Approximations and Endomorphism Algebras of Modules

xm rnm = 0

(n < q),

(5.2.4)

m


xm slm = y(al )

(l < t).

(5.2.5)

m
By Lemma 1.2.13, there is a solution, (b0 , . . . , bp−1 ) of this system in B  . Define a map w : Rp → B  by w(1m ) = bm (m < p). Then w  K = 0 by (5.2.4), so w induces z ∈ HomR (Ai+j , B  ) such that w = zρ. Finally, y = zf by (5.2.5). 2 Now we can prove Theorem 5.2.16. Let R be a ring, A a countably presented module and B = A⊥ . Assume that B (ω) ∈ B for each B ∈ B. Then B is closed under pure submodules.

Proof. Consider the presentation of A from (5.2.1). Let B ∈ B and B  be a pure submodule of B. By assumption, Ext1R (A, B (ω) ) = 0, so ψ clearly has B– factorization property. By Remark 5.2.12 and Lemmas 5.2.14 and 5.2.15, this implies that the inverse system induced by the tower (HomR (Ai , B  ), HomR (hi , B  ) | i < ω) is Mittag–Leffler. Consider the pure–exact sequence ρ

→B− → B/B  → 0. X : 0 → B − Since all modules Ai (i < ω) are finitely presented, an application of HomR (Ai , −) (i < ω) to X yields an inverse system of short exact sequences → HomR (Ai , B) − → HomR (Ai , B/B  ) → 0 (i < ω). 0 → HomR (Ai , B  ) − However, (HomR (Ai , B  ), HomR (hi , B  ) | i < ω) satisfies the Mittag–Leffler condition, so Lemma 5.2.11 gives the exactness of the sequence → lim HomR (Ai , B) − → lim HomR (Ai , B/B  ) → 0 0 → lim HomR (Ai , B  ) − ←− ←− ←− i<ω

i<ω

i<ω

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213

and hence of HomR (A,ρ)

→ HomR (A, B) −−−−−−−→ HomR (A, B/B  ) → 0. 0 → HomR (A, B  ) − In particular, HomR (A, ρ) is surjective, and Ext1R (A, B) = 0 by assumption, so 2 we conclude that Ext1R (A, B  ) = 0. As an immediate corollary, we obtain: Corollary 5.2.17. (Definability of tilting classes) Let R be a ring, T be a tilting module and (A, B) be the cotorsion pair induced by T . Then B is definable.

Proof. Clearly B is coresolving and closed under direct products. By Proposition 5.1.9 (b), B is closed under direct sums. Since the canonical map of a direct sum onto a direct limit is a pure–epimorphism (see Lemma 1.2.7), it suffices to prove that B is closed under pure submodules. However, B is of countable type by Theorem 5.2.10, so the closure of B under pure submodules follows by Theorem 5.2.16. 2

Finite type and resolving subcategories In order to refine Theorem 5.2.10 further to finite type, we will use the following criterion: Lemma 5.2.18. Let R be a ring and T be a tilting module. Let (A, B) be the tilting cotorsion pair induced by T . Then T is of finite type, iff A≤ω ⊆ lim A<ω . −→

Proof. If T is of finite type, then B = (A<ω )⊥ by 5.2.1, so A ⊆ lim A<ω by −→ Theorem 4.5.6. Conversely, let T = (A<ω )⊥ . Then B ⊆ T , and both B and T are definable (by Corollary 5.2.17 and Theorem 5.2.2, respectively). By Lemma 3.1.10, a module belongs to a definable class, iff its pure–injective envelope does. So it remains to show that B and T contain the same pure–injective modules. Let M ∈ T be pure–injective and let A ∈ A. By Theorem 5.2.10, A is A≤ω -filtered. By assumption, A≤ω ⊆ lim A<ω . Since lim A<ω is closed un−→ −→ der extensions and direct limits by Theorem 4.5.6, by induction on the length of a A≤ω -filtration of A, we infer that A ∈ lim A<ω . −→ So there is a direct system (Ai , fji | i ≤ j ∈ I) of modules in A<ω such that A = limi∈I Ai , and ExtjR (Ai , M ) = 0 for all i ∈ I and j > 0. Since M is −→ pure–injective, Lemma 3.3.4 gives ExtjR (A, M ) ∼ ExtjR (Ai , M ) = 0 for = lim ←−i∈I all j > 0, hence M ∈ B.

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This proves that T contains the same pure–injective modules as B.

2

Already at this point, Lemmas 4.5.14 and 5.2.18 yield that each 1–tilting module is of finite type. In order to extend this fact by induction to n–tilting modules, we need one more lemma: Lemma 5.2.19. Let R be a ring, n > 0 and T be a tilting module of projective dimension n. Let C = (A , B  ) be the cotorsion pair defined by B = (Ω(T ))⊥∞ , where Ω(T ) is the first syzygy module of T . Then C is a tilting cotorsion pair induced by an (n − 1)–tilting module T  .

Proof. Since Ω(T ) has projective dimension n−1, by Corollary 5.1.16, it suffices to prove that the class B  is closed under direct sums. For this purpose, it suffices to show that B coincides with the class of all modules M such that there is an exact sequence 0 → M − → B − → C → 0, where ⊥ ∞ and C ∈ Add(T ). B∈T The existence of the exact sequence gives ExtiR (Ω(T ), M ) ∼ = Exti+1 R (T, M ) = i+1 i 0 for each i > 0, because 0 = ExtR (T, C) → ExtR (T, M ) → Exti+1 R (T, B) = 0 is exact. Conversely, if M ∈ B , then the special T ⊥∞ –preenvelope of M yields an exact sequence 0 → M − → B − → C → 0, where B ∈ T ⊥∞ (⊆ B ) and C ∈ ⊥ (T ⊥∞ ) ∩ B  . Moreover, 0 = Ext1 (T, B) → Ext1 (T, C) → Ext2 (T, M ) = R R R Ext1R (Ω(T ), M ) = 0 is exact, so C ∈ ⊥ (T ⊥∞ ) ∩ T ⊥∞ = Add(T ) by Lemma 5.1.8 (c). 2 We arrive at the main result of this chapter: Theorem 5.2.20. (Finite type of tilting modules) Let R be a ring, T a tilting module and (A, B) the cotorsion pair induced by T . Then (a) T and B are of finite type. (b) T is equivalent to a tilting module Tf in such that Tf in is A<ω –filtered.

Proof. (a) The proof is by induction on n = proj dim T . There is nothing to prove for n = 0, since T is then equivalent to R. Assume n > 0 and consider the cotorsion pair C = (A , B  ) defined by B  = (Ω(T ))⊥∞ , where Ω(T ) is the first syzygy module of T . By Lemma 5.2.19, C is a tilting cotorsion pair induced by a tilting module of projective dimension n − 1, so B is of finite type by the inductive premise. Let A ∈ A≤ω . By Lemma 5.2.18, in order to prove that T is of finite type, it suffices to show that A ∈ lim A<ω . −→

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215

By assumption, there is an exact sequence 0 → K − → F − → A → 0 with F ≤ω . For each j > 0 and B  ∈ B , countably generated projective and K ∈ Pn−1 j   we have Extj+1 R (T, B ) = ExtR (Ω(T ), B ) = 0. Since A ∈ A, A is a direct summand in a module filtered by R and by the syzygies of T (see Corollary 3.2.3 j  ∼  and Lemma 5.1.8 (a)). By Lemma 3.1.2, 0 = Extj+1 R (A, B ) = ExtR (K, B ) for all j > 0, so K ∈ A , and K ∈ (A )≤ω . Let C = (A )<ω (⊆ A<ω ). Since B is of finite type, K is a direct summand in a C–filtered module P by Corollary 3.2.3, so K ⊕ L = P for a module L. By Theorem 4.2.6 (for κ = ω) and Lemma 4.3.6, we can, moreover, assume that P is countably presented. Let G = P (ω) . Then K ⊕ G ∼ = (K ⊕ L)(ω) ∼ = G, and there is an exact sequence ⊆

→H− → A → 0, 0→G− where G and H ∼ = F ⊕ G are countably presented and C-filtered. Again by Theorem 4.2.6, there exist strictly increasing C–filtrations (Gi | i < ω) of G, and (Hi | i < ω) of H. Possibly taking a subfiltration of the latter filtration, we can, moreover, assume that Gi ⊆ Hi for each i < ω. Then A ∼ A where = lim −→i<ω i Ai = Hi /Gi . It remains to prove that Ai ∈ A<ω for each i < ω. First we show that Ai ∈ A: since Hi ∈ A, if suffices to extend an arbitrary homomorphism f ∈ HomR (Gi , B) with B ∈ B to some g ∈ HomR (Hi , B). However, G/Gi is C-filtered, so G/Gi ∈ A ⊆ A. Hence Ext1R (G/Gi , B) = 0, and f can be extended to h ∈ HomR (G, B). Similarly, since A ∈ A, h extends to some k ∈ HomR (H, B). Now it suffices to take g = k  Hi . This gives Ai ∈ A. Finally, we prove that Ai ∈ A<ω . Since A is resolving, it suffices to show that any finitely generated module M ∈ A is finitely presented. But M is A≤ω –filtered by Theorem 5.2.10, so Theorem 4.2.6 and Lemma 4.3.6 yield that M is countably ⊆ → R(m) − → M → 0, where presented. Hence  there is an exact sequence 0 → N − m < ω and N = j<ω Nj , where Nj are finitely generated submodules of N .  : N → j<ω Ej by Let Ej denote the injective envelope of N/Nj . Define f f (n) = (n + Nj )j<ω . Then the image of f is contained in j<ω Ej ∈ B. Since  M ∈ A, there is g ∈ HomR (R(m) , j<ω Ej ) such that g  N = f . However, the image of g is finitely generated, so there exists j < ω such that Nj = N proving that M is finitely presented. This proves that A ∈ lim A<ω , and hence that T is of finite type. −→ (b) By part (a), B = (A<ω )⊥ . By Corollary 3.2.3 and Lemma 5.1.8 (c), there are a A<ω -filtered module Tf in and a module Q ∈ Add(T ) such that Tf in = Q ⊕ T . Then Tf in is a tilting module with T ⊥∞ = Tf⊥in∞ , so Tf in is equivalent to T. 2

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Remark 5.2.21. There is an explicit construction of the tilting module Tf in available: as remarked in 5.1.15, the proof of the implication (b) implies (a) in Theorem 5.1.14 shows that any iteration of special B–preenvelopes: ϕ0 : R → T0 of the ring R, ϕ1 of the cokernel of ϕ0 etc., yields a long exact sequence ϕ0

0 → R → T0 → T1 → . . . → Tn−1 → Tn → 0  such that T  = i≤n Ti is a tilting module equivalent to T . By part (a), B = C ⊥∞ , where C = A<ω . By Theorem 3.2.1 (a), each of the special B–preenvelopes ϕi above can be taken so that its cokernel is C–filtered. But then also each Ti (i ≤ n) is C–filtered, and so is T  . In contrast with Theorem 5.2.10, T itself need not in general possess an A<ω – filtration. For example, if T = R ⊕P , where P is a countably generated projective module which is not a direct sum of finitely generated projective modules (see Remark 4.2.13), then T is not P0<ω –filtered. The fact that tilting classes coincide with the classes of finite type makes it possible to classify all tilting classes (over a fixed, but arbitrary, ring R) by the resolving subcategories of mod–R. A class of modules S is called a resolving subcategory of mod–R, if P0<ω ⊆ S ⊆ mod-R, S is closed under extensions and direct summands, and A ∈ S, whenever there is an exact sequence 0 → A − →B− → C → 0 with B, C ∈ S (cf. with Definition 2.2.8 (i)). Before characterizing tilting classes by means of resolving subcategories of bounded projective dimension, we note that the property of being a resolving subcategory can always be tested in a simplified form: Lemma 5.2.22. Let R be a ring and S ⊆ mod–R. Then S is resolving, if and only if R ∈ S, S is closed under extensions and direct summands, and A ∈ S, whenever there is an exact sequence 0 → A − → P − → C → 0 with P finitely generated projective and C ∈ S. In particular, if S ⊆ P1 , then S is resolving, if and only if R ∈ S and S is closed under extensions and direct summands.

Proof. The only–if part is clear. Conversely, if R ∈ S and S is closed under extensions and direct summands, then clearly P0<ω ⊆ S. Consider an exact sequence 0 → A − → B − → C → 0 with B, C ∈ S. By assumption, there is an exact sequence 0 → K − →P − → C → 0 with P finitely

5.2 Classes of finite type

217

generated projective and K ∈ S. Consider the pullback 0 ⏐ ⏐ 

0 ⏐ ⏐ 

K ⏐ ⏐ 

K ⏐ ⏐ 

L −−−−→ ⏐ ⏐ 

P −−−−→ 0 ⏐ ⏐ 

0 −−−−→ A −−−−→ B −−−−→ ⏐ ⏐ 

C −−−−→ 0 ⏐ ⏐ 

0 −−−−→ A −−−−→   

0

0.

Since S is closed under extensions, the left–hand column gives L ∈ S. Since P is projective, the upper row splits, A is a direct summand in L, and hence A ∈ S. 2

Theorem 5.2.23. (A characterization by resolving subcategories of mod–R) Let R be a ring and n < ω. There is a bijective correspondence between n–tilting classes of right R–modules, and resolving subcategories S of mod-R such that S ⊆ Pn<ω . The correspondence is given by the mutually inverse assignments C → (⊥ C)<ω and S → S ⊥ .

Proof. Let C be an n–tilting class. By Theorem 5.2.20, C is of finite type, so there exists T ⊆ Pn<ω such that C = T ⊥∞ . Then clearly (⊥ C)<ω is a resolving subcategory of mod–R. Conversely, let S be a resolving subcategory of mod–R such that S ⊆ Pn<ω . Then C = S ⊥ is a class of finite type, so C is n–tilting by Theorem 5.2.2. Let C be an n–tilting class, so C = T ⊥∞ for class T ⊆ Pn<ω . Let S = (⊥ C)<ω . Then C = (⊥ C)⊥ ⊆ S ⊥ . Conversely, T ⊆ S, so S ⊥ ⊆ T ⊥∞ = C. Let S be a resolving subcategory of mod–R such that S ⊆ Pn<ω . Clearly S ⊆ (⊥ (S ⊥ ))<ω . By Theorem 4.5.6, ⊥ (S ⊥ ) ⊆ (S ) = lim S. By Lemma 1.2.9, −→ 2 S = (lim S)<ω , so we conclude that (⊥ (S ⊥ ))<ω = S. −→ Another consequence of Theorem 5.2.20 is the good behavior of tilting modules and classes with respect to classical localization. Recall that given a commutative ring R, a multiplicative set S ⊆ R, a module T ∈ Mod–R, and a class T ⊆ Mod–R, then S −1 R denotes the localization of

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R at S, S −1 M ∼ = M ⊗R S −1 R the localization of M at S, and TS = {N ∈ −1 Mod–S R | N ∼ = S −1 M for some M ∈ T }. For a maximal ideal m of R and S = R \ m, we also use the notation T(m) = S −1 T and T(m) = TS . Proposition 5.2.24. Let R be a commutative ring, n < ω, T be an n-tilting module and T = T ⊥∞ be the n–tilting class induced by T . (a) Let S be a multiplicative subset of R. Then S −1 T is an n–tilting S −1 Rmodule, the corresponding n–tilting class being TS = (S −1 T )⊥∞ = T ∩ Mod–S −1 R. (b) Let M ∈ Mod–R. Then M ∈ T , if and only if M(m) ∈ T(m) for all maximal ideals m of R.

Proof. By Theorem 5.2.23, there is a resolving class S ⊆ Pn<ω such that T ⊥∞ = S ⊥. Let P denote a projective resolution of T in Mod–R. By Corollary 3.2.3, each syzygy of T in P is isomorphic to a direct summand of an S-filtered module, and conversely, each module in S is isomorphic to a direct summand of a module filtered by R and by the syzygies of T . (a) Applying the exact functor − ⊗R S −1 R to the long exact sequences in (T1) and (T3) for T , we obtain conditions (T1) and (T3) for S −1 T . Further, each element of SS has a projective resolution of length ≤ n consisting of finitely generated S −1 R–modules. Localizing P at S, we obtain a projective resolution Q = P ⊗R S −1 R of S −1 T in Mod–S −1 R. Since S −1 R is a flat module, each syzygy of S −1 T in Q is isomorphic to a direct summand of an SS –filtered S −1 R–module, and conversely, each S −1 R–module in SS is isomorphic to a direct summand of an S −1 R-module filtered by S −1 R and by the syzygies of S −1 T . By Lemma 3.1.2, this yields (S −1 T )⊥∞ = (SS )⊥ in Mod–S −1 R. Let i ≥ 1 and let I be a set. Condition (T2) for T and classical identities for localizations of Ext–groups (see e.g. [155, 3.2.5]) give 0 = Ext1R (C, T (I) ) ⊗R S −1 R ∼ = Ext1S −1 R (S −1 C, S −1 T (I) ) for each C ∈ S. It follows that (S −1 T )(I) ∈ (S −1 T )⊥∞ , hence also condition (T2) holds for the S −1 R–module S −1 T . Similarly, TS ⊆ SS⊥ (= (S −1 T )⊥∞ ), and since N ∼ = N ⊗R S −1 R for any −1 module N ∈ Mod–S R (⊆ Mod–R), we also get N ∈ SS⊥ , iff N ∈ S ⊥ . It follows that TS ⊆ (S −1 T )⊥∞ = T ∩ Mod–S −1 R, and clearly T ∩ Mod–S −1 R ⊆ TS .

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Injectivity properties of tilting modules

219

(b) By [155, 3.2.15], we have Ext1R (C, M ) ⊗R R(m) ∼ = Ext1R (C, M(m) ) for each C ∈ S. Since T = S ⊥ , M ∈ T , iff M(m) ∈ S ⊥ for all maximal ideals ⊥ =T m of R. The latter is equivalent to M(m) ∈ S(m) (m) for all maximal ideals m of R. 2

5.3 Injectivity properties of tilting modules Now we consider further finiteness and injectivity properties of tilting modules. We will start by showing that any tilting module T over any ring R is finendo, that is, T is finitely generated over its endomorphism ring End(T ). First we need a lemma: Lemma 5.3.1. Let R be a ring and M be a module. The following assertions are equivalent: (a) R has an Add(M )–preenvelope. (b) Gen(M ) is a preenveloping class. (c) M is finendo.

Proof. (a) implies (b): let f : R → A be an Add(M )–preenvelope of R. First we prove that f is also a Gen(M )–preenvelope of R. Indeed, if x ∈ HomR (R, G) for some G ∈ Gen(M ), then x factors through the canonical morphism ϕ : M (HomR (M,G)) → G, so there is y ∈ HomR (R, M (HomR (M,G)) ) with x = ϕy. Since f is an Add(M )–preenvelope, there exists an R–homomorphism z : A → M (HomR (M,G)) such that y = zf . It follows that x = ϕzf . Note that f (λ) : R(λ) → A(λ) is a Gen(M )-preenvelope of R(λ) for any λ. Let N ∈ Mod–R, so there is an epimorphism π : R(λ) → N . Consider the pushout of π and f (λ) : f (λ)

R(λ) −−−−→ A(λ) ⏐ ⏐ ⏐ ⏐ ρ π ν

N −−−−→ Then ρ is surjective, so B ∈ Gen(M ).

B

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We prove that ν is a Gen(M )–preenvelope of N : if x ∈ HomR (N, G) for some G ∈ Gen(M ), then xπ factors through the Gen(M )–preenvelope f (λ) . The pushout property then gives a factorization of x through ν. (b) implies (c): first we show that M λ ∈ Gen(M ) for any λ. For each α < λ, denote by πα the α–th canonical projection of M λ onto M . Consider an epimorphism g : F → M λ , where F is a free module. Let f : F → A be a Gen(M )–preenvelope of F . Then foreach α < λ, there is hα : A → M such that πα g = hα f . It follows that h = α<λ hα : A → M λ satisfies g = hf . In particular, h is surjective, so M λ ∈ Gen(A) ⊆ Gen(M ). So there is an epimorphism φ : M (κ) → M M for a cardinal κ. For α < κ, denote by να : M → M (κ) the canonical embedding. For m ∈ M , let πm : M M → M be the canonical projection. (κ) Let x ∈ M be such that φ(x) = idM . Let F be a finite subset of κ such that x = α∈F να (xα ). For each m ∈ M , we have   m = πm (idM ) = πm (φ(x)) = πm φ να (xα ) = eαm (xα ), α∈F

α∈F

where eαm = πm φνα ∈ End(M ) = S. So {xα | α ∈ F } is a finite generating set of the left S–module M . (c) implies (a): let m0 , . . . , mn be an End(M )–generating subset of M . We will prove that the map f : R → M (n) defined by f (1) = (m0 , . . . , mn ) is an Add(M )–preenvelope of R. Since R is finitely generated, it suffices to prove that f is an add(M )–preenvelope of R: Let g ∈ HomR (R, M (p) ). For each projectionof g, gj : R → M (j ≤ p), there exist αji ∈ End(M ) (i ≤ n) such that gj = i≤n αji fi . It follows that the matrix (αji | j ≤ p, i ≤ n) defines a morphism h ∈ HomR (M (n) , M (p) ) such that g = hf . 2 Theorem 5.3.2. Let R be a ring and T be a tilting module. Then T is finendo. ϕ0

Proof. By Lemma 5.1.8 (c), there is an exact sequence 0 → R −→ T0 − → C → 0, where ϕ0 is a special B–preenvelope of R, T0 ∈ Add(T ) and C ∈ A. Then ϕ0 is also an Add(T )–preenvelope of R, and Lemma 5.3.1 applies. 2 Even though M is finendo, in general, there may exist no Add(M )–envelopes and no Gen(M )–envelopes of the free module R — see Corollary 6.3.18 below (and the proof that (a) implies (b) in Lemma 5.3.1). Lemma 5.3.1 suggests the question of when Add(M ) is a preenveloping class for a module M . This turns out to be a much stronger condition than finendo; in particular, it already implies the existence of minimal approximations:

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Injectivity properties of tilting modules

221

Proposition 5.3.3. Let R be a ring and M be a module. The following assertions are equivalent: (a) Add(M ) is an enveloping class. (b) Add(M ) is a preenveloping class. (c) M is product–complete.

Proof. (a) implies (b): this is trivial. (b) implies (c): let κ be a cardinal and f : M κ → A be an Add(M )– preenvelope of M κ . For each α < κ, let πα : M κ → M be the canonical projection. Then there exist σα : A → M such that πα = σα f . So there is g : A → M κ such that πα g = σα for all α < κ. It follows that gf = idM κ , hence M κ ∈ Add(M ). (c) implies (a): let N be a module and κ be an infinite cardinal such that N is ≤ κ–generated. Clearly for each homomorphism f : N → M (λ) there is a subset C ⊆ λ with |C| ≤ κ and Im f ⊆ M (C) . In particular, each f : N → A with A ∈ Add(M ) factors through B = M (κ) . Now denote by h : N → B HomR (N,B) the canonical morphism. Then each f ∈ HomR (N, B) factors through h, and B HomR (N,B) ∈ Add(M ) by the product completeness of M . It follows that h is an Add(M )–preenvelope of N .  Finally, M is –pure–injective, so minimal Add(M )–approximations exist by Proposition 2.1.11. 2 Next we will examine (pure–) injectivity properties of tilting modules, following [16] and [19]. These results will be useful later on in Chapter 7. We will call a module T Σ–pure–split, provided that any pure embedding N → M with M ∈ Add(T ) splits. For example, any Σ–pure–injective module is Σ– pure–split (see Lemma 1.2.23). Proposition 5.3.4. Let R be a ring, T be a tilting module and (A, B) be the tilting cotorsion pair induced by T . Then A is closed under direct limits, if and only if T  is –pure–split.

Proof. First, by Theorems 4.5.6 and 5.2.20, A ⊆ lim A<ω . −→ If A is closed under direct limits, then A = lim A<ω . In particular, A is closed −→ under pure–epimorphic images by Lemma 1.2.9. Moreover, the tilting class B is closed under pure submodules. So, if 0 → N − →M − → P → 0 is a pure–exact sequence with M ∈ Add(T ), then P ∈ A and N ∈ B, and the sequence splits. For the converse, we first claim that Add(T ) = (lim A<ω ) ∩ B. The inclusion −→ ⊆ is clear, since Add(T ) = A ∩ B by Lemma 5.1.8 (c). Let N ∈ (lim A<ω ) ∩ B. −→

222

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Then there is an exact sequence E : 0 → B − →A− → N → 0 with A ∈ A and B ∈ B (induced by a special A–precover of N ). Then A ∈ A ∩ B = Add(T ). Moreover, since N ∈ lim A<ω is a pure–epimorphic image of an element of A, −→ the sequence E is pure–exact. So E splits by assumption, proving our claim. Finally, take an arbitrary module N ∈ lim A<ω . Its special B–preenvelope −→ → A → 0 with A ∈ A and B  ∈ B. yields an exact sequence 0 → N − → B − Then A , and hence also B  , belongs to lim A<ω (by Theorem 4.5.6). So, by the −→ claim above, B  ∈ Add(T ), which yields N ∈ A (as A is resolving). This proves that A = lim A<ω . 2 −→ In the proof of Proposition 5.3.4 we have seen that, if A is closed under direct limits, then A = lim A<ω , hence A is also closed under pure submodules by −→ Lemma 1.2.9. So it is natural to ask, when A is a definable class. Surprisingly, the property of A being closed under direct products is strong enough to guarantee definability: Proposition 5.3.5. Let R be a ring, T be a tilting module and (A, B) be the tilting cotorsion pair induced by T . Then the following are equivalent: (a) A is closed under direct products. (b) A is a definable class. (c) T is product–complete.

Proof. (a) implies (c): since B is always closed under direct products, so is Add(T ) = A ∩ B (see Lemma 5.1.8 (c)), and T is product–complete. (c) implies (b): by Lemma 5.1.9, the class A consists of all Add(T )–coresolved modules of Add(T )–coresolution dimension ≤ n. If T is product–complete, then Add(T ) is closed under direct products, hence so is A. Since any product– complete module is Σ–pure–injective (see Lemma 1.2.25), it is also Σ–pure–split. By Proposition 5.3.4, A = lim A<ω , so A is closed under direct limits and pure −→ submodules by Lemma 1.2.9. (b) implies (a): this is obvious. 2 Remark 5.3.6. Let R be a ring and C = (A, B) be a hereditary cotorsion pair such that A ⊆ P, B is closed under direct sums, and A ∩ B consists of pure–injective modules. By Corollary 5.1.16, C is a tilting cotorsion pair induced by a tilting module T ; since A ∩ B = Add(T ), T is –pure–injective, and A = lim A<ω by Proposition −→ 5.3.4.

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Injectivity properties of tilting modules

223

The module T can be obtained (up to equivalence) as the direct sum of a representative set of all indecomposable modules in A ∩ B. This  follows from the Krull–Schmidt–Azumaya Theorem, and from the fact that each -pure-injective module is a direct sum of indecomposable direct summands with local endomorphism rings (cf. Lemma 1.2.23).

Open problems 1. Characterize the tilting modules T such that the induced tilting class T = T ⊥∞ is enveloping. By Corollary 2.3.7, a sufficient condition for T to be enveloping is that ⊥ T is closed under direct limits. 2. Characterize the tilting modules T such that the regular module R has a minimal T ⊥∞ –coresolution. In other words, this asks, whether/when the sequence (T3) can be obtained by an iteration of T ⊥∞ –envelopes. For example, if R is a domain, S a multiplicative subset of R, and T = δS the Fuchs tilting module from Example 5.1.2 (hence T ⊥ = DS is the class of all S–divisible modules), then the minimal DS –coresolution of R exists, iff the localization S −1 R has projective dimension ≤ 1 as R–module (see Theorem 6.3.16 below).

Chapter 6

1–tilting modules and their applications

In this chapter we restrict ourselves to the particular case of tilting modules of projective dimension 1. In Section 6.1 we will see that in this case there are substantial simplifications of the general theory, partly thanks to a relatively good general understanding of modules of projective dimension 1, and partly to the fact that all 1–tilting classes are torsion classes. In Section 6.2 we will provide an explicit description of 1–tilting classes and modules over particular rings. We will start with the case of artin algebras, but our main concern will be the commutative case: we will completely characterize all tilting modules over Prüfer, valuation and Dedekind domains. In Section 6.3 we will present an application of infinite–dimensional tilting theory to a proof of the following result (Theorem 6.3.16): for any commutative ring R and any multiplicative set S consisting of (some) non–zero divisors, the localization S −1 R has projective dimension ≤ 1, if and only if S −1 R/R is a direct sum of countably presented modules.

6.1

Tilting torsion classes

0–tilting modules are easily seen to coincide with the projective generators. Now we consider in more detail the 1–tilting modules. Some of their examples have already been presented in Section 5.1. First we note a number of simplifications: Lemma 6.1.1. Let R be a ring, T be a 1–tilting module and T = (A, B) be the cotorsion pair induced by T . Then: (a) T is the cotorsion pair generated by T . → T1 → 0 with T0 , T1 ∈ (b) There is a short exact sequence 0 → R − → T0 − Add(T ) (so we can assume r = 1 in condition (T3)).

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225

Moreover, each module M has a T –coresolution of the form 0→M − →G− → T (λ) → 0 for a cardinal λ and G ∈ Gen(T ).

Proof. (a) Since proj dim T ≤ 1, we have T ⊥∞ = T ⊥ and ⊥∞ (T ⊥∞ ) = ⊥ (T ⊥ ). Now Lemma 5.1.8 applies. (b) The first assertion follows by Lemma 5.1.8. For the second, let ϕ : M → G be the special B–preenvelope of M given by Theorem 3.2.1. Then Coker(ϕ) is isomorphic to a direct sum of copies of T by condition (T2). 2 If T is 1–tilting and finitely presented, there are just two category equivalences of the form (5.1.1), namely Ker Ext1R (T, −) and Ker HomR (T, −)

HomR (T,−)



−⊗S T

Ext1R (T,−)



TorS 1 (−,T )

Ker TorS1 (−, T )

Ker(− ⊗S T ).

Moreover, (T ⊥ , Ker HomR (T, −)) and (Ker(− ⊗S T ), T ) are (non–hereditary) torsion pairs in Mod–R and Mod–S, respectively. The pair of equivalences above is called the torsion theory counter equivalence (see [86, §3]). We turn to a characterization of (infinitely generated) 1–tilting modules in terms of the classes they generate: Lemma 6.1.2. Let R be a ring. A module T is 1–tilting, iff Gen(T ) = T ⊥ . In this case Pres(T ) = Gen(T ).

Proof. Assume that T is 1–tilting. Then T ⊥ is closed under homomorphic images. Since (T2) says that T (κ) ∈ T ⊥ , we get Gen(T ) ⊆ T ⊥ . By Lemma 5.1.8 (b), T ⊥ ⊆ Gen(T ), so Gen(T ) = T ⊥ . Conversely, assume that Gen(T ) = T ⊥ . Let N be a module and E be its → E − → E/N → 0, we get injective hull. Applying HomR (T, −) to 0 → N − 0 = Ext1R (T, E/N ) → Ext2R (T, N ) → Ext2R (T, E) = 0, since T ⊥ is closed under homomorphic images. So Ext2R (T, −) = 0, i.e. T ∈ P1 . Condition (T2) is clear by assumption.

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Now we prove that Pres(T ) = Gen(T ). Let M ∈ Gen(T ). Then the canonical map ϕ ∈ HomR (T (HomR (T,M )) , M ) is surjective, so there is an exact sequence ϕ → M → 0. Applying HomR (T, −), we get 0→K− → T (HomR (T,M )) − 0 → HomR (T, K) → HomR (T, T (HomR (T,M )) )

HomR (T,ϕ)

→

HomR (T, M )

→ Ext1R (T, K) → Ext1R (T, T (HomR (T,M )) ) = 0. By definition, HomR (T, ϕ) is surjective, so Ext1R (T, K) = 0 and K ∈ Gen(T ). It remains to verify condition (T3). By condition (T2) and Theorem 3.2.1, there is a special T ⊥ –preenvelope, ψ : R → T0 , of R with T1 = Coker(ψ) isomorphic to a direct sum of copies of T . Since R ∈ ⊥ (T ⊥ ), also T0 ∈ ⊥ (T ⊥ ). Since Gen(T ) = Pres(T ), there are a cardinal λ and an exact sequence 0 → K − → T (λ) − → T0 → 0 with K ∈ T ⊥ . It follows that the sequence splits, and T0 ∈ Add(T ). So (T3) holds for r = 1. 2 In particular, if T is any 1–tilting module, then the 1–tilting class T ⊥ = Gen(T ) is a torsion class of modules, called the tilting torsion class generated by T . The corresponding torsion–free class is Ker HomR (T, −), so (T ⊥ , Ker HomR (T, −)) is a (non–hereditary) torsion pair in Mod–R called the tilting torsion pair generated by T . Example 6.1.3. Let R be a finite–dimensional algebra over a field k. The finite– dimensional 1–tilting modules  were characterized by Bongartz: they coincide with the modules of the form i≤r Mini , where ni < ω, Mi is an indecomposable splitter of projective dimension ≤ 1, Ext1R (Mi , Mj ) = 0 and Mi  Mj for all i = j ≤ r, where r is the rank of the Grothendieck group of R (see [57]). For a generalization to right artinian rings, we refer to [86, §3.7]. Now we will characterize tilting torsion classes among all torsion classes of modules in terms of approximations. We will work in a slightly more general setting of pretorsion classes: a class C of modules is a pretorsion class, provided that C is closed under direct sums and homomorphic images. Theorem 6.1.4. Let R be a ring and T be a class of modules. The following conditions are equivalent: (a) T is a tilting torsion class. (b) T is a special preenveloping torsion class. (c) T is a pretorsion class such that R has a special T –preenvelope.

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227

Proof. (a) implies (b): by Lemma 6.1.2, T = T ⊥ for a 1–tilting module T . Since proj dim T ≤ 1, T is closed under homomorphic images. By Proposition 5.1.9, T is closed under direct sums. Now (b) follows by Theorem 3.2.1 (b). (b) implies (c): this is trivial. (c) implies (a): let 0 → R − → T0 − → T1 → 0 be a special T –preenvelope of R. We will prove that T = T0 ⊕ T0 is a 1–tilting module such that Gen(T ) = T . Since T is a pretorsion class, we have T ∈ T , and Gen(T ) ⊆ T . Let M ∈ ⊥ T (= T1⊥ ). The pushout argument from the proof of Lemma 5.1.8 (b) (for r = 1) shows that M ∈ Gen(T ). Finally, the T –preenvelope of R is special, so T1 ∈ ⊥ T , and T ⊆ (⊥ T )⊥ ⊆ T1⊥ = T ⊥ . This proves that T ⊥ = Gen(T ) = T , so T is 1–tilting by Lemma 6.1.2. 2 Notice that the equivalence of parts (a) and (b) is in fact a particular case of Theorem 5.1.14 for n = 1. Indeed, it suffices to verify that ⊥ T ⊆ P1 in case T is a special preenveloping pretorsion class. But such T contains all homomorphic images of injective modules. If M ∈ ⊥ T , N ∈ Mod–R and E is the injective hull of N , then 0 = Ext1R (M, E/N ) → Ext2R (M, N ) → Ext2R (M, E) = 0, and Ext2R (M, −) = 0. So ⊥ T ⊆ P1 as required. Corollary 6.1.5. Let R be a ring. A pretorsion class T is a tilting torsion class, iff T = T ⊥ for a splitter T .

Proof. By Theorems 3.2.1 and 6.1.4.

2

We finish by a restatement of Corollary 5.1.16 and Theorem 5.2.23 in the particular setting of modules of projective dimension ≤ 1: Corollary 6.1.6. Let R be a ring and C = (A, B) be a cotorsion pair. Then the following assertions are equivalent: (a) C is a 1–tilting cotorsion pair. (b) A ⊆ P1 , and B is closed under direct sums. Corollary 6.1.7. Let R be a ring. Then 1–tilting torsion classes C correspond bijectively to the classes S such that R ∈ S, S ⊆ P1<ω and S is closed under extensions and direct summands. The correspondence is given by the mutually inverse assignments C → (⊥ C)<ω and S → S ⊥ .

Proof. This follows immediately from Lemma 5.2.22 and Theorem 5.2.23.

2

228

6.2

6

1–tilting modules and their applications

The structure of tilting modules and classes over particular rings

Theorem 5.2.23 gives a general classification of all tilting classes in Mod–R via resolving subcategories of mod–R of bounded projective dimension. Over particular rings there are more convenient descriptions available, making use of special properties of modules in the given setting. In this section we present several results of this kind. We give an explicit description of all 1–tilting classes over artin algebras, and of all tilting classes and modules over Prüfer, valuation and Dedekind domains.

1–tilting classes over artin algebras It turns out that in the particular setting of artin algebras R, torsion classes in mod–R are sufficient to classify all 1–tilting classes in Mod–R. Our presentation of this fact follows [295]. The tool making the classification possible is an infinitary version of a well– known formula by Auslander and Reiten. For its proof we refer to [108] and [379]. (Here, for M, N ∈ Mod–R, HomR (M, N ), and HomR (M, N ), denotes the group HomR (M, N ) modulo the subgroup of all homomorphisms that factor through an injective module, and a projective module, respectively.) Lemma 6.2.1. Let R be an artin algebra, D be the standard duality and τ = DT r, τ − = T rD be the Auslander–Reiten translations in mod–R. Let M ∈ mod–R and N ∈ Mod–R. (a) D(Ext1R (M, N )) ∼ = HomR (N, τ M ). Moreover, if M ∈ P1 then D(Ext1R (M, N )) ∼ = HomR (N, τ M ). (b) Ext1R (N, M ) ∼ = D(HomR (τ − M, N )). Moreover, if M ∈ I1 then Ext1R (N, M ) ∼ = D(HomR (τ − M, N )). A tilting torsion class C in Mod–R is called classical, provided there is a finitely generated 1–tilting module T such that C = Gen(T ). Classical tilting torsion classes C over artin algebras were characterized by Assem and Hoshino (see [24, §VI. 6]): the classes C <ω are exactly the torsion classes in mod–R finitely generated by a single module in mod–R and containing all finitely generated injectives. Surprisingly, the characterization of general tilting torsion classes is even simpler – we just remove the condition of single generation:

6.2

The structure of tilting modules and classes over particular rings

229

Theorem 6.2.2. Let A be an artin algebra. (a) There is a bijective correspondence between 1–tilting classes C ⊆ Mod–R and torsion classes T ⊆ mod–R such that T contains all finitely generated injective modules. The correspondence is given by the mutually inverse maps f : C → C <ω and g : T → Ker HomR (−, F) where (T , F) is a torsion pair in mod–R. (b) In the correspondence above, classical tilting torsion classes correspond to the torsion classes T in mod–R such that T = gen(M ) for a module M ∈ mod–R.

Proof. (a) If C is a torsion class in Mod–R, then f (C) is a torsion class in mod–R, since R is right noetherian (see Lemma 4.5.2 (a)). So f is well–defined. Let T be a torsion class in mod–R containing I0<ω with the corresponding torsion pair (T , F) in mod–R. Then T contains all finitely generated cosyzygies of all simple modules, hence ⊥ T consists of modules of projective dimension ≤ 1. (Indeed, if S is a simple module with injective hull E(S), consider the short exact sequence 0 → S → E(S) → X → 0. For M ∈ ⊥ T , we have 0 = Ext1R (M, X) ∼ = Ext2R (M, S). Since Ext2R (M, S) = 0 holds for all simple modules S, we get proj dim M ≤ 1 by Lemma 4.1.8.) By Lemma 6.2.1 (a), for each M ∈ mod–R, M ∈ ⊥ T , iff HomR (T , τ M ) = 0, iff τ M ∈ F. Put τ − F = {M ∈ mod–R | τ M ∈ F}. Since F contains no non–zero injective modules, we have τ (τ − F ) = F for each F ∈ F. As τ − F consists of modules of projective dimension ≤ 1, Lemma 6.2.1 (a) yields g(T ) = Ker HomR (−, τ (τ − F)) = (τ − F)⊥ , so g(T ) is a class of finite type, hence 1–tilting, in Mod–R, and g is well–defined. Clearly T = {M ∈ mod–R | HomR (M, F ) = 0 for all F ∈ F } = f g(T ). Conversely, let C be a 1–tilting class in Mod–R. Let T = f (C), (T , F) be a torsion pair in mod–R and D = gf (C). Then f (D) = f gf (C) = f (C), that is, the finitely generated modules in C and D coincide. We claim that also the pure–injective modules in C and D coincide. To see this, let M be a module and (fi : M → Fi | i ∈ I) a representative set (up to isomorphism) of all epimorphisms from M onto a finitely generated module. Then any homomorphism from M to a finitely generated module has a factorization

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 through f : M → i∈I Fi . By Lemma 1.2.14, the map f is a pure embedding. Since C is a torsion class in Mod–R, we infer that a pure–injective module M belongs to C, iff M is a direct summand in a (possibly infinite) direct product of elements of f (C), and similarly for D. However, f (C) = f (D), and the claim follows. Since D = (τ − F)⊥ , the classes C = S ⊥ and D are of finite type, so they are definable by Theorem 5.2.2. In particular, a module belongs to C, if and only if its pure–injective envelope does (see Lemma 3.1.10), and similarly for D. It follows that C = D, that is, C = gf (C). (b) If C is classical, then C = Gen(M ) for a 1–tilting module M ∈ mod–R, so T = Gen(M ) ∩ mod–R = gen(M ). Conversely, we assume that T is finitely generated by a single module M ∈ mod–R. Clearly M is finendo, that is, the dimension of M over its endomorphism ring is n < ω. By (the proof of) Lemma 5.3.1, there is an Add(M )–preenvelope f : R → M (n) of R which is also its Gen(M )–preenvelope. Since E(R) ∈ gen(M ), f is monic. By Lemma 2.1.3 and Corollary 2.1.10, there exist a direct summand N in M (n) and a monic Gen(M )–envelope g : R → N . Now T = gen(M ), so N ∈ T , and since T is closed under extensions, the version of Lemma 2.1.13 in mod–R gives that P = Coker g ∈ (⊥ T ∩ mod–R). Let C be the tilting torsion class in Mod–R corresponding to T , so T = C ∩ mod–R. Since P ∈ ⊥ T , the argument above shows that also P ∈ ⊥ E where E is the class of all pure–injective modules in C. Since P ⊥ and C are definable classes, Lemma 3.1.10 gives P ∈ ⊥ C, so g is a special C–preenvelope of R. Then T = N ⊕ P ∈ mod–R is a 1–tilting module generating the class C (see Remark 5.1.15), so C is classical. 2 As an example, we will consider the correspondence from Theorem 6.2.2 in the particular setting of Examples 5.1.4 and 5.1.5. Example 6.2.3. Let R be a connected tame hereditary algebra over a field k, P be a non–empty set of tubes, and TP be the corresponding Ringel tilting module (see Example 5.1.4). Then the corresponding torsion class in mod–R is TP⊥ ∩ mod–R which consists of all preinjective modules and all regular modules in the tubes not in P . If R is a connected hereditary algebra of infinite representation type and TP is the Lukas tilting module from Example 5.1.5 (b), then the corresponding torsion class in mod–R consists of all regular modules and all preinjective modules. If R is a connected wild hereditary algebra and TM is the Lukas divisible module from Example 5.1.5 (a), then the corresponding torsion class in mod–R is the class of all preinjective modules.

6.2

The structure of tilting modules and classes over particular rings

231

In particular, by Theorem 6.2.2 (b), neither of the tilting modules TP and TM above is equivalent to a finitely generated tilting module.

Tilting modules and classes over Prüfer domains In this section we will give a complete description of all tilting modules and classes over Prüfer domains based on [41] and [352]. We start with the more general setting of semihereditary rings: a ring R is right hereditary (semihereditary), if all (finitely generated) right ideals of R are projective. Notice that a domain is semihereditary, iff R is a Prüfer domain, and R is hereditary, iff R is a Dedekind domain. Moreover, any right semihereditary ring is right coherent. Recall that a ring R is von Neumann regular, if for each x ∈ R there is y ∈ R such that xyx = x. Any von Neumann regular ring is semihereditary: indeed, a finitely generated submodule of a projective module is always a direct summand. There is an immediate corollary of Theorem 5.2.20 for tilting modules over semihereditary rings: Corollary 6.2.4. Let R be a right semihereditary ring. Then all tilting R–modules have projective dimension ≤ 1. Moreover, if R is von Neumann regular, then tilting modules coincide with the projective generators.

Proof. Let R be right semihereditary. By a classical result of Kaplansky, all finitely generated submodules of projective modules are projective, so all finitely presented modules have projective dimension ≤ 1. By Theorem 5.2.20, all tilting modules have projective dimension ≤ 1. If R is von Neumann regular, then finitely presented modules are projective, so the last claim follows again from Theorem 5.2.20. 2 We also recall the following result of Fuchs on the structure of modules of projective dimension ≤ 1 over a Prüfer domain (for a proof, we refer to [181, VI.6]): Lemma 6.2.5. Let R be a Prüfer domain and M be a module. Let Q = {R/I | I ⊆ R, I finitely generated}. Then M ∈ P1 , iff M is Q–filtered. In fact, if R is a Prüfer domain and M is a module of projective dimension ≤ 1, then any finitely generated submodule N of M is finitely presented (that is, M is

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coherent), and N satisfies proj dim M/N ≤ 1 (that is, N is tight in M ). If N is moreover cyclic, then N ∼ = R/I, where I is finitely generated (hence projective) (see [181, VI.6]). As observed in Example 5.1.2, if R is a Prüfer domain, then C = (P1 , DI) is a tilting cotorsion pair generated by the Fuchs tilting module δ. Notice that Lemma 6.2.5 says that the deconstruction of C is possible not only to countable type (as follows from the general Theorem 5.2.10), but to “cyclic finitely presented” type. In Theorem 6.2.11 below we will show that the same holds for any tilting cotorsion pair over a Prüfer domain. This will be instrumental for characterizing tilting torsion classes as the (generalized) divisibility classes in the sense of the following definition: Definition 6.2.6. Let I be an ideal of a Prüfer domain R. A module M is I–divisible, if M I = M . We will need a homological characterization of I–divisible modules in the case, when I is a finitely generated ideal: Lemma 6.2.7. Let R be a Prüfer domain. Let I be a non–zero finitely generated ideal of R and M be a module. Then M is I–divisible, iff Ext1R (R/I, M ) = 0.

Proof. Let E = Ext1R (R/I, M ). Then E = 0, iff E(p) = 0 for all p ∈ spec R. Let p ∈ spec R. Since R/I ∈ mod–R, we have   E(p) ∼ = Ext1R(p) (R/I)(p) , M(p) (see [155, §3.2]). Moreover, (R/I)(p) ∼ = R/I ⊗R R(p) ∼ = R(p) /I(p) as R(p) – modules. Since I(p) is a finitely generated (= principal) non–zero ideal of the valuation domain R(p) , we have E(p) = 0, iff M(p) I(p) = M(p) . The latter says that 0 = M(p) ⊗R(p) R(p) /I(p) ∼ = (M ⊗R R/I)(p) = 0. Altogether we have E = 0, iff E(p) = 0 for all p ∈ spec R, iff (M ⊗R R/I)(p) = 0 for all p ∈ spec R, iff M ⊗R R/I = 0, iff M I = M . 2 Let S denote the set of all non–zero finitely generated ideals I of the Prüfer domain R such that Ext1R (R/I, T ) = 0, and let G be the filter of ideals in R generated by S. Since G has a basis of finitely generated ideals, G is a Gabriel filter  in the sense R: J = of [377, IX], and the ring of quotients RG of R coincides with the set J∈G

{x ∈ Q | xJ ⊆ R for some J ∈ G}, where Q is the quotient field of R. Since

6.2

The structure of tilting modules and classes over particular rings

233

R is a Prüfer domain, every torsion–free module is flat by Theorem 4.4.10. In particular RG is a flat R–module and, moreover, being an overring of R, RG is a Prüfer domain (see [181, III.1]). Assume that T is a 1–tilting R–module and (A, B) is the cotorsion pair generated by T . For a non–zero ideal I of R, R/I ∈ A implies I ∈ S, since T ∈ B and A ⊆ P1 . Conversely, if I ∈ S, then R/I is a finitely presented module in P1 . Thus (R/I)⊥ is closed under direct sums and epimorphic images; so T ∈ (R/I)⊥ implies (R/I)⊥ ⊇ Gen(T ) = B. Hence R/I ∈ A<ω . We start with a couple of lemmas: Lemma 6.2.8. Let R be a Prüfer domain and T be a module with proj dim T ≤ 1 and Ext1R (T, T ) = 0. If I is the annihilator of some element x ∈ T , then I is projective and Ext1R (R/I, T ) = 0.

Proof. Consider the submodule xR ∼ = R/I of T . Since xR is tight in T , we have proj dim T /xR ≤ 1, and proj dim xR ≤ 1. From the exact sequence 0 → xR − → T − → T /xR → 0 we obtain 0 = Ext1R (T, T ) → Ext1R (xR, T ) → Ext2R (T /xR, T ) = 0. So Ext1R (R/I, T ) = 0.

2

It follows that the torsion submodule t(T ) of T coincides with the torsion submodule tF (T ) of T with respect to the torsion pair associated with the Gabriel filter G. Lemma 6.2.9. Let T be a tilting module over a Prüfer domain R and t(T ) be its torsion submodule. For every non–zero projective ideal I of R, the following are equivalent: (a) Ext1R (R/I, T ) = 0. (b) Ext1R (R/I, T /t(T )) = 0.

Proof. (a) implies (b): this is clear since proj dim R/I ≤ 1. (b) implies (a): let N = T /t(T ) and let G be the Gabriel filter defined above. By the consequence above it follows that N is an RG –module in a canonical way. Moreover, since G has a basis of invertible ideals, TG , the module of quotients of T with respect to the filter G, is isomorphic to T ⊗R RG (see e.g. [377, IX.5.3]). By Lemma 6.2.8, t(T ) = tG (T ), hence t(T ) ⊗R RG = 0 by [377, IX.1.6, 5.3]. We infer that N is isomorphic to T ⊗R RG . We will show that N is a 1–tilting RG –module. In fact, for each cardinal α, N (α) ∈ T ⊥ , so Ext1R (T, N (α) ) ∼ = Ext1RG (T ⊗R RG , N (α) ), since RG is a flat

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overring of R. Tensoring a free presentation of T by the flat module RG , we obtain proj dimRG N ≤ 1. Condition (T3) is easily seen to be satisfied by N , this time tensoring by the flat R–module RG an exact sequence satisfying condition (T3) for the R–module T . Let (A , B  ) be the cotorsion pair generated by the 1–tilting RG –module N . Since N is a flat RG -module, A consists of flat modules. Condition Ext1R (R/I, N ) = 0 is equivalent to Ext1RG (R/I ⊗R RG , N ) = 0 by [78, §VI.4]. The finitely presented RG –module M = R/I ⊗R RG ∼ = RG /IRG is in A ; hence IRG = RG and M = 0. By [377, IX. 5.3], I ∈ G and Ext1R (R/I, T ) = 0. 2 By Theorem 5.2.20, all tilting modules over Prüfer domains are of finite type. In fact, we will prove a stronger result, saying that in this case all tilting modules are A<ω –filtered. The following lemma is the key for the passage from A≤ω – filtrations to A<ω –filtrations: Lemma 6.2.10. Let R be a Prüfer domain, T a tilting R–module and (A, B) the cotorsion pair generated by T . Then C ∈ A≤ω , if and only if C admits a filtration (Cn | n < ω) consisting of finitely presented submodules such that Cn+1 /Cn is cyclic and Cn+1 /Cn ∈ A<ω .

Proof. The if–part follows by Lemma 3.1.2. For the converse, let C ∈ A≤ω and consider a projective resolution of C, 0 →  φ P − → F − → C → 0, F= i∈ω xi R, where P = Ker φ is projective. For every n ≥ 0, let Fn = i≤n xi R and let φn be the restriction  of φ to Fn . Let Cn = Im gφn and Pn = Ker φn = P ∩ Fn . Then P = n∈ω Pn and every Pn is finitely generated, since C is coherent; hence proj dim Cn ≤ 1, and Pn is projective. We have Cn+1 /Cn ∼ = Fn+1 /(Fn + Pn+1 ), hence Cn+1 /Cn ∼ = R/Jn+1 , where Jn+1 is the ideal of R such that xn+1 Jn+1 is the image of Pn+1 under the canonical projection of F onto xn+1 R. Since proj dim Cn+1 /Cn ≤ 1, the ideal Jn+1 is projective. pn+1 → Pn+1 −−−→ Jn+1 → 0, where pn+1 is Consider the exact sequence 0 → Pn − the restriction to Pn+1 of the composition of the canonical (n + 1)–projection with the isomorphism xn+1 R ∼ = R. By the projectivity of Jn+1 , there exists a split map νn+1 : Jn+1 → Pn+1 whose composition with pn+1 is the identity on Jn+1 . Now for every non–zero ideal J of R, every element of HomR (J, R) is multiplication −1 , 0 ≤ i ≤ n + 1 such by an element in J −1 . Thus there are elements qin+1 ∈ Jn+1  n+1 n+1 that, for every a ∈ Jn+1 , νn+1 (a) = i≤n+1 xi (qi a); qn+1 = 1, since νn+1 is

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235

a split map for pn+1 . Let J0 be the finitely generated ideal defined by P0 = x0 J0 and let ν0 : J0 → P0 be the isomorphism defined by ν0 (a) = x0 a.  Clearly P ∼ = i<ω Ji , where we can assume that Ji = 0 for every i ∈ ω, and  μ  there is an exact sequence 0 → i<ω Ji − → i<ω xi R − → C → 0 such that the restriction of μ to Jn is the composition of νn with the inclusion of Pn ⊆ Fn . Let E = {m ≥ 0 | R/Jm ∈ / A}. The condition R/Jm ∈ / A is equivalent to Ext1R (R/Jm , T ) = 0, hence also to Ext1R (R/Jm , T /t(T )) = 0, by Lemma 6.2.9. cannot be exThus, for every m ∈ E we can choose a map fm : Jm →T /t(T ) that tended to R; we let fi = 0, if i ∈ / E. Define a map f : J → i<ω i i<ω T /t(T ) ⊥ is closed under direct sums, there exf . Since C ∈ A and T by f = i<ω i ists a map g : F → (T /t(T ))(ω) such that gμ = f . To g we associate  a matrix M = (ti,j )i,j<ω , where the entries ti,j ∈ T /t(T ) satisfy g(xj ) = i<ω ti,j ∈  T /t(T ), for every j ∈ ω. i<ω We will prove, by induction on j, that ti,j = 0 for every i > j. Writing the elements as column vectors, for every a ∈ Jn we have g(μ(a)) = Mv n = f (a), where v n denotes the column vector element μ(a). If n = 0, μ(a) = x0 a, and then Mv 0 = f (a). Since f (a) has zero component in (T /t(T ))i for every i > 0, we conclude that ti0 is annihilated by J0 ; hence it is zero.  Assuming the claim n+1 a). verified for every m ≤ n, consider a ∈ Jn+1 and μ(a) = j≤n+1 xj (qj For i > n + 1, the of the element Mv n+1 = f (a) is zero and it i–th component ti,j qjn+1 a. By induction, the elements ti,j appearing in the coincides with j≤n+1  sum are zero for every j ≤ n, since i > n + 1 ≥ j. Thus 0 = ti,j qjn+1 a = j≤n+1

n+1 ti,n+1 qn+1 a

n+1 qn+1

= ti,n+1 a since = 1. But then, since T /t(T ) is torsion–free, ti,n+1 = 0. This proves that the matrix M is upper triangular; thus for a ∈ Jn the element g(μ(a)) has zero components for i > n and its n–th component is tn,n a. If ρn is the canonical n–th projection of (T /t(T ))(ω) onto T /t(T ), then ρn gμ = ρn f = fn is the multiplication by the element tn,n ∈ T /t(T ), hence fn extends to 2 R. This proves that E is empty, so Cn+1 /Cn ∼ = R/Jn+1 ∈ A for every n. From Theorem 5.2.10 and Lemma 6.2.10 we immediately obtain

Theorem 6.2.11. Let R be a Prüfer domain, T a tilting module and (A, B) the cotorsion pair generated by T . Then A coincides with the class of all A<ω –filtered modules. In particular, T is A<ω –filtered. In order to characterize all tilting classes over Prüfer domains, we need the notion of a finitely generated localizing system of ideals:

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Definition 6.2.12. Let R be a Prüfer domain. A filter L of non–zero ideals of R is called a finitely generated localizing system provided that (i) L contains a basis consisting of finitely generated ideals, and (ii) J ∈ L, whenever J is an ideal of R such that there is I ∈ L with {r ∈ R | ir ∈ J} ∈ L for all i ∈ I. Remark 6.2.13. Finitely generated localizing systems of ideals of a Prüfer domain R correspond bijectively to overrings of R. Given such localizing system L, the corresponding overring (called the ring of fractions w.r.t. L) is defined by S = {q ∈ Q | Iq ⊆ R for some I ∈ L}. For more details on this correspondence, we refer to [167, §5.1]. In our setting, condition (ii) just says that L is multiplicatively closed: Lemma 6.2.14. Let R be a Prüfer domain. Let L be a filter of non–zero ideals of R such that L satisfies condition (i) of Definition 6.2.12. Then L satisfies condition (ii), iff L is multiplicatively closed.

Proof. Assume (ii). Let I1 , I2 ∈ L. Then for each i1 ∈ I1 , I2 ⊆ {r ∈ R | i1 r ∈ I1 I2 }, so I1 I2 ∈ L by condition (ii). (Note that condition (i) on L is not needed for the proof of this implication.) Assume L is multiplicatively closed. Let J be an ideal of R such that there is I ∈ L with {r ∈ R | ir∈ J} ∈ L for all i ∈ I. By (i), we can assume that I is finitely generated, I = k≤n ik R. For each k ≤ n let Ik = {r ∈ R | ik r ∈ J} ∈ L. Then II1 . . . In ⊆ J, so J ∈ L. 2 Now we can prove the main result of this section: Theorem 6.2.15. (Tilting classes over Prüfer domains) Let R be a Prüfer domain. (a) Let C be a class of modules. Then C is tilting, iff there is a finitely generated localizing system L of ideals of R such that   C = M ∈ Mod–R | M I = M for all I ∈ L . (b) There is a bijective correspondence between tilting classes and finitely generated localizing systems. The correspondence is given by the assignments   f : C → L = J ⊆ R | ∃ I : (0 = I ⊆ J and R/I ∈ (⊥ C)<ω ) and

  g : L → C = M ∈ Mod–R | M is I–divisible for all I ∈ L .

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237

Proof. (a) Let C be a tilting class. So C = T ⊥ for a 1–tilting module T (see Corollary 6.2.4), and C = (A<ω )⊥ , where A = ⊥ C. Let S = {R/I | R/I ∈ A<ω }. By Lemma 6.2.10, C = S ⊥ , so by Lemma 6.2.7, C = {M ∈ Mod–R | M I = M for all 0 = I such that R/I ∈ S}. Let B be the set of all non–zero (finitely generated) ideals I of R such that R/I ∈ S. By Lemma 6.2.7, B is multiplicatively closed, so B is a basis of a finitely generated localizing system L by Lemma 6.2.14. Clearly C = {M ∈ Mod–R | M I = M for all I ∈ L}. Conversely, given a finitely generated localizing system I such that C = {M ∈ Mod–R | M I = M for all I ∈ L}, we let B be the set of all finitely generated ideals in L and S = {R/I | I ∈ B} (⊆ P1<ω ). By Lemma 6.2.7, S ⊥ = C, so C is tilting. (b) It remains to verify that f and g are mutually inverse. By Lemma 6.2.7, C ⊆ gf (C). Conversely, if M ∈ gf (C), then M I = M for all I ∈ f (C). As above, we have C = S ⊥ where S = {R/I | R/I ∈ (⊥ C)<ω }. By Lemma 6.2.7, M ∈ S ⊥ , so M ∈ C. This proves that C = gf (C). Let L be a finitely generated localizing system. Let B be the (multiplicatively closed) set of all finitely generated ideals in L and C = g(L). By Lemma 6.2.7, B ⊆ f (C), so L ⊆ f g(L). Moreover, C = S ⊥ , where S = {R/I | I ∈ B}. Let J ∈ f (C). Then there is a non–zero finitely generated ideal I ⊆ J such that R/I ∈ ⊥ C. By Corollary 3.2.4, R/I is a direct summand of an S ∪ {R}–filtered module N . By Theorem 4.2.6, we can w.l.o.g. assume that N is finitely S ∪ {R}– filtered. By induction on the length of the filtration, we will prove that, if C is a cyclic submodule of N such that C  R, then C ∼ = R/L for some L ∈ L. This will finish the proof, since then R/I ∈ S, so I ∈ B and J ∈ L, hence f g(L) = L.  Let N = i≤n Ni be a finite S ∪ {R}–filtration of N , and let R/L → N with 0 = L = R. If n = 1, then N ∼ = R/I for some I ∈ B, so I ⊆ L, and L ∈ L. If n = k + 1 and R/L → Nk , then Nn /Nk ∼ = R/I for some I ∈ B. Let {ij | j ≤ m} be an R–generating subset of I. For each j ≤ m, (Rij + L)/L is a cyclic submodule of INn ⊆ Nk , so by inductive premise, there is Lj ∈ L with 2 Lj ij ⊆ L. Then IL0 . . . Lm ⊆ L, so L ∈ L. Next we will describe all tilting modules over Prüfer domains up to equivalence. The description, due to Salce [352], makes use of a particular generalization of the notion of the Fuchs tilting module (see Example 5.1.2), where non–zero (= invertible) elements of the Prüfer domain are replaced by its non–zero finitely generated (= invertible) ideals: Definition 6.2.16. (Salce tilting modules) Let R be a Prüfer domain and L be a finitely generated localizing system of ideals of R. Let L0 be the set of all finitely

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generated ideals in L. Denote by Λ the set of all finite sequences of elements of L0 (including the empty sequence ∅). For ∅ = λ = (I0 , . . . , Ik ) ∈ Λ, we define λ− = (I0 , . . . , Ik−1 ) ∈ Λ for k > 0 and λ− = ∅ for k = 0, and Gλ = I0−1 . . . Ik−1 ⊆ Q. We also define G∅ = R. Note that G∅ ⊆ Gλ− ⊆ Gλ for all ∅ = λ ∈ Λ.  Put G = λ∈Λ Gλ , and let K be the submodule of G generated by the elements (xσ | σ ∈ Λ) ∈ G such that there is ∅ = λ ∈ Λ with xλ− = −xλ and xσ = 0 for all σ = λ, λ− . Define the Salce tilting module δL by δL := G/K. Remark 6.2.17. Let R be a Prüfer domain. Recall that for a finitely generated ideal I of R, I −1 denotes the unique (finitely generated) submodule of Q satisfying II −1 = R. In fact, I −1 = {q ∈ Q | qI ⊆ R}. If I, J ∈ L0 , then IJ ∈ L0 and (IJ)(I −1 J −1 ) = R, so (IJ)−1 = I −1 J −1 . Assume, moreover, that J ⊆ I. Then L = I −1 J is the (unique, finitely generated) ideal of R such that IL = J; since J ⊆ L, we have L ∈ L0 (see [181, III.1]). Let M be a module. Then Ext1R (J −1 /I −1 , M ) = 0, iff for each maximal ideal p of R, Ext1R(p) ((J −1 /I −1 )(p) , M(p) ) = 0 (see [155, §3.2]). However, (J −1 /I −1 )(p) ∼ = (J −1 )(p) /(I −1 )(p) = (I(p) )−1 (L(p) )−1 /I −1 ∼ = R(p) /L(p) in (p)

Mod–R(p) , because (I −1 )(p) = (I(p) )−1 by [181, I.2], and I(p) and L(p) are principal ideals of the valuation domain R(p) . It follows that Ext1R (J −1 /I −1 , M ) = 0, iff Ext1R (R/L, M ) = 0.

Lemma 6.2.18. Let R be a domain. Then δL is a 1–tilting module generating the class {M ∈ Mod–R | M I = M for all I ∈ L}.

Proof. The proof is carried out in several steps. (Step I) We prove condition (T1) from Definition 5.1.1 for δL , that is, we prove proj dim δL ≤ 1. We define a strictly increasing chain (δn | n < ω) of submodules of δL as follows: For each n < ω, let Λn be the subset of Λ consisting  of all sequences of length = ( ≤ n (including the empty one), and put δ n λ∈Λn Gλ + K)/K. Clearly  δL = n<ω δn , so it suffices to prove that proj dim(δn+1 /δn ) ≤ 1. ∼ It is easy to see that K for n < ∩ G∅ = 0, so δ0 =G∅ = R. Moreover, ∼ /δ = ( G + K)/( G + K) H ω, we have δ = n+1 n n /(Hn ∩ λ∈Λn+1 λ∈Λn λ  λ  ( λ∈Λn Gλ + K)), where Hn = λ∈Λn+1 \Λn Gλ (⊆ G).   However, Hn ∩( λ∈Λn Gλ +K) = λ∈Λn+1 \Λn Gλ− , so we have δn+1 /δn ∼ =  −1 −1 −1 −1 λ∈Λn+1 \Λn Gλ /Gλ− . Since Gλ /Gλ− = I0 . . . Ik /I0 . . . Ik−1 is finitely

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239

presented for each λ = (I0 , . . . , Ik ), we infer that proj dim(Gλ /Gλ− ) ≤ 1, and hence proj dim(δn+1 /δn ) ≤ 1. (Step II) We prove that δL is I–divisible for each I ∈ L. It suffices to prove this for each I ∈ L0 . However, for each λ ∈ Λ, we can define λ = (λ, I). Then I(Gλ + K)/K ⊇ (Gλ + K)/K by the definition of δL , so IδL = δL . (Step III) We prove that Ext1R (δL , M ) = 0, and δL generates M , whenever M is I–divisible for each I ∈ L. The first claim follows from Lemma 3.1.2 by Step I: Indeed, Gλ /Gλ− ∼ = J −1 /I −1 , where I = I0 · · · Ik−1 ∈ L0 and J = I · Ik ∈ L0 , and Ext1R (J −1 /I −1 , M ) = 0 because Ext1R (R/L, M ) = 0 for L = I −1 J, by Remark 6.2.17 and Lemma 6.2.7. Since

δn+1 /δn ∼ Gλ /Gλ− , = λ∈Λn+1 \Λn

we infer that Ext1R (δn+1 /δn , M ) = 0 for all n < ω, and hence Ext1R (δL , M ) = 0. For the second claim it suffices to prove that for each a ∈ M there is a homomorphism η : δL → M such that a ∈ Im(η). We use the fact that δL = n<ω δn (see Step I) and define η by induction: since δ0 ∼ = R, we can define η on δ0 so that its image contains a. If η is already defined on δn for some n < ω, then, since Ext1R (δn+1 /δn , M ) = 0 by the previous paragraph, η extends from δn to δn+1 , and finally to δ. Notice that by Steps II and III, condition (T2) from Definition 5.1.1 holds for δL , that is, Ext1R (δL , δL (κ) ) = 0 for any cardinal κ. (Step IV) We prove that the module δL /δ0 is isomorphic to a direct summand in δL . We fix a finitely generated ideal J such that R = J ∈ L0 and define ξ : G → δL by ξ  G∅ = 0 and, for λ = (I0 , . . . , Ik ) ∈ Λ and x ∈ Gλ , by ξ(x) = (xκ | κ ∈ Λ) + K, where xκ1 = x for κ1 = (R, I0 , . . . , Ik ), xκ2 = −x for κ2 = (J, I0 , . . . , Ik ), and xκ = 0 otherwise. Then ξ  K = 0, so ξ  (G∅ + K) = 0, and ξ induces an endomorphism ϕ of δL that has a factorization ψ : δL /δ0 → δL through the projection π : δL → δI /δ0 , that is, ϕ = ψπ. Now we will define a homomorphism φ : δL → δL /δ0 such that φϕ = π. Then φψπ = π, so φψ = idδI /δ0 , proving that δL /δ0 is isomorphic to a direct summand in δL . In order to define φ, it suffices to construct a homomorphism η : G → δL /δ0 such that η  K = 0, and φη(g) = π(g + K) for each g ∈ G. It is easy to verify that such η is defined by setting η  Gλ = 0, whenever λ ∈ Λ1 or λ = (I0 , . . . , Ik )

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with I0 = R, and by setting η(x) = ((xκ | κ ∈ Λ) + K) + δ0 , where x− λ = x and xκ = 0 for κ = − λ, when λ = (R, I1 , . . . , Ik ), x ∈ Gλ , and − λ = (I1 , . . . , Ik ). Finally, by Step IV, condition (T3) from Definition 5.1.1 holds (for r = 1) for δL , so δL is a 1–tilting module. By Step III, δL generates the class   M ∈ Mod–R | M I = M for all I ∈ L . 2

Theorem 6.2.19. (Tilting modules over Prüfer domains) Let R be a Prüfer domain. Then the set of Salce tilting modules   D = δL | L is a finitely generated localizing system of ideals of R is a representative set (up to equivalence) of the class of all tilting modules.

Proof. First each module δL ∈ D is tilting by Lemma 6.2.18. Conversely, let T be a tilting module. Then, by Theorem 6.2.15 (a), T ⊥ = {M ∈ Mod–R | M I = M for all I ∈ L} for a finitely generated localizing system L of ideals of R. By Lemma 6.2.18, T ⊥ = δL⊥ , that is, T is equivalent to δL . By Theorem 6.2.15 (b), δL is not equivalent to δL whenever L and L are distinct finitely generated localizing systems of ideals of R. 2

The case of valuation and Dedekind domains Now we focus on two particular instances of Prüfer domains, where the general description of tilting modules and classes takes a much simpler form. We start with the case when R is a valuation domain. In this case, tilting modules are (up to equivalence) exactly the modules δS from Example 5.1.2, where S ranges over all saturated multiplicative subsets S of R, and tilting classes are exactly the classes of all S–divisible modules. (A multiplicative subset S ⊆ R is called saturated, provided that s.s ∈ S implies s ∈ S and s ∈ S for all s, s ∈ R.) In other words, tilting classes correspond to prime ideals of the valuation domain R: Lemma 6.2.20. Let R be a valuation domain and S ⊆ R. Then S is a saturated multiplicative subset of R, if and only if R \ S is a prime ideal of R.

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Proof. The complement of a prime ideal is clearly a saturated multiplicative subset in R (for any ring R). Conversely, if S is multiplicative and saturated, let I be the ideal of R maximal with the property that I ∩ S = ∅. Since S is multiplicative, I is a prime ideal of R. If x ∈ / I, then I  xR, since R is a valuation domain, so x.r ∈ S for some r ∈ R, and x ∈ S because S is saturated. So I ∪ S = R. 2 Theorem 6.2.21. Let R be a valuation domain. (a) There is a bijective correspondence between the set of all tilting classes in Mod–R and the spectrum spec R. The correspondence is given by the mutually inverse assignments   C → PC = s ∈ R | M s = M for some M ∈ C and

  /P . P → CP = M ∈ Mod–R | M s = M for all s ∈

(b) The set of Fuchs tilting modules {δR\P | P ∈ spec R} is a representative set (up to equivalence) of the class of all tilting modules.

Proof. (a) Since all finitely generated ideals are principal, finitely generated localizing systems L correspond bijectively to the saturated multiplicative subsets S of R: Given L, we take S = {s ∈ R | sR ∈ L}, and given S, we let L consist of all ideals I of R such that there exists s ∈ S with sR ⊆ I. In this correspondence, the class CL of all modules I–divisible for each I ∈ L coincides with the class of all S–divisible modules. So the claim follows by Theorem 6.2.15 (b) and Lemma 6.2.20. (b) By Example 5.1.2, the Fuchs tilting module δS generates the class of all S–divisible modules, for any multiplicative subset S in R. So the claim follows by Lemma 6.2.20 and part (a). 2 Now assume that R is a Dedekind domain. In this case we will combine Theorem 6.2.15 with Example 5.1.3. Recall from Example 5.1.3 that given a Dedekind domain  R and a set of maximal ideals P ⊆ mspec R, RP denotes the module π −1 ( p∈P E(R/p)), where  π : Q → Q/R is the canonical projection. By 5.1.3, TP = RP ⊕ p∈P E(R/p) is a 1–tilting module with the corresponding tilting class TP⊥ =



p∈P

R/p

⊥

  = M ∈ Mod–R | M p = M for all p ∈ P .

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Theorem 6.2.22. Let R be a Dedekind domain. (a) There is a bijective correspondence between the set of all tilting classes in Mod–R and the power set of the maximal spectrum P(mspec R). The correspondence is given by the mutually inverse assignments   C → PC = p ∈ mspec R | M p = M for all M ∈ C and

  P → CP = M ∈ Mod–R | M p = M for all p ∈ P .

(b) The set of Bass tilting modules {TP | P ⊆ mspec R} is a representative set (up to equivalence) of the class of all tilting modules.

Proof. (a) Since R is noetherian and each non–zero ideal is uniquely a finite product of maximal ideals (see [322, §11]), a finitely generated localizing system L is uniquely determined by its subset P = L ∩ mspec R: if I is an ideal, then  I ∈ L, iff I = i≤n pi for some n < ω and pi ∈ P (i ≤ n). Theorem 6.2.15 (b) then gives the claim. (b) We show that any tilting module T is equivalent to TP for some P ⊆ mspec R. By part (a), T ⊥ = {M ∈ Mod–R | M p = M for all p ∈ P } for a set of maximal ideals P . But the latter class equals TP⊥ , so T is equivalent to TP . Finally, if P = P  , then TP is not equivalent to TP  (by part (a), or simply because R/p ∈ TP⊥ \ TP⊥ whenever p ∈ P \ P  ). 2 Remark 6.2.23. Theorem 6.2.22 was first obtained in the special cases of R = Z, and R a small Dedekind domain, under the assumption V = L, in [244] and [390], respectively. The general result was then proved in ZFC in [41] using Theorem 5.2.10 and a modification of the proof in [390]. The valuation domain case (Theorem 6.2.21) was first obtained under V = L for small valuation domains in [351]. The general Prüfer domain case (Theorems 6.2.15 and 6.2.19) comes from [41] and [352].

6.3

Matlis localizations

In this section we will employ infinitely generated 1–tilting modules in developing structure theory of Matlis localizations, that is, the localizations of commutative rings R at multiplicative sets S consisting of non–zero–divisors such that

6.3 Matlis localizations

243

proj dim S −1 R ≤ 1 (see Definition 1.5.2). As an application, we will then consider the existence of minimal versions of tilting approximations. Our approach is based on [15] and [381]. In order to characterize Matlis localizations in terms of decomposition theory, we will eventually apply the Hill Lemma. However, we will first need a number of preliminary definitions and results. We start with the ones related to the spectrum, spec R: Let R be a commutative ring, and S a multiplicative subset of R. We write V (S) = {P ∈ spec R | P ∩ S = ∅}. Recall that V (S) is canonically isomorphic to spec S −1 R. Clearly, if S ⊆ S  , then V (S  ) ⊆ V (S). Denote by Σ the multiplicative subset consisting of all non–zero–divisors in R. Let S be a multiplicative subset of R. If s and s are elements of R such that ss ∈ S, then s and s are invertible in the localization S −1 R. Hence S −1 R = (S  )−1 R, where S  = {t ∈ R|s = tt for some s ∈ S and t ∈ R} is a saturated multiplicative subset containing S. S  is called the saturation of S. Remark 6.3.1. If S is a multiplicative subset of R, and I is an ideal of R such that I ∩ S = ∅, then it is well–known that the set C = {J ≤ R | I ⊆ J and J ∩ S = ∅} has maximal elements and that any such maximal element is a prime ideal of R. Let now S ⊆ Σ be a saturated multiplicative subset. Then any x ∈ R \ S satisfies  xR ∩ S = ∅ and thus is contained in a prime ideal from V (S), so S = R \ P ∈V (S) P . Similarly, if S is a multiplicative subset of Σ, then its saturation is easily seen  to coincide with S  = R \ P ∈V (S) P . Observe that Σ is an example of a saturated multiplicative subset of R. Hence R \ Σ is a union of prime ideals of R, and it can be proved that the minimal primes of R are in this union. More generally, let M be a non–zero R–module. The set of elements in R such that multiplication by them induces an injective endomorphism of M is a saturated multiplicative subset of R. Dually, the set of elements in R such that multiplication by them is a surjective endomorphism of M is also a saturated multiplicative subset of R. We now present some results concerning the module S −1 R/R. In particular, we study its endomorphism ring and show that the direct sum decompositions of S −1 R/R have quite nice properties. Proposition 6.3.2. Let R be a commutative ring. Let S be a multiplicative subset of Σ. Then R is canonically a subring of S −1 R and the following holds true:

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(a) If f ∈ EndR (S −1 R/R) and x ∈ S −1 R/R, then f (x) ∈ xR. In particular EndR (S −1 R/R) is a commutative ring.   (b) If S −1 R/R = i∈I Ai and B ⊆ S −1 R/R, then B = i∈I (B ∩ Ai ). (c) If S −1 R/R = A ⊕ B = A ⊕ B  , then B = B  . Moreover, if A and A are direct summands in S −1 R/R, then also A ∩ A and A + A are direct summands in S −1 R/R.  (d) Assume S −1 R/R = i∈I Ai . If i = j, then HomR (Ai , Aj ) = 0.

Proof. (a) Let f ∈ EndR (S −1 R/R). Let x ∈ S −1 R/R, and let s ∈ S. Then sx = 0, if and only if x ∈ ( 1s + R)R. Hence f ( 1s + R) ∈ ( 1s + R)R for any s ∈ S, and then f (x) ∈ xR for any x ∈ S −1 R/R. (b) For any i ∈ I, let πi : S −1 R/R → Ai ⊆ S −1 R/R denote the projection onto Ai . By part (a), for any x ∈ S −1 R/R, πi (x) ∈ xR ∩ Ai . This shows that −1 xR = i∈I (xR ∩ Ai ) for any x ∈ S R/R. Then the same is true for any −1 submodule B of S R/R. (c) For the first statement, apply (b) to see that B  = (A ∩ B  ) ⊕ (B ∩ B  ) = B ∩ B  ⊆ B. By symmetry B = B  . For the second statement, let S −1 R/R = A ⊕ B = A ⊕ B  . Then (b) yields A + A = (A ∩ A ) ⊕ (A ∩ B  ) ⊕ (A ∩ B). So (A + A ) ∩ (B ∩ B  ) = 0, and (A + A ) ⊕ (B ∩ B  ) = S −1 R/R. (d) For any i ∈ I, let πi : S −1 R/R → Ai denote the canonical projection. Let i, j ∈ I, i = j, and let f ∈ HomR (Ai , Aj ). We can apply part (a) to f ◦ πi ∈ EndR (S −1 R/R) to deduce that f (x) ∈ xR ∩ Aj ⊆ Ai ∩ Aj = 0 for any x ∈ Ai . Hence f = 0. 2 We arrive at the important notion of a restriction due to Hamsher [264]: Definition 6.3.3. Let R be a commutative ring and M a module. A submodule N ⊆ M is a restriction of M , if for each prime (equivalently, maximal) ideal p of R, the localization N(p) of N at p satisfies N(p) = 0 or N(p) = M(p) . Remark 6.3.4. An example of restriction is a direct summand of a cyclic module, because a cyclic module over a local ring is indecomposable. Proposition 6.3.6 gives another example of restriction. Both of them will be needed in Theorem 6.3.10. Lemma 6.3.5. Let R be a commutative ring. Let N ⊆ M be R–modules such that N is a restriction of M . Let s ∈ Σ. If the multiplication by s is an onto endomorphism of M , then it is also an onto endomorphism of N . Equivalently, if M is {s}–divisible, then so is N .

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Proof. By the definition of a restriction, for any p ∈ spec R, multiplication by s is an onto endomorphism of N(p) . Thus it is an onto endomorphism of N . 2

Proposition 6.3.6. Let R be a commutative ring and S a multiplicative subset of Σ. (a) If R is local, then S −1 R/R is indecomposable. (b) Let A be an R–submodule of S −1 R such that R ⊆ A. Assume that A/R is a direct summand in S −1 R/R. Then A/R is a restriction of S −1 R/R. (c) Let A be an R–submodule of S −1 R such that R ⊆ A. Assume that A/R is a restriction of S −1 R/R. Then A is a subring of S −1 R.

Proof. (a) Since R is local, all cyclic modules are indecomposable (see Remark 6.3.4), hence S −1 R/R is indecomposable (see e.g. [321, Theorem 4.7]). (b) If A/R is a direct summand of S −1 R/R, then, for any p ∈ spec R, (A/R)(p) is a direct summand of (S −1 R/R)(p) . By part (a) either (A/R)(p) = 0 or (A/R)(p) = (S −1 R/R)(p) . Then A/R is a restriction of S −1 R/R, and (b) is proved. To prove (c), let X = {p ∈ spec R | (A/R)(p) = 0} be the support of A/R. For each p ∈ spec R, denote by αp : S −1 R → (S −1 R/R)(p) the canonical ring homomorphism. Assume that xs is an element of S −1 R such that αp ( xs ) = 0 for all p ∈ spec R \ X. Then, since (S −1 R/R)(p) = (A/R)(p) for all p ∈ X, we deduce that αp ( xs ) ∈ (A/R)(p) for all p ∈ spec R. This implies that for each p ∈ spec R there is an element t ∈ R \ p such that t xs ∈ A and therefore proves that xs ∈ A. Conversely, it is clear that αp (A) = (A/R)(p) = 0 for all p ∈ spec R \ X. We 2 thus conclude that A = p∈X / Ker αp , so A is a ring. We will next investigate direct sum decompositions of S −1 R/R under the assumption of proj dim S −1 R ≤ 1. Remark 6.3.7. Note that localizations of projective dimension at most one are rather frequent. For instance, if R is any commutative ring and S = {1 = s0 , s1 , . . . } is a countable multiplicative subset of Σ, then pd(S −1 R) ≤ 1. This 1 R | i < ω) of S −1 R = follows by Lemma 3.1.2 applied to the filtration ( s0 ...s i  1 i<ω s0 ...si R. The next proposition goes back to Hamsher:

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Proposition 6.3.8. Let R be a commutative ring. Let S be a multiplicative subset of Σ such that S −1 R has projective dimension  −1 at most  1. Let A be a non–zero R– −1 submodule of S R such that proj dim S R/A ≤ 1. Then A/sA is R/sR– projective for any s ∈ S. If, in addition, R is local and A is divisible by a non–unit in S, then A = S −1 R.

Proof. Let Q = S −1 R and s ∈ S. By assumption, proj dim Q ≤ 1 and proj dim Q/A ≤ 1, so proj dim A ≤ 1. Then it is well–known that the projective dimension of A/sA viewed as R/sR–module is also ≤ 1. Since Q/sA ∼ = Q/A, the exact sequence 0 → A/sA − → Q/sA − → Q/A → 0 yields that proj dim A/sA ≤ 1. Now, if the projective dimension of A/sA viewed as R/sR– module equals 1, then proj dim A/sA = 2, a contradiction (see e.g. [290]). So A/sA is R/sR–projective. For the second claim it suffices to show that A = tA for each t ∈ S. By the first part, A/tA is a projective R/tR–module, so, since R is local, A/tA is a free R/tA–module divisible by a non–unit in R/tR, hence A/tA = 0. 2 As proved in Proposition 6.3.6 the direct sum decompositions of S −1 R/R are parameterized by certain subsets of spec R. In the next lemma, we describe the support of S −1 R/R. (Recall that the support of a module M is defined as the set of all p ∈ mspec R such that M(p) = 0; the support is denoted by supp M .) Lemma 6.3.9. Let R be a commutative ring. Let S be a multiplicative subset of Σ, and p ∈ spec R. Then (S −1 R/R)(p) = 0, if and only if p ∈ V (S). In particular, supp (S −1 R/R) = {p ∈ mspec R | p ∩ S = ∅}.

Proof. If p ∈ V (S), then S ⊆ R \ p. Thus (S −1 R/R)(p) ∼ = R(p) /R(p) = 0. Let p be a prime ideal of R such that p ∈ V (S). Let s ∈ p ∩ S. Then 1s + R(p) is 2 a non–zero element of S −1 (R(p) )/R(p) = (S −1 R/R)(p) . The following theorem characterizes direct summands in S −1 R/R: Theorem 6.3.10. Let R be a commutative ring and S be a multiplicative subset of Σ such that proj dim S −1 R = 1. Let M1 = A1 /R be a submodule of S −1 R/R. Put X1 = supp S −1 R/R and X2 = supp M1 . Let  ϕ : S −1 R/R → (S −1 R/R)(p) p∈spec R\V (S)

 be the canonical inclusion. Put M2 = ϕ−1 ( p∈X2 (S −1 R/R)(p) ) = A2 /R. Then the following conditions are equivalent: (a) M1 is a direct summand of S −1 R/R.

6.3 Matlis localizations

247

(b) proj dim(S −1 R/A1 ) ≤ 1 and M1 is a restriction of S −1 R/R. (c) S −1 R/R = M1 ⊕ M2 .

Proof. Since proj dim(S −1 R) = 1, Proposition 6.3.6 (b) yields the implication (a)⇒(b), and clearly (c) implies (a). Assume (b). By the definition of M1 and M2 , it follows that M1 ∩ M2 = 0 and also that, for i = 1, 2, (Mi )(p) = 0 for each maximal ideal p ∈ Xi . We shall show that S −1 R = A1 + A2 . Take s ∈ S. It is enough to prove that 1s ∈ A1 + A2 . By Lemma 6.3.5, s induces an onto map on M1 = A1 /R, hence sA1 +R = A1 . This implies that A1 /sA1 is cyclic and that the map π1 : R/sR → A1 /sA1 defined by r + sR → r + sA1 , for r ∈ R, is surjective. By Proposition 6.3.8, A1 /sA1 is a projective R/sR–module, so π1 splits. Thus there exists a ∈ R such that a − a2 ∈ sR, A1 = aR + sA1 and 1 − a ∈ sA1 . So 1−a s ∈ A1 . We show that a a −1 R/R) , if and only ∈ A . This occurs if and only if ϕ( + R) ∈ (S 2 (p) p∈X2 s s if ((a + sR)R/sR)(p) = 0 for each p ∈ X1 ∪ (spec R \ X). As R/sR = (a + sR)R/sR ⊕ (1 − a + sR)R/sR, (a + sR)R/sR and (1 − a + sR)R/sR are restrictions of R/sR (cf. Remark 6.3.4). For any p ∈ spec R \ X, as s ∈ S, (R/sR)(p) = 0, hence also ((a + sR)R/sR)(p) = 0. Let p ∈ X1 . As (A1 )(p) = (S −1 R)(p) and s ∈ S, 0 = (A1 /sA1 )(p) ∼ = ((a + sR)R/sR)(p) . This finishes the proof of (c). 2 Proposition 6.3.11. Let R be a commutative ring. Let S1 ⊆ S be multiplicative subsets of Σ such that proj dim(S −1 R) = 1 and proj dim(S −1 R/S1−1 R) ≤ 1. Then S1−1 R/R is a restriction of S −1 R/R. More precisely, (a) If p ∈ V (S1 ), then (S1−1 R/R)(p) = 0. (b) If p ∈ spec R \ V (S1 ), then (S1−1 R/R)(p) = (S −1 R/R)(p) .

Proof. (a) follows from Lemma 6.3.9. If p ∈ spec R \ V (S1 ), then there exists a non–unit s ∈ S1 ∩ p ⊆ S. Then Proposition 6.3.8 yields     (S1−1 R/R)(p) = S1−1 R(p) /R(p) = S −1 (R(p) /R(p) = (S −1 R/R)(p) . 2 In view of Theorem 6.3.10 we immediately obtain Corollary 6.3.12. Let R be a commutative ring and S1 ⊆ S a multiplicative subset of Σ such that proj dim S −1 R = 1. Then proj dim(S −1 R/S1−1 R) ≤ 1, if and only if S1−1 R/R is a direct summand in S −1 R/R.

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Example 6.3.13. By Proposition 6.3.6, the complement of the module S1−1 R/R in the statement of Corollary 6.3.12 is of the form A/R for a subring A of S −1 R containing R. However, A is not a localization in general: consider the case when R is a Dedekind domain, S = R \ {0}, and R ⊆ A ⊆ Q = RS −1 is defined by A/R ∼ = E(R/p), where p is a maximal ideal contained in the union of all the maximal ideals q = p (such a p exists iff the class group of R contains torsion– free elements). of the form (S  )−1 R for a multiplicative Then A⊕R(p) /R = Q/R, but A is not   subset S ⊆ S: otherwise, since A = q =p R(q) , we have S ⊆ q =p (R \ q) =  R \ ( q =p q) = R∗ , where R∗ is the set of all units of R, so (S  )−1 R = R = A, a contradiction. We will also need Griffith’s notion of a G(ℵ0 )–family of submodules: Definition 6.3.14. Let R be a commutative ring and M a module. A family S of submodules of M is a G(ℵ0 )–family provided that (G1) 0, M ∈ S, (G2) S is closed under unions of chains, and (G3) if N ∈ S and X is a countable subset of M , then there exists N  ∈ S such that N ∪ X ⊆ N  and N  /N is countably generated. Lemma 6.3.15. Let R be a commutative ring, S a multiplicative subset of Σ and Q = S −1 R. Assume proj dim Q ≤ 1. Then the set S of all restrictions of Q/R is a G(ℵ0 )–family of submodules of Q/R.

Proof. Property (G1) is clear, and (G2) follows from Definition 6.3.3, because N(p) = N ⊗R R(p) , where R(p) is a flat R–module. Property (G3) is a consequence of the following claim: for each N ∈ S and each countable subset X ⊆ Q/R, there is a countable multiplicative subset S1 ⊆ S such that X ⊆ S1−1 R/R and proj dim Q/S1−1 R ≤ 1. Indeed, Proposition 6.3.11 then shows that S1−1 R/R is a restriction of Q/R, and so is N  = N + S1−1 R/R. Since N  /N is countably generated, (G3) follows. In order to prove the claim, we consider a projective resolution of Q ⊆

f

→F − → Q → 0, 0→K− where F = R(S) , f (1s ) = s−1 for each s ∈ S (where (1s | s ∈ S) is the canonical basis of F ).

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6.3 Matlis localizations

Notice that, if T is a countable multiplicative subset of S, then K ∩ R(T ) is generated by the countable set KT = {1t − t 1t.t | t, t ∈ T }: indeed, these elements clearly belong to K. If x ∈ K ∩ R(T ) , x = 1t0 r0 + · · · + 1tk rk , then we can w.l.o.g. assume that t0 = 1 and, for each m < k, tm+1 = tm .um+1 for some −1 um+1 ∈ T . Since f (x) = t−1 0 r0 + · · · + tk rk = 0, we have (u1 . . . uk )r0 + · · · + uk rk−1 + rk = 0. So x = 1t0 r0 + · · · + 1tk−1 rk−1 + 1tk rk − 1tk (u1 . . . uk r0 ) − · · · − 1tk uk rk−1 − 1tk rk = (1t0 − 1tk (u1 . . . uk ))r0 + · · · + (1tk−1 − 1tk uk )rk−1 , which shows that x is generated by the elements of KT .  Since K is projective, K = j∈J Kj where Ki is countably generated (see Corollary 4.2.12). By induction, we define an increasing sequence (Ti | i < ω) of countable subset multiplicative subsets of S as follows: T0 is any countable multiplicative  −1 of S such that X ⊆ RT0 /R. If Ti is defined, then K ∩ Ti ⊆  j∈Ai Kj for a countable set Ai ⊆ J. Let Bi be a countable subset of S such that j∈Ai Kj ⊆ R(Bi ) . Define Ti+1  as a countable multiplicative subset of S containing Ti ∪ Bi . Finally, put S1 = i<ω Ti . Then S1 is a countable multiplicative subset of S such that X ⊆ RS1−1 /R. Moreover, we have the following commutative diagram with exact rows and columns: 0 ⏐ ⏐ 

0 ⏐ ⏐ 

⊆

0 −−−−→ K ∩ R(S1 ) −−−−→ K −−−−→ K/(K ∩ R(S1 ) ) −−−−→ 0 ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ ⊆ ⊆  ⊆

0 −−−−→

R(S1 ) ⏐ ⏐ 

−−−−→ F −−−−→ ⏐ ⏐ 

0 −−−−→

RS1−1

R(S\S1 ) ⏐ ⏐ 

−−−−→ 0

−−−−→ Q −−−−→ ⏐ ⏐ 

Q/RS1−1 ⏐ ⏐ 

−−−−→ 0

0

0.

⊆

By construction, K ∩ R(S1 ) = the right hand column gives





Ai Kj is −1 proj dim Q/S1 R ≤ 1. j∈

i<ω

a direct summand in K, so 2

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Recall that given a commutative ring R and a set S ⊆ R consisting of non–zero divisors, then the class of all S–divisible modules, DS = {M ∈ Mod–R | M s = M for all s ∈ S}, satisfies DS = S ⊥ , where S = {R/sR | s ∈ S} ⊆ P1<ω . By Theorem 5.2.2, DS is a 1–tilting torsion class. Now we are in a position to prove the main result of this section: Theorem 6.3.16. (The structure of Matlis localizations) Let R be a commutative ring and S a multiplicative subset in R consisting of non–zero–divisors. Then the following conditions are equivalent: (a) S −1 R is a Matlis localization of R. (b) T = S −1 R ⊕ S −1 R/R is a 1–tilting R–module. (c) S −1 R/R decomposes into a direct sum of countably presented R–submodules. (d) R has a DS –envelope. Moreover, if T is 1–tilting then T generates the 1–tilting class DS .

Proof. Let Q = S −1 R. First we prove the equivalence of (a)–(c): Assume (a). We will verify conditions (T1)–(T3) for T , thus proving (b). First the projective dimension of Q, Q/R, and hence of T , is ≤ 1 by the assumption, so (T1) holds. (T3) holds, since there is the exact sequence 0 → R → Q → Q/R → 0. In order to prove (T2), in view of (T1), it suffices to show that Ext1R (Q/R, Q(κ) ) = 0 for each cardinal κ. However, Ext1R (Q, Q(κ) ) ∼ = Ext1Q (Q, Q(κ) ) = 0, since Q is a localization of R (see Example 1.1.11). So in order to prove that Ext1R (Q/R, Q(κ) ) = 0, it suffices to show that any f ∈ HomR (R, Q(κ) ) extends to some g ∈ HomR (Q, Q(κ) ) = HomQ (Q, Q(κ) ). But we can simply define g(q) = f (1)q for all q ∈ Q. Next, assume (b). Consider the cotorsion pair (A, B) generated by T . By Theorem 5.2.10, each module in A is A≤ω –filtered. In particular, this holds for Q/R ∈ A. Let F be a family corresponding to a A≤ω –filtration of Q/R by Theorem 4.2.6 (for κ = ℵ1 ). Let G = F ∩ S where S is the G(ℵ0 )–family of all restrictions of Q/R coming from Lemma 6.3.15. We claim that there is a filtration (Gα | α ≤ σ) of Q/R such that Gα ∈ G for all α ≤ σ and  Gα+1 /Gα is countably presented. Indeed, we can define G0 = 0 and Gα = β<α Gβ for limit ordinals α. Assume Gα ∈ G is defined and there is x ∈ (Q/R) \ Gα . Let F0 = S0 = Gα . By Theorem 4.2.6, there is F1 ∈ F such that F0 ∪ {x} ⊆ F1 and F1 /F0 is countably presented. Clearly S0 ⊆ F1 . Let C1 be a countable subset of F1 such that F0 + C1  = F1 . Since S is a G(ℵ0 )–family,

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251

there S1 ∈ S such that S0 ∪ C1 ⊆ S1 and S1 /S0 is countably generated. Then F1 ⊆ S1 . Let D1 be a countable subset of S1 such that S0 + D1  = S1 . Then there is F2 ∈ F such that F1 ∪ D1 ⊆ F2 and F2 /F1 is countably presented. Then S1 ⊆ F2 . Proceeding in this way, we obtain a chain Gα = F0 = S0 ⊆ F1 ⊆ S1 ⊆ F2 ⊆ · · · ⊆ Sn ⊆ Fn+1 ⊆ Sn+1 ⊆ . . .   We define Gα+1 = n<ω Fn = n<ω Sn . Then Gα+1 ∈ G, and since Fn+1 /Fn is countably presented for each n < ω, so is Gα+1 /Gα by Lemma 4.3.6. This proves the claim. Now each Gα is a restriction of Q/R = Gσ such that (Q/R)/Gα ∈ A, so (Q/R)/Gα has projective dimension ≤ 1. By Theorem 6.3.10, Gα is a direct summand in Q/R, and hence in Gα+1 , for each α < σ. This yields a decomposition of Q/R into a direct sum of copies of the countably presented modules Gα+1 /Gα (α < σ), so (c) holds.  Assume (c), so Q/R = i∈I Mi where Mi is a countably presented R–module for each i ∈ I. First we claim that for each i ∈ I, Mi is a direct summand of a module of the form Ti −1 R/R, where Ti is a countable multiplicative subset of S. To prove the claim set J0 = {i}. By induction, we can construct an ascending chain {Jn }n≥0 of countable subsets of I and an ascending chain−1{Sn }n≥0 of countable multiplicative subsets of S such that ⊆ Sn R/R ⊆ j∈Jn Mj    M for all n ≥ 0. Let J = J and T = j i n≥0 n n≥0 Sn . Then j∈Jn+1 −1 j∈J Mj = Ti R/R, and Ti is a countable multiplicative subset of S. This proves the claim. Now, since Ti is countable, Ti −1 R is a countably generated flat R–module, hence Ti −1 R is countably presented and has projective dimension ≤ 1 (see [181, VI.9]). Then we also have proj dim Ti −1 R/R ≤ 1, hence proj dim Mi ≤ 1 for each i ∈ I, and proj dim S −1 R/R ≤ 1, that is, proj dim S −1 R ≤ 1 and (a) holds. Notice that T = Q ⊕ Q/R satisfies Gen(T ) = Gen(Q) ⊆ DS , since Q ∈ DS and DS is a torsion class. If moreover T is 1–tilting, then Gen(T ) = T ⊥ (see Lemma 6.1.2), so by the reasoning above, Gen(T ) = (Q/R)⊥ = i∈I Mi⊥ ⊇ −1 ⊥ i∈I (Ti R/R) , where Ti (i ∈ I) are countable multiplicative subsets of S. However, if countable multiplicative subset of S, then  T = {tk | k−1< ω} is a −1 −1 −1 −1 ∼ . . . t )R and (t T −1 R = k<ω (t−1 0 0 . . . tk+1 )R/(t0 . . . tk )R = R/tk+1 R k −1 R/R)⊥ ⊇ D by Lemma 3.1.2. ∼ for each k < ω, and t−1 S 0 R/R = R/t0 R, so (T ⊥ This proves that (Q/R) ⊇ DS , and hence Gen(T ) = DS . Finally, we prove that (d) is equivalent to (a)–(c). Indeed, if (a)–(c) hold, then the embedding μ : R → Q is a special DS –preenvelope of R. Since Q

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is a localization of R, we have HomR (Q, Q) = HomQ (Q, Q), so the only R– endomorphism of Q fixing 1 is the identity. Thus, μ is left minimal, and μ is a DS –envelope of R. Conversely, assume (d) and let f : R → D be the S–divisible envelope of R. First we show that D is S–torsion–free. To this end, we take s ∈ S and show that the multiplication ψ : D → D by the element s is injective. We know that there is d ∈ D such that sd = f (1). Define an R–homomorphism g : R → D by g(1) = d. Since f is a DS –preenvelope, there is a map φ : D → D such that φf = g. So φψf (1) = s(φf (1)) = sg(1) = sd = f (1). By the left minimality of f we conclude that φψ is an isomorphism, hence ψ is injective. It follows that D is an S–torsion–free, S–divisible module, hence a Q–module. In particular D ∈ Gen(Q). Moreover, since f is an S–divisible preenvelope, each epimorphism R(λ) → X with X S–divisible factors through f (λ) : R(λ) → D(λ) . So D generates DS . This shows that DS is contained in, and therefore coincides with, Gen(Q). Let T  be a 1–tilting module generating the class DS . Since Gen(Q) = Gen(T  ), there exist cardinals κ and λ and R–epimorphisms f : (T  )(κ) → Q and g : Q(λ) → (T  )(κ) . Then f g : Q(λ) → Q is an R– epimorphism and hence a Q–epimorphism. So f g, and also f , splits, and the R–module Q is isomorphic to a direct summand in (T  )(κ) . Then Q has projective dimension ≤ 1, and (a) holds. 2

Remark 6.3.17. (i) Particular cases of Theorem 6.3.16 include the classical result of Kaplansky characterizing Matlis valuation domains as the valuation domains with Q countable (this follows by taking R a valuation domain and S = R \ {0}), the theorem of Lee [308] characterizing Matlis domains by the existence of a direct decomposition of Q/R into countably generated summands (take R a domain and S = R \ {0}), and the theorem of Fuchs and Salce [180] (= the particular case when R is a domain and S a multiplicative subset of R). (ii) By Theorem 6.1.4, the class of all S–divisible modules is a 1–tilting class for any commutative ring R. It is possible to generalize the notion of a Fuchs tilting module defined for domains in Example 5.1.2 to the setting of general commutative rings, so that the resulting 1–tilting module δS generates the class of all S–divisible modules (see [15, §5]). Of course, in the case of Matlis localizations, δS is equivalent to the module S −1 R ⊕ S −1 R/R by Theorem 6.3.16. Theorem 6.3.16 is helpful in answering the question of the existence of minimal versions of preenvelopes in particular cases. We start with its immediate corollary:

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253

Corollary 6.3.18. Let R be a domain. Then R has a DI-envelope, if and only if R is a Matlis domain. Notice that since DI is a tilting torsion class over any domain, tilting approximations need not have minimal versions (this contrasts with the dual setting of cotilting approximations, where the minimal versions always exist, see Theorem 8.1.7). By Theorem 4.1.12, (Pn , Pn⊥ ) is a complete cotorsion pair. It appears open to determine when the Pn⊥ –preenvelopes have minimal versions (= envelopes) for n ≥ 1. Our next corollary will give an answer for n = 1 in the particular case of Prüfer domains that are not Matlis. In this case, also fp–injective envelopes do not exist in general: Corollary 6.3.19. Assume R is a Prüfer domain. (a) The cotorsion pair generated by δ is (P1 , DI) and DI = FI. (b) If proj dim Q ≥ 2, then R does not have an FI–envelope.

Proof. (a) The first equality has already been considered in Example 5.1.2 (see also Lemma 6.2.5). Since any Prüfer domain is coherent, each finitely generated submodule of a finitely presented module is finitely presented. So, if F is finitely presented, then there exist n < ω and a chain of submodules 0 = F0 ⊂ F1 ⊂ · · · ⊂ Fn = F such that Fi+1 /Fi is cyclic and finitely presented for all i < n. So 

⊥ FI = R/I = DI I∈F

where F denotes the set of all finitely generated ideals of R. (b) By part (a) and Corollary 6.3.18.

2

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Open problems 1. Characterize all tilting modules and classes over Matlis domains. For the case when R is Prüfer, we refer to Theorems 6.2.15 and 6.2.19. 2. Let R be a commutative 1–Iwanaga–Gorenstein ring. Is each tilting module equivalent to the Bass tilting module TP constructed in Example 5.1.3? This is true, when R is a Dedekind domain by Theorem 6.2.22.

Chapter 7

Tilting approximations and the finitistic dimension conjectures

In this chapter we present applications of tilting approximations to computing finitistic dimensions of rings and algebras. The simple, but key fact is that the little finitistic dimension of a right noetherian ring is finite, if and only if there is a (possibly infinitely generated) tilting module Tf such that Tf⊥∞ = (P <ω )⊥ (see Theorem 7.1.10 below). The surprising phenomenon here is that even in the artin algebra case, we cannot in general take Tf finitely generated, so the infinite–dimensional tilting theory developed above comes up as a natural tool. Our first application concerns (non–commutative) Iwanaga–Gorenstein rings. In Theorem 7.1.12 we prove that, if R is n–Iwanaga–Gorenstein, then fin dim R = Fin dim R = n. In the second application (Theorem 7.2.4), for a right artinian ring R, we provide a formula for computing fin dim R involving only approximations of the (finitely many) simple modules. Our third application yields a simple proof of a result going back to Auslander– Reiten, Huisgen–Zimmermann and Smalø saying that fin dim R = Fin dim R < ∞ in case R is an artin algebra such that P <ω is contravariantly finite. This chapter is based on [14] and [18].

7.1 Finitistic dimension conjectures and the tilting module Tf In this section we will introduce the finitistic dimension conjectures going back to Bass [35], and then deal with their relations to tilting approximations. In particular, we will introduce the tilting module Tf and study its properties. An explicit computation of Tf for Iwanaga–Gorenstein rings will yield a proof of the Bass conjectures in that particular case (see Theorem 7.1.12).

256

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Tilting approximations and the finitistic dimension conjectures

Recall that the (right) global dimension of R, gl dimR, is the supremum of the projective dimensions of all (right R–) modules. Definition 7.1.1. Let R be a ring. Denote by Fin dim R the big finitistic dimension of R, that is, the supremum of the projective dimensions of arbitrary modules of finite projective dimension. Similarly, fin dim R will denote the little finitistic dimension of R, that is, the supremum of the projective dimensions of all finitely generated modules of finite projective dimension. Obviously, fin dim R ≤ Fin dim R ≤ gl dim R for any ring R. We recall a couple of simple and well–known facts: Lemma 7.1.2. Let R be a ring such that gl dim R < ∞. (a) fin dim R = Fin dim R = gl dim R = max{proj dim R/I | I ⊆ R}. (b) If R is right semiartinian, then all these dimensions are also equal to max{proj dim S | S ∈ simp R}.

Proof. This is an easy consequence of Lemma 3.1.2.

2

So the little and the big finitistic dimensions provide a refinement of the homological dimension theory in the case when gl dim R = ∞. The following example shows that such refinement is needed even in very simple cases: Example 7.1.3. Let R be a quasi–Frobenius (= 0–Iwanaga–Gorenstein) ring which is not completely reducible. (For example, let p be a prime integer, n > 1 and R = Zpn .) Since all projective modules are injective, there is no module of projective dimension 1, hence no module of projective dimension m for any m ≥ 1. By assumption, there is a non–projective simple module M , so proj dim M = ∞. It follows that fin dim R = Fin dim R = 0, while gl dim R = ∞. (Since R = Zpn is of finite representation type, it is certainly the finitistic dimension rather than the global one that reflects better the transparent structure of the module category Mod–Zpn .) Notice that, if R is a right ℵ0 –noetherian ring, then the possible difference between Fin dim R and fin dimR comes from (a representative set of) countably infinitely generated modules of finite projective dimension: Lemma 7.1.4. Assume that each right ideal of R is countably generated. Then Fin dim R = sup{proj dim M | M ∈ P ≤ω }.

7.1 Finitistic dimension conjectures and the tilting module Tf

Proof. This follows by Theorem 4.1.12.

257 2

Example 7.1.5. Let R be a commutative noetherian ring. Then the little and the big finitistic dimensions are known to be closely related to other dimensions of the ring. Bass, Gruson and Raynaud proved that Fin dim R coincides with the Krull dimension of R. Auslander and Buchsbaum proved that, if R is moreover local, then fin dim R = depth R, where depth R denotes the length of a maximal regular sequence in Rad R. So in the local case, both dimensions are finite, but they coincide, if and only if R is Cohen–Macaulay. Examples of commutative noetherian rings with Fin dim R = fin dim R = ∞ were constructed by Nagata: Let R = K[xi | i < ω] be the polynomial ring in countably many variables over a field K. Let (di | i < ω) ⊆ ω be a strictly increasing sequence of natural numbers # such that di+1 $− di > di − di−1 for all i > 0. For each i < ω, let Pi = xdi +1 , . . . , xdi+1 be theprime ideal in R generated by the variables xj (di < j ≤ di+1 ). Let U = R \ i<ω Pi and let S be the localization of R at U . Then S is noetherian, the Krull dimension of S is ∞ and fin dim S = Fin dim S = ∞. For more details we refer to [25, §11]. If R is an arbitrary ring, then the statements (I) Fin dim R = fin dim R, (II) fin dim R < ∞ are known as the first and the second finitistic dimension conjectures for R, respectively. In the case of artin algebras, (I) was disproved by Huisgen–Zimmermann: for each n ≥ 2, she constructed a finite–dimensional monomial algebra Λn such that fin dim R = n and Fin dim R = n + 1 (see [277]). Examples with arbitrarily big differences between the two dimensions were later constructed by Smalø: for each n ≥ 1 there is a finite–dimensional algebra Rn over a field such that fin dim R = 1 and Fin dim R = n (see Example 4.5.13). The second finitistic dimension conjecture has been proved for all finite–dimensional monomial algebras [255], all algebras with representation dimension ≤ 3 [283] et al., but it remains open for general artin algebras. Now we fix our notation for the rest of this section: Let R be a ring. Denote by Cf = (Af , Bf ) the cotorsion pair generated by the class P <ω . Recall that P <ω = P ∩ mod–R, and P <ω is just the class of all finitely presented modules of finite projective dimension in case R is right coherent. Since P <ω always has a representative set of elements, Cf is complete. So Af is a special precovering class, but — unlike P <ω — Af always contains

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infinitely generated modules. However, Af ⊆ lim P <ω by Theorem 4.5.6, so −→ Af ∩ mod–R = P <ω by Lemma 1.2.9. Remark 7.1.6. Let R be an artin algebra. Then there is a (Morita) duality between finitely generated left R–modules and finitely generated right R–modules which takes, for each n < ω, the class of all finitely generated left R–modules of injective dimension ≤ n to Pn<ω . So fin dim R = n, iff n is the maximum of injective dimensions of all finitely generated left R-modules of finite injective dimension. In particular, the finitistic dimension conjectures for artin algebras can be investigated either by means of Af –approximations of right R–modules, or, equivalently, by (⊥ I <ω )⊥ –approximations of left R–modules (see [74]). In the following, we will stick to the first option. Tilting theory relates to the finitistic dimension conjectures by means of the following simple observation: Lemma 7.1.7. Let n < ω. Let R be a right coherent ring and S be a syzygy closed class of finitely presented modules. Let (U, V) be the cotorsion pair generated by S. Then the following assertions are equivalent: (a) U ⊆ Pn . (b) There exists a tilting module T of projective dimension ≤ n such that V = T ⊥∞ .

Proof. Assume (a). By assumption, S ⊆ Pn<ω and S ⊥ = S ⊥∞ , so V is a class of finite type, hence n–tilting by Theorem 5.2.2. 2 If (b) holds, then U = ⊥ (T ⊥∞ ) ⊆ Pn . Varying the class S, we get a rich supply of (infinitely generated) tilting modules: Lemma 7.1.8. Let R be a right coherent ring and n < ω. Denote by (An , Bn ) the cotorsion pair generated by Pn<ω . Then there is a tilting module Tn of projective dimension at most n such that Bn = Tn⊥∞ . If R is right noetherian and fin dim R ≥ n, then Tn has projective dimension n.

Proof. The first assertion follows by Lemma 7.1.7 for S = Pn<ω . Assume that R is right noetherian with fin dim R ≥ n. So there is a finitely presented module M of projective dimension n. Assume that there is a tilting module T ∈ Pn−1 with Bn = T ⊥∞ . On the one hand, by Lemma 7.1.7 we then

7.1 Finitistic dimension conjectures and the tilting module Tf

have An ⊆ Pn−1 . On the other hand, Pn<ω ⊆ that M ∈ Pn−1 , a contradiction.

⊥ ((P <ω )⊥ ) n

259

= An . This implies 2

Example 7.1.9. Let R be a right coherent ring. For all 0 < i ≤ n < ω, denote by (Ani , Bni ) the cotorsion pair generated by the class Ωi−1 (Pn<ω ). By Theorem 3.2.1 and Lemma 3.1.4, this cotorsion pair is complete and Ani ⊆ Pn−i+1 . By dimension shifting, we have Bni = (Pn<ω )⊥i for all 0 < i ≤ n < ω. Note that Bni ⊆ Bnj for i ≤ j, Bni ⊆ Bmi for m ≤ n and Bni ⊆ Bn+k,i+k for each k < ω. Clearly An = An1 , Bn = Bn1 for all 1 ≤ n < ω, and Bf = n<ω Bn . Let 0 < i ≤ n < ω. Then Lemma 7.1.7 for S = Ωi−1 (Pn<ω ) yields a tilting ⊥∞ . module Tni of projective dimension at most n − i + 1 such that Bni = Tni In particular, the classes Bnn (n < ω) form an increasing chain of 1–tilting torsion classes. If R is right noetherian and fin dim R ≥ n, then, as in Lemma 7.1.8, we see that the projective dimension of Tni equals n − i + 1. It is easy to see that there is a single tilting module that controls the global dimension of the category of all finitely generated modules in the case when R is right noetherian and the latter dimension is finite: Theorem 7.1.10. (The tilting module Tf ) Assume R is a right noetherian ring. Then fin dim R < ∞, if and only if there is a tilting module Tf such that Bf = Tf⊥∞ . In this case fin dim R = proj dim Tf and Tf can be taken P <ω –filtered.

Proof. Assume that fin dim R = n < ∞. Then Bf = Bn . By Lemma 7.1.8 there is a tilting module Tf of projective dimension n such that Bf = Tf⊥∞ . The reverse implication follows by Lemma 7.1.7 for S = P <ω . 2 Finally, Tf can be taken P <ω –filtered by Theorem 5.2.20 (b). Clearly the module Tf is unique up to the equivalence of tilting modules. In Theorem 7.3.6, we will see that even for finite–dimensional algebras with little finitistic dimension = 1, it need not be possible to select Tf finitely generated. The proof of Theorem 5.1.14 provides an explicit construction of Tf : By a finite iteration of special (P <ω )⊥ –preenvelopes (of R etc.), we obtain a (P <ω )⊥ –coresolution of R. The tilting module Tf can simply be taken as the direct sum of the members of this finite coresolution (cf. Remark 5.2.21). The problem is that, in general, it is rather difficult to compute these special (P <ω )⊥ –preenvelopes, and hence determine T in this way. However, there are a number of cases when T can be computed explicitly. We start with a very simple one:

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Example 7.1.11. Let R be a right noetherian ring of finite global dimension. Let 0 → R → I 0 → I1 → . . . → In → 0 be the minimal injective coresolution of R. Then Tf = i≤n Ii is a tilting module with proj dim Tf = gl dimR = fin dimR such that Tf⊥∞ = (P <ω )⊥ . Indeed, in this case P <ω = mod–R, so (P <ω )⊥ – envelopes coincide with the injective envelopes. There is another, non–trivial case where the tilting module Tf can be taken as the direct sum of the terms of the minimal injective coresolution of R. This is the case of (non–commutative) Iwanaga–Gorenstein rings considered in Examples 4.1.14 and 5.1.6. The explicit knowledge of Tf here yields a proof of the first finitistic dimension conjecture: Theorem 7.1.12. (Finitistic dimension conjectures for Iwanaga–Gorenstein rings) Let n ≥ 0 and R be an n–Iwanaga–Gorenstein ring. Then (a) fin dim R = Fin dim R = n.  (b) Tf = i≤n Ii where 0 → R → I0 → I1 → . . . → In → 0 is the minimal injective coresolution of R. (c) Af = P = Pn = In = I, Bf = GI is the class of all I0 –resolved modules and Add(Tf ) = I0 .

Proof. First note that by  Example 4.1.14, P = Pn = In = I, so Fin dim R = n. By Example 5.1.6, T = i≤n Ii is an n–tilting module. Denote by (A, B) the tilting cotorsion pair induced by T . Since T is injective and R is right noetherian, clearly Add(T ) ⊆ I0 . We will prove that Add(T ) = I0 . Since Add(T ) = A ∩ B by Lemma 5.1.8 (c), it suffices to prove that I0 ⊆ A. By Proposition 5.1.9 (b), each module B ∈ B is Add(T )–resolved. Denote by B  the kernel of the n–th map in a fixed Add(T )– resolution of B. Since Add(T ) ⊆ I0 , dimension shifting gives Ext1R (I, B) ∼ =  ) for each I ∈ I . However, I ⊆ P , so the latter Ext–group is (I, B Extn+1 0 0 n R zero, proving that I ∈ A. Since Add(T ) = I0 , Proposition 5.1.9 yields that A = In (= P), and B is the class of all I0 –resolved modules. However, by Example 4.1.14, B = P ⊥ = GI is the class of all Gorenstein injective modules. Finally, since T is a tilting module, T is of finite type, so B = (A<ω )⊥ (see Theorem 5.2.20). However, A<ω = P <ω , so (A, B) = (Af , Bf ) and T = Tf . By Corollary 3.2.4, any module P ∈ P is a direct summand of a P <ω –filtered module, so we conclude that fin dim R = Fin dim R. 2

7.1 Finitistic dimension conjectures and the tilting module Tf

261

We have an immediate corollary (its part (a) is a kind of Baer Criterion of Gorenstein injectivity; part (b) is due to Enochs): Corollary 7.1.13. Let R be an Iwanaga–Gorenstein ring and N be a module. (a) N is Gorenstein injective, if and only if Ext1R (P, N ) = 0 for each module P which is finitely generated and has finite projective dimension. (b) If R is commutative, S is a multiplicative subset of R and N is Gorenstein injective, then the localized module S −1 N is Gorenstein injective both as S −1 R–module and as R–module.

Proof. (a) Since (P, GI) is a tilting cotorsion pair, GI is of finite type and the claim follows. (b) Consider an I0 –resolution of N . Its localization is an injective resolution of S −1 N in Mod–S −1 R witnessing that S −1 N is Gorenstein injective as S −1 R– module. However, by Example 1.1.11, the latter resolution is also an injective resolution of S −1 N in Mod–R, so S −1 N is a Gorenstein injective R–module. 2 If R is an arbitrary right noetherian ring with fin dim R < ∞, then Tf is neither injective nor pure–injective in general. The criteria obtained for arbitrary tilting modules in Propositions 5.3.4 and 5.3.5 apply here (recall that for each module M , we have the following implications: M product–complete ⇒ M Σ–pure–injective ⇒ M Σ–pure–split, cf. Lemma 1.2.23): Corollary 7.1.14. Let R be a right noetherian ring with fin dim R < ∞. Then (a) The tilting module Tf is Σ–pure–split, iff the class Af is closed under direct limits. (b) The tilting module Tf is product–complete, iff the class Af is closed under direct products, iff Af is a definable class. Let us notice a case when conditions (a) and (b) of Corollary 7.1.14 are equivalent: Corollary 7.1.15. Let R be a right noetherian ring with fin dim R < ∞. Assume that P <ω is covariantly finite. Then Af is definable, if and only if Af is closed under direct limits. In other words, Tf is product–complete, if and only if Tf is  –pure–split.

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Proof. Indeed, under the given assumptions, if Af is closed under direct limits, then Af = lim P <ω , so Af is closed under direct products by Lemma 4.5.1. 2 −→ Notice that P <ω is covariantly finite in case R is a right artinian ring such that P = lim P <ω (in particular, such that Fin dim R = 1) – see Lemmas 4.5.1 and −→ 4.5.14. Let R be a right noetherian ring with fin dim R < ∞. Then clearly, a sufficient condition for fin dim R = Fin dim R to hold is (III) Af = P. Notice that, if P is closed under direct limits, then (III) implies that Tf is Σ– pure–split by Corollary 7.1.14(a). All our proofs of the first finitistic dimension conjecture for a ring R proceed by proving (III) (see Theorems 7.1.12 and 7.3.8). However, (III) is not necessary for the first finitistic dimension conjecture to hold, even in the case of artin algebras. The relevant example goes back to Igusa, Smaløand Todorov, [282]: Example 7.1.16. Let k be an algebraically closed field and R the finite–dimensional monomial algebra given by the quiver γ 



α 



? 1  2 β

6 6

with the relations αγ = βγ = γα = 0. Then Fin dim R = fin dim R = 1, so lim P <ω = P by Corollary 4.5.12. −→ However, using the fact that R has a factor algebra isomorphic to the Kronecker algebra Λ, and the representation theory of Λ–modules developed by Ringel [340], one can show that Af = P, that is, Af is not closed under direct limits. Hence Tf (∈ P1 ) is an (infinite–dimensional) module which is not Σ–pure–split (for more details, we refer to [19]; an explicit computation of Tf appears in [379]). The fact that Tf is not finite–dimensional can also be seen using another important property of this example proved in [282], namely that P <ω is not contravariantly finite, and applying Theorem 7.3.6 below. Example 7.1.17. If R is a right artinian ring with fin dim R = 1 < Fin dim R (see Example 4.5.13), then Af  P1 , so again, proj dim Tf = 1, but Tf is not Σ–pure–split. Indeed, there is M ∈ P2≤ω \ P1 . Let N = Ω(M ) be countably generated. Then N ∈ P1 \ Af (otherwise, using Eilenberg’s trick 1.3.19 and the Hill Lemma 4.2.6, we can express N as a factor of a countably P1<ω –filtered module A by a countably P1<ω –filtered submodule B ⊆ A, and hence as a direct limit of modules in P1 , since P2<ω ⊆ P1 ).

7.2 A formula for the little finitistic dimension of right artinian rings

263

7.2 A formula for the little finitistic dimension of right artinian rings Before proving the finitistic dimension conjectures for all artin algebras with P <ω contravariantly finite, we will derive a formula for computing fin dim R for right artinian rings, following [386]. The Af –approximations of the simple modules will play a crucial role here. We start with a more general lemma: Lemma 7.2.1. Let R be a right coherent ring and F be a syzygy closed class consisting of finitely presented modules. Let (A, B) be the (complete) cotorsion pair generated by F. For each S ∈ simp R, take a special A–precover of S, fS : XS → S. Put C = {XS | S ∈ simp R}. Let M be a semiartinian module and (Mβ | β ≤ α) be a (transfinite) composition series of M . For each β < α, denote by σβ the inclusion of Mβ into Mβ+1 . Then there exists a continuous well–ordered direct system of short exact sequences fβ

0 → Yβ − → Xβ −→ Mβ → 0 (β ≤ α) with monomorphisms (μβ , νβ , σβ ) (mapping the β–th sequence to the (β + 1)–th one for each β < α) such that Xβ is C–filtered, Yβ ∈ B for each β ≤ α and Coker(νβ ) ∼ = XS for some S ∈ simp R for each β < α.

Proof. The proof is by induction on the composition length of M , l(M ) = α. For α = 0, we take Yα = Xα = 0. gα σα Mα+1 −→ Assume l(M ) = α + 1. We have the exact sequence 0 → Mα −→ S → 0 with S ∈ simp R, Mα+1 = M and l(Mα ) = α. Consider the pullback of fS and gα :

0 −−−−→ Mα −−−−→   

0 ⏐ ⏐ 

0 ⏐ ⏐ 

YS ⏐ ⏐ 

YS ⏐ ⏐ 

B ⏐ ⏐ hα 

−−−−→ XS −−−−→ 0 ⏐ ⏐ fS  gα

σ

0 −−−−→ Mα −−−α−→ Mα+1 −−−−→ ⏐ ⏐  0

S −−−−→ 0 ⏐ ⏐  0.

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By inductive premise, we have a direct system of exact sequences, 0 → Yβ − → fβ

Xβ −→ Mβ → 0 (β ≤ α) with monomorphisms (μβ , νβ , σβ ) (β < α). We will prove that Ext2R (XS , Yα ) = 0, that is, Ext1R (XS , Ω−1 (Yα )) = 0. Since XS ∈ A, Corollary 3.2.3 shows that it is sufficient to prove that Ext1R (F, Ω−1 (Yα )) = Ext2R (F, Yα ) = 0 for any F ∈ F. But Yα ∈ B, and Ω(F ) ∈ A by assumption, so Ext2R (F, Yα ) = Ext1R (Ω(F ), Yα ) = 0. It follows that the homomorphism Ext1R (XS , fα ) is surjective, so there is a commutative diagram with exact rows and columns 0 ⏐ ⏐ 

0 ⏐ ⏐ 

Yα ⏐ ⏐ 

Yα ⏐ ⏐  ν

0 −−−−→ Xα −−−α−→ Xα+1 −−−−→ ⏐ ⏐ ⏐ ⏐ fα  kα  0 −−−−→ Mα −−−−→ ⏐ ⏐ 

B ⏐ ⏐ 

XS −−−−→ 0   

−−−−→ XS −−−−→ 0

0 0. Combining the two diagrams above and using the 3 × 3 lemma from classical homological algebra (see [395, §1.3]), we get a commutative diagram with exact rows and columns: 0 0 0 ⏐ ⏐ ⏐ ⏐ ⏐ ⏐    μα

0 −−−−→ Yα −−−−→ Yα+1 −−−−→ ⏐ ⏐ ⏐ ⏐  

YS −−−−→ 0 ⏐ ⏐ 

ν

0 −−−−→ Xα −−−α−→ Xα+1 −−−−→ XS −−−−→ 0 ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ fS  fα  hα kα  gα

σ

0 −−−−→ Mα −−−α−→ Mα+1 −−−−→ ⏐ ⏐ ⏐ ⏐   0

0

S −−−−→ 0 ⏐ ⏐  0.

7.2 A formula for the little finitistic dimension of right artinian rings

265

fβ

Put fα+1 = hα kα . Then 0 → Yβ − → Xβ −→ Mβ → 0 (β ≤ α + 1) and (μβ , νβ , σβ ) (β ≤ α) induce a continuous direct system of exact sequences with the required properties. Assume l(M ) = α is a limit ordinal. By induction on β < α, we obtain a fβ

→ Xβ −→ continuous well–ordered direct system of exact sequences 0 → Yβ − → Mβ → 0 (β < α) with monomorphisms (μβ , νβ , σβ ) (β < α). Let 0 → Yα − fα

Xα −→ M → 0 be the direct limit of the system. Since all elements of F are finitely presented, B is closed under direct limits by Lemma 3.1.6. In particular, Yα = limβ<α Yβ ∈ B. For each β < α there is S ∈ simp R with Coker(νβ ) ∼ = −→ fβ

→ Xβ −→ Mβ → XS , so Xα = limβ<α Xβ is C–filtered. It follows that 0 → Yβ − −→ 0 (β ≤ α) and (μβ , νβ , σβ ) (β < α) induce a continuous direct system of exact sequences with the required properties. 2 For right semiartinian rings we obtain a more precise description of the special precovering class A: Corollary 7.2.2. Let R be a right semiartinian right coherent ring and F be a syzygy closed class of finitely presented modules. Let (A, B) be the complete cotorsion pair generated by F. For each S ∈ simp R, take a special A–precover of S, fS : XS → S. Put C = {XS | S ∈ simp R}. Then any module M has a special A–precover f : A → M such that A is C– filtered. The class A consists of all direct summands of C–filtered modules.

Proof. By assumption, M is semiartinian. Let (Mβ | β ≤ α) be a composition series of M . Then the map fα : Xα → M constructed in Lemma 7.2.1 is a special A–precover of M , and Xα is C–filtered. By Lemma 3.1.2, any (direct summand of a) C–filtered module is in A. Conversely, if M ∈ A, then the special A–precover of M constructed above splits. So M is a direct summand of a C–filtered module. 2 Recall that Cf = (Af , Bf ) denotes the cotorsion pair generated by P <ω . Also, for each n < ω, we will use the notation Cn = (An , Bn ) for the (complete) cotorsion pair generated by Pn<ω . The classes P and Pn (n < ω) are syzygy closed, and the same is true of the classes P <ω and Pn<ω (n < ω), provided that R is right coherent (cf. the remarks preceding Definition 2.2.8). We will show that special Af –precovers can be computed by an iteration of special An –precovers, as follows: Let M ∈ Mod–R. Take any special A0 –precover, π0 , of M , so there is an ν0 π0 A0 −→ M → 0 (for example, take A0 free and B0 the exact sequence 0 → B0 −→

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corresponding first syzygy module of M ). Assume a special An precover πn , of M νn πn is already constructed, so there is an exact sequence 0 → Bn −→ An −→ M →0 with Bn ∈ Bn . Let μn : Bn → Bn+1 be a special Bn+1 –preenvelope of Bn (see Theorem 3.2.1). Consider the pushout of the monomorphisms νn and μn : 0 −−−−→ Bn ⏐ ⏐ μn 

ν

−−−n−→ An ⏐ ⏐ σn  νn+1

π

−−−n−→ M −−−−→ 0    πn+1

0 −−−−→ Bn+1 −−−−→ An+1 −−−−→ M −−−−→ 0. Since Coker(μn ) ∼ = Coker(σn ) ∈ An+1 , also An+1 ∈ An+1 . So πn+1 is a special An+1 –precover of M . π

→ Af −−M → M → 0 the direct Lemma 7.2.3. Let R be a ring. Denote by 0 → Bf − νn πn M → 0 limit of the direct system of the exact sequences 0 → Bn −→ An −→ (n < ω), with the monomorphisms (μn , σn , idM ) constructed above. Then πM is a special Af –precover of M .

Proof. For each n < ω, An ∈ An ⊆ Af and Coker(σn ) ∈ An+1 ⊆ Af . So Af is Af –filtered. By Lemma 3.1.2, Af ∈ Af . Clearly Bf = n<ω Bn . Since Cn is generated by a set of FP2 –modules, Bn is closed under direct limits by Lemma B ∈ Bn for each n < ω. It follows that Bf ∈ Bf , so 3.1.6. So Bf ∼ = lim −→m≥n m πM is a special Af –precover of M . 2 Of course, if R is right noetherian and fin dim R = n < ∞, then in the conνn An − → M → 0 is already a special Af –precover of struction above 0 → Bn −→ M. If R is right noetherian and fin dim R =∞ (for example, if R is as in the Nagata examples 7.1.5), then taking M = n<ω Mn where, for each n < ω, Mn is finitely generated with proj dim Mn = n, we see that none of sequences νn An − → M → 0 splits while each special Af –precover does split. So 0 → Bn −→ the construction requires performing all of the ω steps. If R is right artinian, we can express fin dim R in terms of the projective dimensions of special Af –precovers of the finitely many simple modules: Theorem 7.2.4. (The little finitistic dimension of right artinian rings) Let R be a right artinian ring. For each S ∈ simp R, take a special Af –precover, fS : XS → S. Then fin dim R = max{proj dim XS | S ∈ simp R}. In particular, fin dim R < ∞, iff XS ∈ P for each S ∈ simp R.

7.3 Artinian rings with P <ω contravariantly finite

267

Proof. Since R is right artinian, simp R is finite, and the syzygy modules of any element of P <ω can be taken in P <ω . So Corollary 7.2.2 applies to our setting. Assume n = fin dim R < ∞. Then P <ω = Pn<ω , so Af = ⊥ ((Pn<ω )⊥ ). By Theorem 4.1.12, Pn = ⊥ (Pn⊥ ), so Af ⊆ Pn . This proves that fin dim R ≥ max{proj dim XS | S ∈ simp R}. Conversely, let n = max{proj dim XS | S ∈ simp R} < ∞. Put C = {XS | S ∈ simp R}, where XS → S is a special Af –precover of S. By Lemma 3.1.4, all C–filtered modules are in Pn . By Corollary 7.2.2, P <ω ⊆ Af ⊆ Pn . This proves 2 that fin dim R ≤ max{proj dim XS | S ∈ simp R}.

7.3 Artinian rings with P <ω contravariantly finite In this section we will employ infinite–dimensional tilting theory (i) in the proof of the second finitistic dimension conjecture for right artinian rings such that P <ω is contravariantly finite, and (ii) in the proof of the first finitistic dimension conjecture for artin algebras with P <ω contravariantly finite. The first result was proved in case of artin algebras by Auslander and Reiten [28], the second by Huisgen–Zimmermann and Smalø [281]. Our approach is different, based on [18] and [386]. An important class of artinian rings where P <ω is contravariantly finite was studied by Huisgen–Zimmermann:  Example 7.3.1. Let R be a right artinian right serial ring (that is, R = i<m ei R, where ei R is a right artinian right uniserial module and ei is a primitive idempotent for each i < m). Then P <ω is contravariantly finite. Moreover, fin dim R = 1 + max{proj dim ei J l | l < n, i < m, proj dim ei J l < ∞}. Here J = Rad(R) is the Jacobson radical of R and n the nilpotency index of J. For more details, we refer to [278]. The following is a criterion for contravariant finiteness of P <ω in terms of the Af –approximations of simple modules: Theorem 7.3.2. Let R be a right artinian ring. Then P <ω is contravariantly finite, iff we can choose XS ∈ mod–R for all S ∈ simp R.

Proof. Assume XS ∈ mod–R for all S ∈ simp R. By Corollary 7.2.2, each finitely generated module F has a special Af –precover XF → F such that XF is finitely C–filtered. Hence XF ∈ P <ω , and P <ω is contravariantly finite.

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Conversely, let gS : YS → S be a P <ω –precover of S in mod–R. By Corollary 2.1.10, we can w.l.o.g. assume that gS is a P <ω –cover. By a version of Lemma 2.1.13 in mod–R (cf. Remark 2.1.14), Ker(gS ) ∈ (P <ω )⊥ . So gS is a special Af –precover of S. 2 As a corollary we obtain a sufficient condition for finiteness of the little finitistic dimension of right artinian rings: Corollary 7.3.3. Let R be a right artinian ring. If P <ω is contravariantly finite, then fin dim R < ∞.

Proof. By Theorem 7.3.2, each simple module S has an Af –approximation XS → 2 S such that XS ∈ P <ω . By Theorem 7.2.4, fin dim R < ∞. The sufficient condition of Corollary 7.3.3 is not necessary even in the case when A is a finite–dimensional monomial algebra over an algebraically closed field: the IST–algebra defined in Example 7.1.16 above satisfies fin dim R = Fin dim R = 1, but P <ω is not contravariantly finite. Using tilting approximations, we will now prove that contravariant finiteness of P <ω for an artin algebra also implies the validity of the first finitistic dimension conjecture. First we will need a lemma making use of an idea of Auslander and Buchweiz [27]: Lemma 7.3.4. Let R be an artin algebra over a commutative artinian ring k, and T be a finitely generated tilting module of projective dimension n. Let C = T ⊥∞ ∩ mod–R. (a) Each module M ∈ mod–R has C–coresolution dimension ≤ n. (b) The class C is covariantly finite, and D = ⊥ C ∩ mod–R is contravariantly finite.

Proof. (a) The claim is trivial for n = 0, since in this case C = mod–R. For n > 0, consider the long exact sequence 0 → M → I0 → . . . → In−1 → N → 0, where all Ii (i < n) are finitely generated injective, hence Ii ∈ C. (T, M ) ∼ Since Extm+n = Extm R (T, N ) = 0 for all m ≥ 1, also N ∈ C. R (b) Let M ∈ mod–R. Let m < ω denote the C–coresolution dimension of M . → By induction on m, we prove that there are two short exact sequences 0 → V1 −

7.3 Artinian rings with P <ω contravariantly finite ϕ1

269

ϕ2

U1 −→ M → 0 and 0 → M −→ V2 − → U2 → 0 such that U1 and U2 have finite add(T )–coresolution dimension and V1 , V2 ∈ C. This is sufficient: since D is resolving in mod–R and add(T ) ⊆ D, we have U1 , U2 ∈ D. So ϕ1 (ϕ2 ) is a special D–precover (C–preenvelope) of the module M in mod–R. Assume m = 0, so M ∈ C. For the first short exact sequence, we take 0 → M − →M − → 0 → 0. For the second, we take an exact sequence 0 → C − →Q− → M → 0 with Q ∈ add(T ) and C ∈ C. The latter sequence exists, as the k–module HomR (T, M ) has a finite k–generating set S, so the canonical map f : T (S) → M has the property that any g ∈ HomR (T, M ) can be factorized through f . Since C ⊆ Gen(T ) by Lemma 5.1.8 (b), f is surjective, and we put C = Ker(f ) and Q = T (S) . This is possible since C ∈ T ⊥∞ : indeed, Ext1R (T, C) = 0, because HomR (T, f ) is surjective by construction and T is tilting; moreover, dimension shifting gives Exti+1 R (T, C) = ExtiR (T, M ) = 0 for each i ≥ 1, since M ∈ C. For the inductive step we split a C–coresolution of M of length m + 1 into two π parts: the exact sequence 0 → M − → C0 − → K → 0 and the long exact sequence of length m: 0 → K → C1 → . . . → Cm → 0. ρ

→ K → 0 with By inductive premise, there is an exact sequence 0 → C − →Q− Q of finite add(T )–coresolution dimension and C ∈ C. Consider the pullback of π and ρ:

σ

0 −−−−→ M −−−−→   

0 ⏐ ⏐ 

0 ⏐ ⏐ 

C ⏐ ⏐ 

C ⏐ ⏐ 

P −−−−→ Q −−−−→ 0 ⏐ ⏐ ⏐ ⏐ ρ  π

0 −−−−→ M −−−−→ C0 −−−−→ ⏐ ⏐  0

K −−−−→ 0 ⏐ ⏐  0.

In the middle row, the module Q is finitely add(T )–coresolved and P ∈ C, so the row provides for the second short exact sequence for M .

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Since P ∈ C, as above, we obtain an exact sequence 0 → V − → U − →P →0  with U ∈ add(T ) and V ∈ C. Consider the pullback of σ and τ : 0 ⏐ ⏐ 

0 ⏐ ⏐ 

V ⏐ ⏐ 

V ⏐ ⏐ 

0 −−−−→ U −−−−→ U  −−−−→ ⏐ ⏐ ⏐ ⏐ τ  σ

0 −−−−→ M −−−−→ ⏐ ⏐ 

Q −−−−→ 0   

P −−−−→ Q −−−−→ 0 ⏐ ⏐ 

0 0. The middle row shows that U has a finite add(T )–coresolution. So the left column provides for the first short exact sequence required for M . 2 The following lemma gives the connection between contravariant finiteness and finite number of generators of the tilting module: Lemma 7.3.5. Let R be an artin algebra. Let S be a syzygy closed subclass of P <ω . Denote by (U, V) the cotorsion pair generated by S. Then the following assertions are equivalent: (a) U <ω is contravariantly finite. (b) There exists a finitely generated tilting module T such that V = T ⊥∞ .

Proof. (a) implies (b): first U <ω ⊆ A<ω = P <ω . Let gS : US → S denote f <ω a U –precover of a simple module S ∈ simp R. By Corollary 2.1.10, we can w.l.o.g. assume that gS is a U <ω –cover. By a version of Lemma 2.1.13 in mod–R, Ker(gS ) ∈ (U <ω )⊥ = V. So gS is a U–cover of S. Let n = max{proj dim US | S ∈ simp R}. Since simp R is a finite set, we have n < ∞ and S ⊆ U ∩ mod–R ⊆ Pn<ω . So U ⊆ Pn , and Lemma 7.1.7 ∞ . Moreover, by Remark provides an n–tilting module T  such that V = (T  )⊥  5.1.15, T is equivalent to the tilting module T = i≤n Ti , where T0 is any special V–preenvelope of R, T1 , any special V–preenvelope of T0 /R etc. By Corollary 7.2.2, each finitely generated module X has a special U–precover gX : UX → X such that UX is finitely {US | S ∈ simp R}–filtered. Using

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271

Remark 2.2.7, we infer that each finitely generated module X has a special V– preenvelope fX : X → VX with VX ∈ mod–R. It follows that all Ti (i ≤ n), and hence T , can be taken finitely generated. (b) implies (a): by (the proof of) Lemma 7.3.4, for any module Y ∈ mod–R there is an exact sequence 0 → V − →U − → Y → 0 such that V ∈ V <ω and U has a finite add(T )–coresolution. Hence U ∈ U <ω , and U <ω is contravariantly finite. 2 Now we can characterize the artin algebras with P <ω contravariantly finite: Theorem 7.3.6. Let R be an artin algebra. The following assertions are equivalent: (a) P <ω is contravariantly finite. (b) There is a finitely generated tilting module Tf such that Bf = Tf⊥∞ .

Proof. By Lemma 1.2.9, P <ω = Af ∩ mod–R. So the assertion follows from 2 Lemma 7.3.5 for S = P <ω . The next lemma is crucial: Lemma 7.3.7. Let R be a ring, T a tilting module and (A, B) the tilting cotorsion pair induced by T . Then the following conditions (a) and (b) are equivalent: (a) Add(T ) is closed under cokernels of monomorphisms. (b) A = P. These conditions imply Fin dim R < ∞. If T ∈ mod–R, then these conditions also imply (c) A = Af . If T ∈ mod–R is Σ–pure–injective, then condition (c) implies (a), so all the three conditions are equivalent. In this case fin dim R = Fin dim R < ∞.

Proof. Assume (a). On the one hand, by Lemma 5.1.8 (b), A ⊆ Pn ⊆ P, where n is the projective dimension of T . On the other hand, if M ∈ P, then the completeness of (A, B) yields an exact sequence 0 → M − →B− → A → 0 with B ∈ B and A ∈ A. Clearly B ∈ P. Let m = proj dim B. By Lemma 5.1.8 (d), there is a long exact sequence 0 → Mm → · · · → M0 → B → 0

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with Mi ∈ Add(T ) for all i ≤ m. Using the assumption and induction, we obtain that B ∈ Add(T ) ⊆ A. Since A is resolving, we get M ∈ A. This proves that P ⊆ A, and hence P = A. Assume (b). Then Add T = P ∩ B by Lemma 5.1.8 (c). So Add T is closed under cokernels of monomorphisms, since P and B share this property. Moreover, (b) implies that P is closed under direct sums, hence there is n < ω with P = Pn , so Fin dim R < ∞. Assume T ∈ mod–R. We will show that (b) implies (c): indeed, we have Af ⊆ A = P = Pn . By assumption, T ∈ Af , so A ⊆ Af . Now assume that T ∈ mod–R is Σ–pure–injective. We will prove that (c) implies (a): First we show that every monomorphism in add(T ) splits. Indeed, let 0 → f

→ B − → C → 0 be a short exact sequence with A and B in add(T ). By A − assumption, add(T ) = mod–R ∩ Add(T ) ⊆ Bf ∩ P <ω . Since P <ω is closed under cokernels of monomorphisms, we have A ∈ Bf and C ∈ P <ω ⊆ Af . Thus Ext1R (C, A) = 0 and f splits. Now let A ⊆ B and A, B ∈ Add(T ). By assumption, each element of Add(T ) is isomorphic to  a direct sum of finitely generated direct summands of T (cf. Remark 5.3.6). Let j∈J xj rij = ai (i ∈ I) be a finite system of R–linear equations with ai ∈ A (i ∈ I) which is solvable in B, by xj = bj (j ∈ J). There is a finitely generated direct summand A ⊆ A such that ai ∈ A for all i ∈ I, and a finitely generated direct summand B  ⊆ B such that A ⊆ B  and bj ∈ B  for all j ∈ J. By the previous paragraph, the embedding A ⊆ B  is pure (even split), so the finite system is also solvable by some xj = aj ∈ A (j ∈ J). This proves that any monomorphism in Add(T ) is pure. Since T is Σ–pure– injective, each monomorphism in Add(T ) splits, and (a) holds. Finally, (b) and (c) give Af = P, so each module of finite projective dimension is a direct summand of a P <ω –filtered module. It follows that fin dimR = Fin dim R. 2 Now we can prove the main result of this section: Theorem 7.3.8. (Finitistic dimension conjectures for artin algebras with P <ω contravariantly finite.) Assume that R is an artin algebra such that P <ω is contravariantly finite. Then every module of finite projective dimension is a direct summand of a P <ω –filtered module, and Fin dim R = fin dim R < ∞.

Proof. By Theorem 7.3.6, there is a finitely generated tilting module Tf such that Bf = Tf⊥∞ . Clearly Tf is Σ–pure–injective and condition (c) of Lemma 7.3.7

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273

holds for T = Tf . Condition (b) then gives Af = P, and the final claim follows again by Lemma 7.3.7. 2

Open problems Let R be a right noetherian ring. 1. What are the characteristic module–theoretic properties of the tilting module Tf whose existence is equivalent to fin dim R < ∞ (see Theorem 7.1.10)? 2. Which additional properties of Tf characterize the case of Fin dim R = fin dim R? By Theorem 7.3.6 and Theorem 7.3.8, if R is an artin algebra, then a sufficient condition for Fin dim R = fin dim R to hold is that Tf is finitely generated. However, this condition is not necessary – the first finitistic dimension conjecture may hold even though Tf is not Σ–pure–split (see Example 7.1.16).

Chapter 8

Cotilting modules

We turn to the dual setting of cotilting left R–modules. By Theorem 8.1.2 below, the duals T d of tilting (right R–) modules T are always cotilting left R–modules. Conversely, any cotilting left R–module C is pure–injective by Theorem 8.1.7, that is, C is a direct summand in a dual module. However, there exist rings with (pure–injective, 1–) cotilting left R–modules which are not equivalent to duals of any tilting (right R–) modules (see Example 8.2.13). So the formal duality of the notions of a tilting and a cotilting module does not always imply the existence of an explicit duality. In other words, in order to develop a general theory of cotilting modules, we have to employ category– theoretic and homological methods for dualizing the results of Chapter 5. An application of an explicit module duality is not sufficient for achieving this goal. Fortunately a large number of cotilting left R–modules are of cofinite type (that is, equivalent to T d for a tilting module T ). Then the corresponding cotilting classes of left R–modules can be characterized by the resolving subcategories of mod–R, see Theorem 8.1.14. We will use this fact to characterize all cotilting modules over particular rings, e.g., over Dedekind domains, up to equivalence (Theorem 8.2.9).

8.1

Cotilting classes and the classes of cofinite type

We start with the definition of a cotilting module: Definition 8.1.1. A left R–module C is cotilting provided that (C1) C has finite injective dimension. (C2) ExtiR (C κ , C) = 0 for all 1 ≤ i < ω and all cardinals κ. (C3) There is r ≥ 0 and a long exact sequence 0 → Cr → . . . → C1 → C0 → W → 0, where Ci ∈ Prod(C) for all i ≤ r and W is an injective cogenerator for R–Mod.

8.1 Cotilting classes and the classes of cofinite type

275

If n < ω and C is a cotilting left R–module of injective dimension ≤ n, then C is called n–cotilting. The class ⊥∞ C (⊆ R–Mod) is the n–cotilting class induced by C. Clearly (⊥∞ C, (⊥∞ C)⊥ ) is a hereditary cotorsion pair in R–Mod, called the n–cotilting cotorsion pair induced by C. If C and C  are cotilting left R–modules, then C  is equivalent to C provided that the induced cotilting classes coincide, that is, ⊥∞ C = ⊥∞ C  . Clearly 0–cotilting modules coincide with injective cogenerators for R–Mod. Since each tilting module is of finite type, the duality (−)d yields an easy way of producing n–cotilting left R–modules from n–tilting (right R–) modules. (Recall that given a right R–module M , the dual left R–module, M d , is defined by M d = HomS (M, E), where E is an injective cogenerator for S–Mod and R is an S–algebra.) Theorem 8.1.2. Let R be a ring, n ≥ 0 and T be an n–tilting module. Then the dual module T d is an n–cotilting left R–module with the induced n–cotilting class   C = T ∞ = M ∈ R–Mod | TorR i (T, M ) = 0 for all i ≤ n . Moreover, if X ⊆ Pn<ω is such that T ⊥∞ = X ⊥∞ , then C = X ∞ .

Proof. Clearly, if M ∈ Mod–R is projective, then M d is an injective left R– module (see Lemma 1.2.11 (b)). Similarly, if M is a generator for Mod–R, then M d is a cogenerator for R–Mod, and M ∈ Add(T ) implies M d ∈ Prod(T d ). This proves conditions (C1) and (C3) for C = T d . Let κ be a cardinal. By Lemma 1.2.11 (b), for each i ≥ 1, ExtiR (C κ , C) = 0, κ ⊥ ⊥∞ ))<ω (⊆ P ). By Theorem 5.2.20, T is of iff TorR n i (T, C ) = 0. Let S = ( (T finite type, so S ⊥ = T ⊥∞ . Let U = T ⊕ Ω1 (T ) ⊕ · · · ⊕ Ωn−1 (T ), the direct sum of the syzygies of T . Then T ∈ ⊥ (U ⊥ ) = ⊥ (S ⊥ ) ⊆ lim S by Corollary 4.5.8. −→ Since the Tor–functor commutes with lim, in order to prove condition (C2) −→ κ it suffices to show that TorR i (S, C ) = 0 (for each i ≥ 1 and each cardinal κ). However, S ⊆ mod–R and C κ = (T (κ) )d , so the latter is equivalent to ExtiR (S, T (κ) ) = 0 by Lemma 1.2.11 (d). This holds since S ⊆ ⊥ (T ⊥∞ ) and T (κ) ∈ T ⊥∞ by condition (T2) defining the tilting module T . Now Lemma 1.2.11 (b) yields ⊥∞ C = T ∞ , which gives the first claim since T has projective dimension ≤ n. For the final claim, let Q be a set containing R and a representative set of elements of X and their syzygies. Then Q⊥ = X ⊥∞ = U ⊥ and Q = X ∞ . By Corollary 3.2.3, Q consists of direct summands of {U, R}–filtered modules, and vice versa, U is a direct summand of a Q–filtered module. By Corollary 3.1.3, Q = U  = T ∞ , so C = X ∞ . 2

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We pause to consider an example: Example 8.1.3. (Bass cotilting modules) Let R be a commutative 1–Iwanaga– Gorenstein ring  (see Example 5.1.3). For i = 0, 1 let Pi = {p ∈ specR | ht(p) = i}, and let Q = p∈P0 E(R/p). As in Example 5.1.3, R has a minimal injective coresolution of the form



π E(R/q) → E(R/p) → 0, 0→R→ q∈P0

p∈P1

and for each subset P ⊆ P1 there is a Bass (1–) tilting module TP = RP ⊕  −1 ( E(R/p), where R = π P p∈P p∈P E(R/p)), and TP generates the tilting 1 ⊥ all p ∈ P }. class TP = {M | ExtR (E(R/p), M ) = 0 for Consider the injective cogenerator E = p∈mspec R E(R/p). By Theorem d 8.1.2, CP = (TP ) = HomR (T P , E) is a (1–) cotilting   module, called the Bass d ∼ cotilting module. Notice that ( p∈P E(R/p)) = p∈P Jp , the product of the p–adic modules  over p ∈ P . Since Q is a flat module and the sequence 0 → RP − →Q− → q∈P1 \P E(R/q) → 0 is exact and its last term is injective, hence of flat dimension ≤ 1, we infer that RP is flat, and hence (RP )d is injective. So the corresponding cotilting class is      Jp = 0 CP = ⊥ CP = M ∈ Mod–R | Ext1R M, 

= M ∈ Mod–R |

p∈P

TorR 1 (E(R/p), M )

 = 0 for all p ∈ P .

There is a dual version of Lemma 5.1.8 for cotilting modules: Lemma 8.1.4. Let R be a ring, n ≥ 0 and C be an n–cotilting left R–module. Denote by C = (A, B) the n–cotilting cotorsion pair induced by C. (a) A ⊆ Cogen(C) and B ⊆ In . (b) Each of the short exact sequences forming the long exact sequence in (C3) is given by a special A–precover of an element of B. The length r in (C3) can be taken ≤ n. (c) The kernel of C equals Prod(C). (d) Each M ∈ A ∩ In has Prod(T )–coresolution dimension ≤ n.

Proof. (a) Since C ∈ In and (⊥ In , In ) is a cotorsion pair (see Theorem 4.1.7), we have B ⊆ In .

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277

The rest of the proof is dual to the one for Lemma 5.1.8 (with the injective cogenerator W in R–Mod replacing the projective generator R). 2 By Lemma 8.1.4 (c), the kernel of the cotorsion pair C equals Prod(C). The proof of Proposition 5.1.9 can be dualized to prove Proposition 8.1.5. Let R be a ring, n ≥ 0, C be an n–cotilting left R–module and C = (A, B) the n–cotilting cotorsion pair induced by C. (a) A coincides with the class of all Prod(C)–coresolved modules. In particular, A is closed under direct products. (b) B coincides with the class of all Prod(C)–resolved modules of Prod(C)– resolution dimension ≤ n. Remark 8.1.6. As in Corollary 5.1.10, we can replace Prod(C)–(co)resolutions in Proposition 8.1.5 by Q–(co)resolutions, where Q = {C κ | κ ≥ 0}; n– cotilting left R–modules can then be characterized by the dual property of ⊥∞ C = Cogn (C). Moreover, dually to Lemma 5.1.12, we obtain that if C1 and C2 are cotilting left R–modules, then C1 is equivalent to C2 , iff Prod(C1 ) = Prod(C2 ), iff Prod(C1 ) ⊆ Prod(C2 ). The crucial property of cotilting modules making a smooth dualization of the theory of tilting modules possible is their pure–injectivity. It was proved in many steps: first for abelian groups [244], and then for modules over Dedekind and Prüfer domains [150], [42], and for 1–cotilting modules over arbitrary rings [37]. The general case of n–cotilting modules was first settled for countable rings in [42]; the case of arbitrary rings is due to Št’ovíˇcek, [378]. Theorem 8.1.7. (Pure–injectivity of cotilting modules) Let R be a ring, n ≥ 0 and C be an n–cotilting left R–module. Denote by C = (A, B) the n–cotilting cotorsion pair induced by C. Then C is pure–injective, A is definable, and C is a perfect cotorsion pair.

Proof. By Lemma 8.1.4 (a), B ⊆ In . By Proposition 8.1.5 (a), A is closed under direct products. Since C is a hereditary cotorsion pair, Theorem 4.3.23 applies and yields that A is definable and C is perfect. In order to prove that C is pure–injective, it suffices to show that for each cardinal κ, Ext1R (C κ /C (κ) , C) = 0. Then the summation map Σ : C (κ) → C extends to C κ , that is, part (e) of Theorem 1.2.19 is satisfied. Clearly the module C (κ) is a direct limit of the direct system consisting of the modules C (F ) , where F runs over all finite subsets of κ. Then C κ /C (κ) =

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limF C κ /C (F ) . Since C κ /C (F ) ∼ = C κ\F ∈ A by condition (C2), and A is closed −→ 2 under direct limits, we conclude that C κ /C (κ) ∈ A. Remark 8.1.8. A bimodule R CS is an n–cotilting bimodule provided that C ∈ R–Mod and C ∈ Mod–S are n–cotilting modules and R CS is faithfully balanced. For example, Morita bimodules (i.e., faithfully balanced bimodules which are injective cogenerators on either side) are exactly the 0–cotilting bimodules. Since cotilting modules are pure–injective, their endomorphism rings S have rather strong properties: S/Rad(S) is von Neumann regular, right self–injective, and idempotents lift modulo Rad(S). This follows from the analogous well– known properties of endomorphism rings of injective objects in Grothendieck categories (cf. the proof of Proposition 2.1.11). These strong properties explain why cotilting bimodules occur rarely as compared to tilting bimodules. If R CS is a Morita bimodule, then C induces a Morita duality between R–mod and mod–S, [8, §24]. Similarly, each 1–cotilting bimodule induces a “generalized Morita duality”. We refer to [86] and [89] for the role of cotilting bimodules in the duality theory for module categories. Now we can characterize n–cotilting classes and n–cotilting cotorsion pairs by their closure properties: Theorem 8.1.9. Let R be a ring, n < ω and C be a class of left R–modules. Then the following assertions are equivalent: (a) C is n–cotilting. (b) C is resolving, covering, closed under direct products and direct summands, and C ⊥ ⊆ In .

Proof. (a) implies (b): this is an immediate consequence of Lemma 8.1.4 (a) and Theorem 8.1.7. (b) implies (a): this is proved dually to the implication (b) implies (a) in Theorem 5.1.14. 2 Corollary 8.1.10. Let n < ω. Let R be a ring and C = (A, B) be a cotorsion pair in R–Mod. Then the following assertions are equivalent: (a) C is an n–cotilting cotorsion pair. (b) C is a hereditary (and perfect) cotorsion pair such that B ⊆ In and A is closed under direct products.

8.1 Cotilting classes and the classes of cofinite type

279

Proof. (a) implies (b): by Theorems 8.1.7 and 8.1.9 for C = A. (b) implies (a): in view of Theorem 8.1.9, we only have to prove that perfectness of C follows from the other assumptions on C. However, this holds by Theorem 4.3.23 (b). 2 By Theorem 8.1.2, the dual module of any tilting module is cotilting. Similarly as all tilting modules and classes are of finite type, the duals of tilting modules and classes are exactly the cotilting modules and classes of cofinite type, defined as follows: Definition 8.1.11. Let R be a ring. Let C be a class of left R–modules. Then C is of cofinite type provided there exist n < ω and S ⊆ Pn<ω such that C = S ∞ . Let C be a left R–module. Then C is of cofinite type provided that the class ⊥∞ C is of cofinite type. Let C be a class of cofinite type and A = C (= ∞ C). Then clearly C = (A<ω )∞ , so S = A<ω is the largest possible choice for S in the Definition 8.1.11. Any class of cofinite type is a cotilting class: Proposition 8.1.12. Let R be a ring and C be a class of left R–modules of cofinite type. Then C is cotilting (and definable).

Proof. By assumption, there are n < ω and S ⊆ Pn<ω such that C = S ∞ . Let S d = {S d | S ∈ S}. Then C = ⊥∞ (S d ) by Lemma 1.2.11 (b), so C is resolving, and it is a covering class by Corollary 3.2.12. Since S ⊆ mod–R, the functor TorR 1 (S, −) commutes with direct products, whenever S ∈ S or S ∈ mod–R is a syzygy of a module in S (see e.g. [155, §3.2]). So C is closed under direct products. Since S d ⊆ In , also C ⊥ ⊆ In . By Corollary 8.1.9, we infer that C is an n–cotilting class. Finally, each cotilting class is definable by Theorem 8.1.7. 2 Notice that if C is of cofinite type, then, by Lemma 1.2.11, the least n such that the class C = ⊥∞ (S d ) is n–cotilting coincides with the least n such that S ⊆ Fn . However, S ⊆ mod–R, and finitely presented flat modules are projective, so this is exactly the least n such that S ⊆ Pn<ω . The essential difference between the tilting and cotilting setting is that the converse of Proposition 8.1.12 does not hold in general. In Example 8.2.13 below, we will construct a 1–cotilting class over a valuation domain which is not of cofinite type. So in general there exist more cotilting modules than just duals of the tilting ones. However, for many important classes of rings, all cotilting modules are of cofinite type, so this surprising phenomenon does not occur.

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Cotilting modules

Theorem 8.1.13. Let R be a ring and n < ω. (a) Let C be an n–cotilting left R–module. Then C is of cofinite type, if and only if there is an n–tilting module TC such that C is equivalent to TCd . (b) If C and C  are n–cotilting modules of cofinite type, then C  is equivalent to C, if and only if the n–tilting modules TC and TC  are equivalent. In particular, TC is uniquely determined by C up to equivalence of tilting modules.

Proof. (a) Assume that C is of cofinite type. By the remark after Definition 8.1.11, there is S ⊆ Pn<ω such that ⊥∞ C = S ∞ and we can w.l.o.g. assume that S = (S ∞ ) ∩ mod–R (so in particular, S a resolving subcategory of mod–R). Since the class S ⊥ is of finite type (see Definition 5.2.1), there is an (n–) tilting module TC such that TC⊥∞ = S ⊥ . By Theorem 8.1.2, TCd is an n–cotilting left R–module inducing the cotilting class TC∞ = S ∞ = ⊥∞ C, so TCd is equivalent to C. Conversely, assume that C is equivalent to T d for an n–tilting module T . Since T is of finite type, there is S ⊆ Pn<ω such that T ⊥∞ = S ⊥∞ , so ⊥∞ C = TC∞ = S ∞ by Theorem 8.1.2, hence C is of cofinite type. (b) Assume that T and T  are tilting modules such that C = T d is equivalent to C  = (T  )d . Then (T  )∞ = T ∞ by Lemma 1.2.11. Let (A, B) and (A , B ) be the tilting cotorsion pairs induced by T and T  . Let S = A<ω and S  = (A )<ω . By Theorem 4.5.6, S = (lim S)<ω = (S ) ∩ mod–R = (T ∞ ) ∩ mod–R, and −→ similarly for S  and T  , so S = S  . By Theorem 5.2.20, S ⊥ = B, and similarly (S  )⊥ = B so B = B  , that is, T and T  are equivalent tilting modules. Conversely, if TC and TC  are equivalent, then TC  ∈ Add(TC ), hence TCd  ∈ 2 Prod(TCd ), so TCd  is equivalent to TCd (see Remark 8.1.6). In particular, if U = {Ui | i ∈ I} is a representative set of all tilting modules up to equivalence, then U d = {Uid | i ∈ I} is a representative set of all cotilting modules of cofinite type up to equivalence. The classes of finite type correspond to resolving subcategories in mod–R. In view of Theorem 8.1.13, it is not surprising that there is a similar correspondence for the classes of cofinite type: Theorem 8.1.14. Let R be a ring and n < ω. There is a bijective correspondence between n–cotilting classes of cofinite type in R–Mod and resolving subcategories S of mod–R such that S ⊆ Pn<ω . The correspondence is given by the mutually inverse assignments C → (C)<ω and S → S .

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281

Proof. Let C be an n–cotilting class of cofinite type in R–Mod. So there is a class T ⊆ Pn<ω such that C = T ∞ . Then clearly (C)<ω is a resolving subcategory of mod–R. Conversely, let S be a resolving subcategory of mod–R such that S ⊆ Pn<ω . Then C = S  is a class of cofinite type, so C is n–cotilting by Proposition 8.1.12. Let C be an n–cotilting class of cofinite type in R–Mod, so C = S ∞ for a class S ⊆ Pn<ω . W.l.o.g., S = (C)<ω , and hence S  = S ∞ = C. Let S be a resolving subcategory of mod–R such that S ⊆ Pn<ω . By Theorem 4.5.6, (S ) = lim S. By Lemma 1.2.9, S coincides with the class of all finitely −→ 2 presented modules in lim S. So S = (lim S)<ω = ((S ))<ω . −→ −→

8.2

1–cotilting modules and cotilting torsion–free classes

In this section we will concentrate on the particular case of 1–cotilting modules and classes. In this case there are further descriptions available. We start with an easy dualization of the first part of Lemma 6.1.1: Lemma 8.2.1. Let R be a ring, C be a 1–cotilting left R–module and C be the cotorsion pair induced by C. Then C is the cotorsion pair cogenerated by C. Moreover, for each left R–module M there are a cardinal λ, a module C  ∈ → C − → M → 0. Cogen(C), and an exact sequence 0 → C λ −

Proof. The first claim is clear, the second follows by Lemma 3.3.9 for A = C, S = Z and B = M . 2 The exact sequence from Lemma 8.2.1 is called the C–torsion–free resolution of M (cf. Lemma 6.1.1). 1–cotilting modules can also be characterized in terms of the classes they cogenerate: Lemma 8.2.2. Let R be a ring. A module C is 1–cotilting, iff Cogen(C) = ⊥ C. In this case Copres(C) = Cogen(C).

Proof. Dual to the proof of Lemma 6.1.2.

2

It follows that if C is 1–cotilting, then Cogen(C) is a torsion–free class of modules, called the cotilting torsion–free class cogenerated by C. We will characterize cotilting torsion–free classes among all torsion–free classes of modules in terms of approximations. Again we will work in a slightly more general setting of pretorsion–free classes: a class of modules, C, is a pretorsion–free class provided that C is closed under submodules and direct products.

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Theorem 8.2.3. Let R be a ring and F be a pretorsion–free class of left R– modules. Let W be an injective cogenerator for R–Mod. Then the following are equivalent: (a) F is a cotilting torsion–free class. (b) F is a covering class. (c) F is a special precovering class. (d) W has a special F–precover.

Proof. (a) implies (b) by Theorem 8.1.9, (b) implies (c) by Lemma 2.1.13 (b), and (c) trivially implies (d). → C2 − → W → 0 be a special F–precover of (d) implies (a): let 0 → C1 − W . A dual proof to that of Theorem 6.1.4 shows that C = C1 ⊕ C2 is a cotilting module such that Cogen(C) = F. 2 Corollary 8.2.4. Let R be a ring. A pretorsion–free class F is a cotilting torsion– free class, iff F = ⊥ C for a splitter C.

Proof. The only–if–part is clear. For the if–part, since F is closed under products, we have Ext1R (C λ , C) = 0, and Lemma 3.3.9 implies that each module has a special F–precover. So F is 1–cotilting by Theorem 8.2.3. 2 Now we can easily show that in the particular case of left noetherian rings, 1– cotilting torsion–free classes are completely determined by their finitely generated modules. This result goes back to Buan and Krause [73]. Theorem 8.2.5. (1–cotilting classes over noetherian rings) Let R be a left noetherian ring. Then there is a bijective correspondence between the 1–cotilting classes C in R–Mod and the torsion–free classes E in R–mod containing R. The correspondence is given by the mutually inverse assignments C → C ∩ R–mod

and

E → lim E. −→

Proof. If C is 1–cotilting then C is a torsion–free class in R–Mod containing R. By (the left–hand version of) Lemma 4.5.2 (a), C ∩ R–mod is a torsion–free class in R–mod. Conversely, given E as in the claim, let C = lim E. By Lemma 4.5.2 (b), C −→ is a torsion–free class in R–Mod. Since R ∈ E, we have C = (E) by (a left

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R-module version of) Theorem 4.5.6. So C is a covering class by Theorem 3.2.15. By Theorem 8.2.3, C is 1–cotilting. Now E = lim E ∩ R–mod by Lemma 1.2.9. Conversely, given a 1–cotilting −→ class C in R–Mod, each M ∈ C is a directed union of the system of all its finitely generated submodules, {Mi | i ∈ I}. Since C is 1–cotilting, Mi ∈ C for each i ∈ I. So C = lim(C ∩ R–mod), and the assignments are mutually inverse. 2 −→ Remark 8.2.6. The corresponding result fails for 1–tilting (torsion) classes: Given a right noetherian ring R and a 1–tilting class T in Mod–R, the class T ∩ mod–R is a torsion class in mod–R. Let C = lim(T ∩ mod–R). By Lemma −→ 4.5.2 (b), C is a torsion class in Mod–R contained in T . However, C is not 1–tilting in general: if T = (P1<ω )⊥ , then C is closed under direct products, iff P1<ω is contravariantly finite in mod–R by Lemma 4.5.1. The latter fails for the IST–algebra from 7.1.16, for example. Nevertheless, if R is an artin algebra, then the dual result to Theorem 8.2.5 does hold: 1–tilting torsion classes in Mod–R correspond bijectively to the torsion classes in mod–R containing an injective cogenerator — see Theorem 6.2.2 above. There is a general criterion for a 1–cotilting class to be of cofinite type: Proposition 8.2.7. Let R be a ring and C be a class of left R–modules. Then C is 1–cotilting of cofinite type, if and only if there is a module M ∈ P1 such that C = M .

Proof. For the only–if–part, consider S ⊆ Pn<ω such that C = S ∞ . Let M be the direct sum of a representative set of all modules in S. Then C = M ∞ = ⊥∞ M d . Since C is 1–cotilting, C ⊥ ⊆ I , so M d ∈ I , hence M ∈ F by Lemma 1 1 1 1.2.11. So S ⊆ F1 ∩ mod–R = P1<ω (since finitely presented flat modules are projective — see Lemma 1.2.16). So M ∈ P1 , and C = M . For the if–part, we consider the cotorsion pair (A, B) generated by M . Since A ⊆ P1 , the claim will follow once we prove that M  = (A<ω ). If N ∈ M , then N d ∈ B, so N ∈ A ⊆ (A<ω ). Conversely, by Theorem 4.5.6 and Lemma 2 4.5.14, M ∈ lim A<ω = ((A<ω )), so M  ⊇ (A<ω ). −→ Now we can prove that in many cases, all 1–cotilting classes are of cofinite type: Theorem 8.2.8. Let R be a left noetherian ring such that F1 = P1 (this occurs when R is right hereditary, or right perfect, or 1–Iwanaga–Gorenstein). Then all 1–cotilting classes are of cofinite type, that is, all 1–cotilting left R– modules are equivalent to duals of 1–tilting (right R–) modules.

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Proof. Let C ⊆ R–Mod be a 1–cotilting class. By a version of Theorem 4.5.6 for left R–modules, and by Theorem 8.2.5, C = D, where D = (C <ω ). Since C is closed under submodules, we have D ⊆ F1 . By Lemma 3.2.13, each module D ∈ D is D≤κ –filtered, where κ = |R| + ℵ0 . Let M be the direct sum of a representative set of all modules in D≤κ . By Corollary 3.1.3 C = M . Finally, since D ⊆ F1 = P1 by assumption, we conclude that C is of cofinite type by Proposition 8.2.7. 2 In particular, if R is left artinian or 1–Iwanaga–Gorenstein, then we can describe 1–cotilting classes either by means of the torsion–free classes in R–mod containing R (as in Theorem 8.2.5), or by means of the subcategories S ⊆ mod–R closed under extensions and direct summands and satisfying P0<ω ⊆ S ⊆ P1<ω (as in Corollary 6.1.7).

Cotilting modules and classes over Dedekind domains As an application we consider in more detail the case of Dedekind domains. Given a Dedekind domain R and a set of maximal ideals P ⊆ mspec R, we define a module QP by QP = Q ⊕



E(R/q) ⊕

q∈mspec R\P



Jp .

p∈P

Given an ideal I of R, we will call a module M ∈ R–Mod I–torsion–free provided that TorR 1 (R/I, M ) = 0. Denote by CP the class of all left R–modules that are p–torsion–free for all p ∈ P , that is, CP = {M ∈ R–Mod | TorR 1 (R/p, M ) = 0 for all p ∈ P }. Theorem 8.2.9. Let R be a Dedekind domain. (a) Let C be a class of modules. Then C is cotilting, iff there is a set of maximal ideals, P , such that C = CP . (b) The set {QP | P ⊆ mspec R} is a representative set (up to equivalence) of the class of all cotilting modules.

Proof. (a) Clearly, given P ⊆ mspec R, CP is a class of cofinite type, hence a cotilting one by Proposition 8.1.12. Conversely, if C is cotilting, then C is of cofinite type by Theorem 8.2.8, so by Theorems 6.2.22 and 8.1.13, there is a subset P ⊆ mspec R such that C = TP . By

8.2 1–cotilting modules and cotilting torsion–free classes

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Example 8.1.3, C = {M ∈ Mod–R | TorR 1 (E(R/p), M ) = 0 ∀p ∈ P }. Since R is hereditary and E(R/p) is {R/p}–filtered for each p ∈ mspec R, the condition R TorR 1 (E(R/p), M ) = 0 is equivalent to Tor1 (R/p, M ) = 0, for each p ∈ P , and the claim follows. (b) In view of Theorems 6.2.22, 8.1.13 and 8.2.8, it suffices to prove that for each P ⊆ mspec R, QP is a cotilting module equivalent to the cotilting module CP defined in Example 8.1.3. Condition (C1) for QP is clear. For (C2), since Q ⊕ JP is flat and JP is pure–injective,  where JP = p∈P Jp , it suffices to prove that Ext1R (Iκ , JP ) = 0, where Iκ = ( q∈mspec R\P E(R/q))κ , for each κ. However, Iκ is an injective module, and it is easy to see that Iκ has no indecomposable direct summands isomorphic to E(R/p) for p ∈ P . Since Jp is q–divisible for all p ∈ P and q ∈ mspec R \ P , we infer that Ext1R (Iκ , JP ) = 0. For condition (C3), consider the exact sequence 0 →  JP − → E(JP ) − → E(JP )/JP → 0. Then (λ) and E(J )/J ∼ (αp ) for some cardinals λ > 0 Q E(R/p) E(JP ) ∼ = P P = p∈P  and αp > 0 (p ∈ P ). Let W = E(JP )/JP ⊕ q∈mspec R\P E(R/q). Then → W is an injective cogenerator for Mod–R, and the exact sequence 0 → JP −  E(JP ) ⊕ q∈mspec R\P E(R/q) − → W → 0 proves (C3). ⊥ ⊥ ⊥ 2 Finally, QP = CP = JP , so QP and CP are equivalent. In fact, for Dedekind domains, cotilting cotorsion pairs correspond bijectively to Tor–pairs. More precisely, they form the image of the canonical embedding of the lattice LT or into LExt (see Lemma 2.2.3). Before proving this, we recall another notion: A torsion pair (T , F) in Mod–R is called hereditary, if T is closed under submodules, or equivalently, F is closed under injective hulls. A 1–cotilting torsion–free class C is called hereditary, provided that it is closed under injective hulls. If C is a 1–cotilting module cogenerating C, then it is easy to see that C is hereditary, iff E(C) ∈ Cogen(C). The following technical lemma requires a deeper insight into the structure flat modules over commutative noetherian rings (cf. Example 1.2.26 (c)): Lemma 8.2.10. Let R be a Dedekind domain. Let C be an Enochs cotorsion module and f : F → C be the flat cover of C. Then F is pure–injective and ⊥C = ⊥F .

Proof. F is pure–injective by Theorem 4.4.12 (a). Let M be a module. Consider the exact sequence 0 → T − →M − → M/T → 0, where T is the torsion part of M . Since R is hereditary and both ⊥ F and ⊥ C contain all torsion–free (= flat) modules (cf. Theorem 4.4.9), we see that M ∈ ⊥ F , iff T ∈ ⊥ F , and similarly M ∈ ⊥ C, iff T ∈ ⊥ C. Since T is a torsion module over a Dedekind domain, T is a direct

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sum of its p–components, where p ranges over mspec R. Since each p–component (being a submodule of E(R/p)(αp ) for an ordinal αp ) is {R/p}–filtered, it remains to prove that R/p ∈ ⊥ F iff R/p ∈ ⊥ C for each p ∈ mspec R. For this purpose, we use Xu’s formula for the dual Bass invariants of Enochs cotorsion modules (see Remark 4.1.15 (ii)): for p ∈ spec R, let k(p) = (R/p)(p) be the residue field of the localization R(p) . Since F is the flat cover of C, the  formula yields F = p∈mspec R Lp ⊕ Q(α0 ) , where Lp is the p–adic completion of the free R(p) –module of rank αp ≥ 0, and αp = dimk(p) k(p) ⊗R(p) HomR (R(p) , C)

(p ∈ spec R)

are the 0–th dual Bass invariants of C. Let p ∈ mspec R. Then F(p) ∼ = Lp ⊕ Q(βp ) for some βp ≥ 0. In particular, F(p) is a divisible R(p) –module, iff αp = 0. Denote by νp the canonical embedding of pR(p) into R(p) . Then we have αp = 0, iff k(p) ⊗R(p) HomR (R(p) , C) = 0, iff Im(νp ⊗R(p) id) = HomR (R(p) , C), iff pR(p) . HomR (R(p) , C) = HomR (R(p) , C). The latter says HomR (R(p) , C) ∼ = HomR(p) (R(p) , C(p) ) ∼ = C(p) is a divisible R(p) –module. This proves that for each p ∈ mspec R, F(p) is a divisible R(p) –module, iff C(p) is one. Since R/p is finitely generated, the classical formulas for Ext–groups of localizations (see e.g. [155, §3.2]) finally yield that R/p ∈ ⊥ F , iff R/p ∈ ⊥ C. 2 The cotilting torsion–free classes over any Dedekind domain correspond bijectively to Tor–pairs: Theorem 8.2.11. (Tor–pairs for Dedekind domains) Let R be a Dedekind domain. Let C ⊆ Mod–R. The following conditions are equivalent: (a) (C, C ) is a Tor–pair. (b) (C, C ⊥ ) is a cotorsion pair such that C ⊇ FL. (c) There is a subset P ⊆ mspec R such that C is the class of all modules which are p–torsion–free for all p ∈ P . (d) C is a hereditary cotilting torsion–free class. (e) C is a cotilting torsion–free class.

Proof. (a) implies (b) by Lemma 2.2.3. (b) implies (c): by assumption, all modules in C ⊥ are Enochs cotorsion. Let F (E) → E be the flat cover of E. Put F = {F (E) | E ∈ C ⊥ }. By Lemma 8.2.10,

8.2 1–cotilting modules and cotilting torsion–free classes

287

F is aclass of pure–injective flat modules and C = ⊥ F. By Example 1.2.26 (c), F ∼ = p∈mspec R Lp ⊕ Q(α) , where Lp is the p–adic completion of the free R(p) – module of rank αp (for some αp ≥ 0) for each p ∈ mspec R and α ≥ 0. Since R(p) is a local principal ideal domain, we have      (α ) Lp ∼ = R(p)p ∼ = HomR(p) Q(p) /R(p) , Q(p) /R(p) ∼ = HomR E(R/p), E(R/p)(αp ) byLemma 4.4.6. So ⊥ Lp = E(R/p) = ⊥ Jp for each p ∈ mspec R, and C = ⊥ p∈P Jp = CP where P = {p ∈ mspec R | αp = 0} (cf. Theorem 8.2.9). (c) implies 8.2.9, C = ⊥ QP for the cotilting module  (d): by (c) and Theorem QP = Q ⊕ q∈mspec R\P E(R/q)) ⊕ p∈P Jp . Since E(QP ) ∼ = Q(α) ⊕



E(R/q)

q∈mspec R\P

for some α > 0, we see that E(QP ) ∈ Cogen(QP ), so C is hereditary. (d) implies (e): this is trivial. (e) implies (a): by Theorem 8.2.8 (for right R–modules), C is of cofinite type, so there is S ⊆ R–mod consisting of left R–modules of projective dimension ≤ 1 such that C = S, and (a) follows. 2 Now we briefly consider the more general case when R is a Prüfer domain. Since all cotilting R–modules are pure-injective, they have injective dimension ≤ 1 by Theorem 4.4.10. Corollary 8.2.12. Let R be a Prüfer domain. (a) There is a bijective correspondence between cotilting classes of cofinite type in R–Mod and finitely generated localizing systems. The correspondence is given by the assignments   C → L = J ⊆ R | ∃I : 0 = I ⊆ J and R/I ∈ (C)<ω and L → CL , where CL is the class of all modules that are I–torsion–free for all I ∈ L ∩ mod–R. (b) Up to equivalence, cotilting modules of cofinite type are the duals of the Salce tilting modules δL from Theorem 6.2.19, where L ranges over all finitely generated localizing systems.

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Proof. This follows by Theorems 6.2.15 (b), 6.2.19, 8.1.2 and 8.1.13.

2

Unfortunately, Corollary 8.2.12 does not classify all cotilting classes and modules over Prüfer domains. There exist valuation domains with cotilting classes that are not of cofinite type. The next example is due to Bazzoni: Example 8.2.13. (A cotilting class which is not of cofinite type) Let R be a maximal valuation domain with a non–principal maximal ideal p (see [181, XIII.5]). Then R is pure–injective, so the class W1 of all Whitehead modules is a covering class by Theorem 3.2.9. Since R has injective dimension ≤ 1 by Theorem 4.4.10, W1 is closed under submodules and extensions. In order to prove that W1 is a cotilting torsion–free class, it suffices to show that M ∈ W1 , iff p annihilates the torsion part t(M ) of M (then if 0 = x is an element in a direct product of Whitehead modules and Ann(x) = 0, then Ann(x) = p, so W1 is closed under direct products, and Theorem 8.2.3 applies). First, since torsion–free modules are flat and R is pure–injective (see Theorem 4.4.10), M ∈ W1 , iff t(M ) ∈ W1 . Assume that pt(M ) = 0. Then t(M ) ∼ = (R/p)(α) , so it suffices to prove that 1 1 ExtR (R/p, R) = 0. Note that ExtR (R/p, R) ∼ = (Q/R)[p]/((Q[p] + R)/R), where M [p] denotes the set of all elements x ∈ M annihilated by p. (This follows by applying HomR (R/p, −) to the exact sequence 0 → R − →Q− → Q/R → 0, and identifying HomR (R/p, N ) with N [p] for N = Q and N = Q/R.) So Ext1R (R/p, R) = 0, iff (Q/R)[p] = 0. Let (R : p) = {x ∈ Q | x.p ⊆ R}. Since p is non–principal, (R : p) = R, so (Q/R)[p] = (R : p)/R = 0. This proves that pt(M ) = 0 implies M ∈ W1 . Conversely, assume that M is a torsion Whitehead module and consider 0 = x ∈ M . Then Ext1R (Rx, R) = 0. Let I = Ann x. Then as above 0 = Ext1R (R/I, R) ∼ = (Q/R)[I]/((Q[I] + R)/R), so (Q/R)[I] = (R : I)/R = 0, and (R : I) = R. Then I is not principal, so I(R : I) = I  where I  is the prime ideal associated with I. But then I = I  , and since (R : J) = R(J) for every prime ideal J, we conclude that I = p. This proves that M ∈ W1 implies pt(M ) = 0, and finishes the proof that W1 is a cotilting torsion–free class. Finally, we show that W1 is not of cofinite type. Assume there is S ⊆ mod–R = P1<ω such that W1 = S . Since R is a valuation domain, finitely presented modules coincide with direct sums of cyclically presented modules. So there is a set of non–zero elements {rα | α < κ} ⊆ p such that W1 = {M | R ∼ TorR 1 (R/rα R, M ) = 0 ∀α < κ}. However, Tor1 (R/rα R, M ) = M [rα ], so R ∼ in particular Tor1 (R/rα R, R/p) = R/p = 0. However, R/p ∈ W1 , a contradiction.

8.2 1–cotilting modules and cotilting torsion–free classes

289

Remark 8.2.14. In [40], Bazzoni proved that all cotilting modules over a valuation domain R are of cofinite type, iff R is strongly discrete (that is, R has no non–zero idempotent prime ideals). Moreover, if C is a cotilting module over a Prüfer domain R, then C(m) is a cotilting R(m) –module for every  maximal ideal m ∈ mspec R, and C is equivalent to the cotilting module m∈mspec R C(m) (cf. Proposition 5.2.24). So the classification of cotilting modules can be reduced to the valuation domain case.

Ext–rigid systems For an arbitrary ring R, it is not so easy to describe all resolving subcategories of mod–R, let alone the fact that, in general, resolving subcategories are not sufficient to classify all cotilting modules (cf. Example 8.2.13). So further characterizations of cotilting modules are desirable. Since each cotilting module is pure– injective, and any pure–injective module has the canonical decomposition as in Lemma 1.2.24, we are naturally led to the question of when cotilting classes can be characterized in terms of the rigid systems of indecomposable pure–injectives: Definition 8.2.15. Let R be a ring and n < ω. Consider a set M = {Mα | α < κ} of left R–modules such that for each Mα (α < κ), Mα is pure–injective, inj dim Mα ≤ n, and ExtiR (Mα , Mβ ) = 0 for all α, β < κ and 1 ≤ i ≤ n. (So in particular, each Mα is a splitter.) Then M is an n–rigid system if all elements of M are indecomposable. M is almost n–rigid if M0 is superdecomposable, and all Mα (0 < α < κ) are indecomposable. Usually, the term “rigid system” of modules refers to non–existence of non– zero homomorphisms between different members of the system. Here we allow non–zero homomorphisms (that is, non–zero Ext0 –groups), but no extensions, including self–extensions (the Exti –groups are required to be zero for all i > 0). Theorem 8.2.16. Let R be a ring, n < ω, and C an n–cotilting left R–module with the induced n–cotilting class C. Then there is an almost n–rigid system M  such that C = M ∈M M is an n–cotilting module equivalent to C. In particular, C = M ∈M,inj dim M >0 ⊥∞ M .

Proof. By Lemma 1.2.24, the pure–injective module C is of the form C = M0 ⊕ E, where M0 is zero or superdecomposable, and E is zero or a pure– injective modules, E =  hull of a direct sum of indecomposable pure–injective  P E( 0<α<κ Mα ). Then E is a direct summand in P = 0<α<κ Mα and P is

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a pure submodule, hence a direct summand, in E κ . Put C  = M0 ⊕ P . Then Prod(C) = Prod(C  ), so C  is an n–cotilting module equivalent to C. It follows 2 that M = {Mα | α < κ} is an almost n–rigid system satisfying the claim. Assume there are no superdecomposable pure–injective modules (i.e., in the terminology of Jensen and Lenzing [286, §8], R has sufficiently many algebraically compact indecomposable modules). Then the system M in Theorem 8.2.16 is n–rigid, and it only remains to determine which of the n–rigid systems yield n– cotilting modules. This occurs when R is a Dedekind domain or a tame hereditary algebra, for example; in fact, in these cases the structure of indecomposable pure–injective modules is well-known, see Example 1.2.26. In the Dedekind domain case, if M is a (1–) rigid system, then M clearly contains no finitely generated modules. So the non–injective elements of M are (some of the) adic modules; Theorem 8.2.16 thus yields a simple alternative proof of Theorem 8.2.9 (a). We will now show that even if R admits superdecomposable pure–injective modules, rigid systems are enough to capture cotilting classes, if R is left noetherian and n = 1. This is really useful: indecomposable pure–injective modules are well understood for many noetherian rings (because they are exactly the points of the so called Ziegler spectrum of R, a topological space reflecting the key model– theoretic properties of R–modules, see [108, §2] and [336]). Theorem 8.2.17. Let R be a left noetherian ring. (a) If C is a 1–cotilting class in R–Mod, then there is a 1–rigid system M such that C = M ∈M,inj dim M =1 ⊥ M . (b) Assume that R is left artinian. Let C be a class of left R–modules. Then C is 1–cotilting, if and only if there exists a 1–rigid system M such that C = M ∈M,inj dim M =1 ⊥ M .

Proof. (a) Let C be a 1–cotilting left R–module such that C = ⊥ C. By a result of Ziegler, C is elementarily equivalent to a pure–injective envelope of a direct sum of indecomposable pure–injective modules (see e.g. [336]), hence to a direct  product of indecomposable pure–injective modules, E = α<κ Mα . In particular, E is a direct summand in an ultrapower of C. Since any ultrapower of C is isomorphic to a direct limit of products of copies of C, we infer that E ∈ C by Theorem 8.1.7. Since R is left noetherian, the Baer Criterion shows that I1 is definable, so E ∈ I1 because E is elementarily equivalent to C ∈ I1 .

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Since A⊥ is definable for each finitely presented module A, we have C ∈ A⊥ , iff E ∈ A⊥ . By Lemma 3.3.4, Ext1R (−, I) takes direct limits into inverse ones for any pure–injective module I. Since R is left noetherian, it follows that C = ⊥ C = ⊥ E. In particular, E is a pure–injective splitter of injective dimension ≤ 1, so the modules Mα (α < κ) form a 1–rigid system. (b) Assume that R is left artinian. Let M be a pure–injective module of injective dimension ≤ 1. We claim that A = ⊥ M is a torsion–free class in R–Mod. Indeed, let F = <ω A . If N ∈ R–mod, then N has a smallest submodule P (N ) ⊆ N such that N/P (N ) ∈ F, namely P (N ) = P ⊆N,N/P ∈F P (here, N/P (N ) ∈ F follows from the fact that N is an artinian module, hence finitely cogenerated, and F is closed under submodules and finite direct products). Let T = {T ∈ R–mod | HomR (T, F) = 0}. For each N ∈ R–mod, P (N ) ∈ T because F is closed under extensions. In particular, (T , F) is a torsion pair in R–mod (because given N ∈ R–mod such that HomR (T , N ) = 0, we have P (N ) ∈ T , so P (N ) = 0 and N ∈ F). By Lemma 4.5.2 (b), (lim T , lim F) is a torsion pair in R–Mod. −→ −→ By Lemma 3.3.4, A is closed under direct limits, so lim F ⊆ A. Since R is left −→ noetherian and A is closed under submodules, also A ⊆ lim F. So A = lim F is −→ −→ a torsion–free class in R–Mod.  To prove the if–part, let M be a 1–rigid system and Q = M ∈M,inj dim M =1 M . Then C = ⊥ Q is a torsion–free class, and C is covering by Theorem 3.2.9, so C is a 1–cotilting class by Theorem 8.2.3. The only–if–part follows by part (a). 2 Remark 8.2.18. A similar result to Theorem 8.2.17 (a) holds true for Prüfer domains. By [40], each (1–) cotilting module C is equivalent to a cotilting module which is a direct product of indecomposable pure–injective modules. In particular, there is a 1–rigid system M such that C = M ∈M,inj dim M =1 ⊥ M .

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Open problems 1. Characterize the rings R such that all n–cotilting left R-modules are of cofinite type (i.e., equivalent to duals of n–tilting right R–modules). By Theorem 8.2.8, for n = 1, these rings include all left artinian rings and all 1–Iwanaga–Gorenstein rings. Among valuation domains, these are exactly the strongly discrete ones by Remark 8.2.14. 2. Is any cotilting left R–module equivalent to a cotilting left R–module which has no superdecomposable direct summands? Is any n–cotilting class of the form M ∈M,inj dim M >0 ⊥∞ M for an n–rigid system M? Of course, these questions are of interest only for rings possessing superdecomposable pure–injective modules (see Theorem 8.2.16). The first question has a positive answer for any Prüfer domain by [40], and trivially, for any von Neumann regular ring. The second question has a positive answer also for n = 1 for any left noetherian ring by Theorem 8.2.17 (a). 3. Let R be a commutative 1–Iwanaga–Gorenstein ring. Is any cotilting module equivalent to the Bass cotilting module CP constructed in Example 8.1.3? By Theorem 8.2.8, the answer is positive in case Open problem 2 from Chapter 6 has a positive answer for R. This occurs when R is a Dedekind domain (see Theorems 6.2.22(b) and 8.2.9(b)).

Chapter 9

The Black Box and its relatives

9.1 Survey of prediction principles using ZFC and more We start with a rather strong prediction principle, Jensen’s Diamond [284]. This principle holds, for example, under the assumption of Gödel’s Axiom of Constructibility (V=L). However, its weaker version, called the Weak Diamond (which was proved to be equivalent to 2ℵ0 < 2ℵ1 by Devlin, Shelah [133]), is often sufficient for applications, especially for proving many realization theorems; it will also be discussed below. We begin with introducing the necessary notations and results before we present three different versions of the Diamond Principle. The closed unbounded filter In Chapter 1, Section 1.4, we introduced λ–complete filters and considered one specific example, namely Fλ . Here we will need another one. The next theorem follows from ‘back–and–forth–arguments’; the proof can be found, for instance, in [284]. Theorem 9.1.1. Let λ be a regular, uncountable cardinal and let Fλ be the set of all subsets of λ containing a cub. Then: (a) Fλ is a filter, the so–called closed unbounded filter of λ. (b) The intersection of < λ cubs is a cub, thus Fλ is λ–complete. (c) Fλ is a normal filter. The boolean algebra B(λ) = P(λ)/Fλ for this filter Fλ on λ is particularly important. Recall that the equivalence of two sets X, Y ⊂ λ modulo Fλ is given by [X] = [Y ] in B(λ) ⇐⇒ X ∼ Y ⇐⇒ ∃ C ⊆ λ cub : X ∩ C = Y ∩ C.

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In particular, for X ⊆ λ we have [X] = 0 in B(λ) ⇐⇒ ∃ C ⊆ λ cub : X ∩ C = ∅. Elements in B(λ) are often used as Γ–invariants of a module. Γ–invariants are defined by filtrations of a module of cardinality λ, and passing from P(λ) to P(λ)/Fλ makes the chosen subset of λ independent of the filtration (see Eklof, Mekler [142]). Definition 9.1.2. A subset E of λ is stationary in λ, if and only if [E] = 0 in B(λ). This is equivalent to E ∩ C = ∅ for all cubs C ⊆ λ. By Theorem 9.1.1 (b), cubs cannot be partitioned into cubs. It is important that stationary sets can be partitioned into stationary sets. Here is a first example: let λ ≥ ℵ2 be regular, let λo = {α ∈ λ : cf(α) = ω}, and let E1 = {α ∈ λ : cf(α) = ω1 }. Then λo and E1 are disjoint and stationary in λ. Solovay proved a more general and useful result: Lemma 9.1.3. [Solovay] Let λ be a regular cardinal. Then any stationary subset E of λ can be partitioned into λ stationary sets. For the non–trivial proof we refer the reader to Jech [284, p. 433].

Three equivalent versions of the Diamond Principle If X is a subset of a cardinal λ, then we may consider the initial segments X ∩ α (α ∈ λ). The Diamond Principle for a stationary subset E of λ provides a list {Wα ⊆ α : α ∈ E} which ‘very often’ predicts these initial segments X ∩ α, independent of the choice of X ⊆ λ. More precisely, the Diamond Principle, ♦λ E, is the following: ♦λE : Let E be a stationary subset of a regular, uncountable cardinal λ. Then there is a family {Wα : α ∈ E} of initial segments Wα of α such that, for any subset X ⊆ λ, the set {α ∈ E : Wα = X ∩ α} is stationary in λ. Note that the family {Wα : α ∈ E} ‘predicts’ stationary many initial segments of any subset X of λ. Definition 9.1.4. Let λ be a regular, uncountable cardinal. A subset E ⊆ λ is said to be non–reflecting or sparse, if for all α ∈ λ with cf(α) > ω, the set E ∩ α is not stationary in α. The following result is due to Ronald Jensen [285]. Recall from above that λo = {α ∈ λ | cf(α) = ω} is a stationary subset of λ.

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Theorem 9.1.5. Assume that Gödel’s constructible universe, V = L, holds. Let λ be a regular, uncountable cardinal and let E be any stationary subset of λ. Then ♦λ E holds. Moreover, there is a stationary non–reflecting subset E  ⊆ λo , if (and only if) λ is not weakly compact. For applications an equivalent version of ♦λ E is often more useful. It requires the notion of a λ–filtration (introduced in Section 4.3 for modules): Definition 9.1.6. Let A be a set of cardinality at most λ. An ascending, continuous chain {Aα : α < λ} of subsets Aα of A is called a λ–filtration of A, if A =  α<λ Aα and if the cardinality of Aα is less than λ for each α < λ. The second version of ♦λ E can now be formulated: Theorem 9.1.7. ♦λ E is equivalent to the following statement: Let A = α<λ Aα be any λ–filtration of a set A of cardinality λ. Then there is a family {Yα | α ∈ E} of subsets Yα of Aα such that, for all X ⊆ A, the set {α ∈ E : Yα = X ∩ Aα } is stationary in λ.

Proof. Without canonical  loss of generality we identify A = λ. Thus A has a filtration A = α<λ α, which we compare with the given filtration A = α<λ Aα . Hence there is a cub C ⊆ λ on which the two filtrations agree: α = Aα for all α ∈ C. Let {Wα ⊆ α : α ∈ E} be as in ♦λ E and put  Wα ⊆ Aα , if α ∈ E ∩ C Yα = ∅, otherwise. Now let X ⊆ A be an arbitrary set. Then     = α ∈ E ∩ C | Wα = X ∩ α α ∈ E ∩ C | Y α = X ∩ Aα   = α ∈ E : Wα = X ∩ α ∩ C is stationary in λ. Hence the second version of ♦λ E follows from the first one. The converse is immediate. 2 The next version of ♦λ E is especially useful in connection with the realization of endomorphism rings.   Theorem 9.1.8. Suppose that ♦λ E holds. Let A = α<λ Aα and B = α<λ Bα be two λ–filtrations. Then there are so–called Jensen–functions gα : Aα −→ Bα

(α ∈ E)

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such that, for any function g : A → B, the set   α ∈ E : gα = g  Aα is stationary in λ. Remark: as before, the converse of Theorem 9.1.8 is easy.

Proof. We apply Theorem 9.1.7 to g ⊆ A × B viewed as a graph and to the  λ–filtration A × B = α<λ Aα × Bα . Thus there is a predicting family 

 Yα ⊆ Aα × Bα : α ∈ E .

We use this family to define the required Jensen-functions: let α ∈ E. If the set Yα ⊆ Aα × Bα turns out to be the graph of a function, then we put gα = Yα : Aα → Bα ; otherwise we choose gα arbitrary, e.g. put gα = ∅. Now the defined family consists, indeed, of maps and, for any arbitrary g : A → B, there is a cub C ⊆ λ such that g  Aα is a function into Bα for any α ∈ C. Again we view g ⊆ A × B as a graph and, similarly, gα ⊆ Aα × Bα for all α ∈ C. Therefore the set   α ∈ E ∩ C : g ∩ (Aα × Bα ) = Yα = gα is stationary in λ by Theorem 9.1.7, as desired.

2

While Jensen’s result on the diamond in Gödel’s universe comes from a complicated and long paper [285], we can prove a relevant part directly from GCH and less, which is our next intention. However, assuming V=L, there are additional non–reflecting stationary subsets of any not weakly compact cardinal (see Theorem 9.1.5). Thus, assuming only GCH, the algebraic assumptions must be robust enough to pass through limit ordinals not cofinal to ℵ0 . (‘Inductive assumptions’ on the class of modules under construction, like Pontryagin’s theorem, will then be of help.) The consequences of GCH will be based on papers by Shelah [367], Gregory [256] and Kunen [304]. The stationary subsets of a cardinal λ involved will include those related to cofinalities; for regular cardinals ℵ0 ≤ κ < λ, we define:   Eλ (κ) = α | α ∈ λ, cf(α) = κ . First we will show that, for successor cardinals λ+ , a weak version of the Diamond Principle is equivalent to the usual (strong) version above. We begin by defining the weak version and then state the equivalence. Definition 9.1.9. Let E be a stationary subset of the regular cardinal λ. Then ♦− λ (E) holds, if there is a family {Wα | α ∈ E} with the following three properties: (i) Wα ⊆ P(α) for all α ∈ E;

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(ii) |Wα | ≤ |α| for all α ∈ E; (iii) Prediction: For all X ⊆ λ, the set {α ∈ E | X ∩ α ∈ Wα } is stationary in λ. It is clear that ♦− λ (E) follows from ♦λ (E). Hence one implication of the following theorem obviously holds. Theorem 9.1.10. Let λ ≥ ℵ0 be any cardinal and let E ⊆ λ+ be stationary. Then (E) holds, if and only if ♦λ+ (E) does. ♦− λ+

Proof. Suppose that ♦− (E) holds and let {Wα | α ∈ E} be the family given by λ+ Definition 9.1.9. Moreover, let f : λ+ → λ × λ+ be a bijection. Then, for the regular cardinal λ+ , there is a cub C = {α ∈ λ+ | λ ≤ α and f (α) = λ × α}, where α, λ × α are considered as sets: we just compare the two natural λ+ – filtrations of λ+ and λ × λ+ , respectively. They agree under f on a cub. We use this cub C to define a new family {Vα | α ∈ E} with Vα ⊆ P(λ × α) as follows:  ∅, if α ∈ /C Vα := f (Wα ), if α ∈ C, where f (Wα ) = {f (Y ) | Y ∈ Wα }. Now we claim:   ∀X ⊆ λ × λ+ , the set α ∈ E ∩ C | X ∩ (λ × α) ∈ Vα is stationary, (9.1.1) that is, ♦− (E) holds for λ × λ+ . (E) Note first that S =: {α ∈ E | f −1 (X) ∩ α ∈ Wα } is stationary by ♦− λ+ and hence S ∩ C is also stationary because C is a cub. If α ∈ S ∩ C, then X ∩ (λ × α) = f f −1 (X) ∩ f (α) = f (f −1 (X) ∩ α) ∈ f (Wα ) = Vα , since f is bijective and thus closed under intersections. Therefore (9.1.1) follows. Now, for α ∈ E ∩ C, we have |Vα | = |Wα | ≤ |α| ≤ λ and so we may enumerate   (9.1.2) Vα = Vαβ | β ∈ λ with Vαβ ⊆ λ × α ⊆ λ × λ+ for all α, β. Moreover, for γ ∈ λ we put   β := ζ | ζ ∈ λ+ , (γ, ζ) ∈ Vαβ ⊆ α. Vα,γ β We now claim that there is a β ∈ λ such that the sequence {Vα,β | α ∈ E ∩ C} is the required family of Jensen sets for ♦λ+ (E).

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Suppose, for contradiction, that this is not the case. Then, for each β ∈ λ, there is a set X β ⊆ λ+ such that  β  is not stationary. (9.1.3) S β := α ∈ E ∩ C | X β ∩ α = Vα,β Therefore there are cubs C β (β < λ) such that C β ∩ S β = ∅.  Put C  = β<λ C β and X = β<λ ({β} × X β ) ⊆ λ × λ+ . Then C  is also a cub and we have X ∩ ({β} × λ+ ) = {β} × X β . We show that X ∩ (λ × α) ∈ / Vα for any α ∈ E ∩ C ∩ C  , which then obviously contradicts (9.1.1). So let α ∈ E ∩C ∩C  and suppose that X ∩(λ×α) ∈ Vα . Then X ∩(λ×α) = β Vα for some β < λ by (9.1.2). Thus we also have X ∩({β}×α) = Vαβ ∩({β}×α) β β and hence {β} × (X β ∩ α) = {β} × Vα,β , respectively X β ∩ α = Vα,β , i.e. β  β α ∈ S . This contradicts the assumption that α ∈ C ⊆ C . Therefore the set {α ∈ E ∩C | X ∩(λ×α) ∈ Vα } is not stationary, contradicting (9.1.1). Hence our β | α ∈ E ∩ C} original claim is correct: there is β < λ such that the sequence {Vα,β is the required family of Jensen sets for ♦λ+ (E). Thus the proof is finished. 2 The next proposition is the link between GCH and this particular weaker Diamond Principle. Proposition 9.1.11. Let κ be regular, ℵ0 ≤ κ < μ and λ = 2μ = μ+ . Then, for any stationary subset E ⊆ Eλ (κ) ∩ (μ, λ), the principle ♦− λ (E) holds, if either of the following conditions is satisfied: (a) μκ = μ, or (b) μ singular with cf(μ) = κ and also |δ|κ < μ for all δ < μ. Note that we are mainly interested in the case κ = ℵ0 .

Proof. Let the cardinals κ, μ, λ be as above. In both cases we need to define a family {Wα | α ∈ E} satisfying the conditions (i), (ii) and (iii) of Definition 9.1.9; the suitable definition of the family is the main idea of the proof. In order to do so we first ‘enumerate’ the set of all bounded subsets of λ by {Aα | α ∈ λ} in such a way that each bounded subset A ⊆ λ appears λ times. This is possible since, for A ⊆ λ bounded, we have |A| < λ = μ+ , hence |A| ≤ μ and thus there are at most 2μ = λ such sets. Case (a) [μκ = μ]: For each α ∈ E, we define Wα as follows:    Wα := Y | |Y | ≤ κ, Y ⊆ P(α) ∩ Aβ | β < α .

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It follows immediately from this definition that Wα ⊆ P(α) for all α ∈ E, that means condition (i) is satisfied. Next we show (ii), that is, |Wα | ≤ |α| for all α ∈ E: by assumption we have μ < α < λ = μ+ and  so |α| = μ. We now consider the set Wα . An element of Wα is of the form Y for some Y ⊆ P(α) ∩ {Aβ | β < α} with |Y | ≤ κ. Hence there is a unique Y  ⊆ α with |Y  | ≤ κ and Y = {Aβ | β ∈ Y  }. Therefore |Wα | ≤ |α|κ = μκ = μ, as required. Finally, we show the prediction (iii). Let X ⊆ λ be an arbitrary set. First we inductively construct an unbounded ascending continuous chain (αρ | ρ < λ) of ordinals αρ in λ such that X ∩ αρ = Aαρ+1 for all ρ < λ. Put α0 = 0. Now suppose that αρ has been defined for all ρ < δ and for some δ ∈ λ. If δ is a limit ordinal, then let αδ := supρ<δ αρ . Clearly, we thus have αρ < λ in this case, since |δ| < λ = cf(λ). If δ = ρ + 1 is a successor ordinal, then we choose αδ > αρ in such a way that X ∩ αρ = Aαδ is satisfied. This is possible because X ∩ αρ ⊆ λ is bounded and hence appears λ often in the list of bounded subsets fixed above. Therefore we have obtained the desired sequence of ordinals. Next let C = {αδ | δ < λ, δ limit ordinal}; obviously C is a cub. It is now immediate that E ∩ C is a stationary subset of λ, and so it remains to show that X ∩ α ∈ Wα for all α ∈ E ∩ C. Recall, E ⊆ (μ, λ) ∩ Eλ (κ) by assumption. Thus, if α ∈ E ∩ C, then α = αδ = supρ<δ αρ for some δ < λ and also cf(α) = κ. Hence we have α = supj<κ αρj with αρj < α and X ∩ αρj = Aαρj +1 by the definition of the above sequence. Therefore   X ∩α= X ∩ αρ j = Aαρj +1 j<κ

j<κ

and so, by the definition of Wα , X ∩ α ∈ Wα . Thus the prediction (iii) is shown.

Case (b) [κ = cf(μ) < μ, |δ|κ < μ ∀ δ < μ]: This case is very similar to (a), but we must take special care of the singularity of μ. + If α ∈ E, then  μ <α α < λ =α μ , i.e. |α| = μ, and thus we may write α as a filtration α = j<μ Vj with |Vj | < μ. Using these sets as well as the above list of bounded subsets of λ, this time we define Wα by     !! Aβ | β ∈ Y ! Aβ ⊆ α, |Y | ≤ κ, Y ⊆ Vjα for some j < μ . Wα := Again we need to show that conditions (i), (ii) and (iii) of Definition 9.1.9 are satisfied. Obviously, Moreover, we calculate the size of Wα for each  (i) holds.  α ≥ μ as |Wα | = μ |Vjα |κ ≤ μ μ = μ = |α| and thus (ii) also holds.

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As before in (a), we finally show the prediction (iii). So let X ⊆ λ be an arbitrary set. Here we fix a strictly increasing (not necessarily continuous) function f : λ → λ such that X ∩ α = Af ({α}) ; clearly, f can be defined inductively. Moreover, we put C := {α ∈ λ | ∀β < α : f ({β}) < α}, which is easily seen to be a cub in λ. Now let α ∈ C ∩ E be arbitrary. Then, in particular, cf(α) = κ, and hence we can choose an increasing sequence αi (i < κ) of ordinals less than α with α = supi<κ αi . For such a fixed α ∈ C ∩ E, we claim: !  !! ! ∃ j < μ such that κ = !Vjα ∩ f ({αi }) | i < κ !.

(9.1.4)

In order to prove the claim we put A = {f ({αi }) | i < κ} and assume the contrary: κj := |Vjα ∩ A| < κ for all j < μ. Since α ∈ C and |A| = κ we deduce A ⊆ α = j<μ Vjα and also κ = supj<μ κj . Therefore, on the one hand,we may choose an unbounded sequence (jα )α<κ in μ with α ∈ κjα . Hence μ = α<κ jα and so cf(μ) ≤ cf(κ). On the other hand, cf(κ) ≤ cf(μ) since κ = j<μ κj and thus cf(μ) = cf(κ). But κ is regular and so κ = cf(μ), contradicting the assumption of part (b) of the proposition. Hence Claim (9.1.4) is proved. Using (9.1.4) for j ∈ μ and the definitions of f and Wα , we finally obtain (for any α ∈ C ∩ E)   X ∩ αi | i < κ, f ({αi }) ∈ Vjα X ∩α =   Af ({αi }) | i < κ, f ({αi }) ∈ Vjα ∈ Wα , = and thus part (b) of the proposition follows.

2

Putting the previous proposition and Theorem 9.1.10 together we immediately get the following interesting Corollary 9.1.12. Let κ, μ, λ be cardinals with κ regular, ℵ0 ≤ κ < μ and λ = 2μ = μ+ . Then the following holds for any stationary subset E ⊆ Eλ (κ): (a) if μκ = μ, then ♦λ (E) holds; (b) if μ is singular but cf(μ) = κ and if also |δ|κ < μ for all δ < μ, then ♦λ (E) holds. Notice that the cardinal conditions μκ = μ, λ = 2μ = μ+ above are the same as the cardinal assumptions (9.2.1) for predictions by the Strong Black Box, which holds in any model of ZFC – of course, 2μ = μ+ must go, but precisely this needs extra work in Section 9.2.

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The Weak Diamond Principle The following prediction principle, which is a weakening of the Diamond Principle ♦ (= ♦λ for all regular cardinals λ) was derived in Devlin, Shelah [133]. It is equivalent to 2ℵ0 < 2ℵ1 for λ = ℵ1 . In [133] it is stated in a form suitable for applications, like Theorem 9.1.8. However, we will begin with a version closely patterned after ♦. The Weak Diamond Principle Φλ E is defined as follows: ΦλE: Let E be a stationary subset of a regular, uncountable cardinal λ. Then, for any family of partitions Pα : P(α) −→ 2 (α ∈ E), there is a weak diamond function F : E −→ 2 such that, for all X ⊆ λ, the set   α ∈ E | Pα (X ∩ α) = F ({α}) is stationary in λ. In this case we also say that E is non–small. Theorem 9.1.13. Φℵ1 (ℵ1 ) holds, if and only if 2ℵ0 < 2ℵ1 . For a proof see [133]. Next we want to present another version of the Weak Diamond Principle. We use the notation map(A, B) = A B for the set of all maps A → B. E holds. Moreover, let Theorem 9.1.14. Let λ, E be as above  and assume Φλ A, B be sets with λ–filtrations A = α<λ Aα , B = α<λ Bα and, for each α ∈ E, let Pα : map(Aα , Bα ) −→ 2 be a partition function. Then there is a ‘global’ partition function ρ : E −→ 2 such that, for all f ∈ map(A, B), the set 

 α ∈ E | Pα (f  Aα ) = ρ(α) is stationary in λ.

The proof is very similar to the proof of Theorem 9.1.8 and hence omitted. There is another combinatorial principle called the Generalized Weak Diamond: cardinal λ and ΨλE: Let E be a stationary subset  of a regular, uncountable  A, B be sets with λ–filtrations A = α<λ Aα , B = α<λ Bα . For each α ∈ E, let 2 ≤ pα < ω and let Pα : map(Aα , Bα ) −→ pα be given. Then there is ρ : E −→ ω such that ρ(α) ∈ pα for all α ∈ E and for all f ∈ map(A, B), the set 

 α ∈ E | Pα (f  Aα ) = ρ(α) is stationary in λ.

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Lemma 9.1.15. Let λ be a regular uncountable cardinal and E be a stationary subset of λ. Then ♦λ (E) =⇒ Ψλ E =⇒ Φλ E. In particular, Ψλ E holds under V = L for each regular uncountable cardinal λ and each stationary subset E ⊆ λ.

Proof. Take ρ(α) = Pα (Yα ) for all α ∈ E for the first implication, and pα = 2 for all α < λ for the second. 2 In general, by [364, XIV], none of the implications from Lemma 9.1.15 can be reversed.

Applications: the existence of almost free R–modules with a prescribed endomorphism ring This section presents a shortcut through Dugas, Göbel [120] using [199], see also [205]. The algebraic part for proving EndR M = A assuming the Diamond Principle ♦λE and the Weak Diamond ΦλE Let λ be a regular uncountable cardinal, let R be an S–ring (cf. Definition 1.1.1), and let A be an R–algebra of cardinality < λ, whose underlying R–module R A is free. Moreover, let S = p = {pn | n < ω} be generated by one element p and  R) = 0. So R and therefore also A are p–cotorsion–free. (The suppose Hom(R, element p only pretends to be a prime.) We want to prove the following Step Lemma, which will be used for applying the Weak Diamond, if |R| < 2ℵ0 , and for the Diamond, otherwise. If we only assume that A is p–cotorsion–free, then we need the stronger hypothesis |A| < 2ℵ0 for the Weak Diamond.  Step Lemma 9.1.16. Let F = n<ω Fn be the union of an ascending chain of free (right) A–modules Fn such that F0 = bA ⊕ D−1 and Fn+1 = Fn ⊕ Dn with Dn = 0 are free A–modules. Moreover, if |A| ≥ 2ℵ0 , we also fix some ϕ ∈ EndR F . Then there exist two extensions F ε ⊇ F (ε ∈ {0, 1}), which depend on ϕ, if |A| ≥ 2ℵ0 , such that the following holds: (a) the A–module F ε is free and either F ε /F ∼ = S−1 A or F ε = F ; (b) for each n < ω, the A–module F ε /Fn is free; (c) the A–module F ε /bA is free;

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(d) if |A| ≥ 2ℵ0 and ϕ extends to both ϕε ∈ EndR F ε for ε ∈ {0, 1}, then bϕ ∈ bA; if |A| < 2ℵ0 and if any ψ ∈ EndR F extends to both ψ ε ∈ EndR F ε for ε ∈ {0, 1}, then also bψ ∈ bA. Remark: if |A| < 2ℵ0 , then the last condition provides two uniform extensions of F , which at the same time control all endomorphisms of EndR F at a fixed element b ∈ F . Fε

Before we prove the lemma, we will introduce some simple notions that are basic at many places. We will meet variations of them again soon (see also Section 1.1).  If 0 = any free A–module y ∈ F is an element of thep–adic completion of  F = i∈I ei A, then we write y = i∈I ei yi with yi ∈ A and call [y] := {i ∈ I | yi = 0} the support of y (with respect to the given direct sum decomposition). Now let F and S = p be as in the above Step Lemma iand let ei ∈ Di (i < ω) be a family of basic elements. Then we call y = i<ω ei p ∈ F a branch element. Moreover, we call πx + y a branch–like element, if y is a branch element, x is a  basic element of F and π ∈ R. We are now ready to prove Step Lemma 9.1.16. Proof. Let the assumptions be as above. First we consider an arbitrary branch element y ∈ F and the extension F  = F, yA∗ ⊆ F of F . We claim: F  /F is p–divisible and F  , F  /Fn (n < ω) are free A–modules.

(9.1.5)

Since F is the p–adic completion of F , the quotient F/F is p–divisible, and so the p–divisibility of F  /F ⊆ F /F follows by purity. Now let the elements y k = k≤i<ω ei pi−k (k < ω) be a ‘divisibility chain’ of the branch element y. Then F  = F, y k A | k < ω and y k − y k+1 p = ek . Without loss of generality we may assume that the family B = {ei | i < ω} is an A–basis of F . Moreover, by a support argument, the set B  = {y k | k < ω} is A–independent. From the displayed formula it also follows that F  = B  A . Hence F  is a free A–module. Similarly, F  /Fn is a free A–module, because we can choose a basis of F/Fn and deal with the elements y k for k ≥ n. Hence (9.1.5) is shown. Next we consider a branch–like element z = y + πx, which is derived from a branchelement y, and show that  the result corresponding to (9.1.5) is also true: let F = i<ω ei A ⊕ xA, y = i<ω ei pi ∈ F, and let x be a basic element of F0 .

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Then y is a branch element and z is a branch–like element.  Again, we show that the A–module F  = F, zA∗ ⊆ F is free. Write y k = k≤i<ω ei pi−k as above   and, similarly, π = i<ω ai pi with ai ∈ R. We also define π k = k≤i<ω ai pi−k and z k = y k + π k x (k < ω). We thus obtain z k − z k+1 p = ek + ak x. As above, F  = F, zA∗ = F, z k A | k < ω is freely generated by the z k s and x. Again, a slight modification shows that F  /Fn is also a free A–module for each n < ω. Now we choose suitable branch–like elements y 0 = y ∈ F and y 1 = z ∈ F as  x ∈ F0 ; we shall specify y 0 , y 1 later. Moreover, above for certain elements π ∈ R, we define the two extensions of F by F ε = F, y ε A∗ ⊆ F for ε = 0, 1. The assertions (a), (b) and (c) of the Step Lemma are already shown; clearly, they are also true for F ε = F . So it remains to prove condition (d). We first consider the case |A| ≥ 2ℵ0 , that is, we deal with a fixed endomorphism ϕ ∈ EndR F . Note that ϕ uniquely extends to an endomorphism of Fˆ , which we also denote by ϕ. If bϕ ∈ bA, then (d) is already satisfied for F ε = F (ε = 0, 1). Otherwise we have bϕ ∈ / bA and thus also pk bϕ ∈ / bA for any k < ω, since b is basic in F0 . Now we fix an arbitrary branch element y 0 = y; recall that F 0 = F, yA∗ . If yϕ ∈ / F 0 , then ϕ violates the hypothesis of (d) for F 0 and 1 F = F and hence (d) holds in this case. The non–trivial case arises, if yϕ ∈ F 0 , i.e. ϕ extends to ϕ0 (= ϕ) ∈ End F 0 : then there are k < ω and a ∈ A such that pk yϕ − ya ∈ F. Hence we deduce 0 = pk bϕ − ba ∈ F from the two above displayed equations.  F ) = 0, and thus R(p  k bϕ − Now we use that F is p–cotorsion–free, so Hom(R,  such that ba) ⊆ F is impossible. Therefore there is an S–pure element π ∈ R π(pk bϕ − ba) ∈ F \ F.

(9.1.6)

We use this element π to specify our branch-like element z: z = y + πb; hence F 1 (for y 1 = z) is now also defined. Contrary to the lemma we assume that ϕ extends

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(uniquely) to ϕ1 (= ϕ) ∈ End F 1 . Hence pl y 1 ϕ − y 1 a ∈ F for some l ≥ k and a ∈ A. If we substitute y 1 , then pl yϕ + pl πbϕ − ya − πba ∈ F. Moreover, we have pl−k (pk yϕ − ya) ∈ F and thus their difference is also in F , that is, y(pl−k a − a ) + π(pl bϕ − ba ) ∈ F. But b, bϕ ∈ F have finite support, while [y] = {ei | i < ω}. Therefore pl−k a−a = 0 and so π(pk bϕ − ba) ∈ F , contradicting (9.1.6). Thus ϕε ∈ EndR F ε for ε = 0, 1 is impossible, proving (d). This shows the Step Lemma in case |A| ≥ 2ℵ0 . Finally, we consider the case |A| < 2ℵ0 . Since A is ‘relatively small’, we are able to control all elements ψ in EndR F by only two branch–like elements y 0 and y 1 , i.e. we can find a uniform π for all ψs.  has transcendence degree 2ℵ0 over A. So we can pick By Theorem 1.1.20, R  such that A ∩ πA = 0, where A is considered as an R–submodule of the π∈R   Thus we also have R–module A. F ∩ πF = 0 (in F ) because F is a free A–module. Now we can choose y 0 = y as before and y 1 = y + πb. The given argument shows that any ψ ∈ EndR F will only extend to both ψ ε ∈ EndR F ε for F ε = F, y ε A∗ ⊆ F , if bψ ∈ bA. Thus the Step Lemma is also shown in this case, and so the proof is finished. 2 Remark: the arguments used in the proof of part (d) will turn up again; they are crucial for working with endomorphism rings. The construction of R–modules M with EndR M = A assuming ♦ The set–theoretic assumptions of our next theorem are justified by Jensen’s Theorem 9.1.5. The algebraic assumptions are from above; they were used in Step Lemma 9.1.16: Recall that A is an R–algebra of cardinality < λ, whose underlying R–module R A is free, λ is a regular, uncountable cardinal and R is S–reduced and S–torsion– free (as in Definition 1.1.1). Moreover, let S = p = {pn | n < ω} be generated  R) = 0, so R, and therefore also A, are by one element p and suppose Hom(R, p–cotorsion–free.

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We also refer the reader to Lemma 9.1.3 for Solovay’s decomposition theorem. Under the strong assumption of ♦λ E such a decomposition (of E) is almost immediate, as we shall see next. Theorem 9.1.17. If ♦λ E holds, then there is a decomposition E = such that ♦λ Eβ holds for all β < λ.



β<λ Eβ

Note ♦λ E implicitly states that λ must be regular and E is stationary; so also the Eβ s are stationary.

Proof. Consider two copies of the canonical λ–filtration λ = gα : α −→ α



ν<λ ν

and let

(α ∈ E)

be the Jensen–functions given by Theorem 9.1.8. For any ν < λ, let   Eν := α ∈ E | gα (0) = ν .  Clearly ν<λ Eν is a partition of E. In order to show ♦λ (Eν ) for some arbitrary (but fixed) ν < λ, we derive new Jensen–functions hα = hνα for α ∈ Eν \ ω from gα by getting rid of gα (0); we shift the first ω ordinals: put hα (n) = gα (n + 1) for n < ω and hα (α \ ω) = gα (α \ ω). (Remark: to be absolutely precise we also need to define hα for α ∈ Eν ∩ ω, but since they play no role for the stationarity we may choose arbitrary functions for these αs.) Now let h : λ → λ be an arbitrary function and define g : λ → λ dually to the above by g(0) = ν and g(n+1) = h(n) (n < ω), as well as g (λ\ω) = h (λ\ω). Then it follows from ♦λ E and from the definition of Eν that     E  = α ∈ E | g  α = gα = α ∈ Eν | g  α = gα = Eν is stationary in λ. Moreover, if α ∈ Eν \ ω, then ν = g(0) = gα (0) and also h(n) = g(n + 1) = gα (n + 1) = hα (n) for each n < ω. Therefore h  α = hα for each α ∈ Eν \ ω and thus   α ∈ Eν | h  α = hα ⊇ Eν \ ω is stationary, i.e. ♦λ (Eν ) holds.

2

Using the result above we are now ready to prove the main theorem of this subsection which can be formulated in a stronger form using the following, generally accepted and by now well–known definition; (see [142, pp. 88, 92]). Definition 9.1.18. Let λ be a regular uncountable cardinal and M be an R–module.

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(i) M is λ–free, if there is a family M of submodules of M that are free and < λ–generated and every subset of M of size < λ is a subset of a member of M. Moreover, M is closed under unions of well–ordered chains of length < λ. (ii) An R–module M is said to be strongly λ–free, if there is a family M of < λ–generated free submodules containing 0 such that for any subset X of M of size < λ and any N ∈ M, there exists N  ∈ M such that N  ⊇ N ∪X and N  /N is free. If |R| < λ, then Definition 9.1.18 is designed to be translated easily into the existence of λ–filtration – as established in the next proof (see [142]). In particular, observe that the two notions are very transparent in case of abelian groups (the category where they come from). We discussed cases of (i) in Chapter 1 (see also the simplified Definition 1.1.12, which is the same for abelian groups, but often good enough). Theorem 9.1.19. Assume ♦λ E for a non–reflecting stationary subset E of λ consisting of limit ordinals cofinal to ω. Moreover, let A be a free R–algebra of cardinality < λ and suppose that R is p–cotorsion–free. Then there exists a strongly λ–free R–module M of cardinality λ with EndR M = A. Remark 9.1.20. If R A in Theorem 9.1.19 is only assumed to be p–cotorsion–free (and not necessarily free), then M can only be shown to be strongly λ–free as an A–module and p–cotorsion–free; the proof is the same.  Proof. We decompose E = β<λ Eβ such that each Eβ satisfies ♦λ Eβ , which is possible by Theorem 9.1.17. The module M will be constructed inductively by defining  a module structure on a given set M of cardinality λ. We fix a λ–filtration M = ν<λ Mν of M such that |Mν | = |ν| + |A| = |Mν+1 \ Mν | for each ν < λ. Moreover, we enumerate M by M = {mν | ν < λ}. Then, for any β < λ, the set {ν < λ | mβ ∈ / Mν } is bounded by some α < λ, since any element m belongs to Mν for all ν < λ large enough. Therefore we may replace Eβ by Eβ \ α, and thus the following additional condition holds without loss of generality: mβ ∈ Mν for all ν ∈ Eβ . Also, for each α < λ with cf(α) = ω, we can choose a strictly increasing sequence αn ∈ α \ E with sup αn = α n<ω

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because E only consists of limit ordinals cofinal to ω and so we may pick the αn s as successor ordinals. We will use the same Greek letter for a converging sequence and its limit, so the elements of the sequence only differ by the suffix. Similarly, as E is non–reflecting by assumption, we may choose a strictly increasing, continuous sequence (αν )ν
for any limit ordinal with cf(α) > ω. This is crucial, since the submodule Mα (for such an α) of the (continuous) λ–filtration of M must be a free A–module in order to proceed the transfinite construction. Note, this case does not occur for λ = ℵ1 . Using Theorem 9.1.8 and Lemma 9.1.16 inductively, we define an A–module structure on Mν . We begin with putting M0 = A. By the continuity of the ascending chain the construction is reduced to an inductive step passing from Mν to Mν+1 . We will carry on our induction hypothesis of the filtration at each step. In particular, the following three conditions must hold for each ν < λ: (i) Mν is a free right A–module. (ii) If ρ ∈ ν \ E, then Mν /Mρ is A–free. (iii) For ρ ∈ Eβ ∩ ν for some β, let (ρn )n<ω be the given sequence with supn<ω ρn = ρ (as displayed above for an ordinal α). If Mρn /mβ A is A–free for some n < ω and the Jensen–function gρ of ♦λ Eβ is an R–homomorphism / mβ A, gρ ∈ EndR Mρ with mβ gρ ∈ then we choose the modules Mρ+1 = F 0 or Mρ+1 = F 1 corresponding to Step Lemma 9.1.16 for b = mβ , F = Mρ and Fn = Mρn such that gρ does not extend to an endomorphism of Mρ+1 . Following these rules we now deal with the step from ν to ν + 1: if the hypotheses of condition (iii) are violated, if ν ∈ / E, for instance, we choose Mν+1 = Mν ⊕ A. The opposite case implies ν ∈ Eβ for some β < λ, and Mνn /mβ A / mβ A is is A–free for some n < ω, and the Jensen–function gν with mβ gν ∈ an R–homomorphism gν ∈ EndR Mν \ A. We apply the Step Lemma 9.1.16 as indicated in (iii) for ρ = ν to obtain Mν+1 . However, next we must check that the conditions (i) to (iii) are, indeed, carried over to ν + 1. If the hypotheses of condition (iii) are violated, this is obvious. In the other case the Step Lemma is designed to guarantee the following conditions:

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Condition (i) is the freeness of F ε in the Step Lemma. Condition (ii) needs that Mν+1 /Mρ is A–free. However, Mρ ⊆ Mνn for a large enough n < ω. Hence (ii) follows from the freeness of Mνn /Mρ (inductively) and Mν+1 /Mνn (by the Step Lemma). In the limit case  γ we have two possibilities: if cf(γ) = ω, then supn<ω γn = γ, hence Mγ = n<ω Mγn and Mγ is A–free with the help of (i) and (ii) by induction. If cf(γ) > ω, then, by our set–theoretic assumption (E non–reflecting), we havea limit supα
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9 The Black Box and its relatives

A–free by construction. On the other hand, Mα+1 /Mα is p–divisible by construction, that is, topologically speaking, Mα+1 is the p–adic closure of Mα in Mγ . But homomorphisms are continuous in the p–adic topology and so, necessarily, Mα+1 ϕ ⊆ Mα+1 , i.e. ϕ ∈ End Mα+1 extends gα , a contradiction. This finishes 2 the proof of EndR M = A.

The existence of R–modules M with EndR M = A assuming 2ℵ0 < 2ℵ1 Similar to the ♦–case we will need a decomposition into non–small stationary sets. This is a result which requires λ to be a successor cardinal. Theorem 9.1.21. If μ is an infinite cardinal such that Φλ (E) holds for λ = μ+ ,  then there is a decomposition E = β<λ Eβ such that Φλ (Eβ ) holds for all β < λ.

Proof. See Eklof, Mekler [142, p. 144].

2

Next we apply the Weak Diamond Principle Φω1 (ω1 ), which is equivalent to 2ℵ0 < 2ℵ1 , to derive a result similar to Theorem 9.1.19. It is interesting to observe that less set–theoretic input must be replaced by stronger algebraic arguments. This effect will occur again, when we dismiss weak CH and work entirely in ZFC. Theorem 9.1.22. Assume ZFC + Φλ (E) for some non–reflecting stationary subset E of limit ordinals cofinal to ω in a successor cardinal λ ≤ 2ℵ0 . If A is a p– cotorsion–free R–algebra of cardinality < λ, then there exists a strongly λ–free A–module M of cardinality λ with EndR M = A. If R A is a free R–module, then M is also a strongly λ–free R–module. Corollary 9.1.23. [ZFC + 2ℵ0 < 2ℵ1 ] Let A be a countable p–reduced and p– torsion–free R–algebra. Then there exists a strongly ℵ1 –free A–module M of cardinality ℵ1 such that EndR M = A. First we note that A in the corollary is also p–cotorsion–free by Corollary 1.1.25. Hence Corollary 9.1.23 is an immediate consequence of Theorem 9.1.22 because E = {α < ℵ1 | cf(α) = ω} is stationary and, obviously, also non–reflecting. From Theorem 9.1.13 follows Φℵ1 (ℵ1 ) and thus also Φℵ1 (E) for E as above.

Proof of Theorem 9.1.22. We modify the proof of Theorem 9.1.19. As before,  we decompose E = β<λ Eβ such that each Eβ satisfies Φλ (Eβ ); this time we apply Theorem 9.1.21 which requires λ = μ+ to be a successor cardinal. The module M will again be constructed by inductively defining a module structure on

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a given λ–filtration M = Theorem 9.1.19.



ν<λ Mν

311

of sets; we mainly run through the proof of

The A–Module structure on Mν can obviously be considered as a certain subset of the set AMν := Mν5 × A and maps Mν → Mν as subsets of Mν2 . Thus the A–module structure and endomorphisms together can be viewed as subsets of Relν := AMν ∪ Mν2 , where the AMν –part describes the A–module Mν . As in the proof of Theorem 9.1.19 we will, for simplicity, suppress the A–module structure and restrict ourselves to P(Relν ). We now define partition functions pβν : P(Relν ) −→ 2 for all ν ∈ Eβ . We put pβν (X) = 0 for X ∈ P(Relν ), if the following holds: (1) Mν is a free right A–module with the A–module structure induced by X. (2) If ρ ∈ ν \ E, then Mν /Mρ is A–free. (3) Let (νn )n<ω be the given sequence with supn<ω νn = ν. Then Mνn /mβ A is A–free for some n < ω. (4) If Mν0 and Mν1 are fixed extensions of Mν given by F 0 and F 1 , respectively, according to Step Lemma 9.1.16, identifying (M, Mνn , mβ | n < ω) and (F, Fn , b : n < ω), then ϕ does not extend to an element of EndR Mν0 . Note that |A| < 2ℵ0 and therefore F 0 and F 1 are chosen independent of ϕ. Otherwise we put pβν (X) = 1. By Φλ Eβ there is a global partition function pβ : Eβ → 2 such that, for all X ∈ P(Rel) = P(AM ∪ M 2 ), the set   ν ∈ Eβ | pβν (X ∩ Relν ) = pβ (ν) is stationary in λ. If we compare this with the prediction of ♦, we see that ♦ predicts more, namely, for any f ∈ map(M, M ), the entire partial map f  Mν is predicted, and not only the class to which it belongs, like here. However, the construction remains the same except for the following change of condition (iii), which now reads as follows: (iii)∗ For ρ ∈ Eβ ∩ ν for some β, let (ρn )n<ω be the given sequence with supn<ω ρn = ρ. If Mρn /mβ A is A–free for some n < ω, then let Mρ+1 be the module β F p (ρ) given by Step Lemma 9.1.16, replacing F, Fn by Mρ , Mρn and b by mβ (the same fixed extensions F 0 and F 1 as under (4) above).

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We would like to mention that this also applies mutatis mutandis to ν ∈ Eβ . Using the same arguments as before shows that M is a strongly λ–free A– module. So it remains to show that EndR M ⊆ A. If ϕ ∈ EndR M \ A, then, by (9.1.7), there is some ν ∈ λ \ E and some m ∈ Mν such that Mν = mA ⊕ D and mϕ ∈ / mA. Let m = mβ and let C = {γ < λ | γ a limit ordinal, Mγ ϕ ⊆ Mγ }; clearly, C is a cub in λ. Moreover, let M and ϕ be represented as above by X ∈ P(Rel). By Φλ (Eβ ) the set   T = ν ∈ Eβ | pβν (X ∩ Relν ) = pβ (ν) is stationary in λ. Therefore we can find ν ∈ C ∩ T and ν < γ < λ such that Mν+1 ϕ ⊆ Mγ and so, similar to the proof of Theorem 9.1.19, ϕ  Mν extends to an endomorphism of Mν+1 . Thus, by (4) and pβν (X ∩ Relν ) = pβ (ν), we have pβ (ν) = 1 and ϕ Mν extends to Mν0 as well as to Mν+1 = Mν1 . Hence it follows from Step Lemma 9.1.16 that mβ ϕ ∈ mβ A, a contradiction; this finishes the proof. 2 We could apply the prediction principle stated in Corollary 9.1.12 that holds under a weak form of GCH (assuming λ = μ+ = 2μ ). In this case a module similar to M in Theorem 9.1.19 can be constructed using a stationary subset E ⊆ λo = {α < λ | cf(α) = ω}. In this case we must circumvent the problem that we cannot assume that E is sparse. Thus the resulting module will not be strongly λ–free. We could replace strongly λ–freeness by the weaker ℵ1 –freeness (in order to advance in the construction through steps of the chain of submodules related to limit ordinals not cofinal to ω). However, under this weaker demand we will also be able to proceed without any additional axioms of ZFC. So we get ready to prove the required prediction principle known as ‘Black Box’, which we will then apply in Chapter 12 to derive theorems parallel to Theorem 9.1.19.

9.2

The Black Boxes

In order to develop ‘Black Box’ prediction principles we define various concepts: these are versions of ‘admissible sequences’ and preliminary ‘traps’ (tuples of partial homomorphisms and trees or submodules and sequences of S–pure elements). Of these concepts only the final ‘canonical homomorphisms’ and ‘traps’ are used later on. The other sequences of homomorphisms, partial and admissible traps are technical tools to find the correct subfamily of canonical homomorphisms and genuine traps, respectively. We consider case by case, beginning with the Strong Black Box. The choice of special (strong) cardinals – certain successor cardinals

9.2 The Black Boxes

313

– allows us to prove a stronger prediction principle; as a bonus we can avoid parts of the algebraic proofs in some applications. Thus the algebraic results are more easily verified but hold only for constructions that rely on these particular cardinals. Our proof [250] is a simple counting argument as in [102]. It is inspired by [142], using a nice trick due to Shelah, but we refrain from set–theoretic models and games. The Strong Black Box is recommended to readers short of time but interested in results concerning endomorphism rings or cotorsion pairs.

The Strong Black Box We consider three variants of the Strong Black Box: for constructing modules with prescribed endomorphism rings, for E–rings and for constructing ultra–cotorsion– free modules. These variants also cover many other possible situations to which we want to apply the Strong Black Box. Our three examples show how to formulate other variants of the Strong Black Box ready to be adapted for different applications without going into the details of the proof anymore. Moreover, we stress that these Black Boxes are clever, yet uncomplicated, counting arguments relying only on an undergraduate course on naive set theory. The Strong Black Box for endomorphism rings The Strong Black Box for endomorphism rings is designed for modules over S– rings R with prescribed endomorphism R–algebra A from a large specific class. In this subsection we formulate and prove the Strong Black Box in full detail. To do so we choose infinite cardinals κ, λ, μ such that |R| ≤ κ ≤ μ, μκ = μ and λ = μ+ .

(9.2.1)

Let A be an R–algebra of cardinality ρ ≤ λ such that there is a free (left) R– submodule

Raε BA = ε<ρ

with A and 1A ∈ BA , w.l.o.g. a0 = 1. BA ⊆ A ⊆∗ B

(9.2.2)

A denotes the completion of BA in the Hausdorff S–topology and As before, B ⊆∗ indicates that A is an S–pure submodule. Both Black Box constructions, the strong and the general one, deal with algebraic objects between direct sums and direct products. For the above described

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situation we formulate the parameters of the Strong Black Box accordingly to arrive at a canonical direct sum as well as a direct product of modules. They are our extremal objects: We substitute the above definition BA into

B= BA eα α<λ

and abbreviate aε eα by eα,ε . Thus we obtain the free R–module



BA eα = Reε,α , B= α<λ

(α,ε) ∈ λ×ρ

which, by (9.2.2), is an S–pure submodule of the free right A–module

eα A. B  := α<λ

 is an R–A–bimodule, in particular, that B  is dense in B. =B  It follows that B We order the product λ × ρ lexicographically; since ρ, λ are ordinals, the index set λ × ρ is also well–ordered.  we write For any element g ∈ B, ⊆ gα,ε eα,ε ∈ B g= (α,ε) ∈ λ×ρ



 α,ε . Re

(α,ε) ∈ λ×ρ

 are (converging) countable sums in the product. Defining Recall that elements in B the support of g by   [ g ] = (α, ε) ∈ λ × ρ | gα,ε = 0 ,  The support of any subset M of B  is defined by we have |[ g ]| ≤ ℵ0 for all g ∈ B.  [ g ]. [M ] = g∈M

Moreover, we define the λ–support of g by   [ g ]λ = α < λ | ∃ ε < ρ : (α, ε) ∈ [ g ] ⊆ λ and, symmetrically, the A–support of g by   [ g ]A = ε < ρ | ∃ α < λ : (α, ε) ∈ [ g ] ⊆ ρ.

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The A–support of g collects the ordinals ε < ρ needed in the sum representation gα,ε eα,ε . g= (α,ε) ∈ λ×ρ

These ordinals ε, or better the elements aε ∈ A, represent free generators of the ⊆ A, while the λ–support refers to the representation of g as an base module BA  = element of B α<λ BA eα . Next we define a λ–norm on λ by putting  {α}  = α + 1 (α < λ). This definition naturally extends to bounded subsets M of λ:  M  = sup  {α}  (M ⊆ λ), α∈M

 and also to elements g in B:    g  =  [ g ]λ , which is  g  = min β < λ | [ g ]λ ⊆ β . Note

 for B  = [ g ]λ ⊆ β ⇐⇒ g ∈ B β β



eα A.

α<β

We also define an A–norm of g by  g A =  [ g ]A . The reader who is only interested in the case A = R, i.e. in realizing the ring R as endomorphism ring of an R–module, can ignore the “ρ–” and the “A–  component” and just work with B = α<λ Reα . Finally, we define canonical homomorphisms; they will play a decisive role in the formulation and the proof of the Black Box. For this we fix once and for all bijections gγ : μ −→ γ for all γ with μ ≤ γ < λ, in particular, we put gμ = idμ . This is possible, since μ ≤ γ < μ+ = λ, hence |γ| = |μ|. For technical purposes we also put gγ = gμ for any γ < μ. Definition 9.2.1. Let gγ (γ < λ) be the above bijections and, for all (α, ε) ∈ λ × ρ, let gα,ε = gα × gε : μ × μ −→ α × ε be the natural product  map. We define P := (α,ε) ∈ I Reα,ε to be a canonical summand of B, if I is a subset of λ × ρ of size at most κ such that the following holds:

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9 The Black Box and its relatives

(i) (α, ε) ∈ I ⇒ (ε, ε) ∈ I. (ii) (α, ε) ∈ I ⇒ (α, 0) ∈ I. (iii) (α, ε) ∈ I, and α < ρ ⇒ (ε, α) ∈ I. (iv) (α, ε) ∈ I ⇒ (I ∩ (μ × μ)) gα,ε = I ∩ α × ε.  is a canonical homomorphism, if P is a canonAn R–homomorphism ϕ : P → B ical summand of B and Im ϕ ⊆ P; we put [ ϕ ] = [ P ], [ ϕ ]λ = [ P ]λ and  ϕ  =  P . Let C denote the set of all canonical homomorphisms. By (i), a canonical summand P satisfies  P A ≤  P . By our choice of cardinals, μκ = μ and thus 2κ ≤ μ and therefore λκ = λ (see Jech [284, Chapter I, (6.18)]). It follows that |C| = λ. We are now ready to state the main theorem of this subsection, i.e. the first variant of the Strong Black Box: The Strong Black Box 9.2.2 (Variant 1). Let R be an S–ring, κ, μ, λ be a cardi as before. Moreover, let E ⊆ λo := {α < λ | cf(α) = nal as in (9.2.1) and B, B ω} be a stationary subset of λ. Then there is a subset C∗ of canonical homomorphisms with the following properties: (1)  ϕ  ∈ E for all ϕ ∈ C∗ . (2) If ϕ, ϕ are two different elements of C∗ with the same norm α, then % &  [ ϕ ]λ ∩ ϕ λ  < α.  and for any subset (3) P REDICTION : For any R–homomorphism ψ : B → B I ⊆ λ × ρ with |I| ≤ κ, the set 

α ∈ E | ∃ ϕ ∈ C∗ :  ϕ  = α, ϕ ⊆ ψ, I ⊆ [ ϕ ]



is stationary. To prove this Strong Black Box we need further definitions and supporting results. We begin by defining an equivalence relation on C. While C has cardinality λ, the set of equivalence classes has only size μ, which is small enough to be used for predictions and still fine enough to provide sufficient information about the represented partial homomorphisms.

9.2 The Black Boxes

317

Definition 9.2.3. Two canonical homomorphism ϕ, ϕ are of the same type or equivalent (notation: ϕ ≡ ϕ ), if % & [ ϕ ] ∩ (μ × μ) = ϕ ∩ (μ × μ) and if there exists an order–isomorphism % & f : [ ϕ ] −→ ϕ with the following property: let eα,ε f := e(α,ε)f ((α, ε) ∈ [ ϕ ]), then f has a unique extension ϕ ϕ −→ Dom f : Dom as an R–homomorphism, and we require (xf)ϕ = (xϕ)f for all x ∈ Dom ϕ. Let P = Dom ϕ and P  = Dom ϕ , then the above equations can be visualized by the commutative diagram: f

[P ] −−−−→ ⏐ ⏐ 

[P  ] ⏐ ⏐ 

f

P −−−−→ P  ⏐ ⏐ ⏐ ⏐ ϕ ϕ  

f P −−−−→ P .

Since [ ϕ ] and [ ϕ ] are well–ordered, f : [ ϕ ] → [ ϕ ] is unique. Thus, if ϕ ≡ ϕ and [ ϕ ] = [ ϕ ], then f = id and so ϕ = ϕ . Clearly any type in (C, ≡) is represented by a subset V ⊆ μ × μ of cardinality at most κ, an order–type of a set of cardinality κ and a homomorphism from a free R– module P of rank κ into its completion P. The possibilities of these representing sets are bounded as follows: |{V ⊆ μ × μ | |V | ≤ κ}| = μκ = ! μ; there are at ! most ! ! κ  2 ≤ μ non–isomorphic well–orderings on a set of size κ; !HomR (P, P )! = 2κ for a fixed P . Therefore there are at most μ different types. Certain sequences of canonical homomorphisms will play an important role:

318

9 The Black Box and its relatives

Definition 9.2.4. Let ϕ0 ⊂ ϕ1 ⊂ . . . ⊂ ϕn ⊂ . . . (n < ω) be an increasing sequence of canonical homomorphisms. Then (ϕn )n<ω is said to be admissible, if [ ϕ0 ] ∩ (μ × μ) = [ ϕn ] ∩ (μ × μ) and  ϕn  <  ϕn+1  for all n < ω. Moreover, two admissible sequences (ϕn )n<ω , (ϕn )n<ω are of the same type or equivalent, if ϕn ≡ ϕn for all n < ω. We denote the set of all possible types of admissible sequences of canonical homomorphisms by T. Also, (ϕn )n<ω is called admissible for a sequence (βn )n<ω of ordinals in λ (for short: (ϕn , βn )n<ω is admissible), if (ϕn )n<ω is admissible with  ϕn  ≤ βn <  ϕn+1  and [ ϕn ] = [ ϕn+1 ] ∩ (βn × βn ) for all n < ω.  Note, if (ϕn )n<ω is admissible, then ϕ = n<ω ϕn is an element of C with  ϕ  = sup  ϕn  ∈ λo . n<ω

By definition, any type in T can be identified with a sequence (τn )n<ω for some types τn in (C, ≡) so that the corresponding subsets of μ × μ agree. Hence we obtain |T| ≤ μℵ0 = μ. If (ϕn )n<ω or (ϕn , βn )n<ω are admissible of type τ , then we also say that they are τ –admissible. Moreover, if τ = (τn )n<ω ∈ T and if (ϕn )n
9.2 The Black Boxes

319

We would like to rephrase the claim of Proposition 9.2.5 in terms of a game between two players, where the player choosing the partial homomorphisms will win. In the first move Player I chooses a type τ (obtained from a friendly kibitzer) and at each turn he chooses a homomorphism ϕn ∈ K, while Player II chooses ordinals βn ≥  ϕn  at his turns. We identify ourselves with Player I and will win, if we are able to continue the game ω steps and remain τ –admissible all the time, hence (ϕn , βn )n<ω will be τ –admissible. As in the first variant of the Strong Black Box we will succeed! In other words: There is a type τ ∈ T such that the following recursion holds. If ϕ0 , β0 , . . . , ϕn−1 , βn−1 , ϕn is a finite admissible sequence of type τ and βn is any ordinal such that  ϕn  ≤ βn , then we can find a partial homomorphism ϕn+1 ∈ K such that ϕ0 , β0 , . . . , βn−1 , ϕn , βn , ϕn+1 is a finite admissible sequence of type τ which extends the old one. Proof. Suppose, for contradiction, that there is no such type. Then, since the above formula is of ‘finite character’, we have, for any type τ ∈ T, (∀ ϕ0 ∈ K) (∃ β0 ≥  ϕ0 ) . . . (∀ ϕn ∈ K) (∃ βn ≥  ϕn ) . . . with (ϕn , βn )n<ω not τ –admissible. Hence this means that there is a k < ω such that the subsequence (ϕn , βn )n
Bα = Reβ,ε (α < λ). (β,ε) ∈ Tα

α We define C to be the set of all ordinals α < λ such that Bα ψ ⊆ B and βn (τ, ϕ0 , . . . , ϕn ) ≤ α for each type τ ∈ T and for any finite sequence (ϕ0 , . . . , ϕn ) with ϕi ∈ K and  ϕi  ≤ α (that is, if and only if [ ϕi ] ⊆ α × α) for all i ≤ n. Then C is unbounded since, given an arbitrary α0 < λ, we inductively define ordinals αk < λ (k < κ+ ≤ μ) by αk = sup{αl | l < k} for k a limit ordinal and αk+1 = αk + 1 ∨  Bαk ψ  ∨ sup Bk , where

  Bk = βn (τ, ϕ0 , . . . , ϕn ) | τ ∈ T, ϕi ∈ K,  ϕi  ≤ αk

is a set of cardinality at most μ, since |T| ≤ μ, |{ϕ ∈ K |  ϕ  ≤ αk }| ≤ |αk |κ ≤ μκ = μ provided that αk < λ. Then, using that |[ ϕ ]| ≤ κ for all ϕ ∈ K, it is easy to see that α = sup{αk | k < κ+ } is an element of C.

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9 The Black Box and its relatives

Now we choose an increasing sequence   max μ,  I ,  I A ≤ α0 < α1 < . . . < αn < . . .

(n < ω)

in C and put α = supn<ω αn (∈ C). Note, since αn ∈ C for all n < ω, we also α . Also, let {εn | n < ω} be an arbitrary but fixed set of elements have Bα ψ ⊆ B of α ∩ ρ. Using these αn s and εn s we inductively define subsets In (n < ω) of α × (α ∩ ρ) = Tα = [ Bα ] of cardinality at most κ by   I0 = I ∪ (αn , εn ) | n < ω and

1

2

3

ψ

g

In+1 = In ∪ In ∪ In ∪ In ∪ In ∪ In , where

  1 In = (β, 0) | ∃ β < λ, ε < ρ : (β, ε) ∈ I ,   2 In = (ε, ε) | ∃ β < λ : (β, ε) ∈ In ,   3 In = (ε, β) | ∃ β ∈ ρ : (β, ε) ∈ In ,  ψ In = [ ei ψ ], i∈In

and

g

In =

 

 (In ∩ (μ × μ)) gi ∪ (In ∩ Im gi ) gi−1 .

i∈In

These In s satisfy the required conditions: for n = 0 we have |I0 | ≤ κ and I0 ⊆ Tα by  I  and  I A ≤ α0 < α. Next suppose |In | ≤ κ and In ⊆ Tα . Then 1 2 3 ψ α and |In+1 | ≤ κ; also In ∪ In ∪ In ⊆ Tα and In ⊆ Tα , since Bα ψ ⊆ B g In ⊆ Tα by the definition of the gi s, and  thus In+1 ⊆ Tα . ∗ ∗ We now put I = n<ω In and P := i∈I ∗ Rei . Then |I | ≤ κ, I ∗ ⊆ Tα = α × (ρ ∩ α) and I ⊆ I ∗ = [ P ]. Moreover, (I ∗ ∩ (μ × μ)) gi = I ∗ ∩ Im gi for g all i ∈ I ∗ by the definition of the In s, and I ∗ also satisfies: if (β, ε) ∈ I ∗ , then (β, 0), (ε, ε) ∈ I ∗ , and if (β, ε) ∈ I ∗ , β ∈ ρ, then (ε, β) ∈ I ∗ (see definition of i the In s). Therefore P is a canonical summand of B (see Definition 9.2.1) with ψ P ψ ⊆ P, where the latter follows from the definition of the In s. Thus ϕ := ψ  P is a canonical homomorphism with I ⊆ [ ϕ ], i.e. ϕ ∈ K. Finally, we put ϕn = ϕ  (P ∩ Bαn ). Using the definitions of the Bαn s and of the cub C ⊆ Cψ and an ∈ C, it is easy to check that ϕn ∈ K (n < ω) and

9.2 The Black Boxes

321

that (ϕn )n<ω is an admissible sequence with  ϕn  ≤ αn <  ϕn+1 . Let τ ∈ T be the type of (ϕn )n<ω . By the definition of C ⊆ Cτ we also have βn ≤ αn , since [ ϕn ] ⊆ αn × αn for any n < ω. Therefore  ϕn  ≤ βn ≤ αn and [ ϕn ] = [ ϕ ] ∩ (αn × αn ) = [ ϕ ] ∩ (βn × βn ) = [ ϕn+1 ] ∩ (βn × βn ), i.e. (ϕn , βn )n<ω is τ –admissible. This contradicts the assumption that it is not so for βn . Hence the original statement holds and the proof is finished. 2 In order to prove the Strong Black Box we also need the following well–known lemma due to Shelah. For example, it also appears in [142]. For an ordinal α, a mapping ηα : ω → α is a ladder on α, if it is strictly increasing and Im α is cofinal in α; a family of such ladders indexed by various αs is called a ladder system. Lemma 9.2.6. Let E ⊆ λo be a stationary subset of λ = μ+ for some μ with μℵ0 = μ. Then there is a ladder system {ηα | α ∈ E} such that, for all cubs C, the set {α ∈ E | Im ηα ⊆ C} is stationary.

Proof. For any α ∈ E, let {ηαi | i < μ} be an enumeration of all ladders on α. By allowing repetitions we may choose μ independent of α. For each i < μ, let η i := {ηαi | α ∈ E}. We claim that there is an i < μ such that η i satisfies the conclusion of the lemma. Suppose not. Then, for any i < μ, there is a cub Ci ⊆ λ such that the set Ti := {α ∈ E | Im ηαi ⊆ Ci } is not stationary, i.e. there is a cub Di with Ti ∩ Di = ∅. Replacing Ci by the cub Ci ∩ Di , we may assume that Ti = ∅ for any i < μ, i.e. Im ηαi ⊆ Ci for all α ∈ E (i < μ). We put C = i<μ Ci . Then C is also a cub in λ (see Theorem 9.1.1 (b)). We choose an ordinal α ∈ C ∩ E which is a limit point of C, i.e. α = supn<ω αn for some αn ∈ C ∩ α with αn < αn+1 . Such limit points always exist since limit points of a cub in λ constitute a cub. Therefore the map ηα : ω → α defined by ηα (n) = αn is a ladder on α with Im ηα ⊆ C. By the above enumeration ηα = ηαi for some i < μ, hence Im ηαi ⊆ Ci . From C ⊆ Ci follows the contradiction Im ηαi ⊆ C. 2 Finally, we prove the main theorem of this subsection.

Proof of the Strong Black Box 9.2.2. First we decompose the given stationary set E into |T| ≤ μ pairwise disjoint stationary subsets, say  E= Eτ . τ ∈T

Applying Lemma 9.2.6 we choose, for each τ ∈ T, a ladder system {ηα | α ∈ Eτ }

322

9 The Black Box and its relatives

such that the set 

α ∈ Eτ | Im ηα ⊆ C



is stationary for any cub C.

For any α ∈ Eτ , we define Cα⊆ C to be the set of all canonical homomorphisms ϕ such that  ϕ  = α and ϕ = n<ω ϕn for some τ –admissible sequence (ϕn )n<ω with [ ϕn ] = [ ϕ ] ∩ (ηα (n) × ηα (n)) (n < ω). Note, for ϕ, ϕ ∈ Cα with Dom ϕ = Dom ϕ (if and only if [ ϕ ] = [ ϕ ]), we obtain ϕn = ϕn for all n < ω, and so ϕ = ϕ (cf. Definition 9.2.3). Now we define C∗ to be the union of all these Cα s, i.e.  C∗ = Cα . α∈E

Condition (1) holds by the definition of the Cα s. Next we show that condition (2) is satisfied. So let ϕ, ϕ ∈ C∗ with  ϕ  =  ϕ  = α. Then ϕ =  ϕ, ϕ  ∈ Cα , where α ∈ Eτ for some τ ∈ T and thus   ) ϕ , ϕ = ϕ for some τ –admissible sequences (ϕ ) , (ϕ n n<ω n n<ω . n<ω n n<ω n   Suppose that  [ ϕ ]λ ∩ [ ϕ ]λ  = α. Then there are αn ∈ [ ϕ ]λ ∩ [ ϕ ]λ with supn<ω αn = α. Let εn , εn ∈ ρ such that (αn , εn ) ∈ [ ϕ ] and (αn , εn ) ∈ [ ϕ ]. We consider two cases. First case: ρ < λ, i.e. ρ ≤ μ. Then gαn ,εn = gαn × gμ = gαn ,εn (n < ω). Since (ϕn )n<ω , (ϕn )n<ω are of the same type τ , we know that % & % & [ ϕ ] ∩ (μ × μ) = [ ϕ0 ] ∩ (μ × μ) = ϕ0 ∩ (μ × μ) = ϕ ∩ (μ × μ). Hence [ ϕ ] ∩ (αn × μ) = [ ϕ ] ∩ Im gαn ,εn = ([ ϕ ] ∩ (μ × μ)) gαn ,εn  % & % & = ϕ ∩ (μ × μ) gαn ,εn = ϕ ∩ Im gαn ,εn % & = ϕ ∩ (αn × μ) for all n < ω because Dom ϕ, Dom ϕ are canonical summands of B. Therefore    % &   % & [ϕ] = [ ϕ ] ∩ (αn × μ) = ϕ ∩ (αn × μ) = ϕ . n<ω

n<ω

Second case: ρ = λ. Then (αn , εn ) ∈ [ ϕ ] implies (εn , αn ) ∈ [ ϕ ] and so (αn , αn ) ∈ [ ϕ ] (see Definition 9.2.1). Similarly, we obtain (αn , αn ) ∈ [ ϕ ] and,

9.2 The Black Boxes

323

as in the first case,       [ ϕ ] ∩ (αn × αn ) = [ ϕ ] ∩ (μ × μ) gαn ,αn [ϕ] = n<ω

=

 %

n<ω

 % & &   % & ϕ ∩ (μ × μ) gαn ,αn = ϕ ∩ (αn × αn ) = ϕ .

n<ω

n<ω

Both cases yield ϕ = ϕ which proves (2).  be an R–homomorphism and let It remains to show (3). So let ψ : B → B I ⊆ λ × ρ with |I| ≤ κ. By Proposition 9.2.5, there is a type τ ∈ T such that: (∃ ϕ0 ∈ K) (∀ β0 ≥  ϕ0 ) . . . (∃ ϕn ∈ K) (∀ βn ≥  ϕn ) . . . : (ϕn , βn )n<ω is τ –admissible, where K = Kψ,I = {ϕ ∈ C | ϕ ⊆ ψ, I ⊆ [ ϕ ]}. α (reWe form a cub C such that, for all α ∈ C, α ≥ μ, α ≥  ϕ0 , Bα ψ ⊆ B call, Bα = (β,ε) ∈ Tα Reβ,ε with Tα = α × (α ∩ ρ)), and, if (ϕ0 , β0 , . . . , ϕn , βn ) is a finite part of one of the above τ –admissible sequences with βn < α, then there is also ϕn+1 ∈ K with [ ϕn+1 ] ⊆ α × α, and (ϕ0 , β0 , . . . , ϕn , βn , ϕn+1 ) is τ –admissible. Since C is a cub, the set   Eτ = α ∈ Eτ | Im ηα ⊆ C is stationary. Now let α ∈ Eτ be arbitrary but fixed and abbreviate αn = ηα (n) (∈ C). By the definition of C, we have  ϕ0  ≤ α0 < α1 , and so there is ϕ1 ∈ K with  ϕ1  ≤ α1 such that (ϕ0 , α0 , ϕ1 ) is τ –admissible. We proceed inductively on n < ω, i.e., whenever we have the τ –admissible sequence (ϕ0 , α0 , . . . , ϕn , αn ) with  ϕn  ≤ αn < αn+1 , we can find ϕn+1 ∈ K with  ϕn+1  ≤ αn+1 such that (ϕ0 , α0 , . . . , ϕn , αn , ϕn+1 ) is τ –admissible. Therefore we obtain an infinite τ –admissible sequence (ϕn , αn )n<ω , i.e.  ϕn  ≤ αn < ϕn+1  and [ ϕn+1 ] ∩ (αn × αn ) = [ ϕn ] (cf. Definition 9.2.4). We put ϕ = n<ω ϕn , then  ϕ  = supn<ω  ϕn  = supn<ω αn = α and   [ ϕk ] ∩ (αn × αn ) = [ ϕn ]. [ ϕ ] ∩ (αn × αn ) = k≥n



Hence ϕ = n<ω ϕn ∈ Cα ⊆ C∗ . Since α ∈ Eτ was arbitrary and Eτ is stationary the proof is finished. 2 We conclude this subsection with an ‘enumerated’ version of the Strong Black Box 9.2.2, which will be used in Chapter 12.

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9 The Black Box and its relatives

Corollary 9.2.7. Let the assumptions be the same as in the Strong Black Box 9.2.2. Then there is a family (ϕβ )β<λ of canonical homomorphisms such that (i)  ϕβ  ∈ E for all β < λ. (ii)  ϕγ  ≤  ϕβ  for all γ ≤ β < λ. (iii)  [ ϕγ ]λ ∩ [ ϕβ ]λ  <  ϕβ  for all γ < β < λ.  and for any subset (iv) P REDICTION : For any homomorphism ψ : B → B I ⊆ λ × ρ with |I| ≤ κ, the set 



α ∈ E | ∃ β < λ :  ϕβ  = α, ϕβ ⊆ ψ, I ⊆ [ ϕβ ]

is stationary.

Proof. By the Strong Black Box 9.2.2 there is a class C∗ of canonical homomorphisms satisfying the conditions (i) and (iv), which are obviously independent of the enumeration (cf. conditions (1) and (3) in Theorem 9.2.2). Moreover, we put an arbitrary well–ordering on the sets Cα = {ϕ ∈ C∗ |  ϕ  = α} (α ∈ E) and define ϕ ∈ Cα to be less than ϕ ∈ Cα if α < α . This defines a well–ordering on C∗ , and hence there is a corresponding ordinal λ∗ such that the condition (ii) is satisfied. From |C∗ | = μ < μ+ = λ it follows that each initial segment of λ∗ has cardinality < λ and thus λ = λ∗ . Condition (iii) also easily follows, since  [ ϕ ]λ ∩ [ ϕ ]λ  <  ϕ  is obvious for  ϕ  <  ϕ , and it coincides with con2 dition (2) in the Strong Black Box 9.2.2 for  ϕ  =  ϕ .

The Strong Black Box for E–rings In this subsection we shall formulate another version of the Strong Black Box which will only be slightly different from the last one. Hence we shall only outline the proofs. Let R be an S–ring as in the previous subsection and suppose κ, μ, λ are cardinals satisfying (9.2.1). The second variant of the Strong Black Box will be used to construct E(R)–algebras over a commutative ring R. As usual, we define a direct sum and a direct product, which are the initial objects of the Strong Black Box. Here we replace the free R–module B and its  of the last subsection by a polynomial ring B which is an S–adic completion B R–algebra over R with the usual multiplicatively closed subset S. Let B = R[ xα | α < λ ]

325

9.2 The Black Boxes

be the polynomial ring in the commuting variables xα and let M be the set of all monomials. Then

Rm B= m∈ M

is a free R–module. For any ⊆ g = (gm m)m∈ M ∈ B



 Rm,

m∈ M

we define the support of g by   [ g ] = m ∈ M | gm = 0 ;  The support of any subset M of B  is defined note that |[ g ]| ≤ ℵ0 for all g ∈ B. by  [M ] = [g]. g∈M

Moreover, we define the X–support of g by   [ g ]X = α ∈ λ | xα occurs in some m ∈ [ g ] ⊆ λ.  by Similarly, the X–support is extended to subsets M of B  [ g ]X . [ M ]X = g∈M

Next we define a norm on λ and let  {α}  = α + 1

(α ∈ λ).

This naturally extends to bounded subsets M of λ:  M  = sup  {α}  α∈M

(M ⊆ λ)

 and also to elements g in B:    g  =  [ g ]X , that is  g  = min β ∈ λ | [ g ]X ⊆ β . Note that

β for Bβ = R[ xα | α < β ]. [ g ]X ⊆ β ⇐⇒ g ∈ B

 the above definitions extend naturally. As before, for a subset M of B Again we need to say what we mean by a canonical homomorphism. For this we fix bijections gγ : μ −→ γ for all γ with μ ≤ γ < λ

326

9 The Black Box and its relatives

where we put gμ = idμ . For technical purposes we also put gγ = gμ for γ < μ. Definition 9.2.8. Let gγ (γ < λ) be the above bijections. We define P = R[xα | α ∈ I] to be a canonical R–subalgebra of B, if the following holds for I ⊆ λ: (i) |I| ≤ κ; (ii) α ∈ I =⇒ (I ∩ μ) gα = I ∩ α.  is a canonical homomorAccordingly, an R–module homomorphism ϕ : P → B phism, if P is a canonical subalgebra of B and Im ϕ ⊆ P; we put [ ϕ ] = [ P ], [ ϕ ]X = [ P ]X and  ϕ  =  P . Let C denote the set of all canonical homomorphisms. Clearly |C| = λ (as before). We are now ready to formulate the second variant of the Strong Black Box: The Strong Black Box 9.2.9 (Variant 2). Let λ = μ+ , μκ = μ, R be an S–ring  as above. Moreover, let E ⊆ λo = {α ∈ λ | cf(α) = ω} be a stationary and B, B subset of λ. Then there is a subset C∗ of canonical homomorphisms with the following properties: (1) If ϕ ∈ C∗ , then  ϕ  ∈ E. (2) If ϕ, ϕ are two different elements of C∗ with the same norm α, then % &  [ ϕ ]X ∩ ϕ X  < α.  and for any subset (3) P REDICTION : For any R–homomorphism ψ : B → B I ⊆ λ with |I| ≤ κ, the set 

α ∈ E | ∃ ϕ ∈ C∗ :  ϕ  = α, ϕ ⊆ ψ, I ⊆ [ ϕ ]

is stationary.



9.2 The Black Boxes

327

Note that, although the above theorem reads exactly like the first variant of the Black Box 9.2.2, the definition of a canonical homomorphism is slightly different from Definition 9.2.1. As mentioned before, we will not give all the details of the proof (again). However, we state all the used definitions and results, even when they coincide with their counterpart in the first version. We begin by adjusting the definition of the equivalence relation on C: Definition 9.2.10. Two canonical homomorphisms ϕ, ϕ are of the same type or equivalent (notation: ϕ ≡ ϕ ), if [ ϕ ]X ∩ μ = [ ϕ ]X ∩ μ and if there exists an order–isomorphism f : [ ϕ ]X → [ ϕ ]X with the following property: we let     k0 α0 , . . . , αn ∈ [ ϕ ]X , xα0 · · · xkαnn f = xkα00 f · · · xkαnn f then f has a unique extension ϕ ϕ −→ Dom f : Dom as an R–homomorphism, and we require (xf)ϕ = (xϕ)f for all x ∈ Dom ϕ. Note, f : [ ϕ ]X → [ ϕ ]X is unique, since [ ϕ ]X , [ ϕ ]X are well–ordered. Thus, if ϕ ≡ ϕ and [ ϕ ]X = [ ϕ ]X , then f = id[ϕ]X and so ϕ = ϕ . As before, it is easy to see that there are at most μ different types (equivalence classes) in (C, ≡). Next we recall the definition of an admissible sequence and of all other related notions: Definition 9.2.11. Let ϕ0 ⊂ ϕ1 ⊂ . . . ⊂ ϕn ⊂ . . . (n < ω) be an increasing sequence of canonical homomorphisms. Then (ϕn )n<ω is admissible, if [ ϕ0 ]X ∩ μ = [ ϕn ]X ∩ μ and  ϕn  <  ϕn+1  for all n < ω. Moreover, two admissible sequences (ϕn )n<ω , (ϕn )n<ω are of the same type or equivalent, if ϕn ≡ ϕn for all n < ω. Let T denote the set of all possible types of admissible sequences of canonical homomorphisms.

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9 The Black Box and its relatives

Also, (ϕn )n<ω is admissible for a sequence (βn )n<ω of ordinals in λ (or (ϕn , βn )n<ω is admissible), if (ϕn )n<ω is admissible with  ϕn  ≤ βn <  ϕn+1  and [ ϕn ]X = [ ϕn+1 ]X ∩ βn for all n < ω.  If (ϕn )n<ω is admissible, then ϕ = n<ω ϕn is an element of C with  ϕ  ∈ λo , and |T| ≤ μℵ0 = μ follows as before. If (ϕn )n<ω or (ϕn , βn )n<ω are admissible of type τ , then we also say that they are τ –admissible. Moreover, if τ = (τn )n<ω ∈ T and (ϕn )n
9.2 The Black Boxes

329

Proof. Suppose, for contradiction, that there is no such type as claimed in Proposition 9.2.12. Then, since the above formula is of ‘finite character’, we have, for any type τ ∈ T, (∀ ϕ0 ∈ K) (∃ β0 ≥  ϕ0 ) . . . (∀ ϕn ∈ K) (∃ βn ≥  ϕn ) . . . with (ϕn , βn )n<ω not τ –admissible. Hence this means that there is a k < ω such that the subsequence (ϕn , βn )n
 

 (In ∩ μ) gi ∪ (In ∩ i) gi−1 .

i∈In

We put I∗ =



In and P = R[ xβ | β ∈ I ∗ ].

n<ω

It is easy to check that P is a canonical subalgebra satisfying  P  = α and P ψ ⊆ P. Hence ϕ = ψ  P is a canonical homomorphism with I ⊆ [ ϕ ]X , i.e. ϕ ∈ K. Finally, we put ϕn = ϕ  (P ∩ Bαn ). Using the same arguments as in the proof of Proposition 9.2.5, we deduce that (ϕn , βn )n<ω is a τ –admissible sequence for some type τ and for βn = βn (τ, ϕ0 , . . . , ϕn ). This contradiction finishes the proof. 2 We have now provided all the necessary definitions and results to prove the main theorem of this subsection.

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9 The Black Box and its relatives

Proof of the Strong Black Box 9.2.9. Exactly as in the proof of Theorem 9.2.2 we decompose the given stationary set E into |T| ≤ μ pairwise disjoint stationary subsets,  E= Eτ , τ ∈T

and, for each τ ∈ T, we choose a ladder system {ηα | α ∈ Eτ } such that the set   α ∈ Eτ | Im ηα ⊆ C is stationary for any cub C, (cf. Lemma 9.2.6). Also as in the first variant of the Strong Black Box, we define  Cα C∗ = α∈E

where, for each α ∈ Eτ , the set Cα consists of all canonical homomorphisms ϕ such that  ϕ  = α and ϕ = n<ω ϕn for some τ –admissible sequence (ϕn )n<ω with [ ϕn ]X = [ ϕ ]X ∩ ηα (n) (n < ω). Note, for ϕ, ϕ ∈ Cα with Dom ϕ = Dom ϕ (if and only if [ ϕ ]X = [ ϕ ]X ) we deduce ϕ = ϕ . Now Condition (1) is immediate by the definition of C∗ . Condition (2) follows from       [ ϕ ]X ∩ αn = [ ϕ ]X ∩ μ gαn [ ϕ ]X = n<ω

=

 

n<ω

n<ω

    [ ϕ ]X ∩ μ gαn = [ ϕ ]X ∩ αn = [ ϕ ]X n<ω

for μ ≤ αn ∈ [ ϕ ]X ∩ [ ϕ ]X with supn<ω αn =  ϕ  =  ϕ  (cf. 9.2.2). Finally, the proof of Condition (3) is the same as the corresponding part of the proof of the Strong Black Box 9.2.2 using Proposition 9.2.12 instead of Proposition 9.2.5. 2 As in Corollary 9.2.7 we also present an ‘enumerated’ version of the Strong Black Box 9.2.9. For the proof we refer to that of Corollary 9.2.7. Corollary 9.2.13. Let the assumptions be the same as in the Strong Black Box 9.2.9. Then there is a family (ϕβ )β<λ of canonical homomorphism such that

9.2 The Black Boxes

331

(i)  ϕβ  ∈ E for all β < λ; (ii)  ϕγ  ≤  ϕβ  for all γ ≤ β < λ; (iii)  [ ϕγ ]X ∩ [ ϕβ ]X  <  ϕβ  for all γ < β < λ;  and for any subset (iv) P REDICTION : For any R–homomorphism ψ : B → B I of λ with |I| ≤ κ, the set   α ∈ E | ∃ β < λ :  ϕβ  = α, ϕβ ⊆ ψ, I ⊆ [ ϕβ ]X is stationary. The Strong Black Box for ultra–cotorsion–free modules In this subsection we will formulate and prove the necessary third variant of the Strong Black Box in full detail. Comparing the version presented here with the ones given in the two previous subsections should enable the reader to understand that the Strong Black Box can be formulated (and proved) in rather different settings provided that the cardinalities in question are bounded by μ (number of types), respectively by λ (e.g. number of canonical summands). This variant is used to construct ‘very’ rigid families of R–modules employed to characterize the lattice of cotorsion pairs. Let R be an S–ring as before and suppose that the cardinals μ, κ, λ satisfy (9.2.1). The following simple but basic definition will only be needed later, namely in the corresponding subsection of Chapter 11. Definition 9.2.14. An R–module M is ultra–cotorsion–free, if the only submodule U of M with |U | = |M | and M/U S–cotorsion–free is M itself. As usual, we must define a suitable direct sum and a related direct product of modules as an initial step for the setting of the third variant of the Strong Black Box and for the construction of ultra–cotorsion–free modules. Again, let B be a  be its S–adic completion; we set B =  free R–module and let B α<λ Reα . For any   α, ⊆ g = (gα eα )α<λ ∈ B Re α<λ

we define the support of g by   [ g ] = α < λ | gα = 0 ⊆ λ  by and note |[ g ]| ≤ ℵ0 . The support is extended to any subset M of B  [M ] = [ g ]. g∈M

332

9 The Black Box and its relatives

Next we define a norm on λ and put  {α}  = α + 1

(α < λ).

This naturally extends to bounded subsets M of λ:  M  = sup  {α}  α∈M

(M ⊆ λ)

 and also to elements g in B:    g  =  [ g ] , that is  g  = min β ∈ λ | [ g ] ⊆ β . Note,

β for Bβ = [ g ] ⊆ β ⇐⇒ g ∈ B



Reα .

α<β

We also need to define canonical summands and other ‘canonical objects’ which shall play a decisive role in the formulation and the proof of the Strong Black Box. Note, in the version presented here we basically want to predict kernels of homomorphisms (i.e. submodules, respectively their elements) not the homomorphisms themselves (see Definition 9.2.14). As before, we fix bijections gγ : μ −→ γ for all γ with μ ≤ γ < λ, where we put gμ = idμ . For technical purposes we also put gγ = gμ for γ < μ. Definition 9.2.15.  Let gγ (γ < λ) be the above bijections. We define P = α∈I Reα to be a canonical summand of B, if the subset I ⊆ λ has size at most κ and if (I ∩ μ)gα = I ∩ α for any α ∈ I. We call (P, v) a canonical pair, if P is a canonical summand of B and v is an S–pure element of P.  satisfying Moreover, an infinite sequence (vn )n<ω of S–pure elements of B  vn  <  vn+1 

(n < ω)

is a Signac–branch. A pair (P, (vn )n<ω ) is called a canonical Signac–pair, if P is a canonical summand of B and (vn )n<ω is a Signac–branch with the following properties: (i) vn ∈ P for all n < ω; and (ii)  P  = supn<ω  vn .

9.2 The Black Boxes

333

Let P denote the set of all canonical pairs and let C denote the set of all canonical Signac–pairs. It follows immediately from the above definition that |P| = |C| = λ. Notice, the way Signac–branches are constructed mathematically is supposed to remind the reader of the painting technique used by the pointillist Signac; we refer, for instance, to Signac’s tree exhibited at [370]. In the definition of a trap presented below we will consider pairs (b, π) from  where the first component b is an S–pure element in B. P = B × R, Definition 9.2.16. Let (P, (vn )n<ω ) be a canonical Signac–pair, b an S–pure ele Then ment of B and π ∈ R.   t = P, (vn )n<ω , b, π is said to be a trap, if b ∈ P (especially  b  <  P ). We put [ t ] = [ P ] and  t  =  P . The reader familiar with the General Black Box 9.2.27 will immediately observe links to Definition 9.2.25 of traps given there. We are now ready to state the desired variation of the Strong Black Box:  be The Strong Black Box 9.2.17 (Variant 3). Let R be an S–ring and let B, B κ as before. Moreover, let |R| ≤ κ ≤ μ ≤ λ be infinite cardinals such that μ = μ and λ = μ+ and let E ⊆ λo = {α ∈ λ | cf(α) = ω} be a stationary subset of λ. Then there is a family C∗ of traps   t = Pt , (vt,n )n<ω , bt , πt with the following properties: (1) If t ∈ C∗ , then  t  ∈ E. (2) If t and t are two different elements of C∗ with the same norm α, then % &  [ t ] ∩ t  < α.  of cardinality λ, for (3) P REDICTION : For any set U of S–pure elements of B  the set any S–pure element b of B and for any π ∈ R,   α ∈ E | ∃ t ∈ C∗ :  t  = α, {vt,n | n < ω} ⊆ U, b = bt , π = πt is stationary.

334

9 The Black Box and its relatives

To prove the above theorem we need further definitions and other results. We begin with an equivalence relation on P. Definition 9.2.18. Two canonical pairs (P, v), (P  , v  ) are of the same type or equivalent (notation: (P, v) ≡ (P  , v  )), if % & [P ] ∩ μ = P ∩ μ and if there exists an order–isomorphism % & f : [ P ] −→ P  with the following property: we let eα f = eαf

(α ∈ [ P ]),

then f has a unique extension  f : P −→ P as an R–homomorphism, and we require v f = v  . Note, f : [ P ] → [ P  ] is unique, since [ P ], [ P  ] are well–ordered. Thus, if (P, v) ≡ (P  , v  ) and [ P ] = [ P  ], then f = id[ P ] and so (P, v) = (P  , v  ). Obviously, any type in (P, ≡) can be represented by a subset V of μ of cardinality at most κ, an order–type of a set M of cardinality at most κ and a countable sequence (αn , rn )n<ω with αn ∈ M , rn ∈ R (which describes v). Therefore there are at most μκ · κκ · κℵ0 · |R|ℵ0 = μ different types in (P, ≡). Next we consider certain infinite sequences of canonical pairs: Definition 9.2.19. A sequence (Pn , vn )n<ω of canonical pairs is said to be admissible, if P0 ⊂ P1 ⊂ . . . ⊂ Pn ⊂ . . . (n < ω), [ P0 ] ∩ μ = [ Pn ] ∩ μ, and  Pn  <  vn+1  (≤  Pn+1 ) for each n < ω. Moreover, two admissible sequences (Pn , vn )n<ω , (Pn , vn )n<ω are of the same type or equivalent, if (Pn , vn ) ≡ (Pn , vn ) for all n < ω. Let T denote the set of all possible types of admissible sequences of canonical pairs.

9.2 The Black Boxes

335

We also say that (Pn , vn )n<ω is admissible for a sequence (βn )n<ω of ordinals in λ, if (Pn , vn )n<ω is admissible such that  Pn  ≤ βn <  Pn+1  and [ Pn ] = [ Pn+1 ] ∩ βn for all n < ω.    Note, if (Pn , vn )n<ω is admissible, then P = n<ω Pn , (vn )n<ω is a canonical Signac–pair, since  P  = supn<ω  Pn  = supn<ω  vn . It follows immediately from the above definition that any type τ in T can be represented by τ = (τn )n<ω , where the τn s are equivalence classes of (P, ≡) with the same underlying subset V of μ. Hence we deduce |T| = μℵ0 = μ. If (Pn , vn )n<ω is an admissible sequence of type τ , then we also use the notion τ –admissible. Moreover, if τ = (τn )n<ω ∈ T and (Pn , vn )n
336

9 The Black Box and its relatives

Proof. Suppose, for contradiction, that there is no such type. Then, for any τ ∈ T, we have: (∀ (P0 , v0 ) ∈ K) (∃ β0 ≥  P0 ) . . . (∀ (Pn , vn ) ∈ K) (∃ βn ≥  Pn ) . . . with (Pn , vn )n<ω not τ –admissible for (βn )n<ω . Hence this means that there is a k < ω such that the subsequence ((Pn , vn ), βn )n
and In+1 = In ∪ We put

I∗

=



 

 (In ∩ μ)gi ∪ (In ∩ i)gi−1 .

i∈In n<ω In .

Then it is easy to check that

Reα P := α∈I ∗

is a canonical summand of B such that b ∈ P , vn ∈ P for all n < ω and  P  =  I ∗  = α = sup  vn , n<ω

i.e. (P, (vn )n<ω ) is a canonical Signac–pair. Finally, let

  Bαn = Pn = P ∩ Bαn Reβ , β<αn

337

9.2 The Black Boxes

i.e. [ Pn ] = [ P ] ∩ αn . Then b ∈ P0 (⊆ Pn ), since  b  ≤ α0 ([ b ] ⊆ α0 ) by the n for any n < ω, since  vn  ≤ αn ([ vn ] ⊆ αn ) by definition of C and vn ∈ P the choice of the vn s and αn s. Hence (Pn , vn ) ∈ K for all n < ω. Moreover, [ P0 ] ∩ μ = [ Pn ] ∩ μ = [ P ] ∩ μ

(n < ω),

μ ≤ α0 < α1 < . . . < αn < . . .

(n < ω).

since Therefore (Pn , vn )n<ω is an admissible sequence, say of type τ ∈ T. By the definition of C we also have that  Pn  ≤ βn = β(τ, P0 , . . . , Pn ) ≤ αn for any n < ω and thus [ Pn+1 ] ∩ βn = [ Pn ] ∩ βn = [ Pn ]. Hence (Pn , vn )n<ω is τ –admissible for (βn )n<ω , contradicting the assumption that it is not for βn = βn (τ, P0 , . . . , Pn ). Therefore the original conclusion holds, and so the proof is finished. 2 We are now ready to prove the main theorem of this subsection.

Proof of the Strong Black Box 9.2.17. Let p(B) denote the set of all S–pure  Then |P| = λ · 2κ = λ. elements of B and put P = p(B) × R. First we partition the given stationary set E into |T| ≤ μ pairwise disjoint stationary subsets, say  E= Eτ . τ ∈T

Moreover, for each τ ∈ T, we decompose Eτ into |P| = λ pairwise disjoint stationary subsets:  Eτ = Eτ,p . p∈P

Note, for p = (b, π) ∈ P and τ ∈ T, we may assume that  b  < α for all α ∈ Eτ,p . For each τ ∈ T and each p ∈ P, we choose a ladder system {ηα | α ∈ Eτ,p } such that the set   α ∈ Eτ,p | Im ηα ⊆ C is stationary for any cub C

338

9 The Black Box and its relatives

(cf. Lemma 9.2.6). Let τ ∈ T, p = (b, π) ∈ P and α ∈ Eτ,p ; note  b  < α. We define Cα to be theset of all traps t = (P, (vn )n<ω , b, π) such that  t  =  P  = α and P = n<ω Pn for some τ –admissible sequence (Pn , vn )n<ω of canonical pairs with [ Pn ] = [ P ] ∩ ηα (n). Note, for t, t ∈ Cα with [ t ] = [ t ] (if and only if P = P  ), we clearly deduce t = t , since bt = b = bt , πt = π = πt and & % [ Pt,n ] = Pt ,n , (Pt,n , vt,n ) ≡ (Pt ,n , vt ,n ) imply (Pt,n , vt,n ) = (Pt ,n , vt ,n ) for all n < ω (cf. Definitions 9.2.16 and 9.2.18). Now we define C∗ to be the union of all these Cα s, i.e.  C∗ = Cα . α∈E

Condition (1) of the Strong Black Box follows from the definition of C∗ . To see (2), let t, t ∈ C∗ with  t  =  t  = α. Then t, t ∈ Cα and thus t = (P  , (vn )n<ω , b , π  )

t = (P, (vn )n<ω , b, π), with and

b = b ,

π = π,

(P, (vn )n<ω ), (P  , (vn )n<ω ) are of the same type τ,

since α ∈ Eτ,(b,π) , where P =



Pn ,

P =

n<ω

and [ P ] ∩ ηα (n) = [ Pn ],



Pn ,

n<ω

%

& % & Pn ∩ ηα (n) = Pn .

Suppose, for contradiction, that  [ t ] ∩ [ t ]  = α (recall: [ t ] = [ P ] and = [ P  ]). Then there are αn ∈ [ P ] ∩ [ P  ] with supn<ω αn = α. Now  %  & % & ([ P ] ∩ μ)gαn = [ P ] ∩ αn and P ∩ μ gαn = P  ∩ αn ,

[ t ]

since P , P  are canonical summands of B (see Definition 9.2.15). Moreover, % & % & [ P ] ∩ μ = [ P0 ] ∩ μ = P0 ∩ μ = P  ∩ μ

9.2 The Black Boxes

and therefore [P ] =

 

339

 % &   % & [ P ] ∩ αn = P  ∩ αn = P  .

n<ω

n<ω

This implies (P, (vn )n<ω ) = (P  , (vn )n<ω ), and so property (2) follows.  of It remains to show (3). To do so, let U be a set of S–pure elements of B  cardinality λ and let p = (b, π) ∈ P = p(B) × R. By Proposition 9.2.20, there is a type τ ∈ T such that (∃ (P0 , v0 ) ∈ K) (∀β0 ≥  P0 ) . . . (∃ (Pn , vn ) ∈ K) (∀ βn ≥  Pn ) . . . with (Pn , vn )n<ω admissible of type τ for (βn )n<ω , where

  K = (P, v) ∈ P | v ∈ U, b ∈ P .

Let C be the set of all ordinals α < λ such that α ≥ μ, α ≥  P0  and, if (Pn , vn )n≤k is a finite part of one of the above τ –admissible sequences for (βn )n≤k with βk < α, then there is (Pk+1 , vk+1 ) ∈ K with [ Pn+1 ] ⊆ α, and (Pn , vn )n≤ k+1 is τ –admissible. Obviously, C is a cub. Therefore  = {α ∈ Eτ,p | Im ηα ⊆ C} Eτ,p is stationary by Lemma 9.2.6.  be fixed, i.e. η (n) ∈ C for all n < ω. By the In the following let α ∈ Eτ,p α definition of C we have  P0  ≤ ηα (0) < ηα (1), and so there is (P1 , v1 ) ∈ K with  P1  ≤ ηα (1) such that (Pn , vn )n≤1 is τ –admissible for (ηα (0), ηα (1)). We proceed like this for each n < ω, that is, whenever we have a sequence (Pn , vn )n≤k which is τ –admissible for (ηα (n))n≤k , we can find (Pk+1 , vk+1 ) ∈ K with  Pk+1  ≤ ηα (k+1) such that (Pn , vn )n≤ k+1 is τ –admissible for (ηα (n))n≤ k+1 . Therefore we obtain an infinite τ –admissible sequence  (Pn , vn )n<ω with  Pn  ≤ ηα (n) and [ Pn+1 ] ∩ ηα (n) = [ Pn ]. We put P = n<ω Pn . Then  P  = sup  Pn  = sup ηα (n) = α n<ω

and [ P ] ∩ ηα (n) =

n<ω

 i≥n

 [ Pi ] ∩ ηα (n) = [ Pn ].

340

9 The Black Box and its relatives

 was arbitrary and E  is stationHence (P, (vn )n<ω , b, π) ∈ Cα . Since α ∈ Eτ,p τ,p ary, the proof is finished. 2

We finish this subsection with an ‘enumerated’ version of the third variant of the Strong Black Box 9.2.17. For the proof we refer the reader to the that of Corollary 9.2.7. Corollary 9.2.21. Let the assumptions in the corollary be the same as in the Strong Black Box 9.2.17. Then there is a family (tβ = (Pβ , (vβ,n )n<ω , bβ , πβ ))β<λ of traps such that (i)  tβ  ∈ E for all β < λ; (ii)  tγ  ≤  tβ  for all γ ≤ β < λ; (iii)  [ tγ ] ∩ [ tβ ]  <  Pβ  for all γ < β < λ;  of cardinality λ, for (iv) P REDICTION : For any set U of S–pure elements of B  the set any S–pure element b of B, and for any π ∈ R,   α ∈ E | ∃ β < λ :  tβ  = α, {vβ,n | n < ω} ⊆ U, b = bβ , π = πβ is stationary.

The more General Black Box Here we pose fewer restrictions on the cardinal λ, hence we must work harder for the combinatorial prerequisites as well as for the final algebraic results. The reader who does not care about the particular size of the constructed objects, but only about the fact that the construction takes place in ZFC, can skip this section and go directly to Section 9.3 for another prediction principle or to the applications by the Strong Black Box in Chapters 11 and 12. The attractive feature of this subsection (compared with the Strong Black Box) is that the combinatorial setting is based on trees, which allow (geometric) pictures of the situation under discussion, e.g. elements of modules are constructed on branches of a tree and there are families of trees which constitute a forest, so that particular submodules have their personal tree. The trees For a non–empty set X we let TX = ω> X = {τ : n → X | n < ω}

9.2 The Black Boxes

341

be the tree on X, where we consider the natural number n as the set {0, . . . , n−1}. The functions τ : n → X are called the nodes of the tree TX and τ has length l(τ ) = Dom τ = n; we also write τ = τ (0)∧ . . . ∧ τ (n − 1). The nodes are ordered by set–theoretical containment: σ ≤ τ , if and only if σ ⊆ τ , this is to say that Dom σ ⊆ Dom τ and τ  Dom σ = σ. If τ ∈ TX , then we consider the associated finite branch, which is the linearly ordered set {τ  i | 0 < i ≤ l(τ )} of l(α) elements, which is naturally often identified with τ . Infinite branches of TX are the (disjoint) elements in Br(TX ) = ω X = {f : ω −→ X} and, for f ∈ Br(TX ), we also identify f with the set of its restrictions: f = {f  n | n < ω}, which is a maximal, infinite, linearly ordered subset of TX . Moreover, recall that a subset of a tree TX is a subtree, if initial segments of branches belong again to the this subset; so a branch f and its initial segments are an example of a quite thin subtree of TX . If X = U × V for sets U, V , then we will deal with three trees, namely TX where each node τ has values τ (i) = (τ1 (i), τ2 (i)) ∈ U × V, and TU , TV which are images of TX . We will consider TV as a nursery and attach a branch f ∈ Br(TU ) to TV to get a subtree of TX having all the desired genes: If τ = τ (0)∧ . . . ∧ τ (n − 1) ∈ TV and f = (f (0), f (1), . . . ) ∈ Br(TU ), then f × τ = (f (0), τ (0))∧ . . . ∧ (f (n − 1), τ (n − 1)) is a branch of TX of length n, and the collection f × TV of all these branches (for τ ∈ TV ) constitutes a subtree of TX . If v ∈ Br(TV ) is an infinite branch v = (v(0), v(1), . . . ), then   f × v = (f (0), v(0))∧ (f (1), v(1))∧ . . . ∈ Br(TX ), and with f ×Br(TV ) := {f ×v | v ∈ Br(TV )} we have Br(f ×TV ) = f ×Br(TV ). If τ ∈ TU ×V , then τ = (τU (0), τV (0))∧ . . . ∧ (τU (n − 1), τV (n − 1)) is a branch of length n, and we can rearrange τ as τ = (τU , τV ) with τU = τU (0)∧ . . . ∧ τU (n − 1) ∈ TU and τV = τV (0)∧ . . . ∧ τV (n − 1) ∈ TV . Thus TU ×V ⊆ TU × TV .

342

9 The Black Box and its relatives

The norm Now we fix three cardinals κ, λ and μ such that μ is regular, λκ = λℵ0 and κ+ ≤ μ ≤ λ; we may choose μ = κ+ . Accordingly, we consider the obvious trees T = Tμ×λ×κ as well as Tμ , Tμ×λ and Tκ , where Tμ is our norm tree and Tκ is our nursery for getting new subtrees of T from infinite branches of Tμ×λ as described above. By the above, any τ ∈ T can be expressed uniquely as τ = (τμ , τλ , τκ ) with τμ ∈ Tμ , τλ ∈ Tλ and τκ ∈ Tκ . We define the norm of τ to be the ordinal  τ  = sup τμ (i) < μ i
which can easily be extended to subsets ∅ = X ⊆ T taking  X  = sup  τ . τ ∈X

Clearly  X  ≤ μ and if |X| < μ, then  X  < μ, since μ = cf(μ). Moreover, we put  ∅  = 0 and  X  = ∞, if |X| = μ. Similarly, we define the norm on Tμ×λ and let the norm on Tκ be trivially 0 for all subsets of Tκ . If f is an infinite branch of T (or of Tμ×λ ), then we will say that the branch f is stretched (by the norm), if  f  n  <  f  n + 1  for all n < ω. We deduce an easy but useful Lemma 9.2.22. Let f ∈ Br(Tμ×λ ) be a stretched branch. Then the following holds for the norms on T and Tμ×λ as above: (a)  f  < μ is a limit ordinal of cofinality ω. (b)  g  =  f  for all branches g ∈ Br(f × Tκ ).

Proof. (a) We identify f with the set {f  n | n < ω} and thus  f  < μ because μ = cf(μ) is uncountable. Visibly, supn<ω  f  n  is a proper limit for a stretched f. (b) From Br(f × Tκ ) = f × Br(Tκ ) follows that g is of the form g = f × v for some v ∈ Br(Tκ ). So obviously,  g  = sup  (f  n, v  n)  = sup | f  n  =  f , n<ω

observing that only the μ–component counts.

n<ω

2

9.2 The Black Boxes

343

Various traps We begin with an observation and define μo as usual: μo = {α < μ | cf(α) = ω}. A subset C ⊆ μ of μ is ω–closed if, for any bounded sequence αn ∈ C (n < ω) with supn<ω αn < μ, also supn<ω αn ∈ C. Moreover, C is an ω–cub of μ, if C is ω–closed and unbounded in μ. The closure of C in μ is the subset C of μ containing C and all suprema in μ of subsets of C; hence C is closed, if it coincides with its closure. Observation 9.2.23. If V is a stationary subset of μo and C ⊆ μ is an unbounded, ω–closed subset of μ, then C ∩ V = ∅.

Proof. Let C be the closure of C in μ. Then C is a cub and thus C ∩ V = ∅. But, obviously, C ∩ μo = C ∩ μo and so the observation follows. 2

Partial traps of length n < ω We choose a fixed set Γ of cardinality at most λℵ0 , which will be our ‘coding set’. It will help us to distinguish the members of a collection of modules. However, often our collection is a singleton and then we will discard Γ, in which case the triples defined next are just pairs. Let R be an S–ring as before (see Definition 1.1.1), let

B= Bτ τ ∈T

 be the S–adic with |B| ≤ λ be our base module as in Section 1.1, and let B completion of B. Partial traps of length n < ω are triples pn = (f n , ϕn , cn ),  with domain Pn = Dom ϕn ⊆ where cn ∈ Γ, ϕn is a partial R–endomorphism of B   B a κ–generated submodule of B, and f n : n −→ μ × λ is a finite branch of length n, i.e. f n is an element of Tμ×λ .

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It is easy to see that the size of the set PT of all partial traps is at most λκ , which equals λℵ0 by assumption. By a counting argument (due to Cantor), we also have |ω> λ| = λ and hence |PT| = λℵ0 and |ω> λ| = λ. Therefore we can find coding maps, that is, we fix two injections δ : PT −→ ω λ and γ : ω> λ −→ λ.

(9.2.3)

These coding maps can be applied to finite sequences (pk )k PT. We obtain a derived coding map: ζ : ω> PT −→ λ, (pk )k
Proof. Let m < ω and take any n < ω with m < n, where ζ((pk )k
ω PT

Admissible traps These special members of ω PT are the admissible traps (see [171] and below).  of cardinality |X| ≤ κ and an R–endomorphism ϕ : B → Given c ∈ Γ, X ⊆ B ω  B, we want to construct inductively special sequences p = (pk )k<ω ∈ PT with ck = c for all k < ω, X ⊆ P0 = Dom ϕ0 , ϕ  X ⊆ ϕ0 and some further properties. We will collect these admissible traps p, until the proof is complete, in a temporary subset PTϕXc ⊆ ω PT which will be defined in detail later on.

9.2 The Black Boxes

345

For the ‘classic Black Box’ as in [102] it would be enough to produce one element in PTϕXc . However, we want to derive a stronger prediction principle under the name ‘General Black Box’ which can be applied to a given stationary set like the ♦–principle (see Theorem 9.1.8). Thus, at the moment, we will need more than one admissible trap p. Recall that any pn = (f n , ϕn , cn ) is a triple with f n ∈ Tμ×λ , hence  f n  < μ, and so we may consider the norm of elements p = (pk )k<ω ∈ ω PT by putting  p  = sup  f n . n<ω

We also let

  CϕXc =  p  | p ∈ PTϕXc ⊆ μ.

So it is often enough to show that CϕXc = ∅. However, we claim for the constructed set PTϕXc that ω ω an ω–cub : CϕXc ⊆ CϕXc . CϕXc ⊆ μo and ∃ CϕXc

(9.2.4)

First, for fixed ϕ, X, c as above, we define PTϕXc using a concrete construction and show that CϕXc ⊆ μo is unbounded. For this purpose we construct, for any α < μ, elements p ∈ PTϕXc with α <  p  ∈ μo . We proceed by induction.  such that X ⊆ P0 , We begin with choosing a κ–generated submodule P0 ⊆ B (α, β1 , β2 ) ∈ [ P0 ] ⊆ T for α as above and for some (β1 , β2 ) ∈ λ × κ, and [ P0 ] is a subtree of T ; this can easily be arranged by ‘filling up’ P0 . Moreover, we put f 0 : 0 → μ × λ to be the empty map and ϕ0 = ϕ  P 0 . Now suppose f 0 ⊆ · · · ⊆ f n , P0 ⊆ · · · ⊆ Pn are defined in such a way that f n : n → μ × λ, f n × n> κ ⊆ [ Pn ] for the tree f n × n> κ defined as above,  κ–generated with Pn ⊆ B  f i+1  =  Pi  <  Pi+1  (i < n),

[ Pn ] a subtree of T, and ϕn = ϕ  Pn .

 is defined by  P  :=  [ P ]  for any P ⊆ B,  and we have Here the norm on B +   P  < μ for any κ–generated P ⊆ B because μ ≥ κ is regular. Therefore pn = (f n , ϕn , cn ) is a partial trap and, obviously, α ≤  P0  <  Pn−1  =  f n  for n > 1. Using the above, we next define f n+1 : n+1 → μ×λ : Extending f n : n → μ×λ we must only determine f n+1 (n); we set   (9.2.5) f n+1 (n) =  Pn , ζ((pk )k
(9.2.6)

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9 The Black Box and its relatives

So clearly f n ⊆ f n+1 and  f n+1  =  Pn . Now let αn < μ with  Pn  < αn ,  let (β1n , β2n ) ∈ λ × κ, and let τn = (αn , β1n , β2n ). Then we choose Pn+1 ⊆ B large enough such that (a) τn ∈ [ Pn+1 ]; (b) Pn ⊆ Pn+1 is κ–generated; (c) Pn ϕ ⊆ Pn+1 ; (d) f n+1 × n+1> κ ⊆ [ Pn+1 ] ⊆ T ; and (e)  Pn  <  Pn+1 . Note that (a), (b), (c) and (d) are easily arranged, where the latter follows from |f n+1 × n+1> κ| = κ. Moreover, condition (e) follows from (a), since this implies  Pn  < αn ≤  Pn+1 . Finally, we put ϕn+1 = ϕ  Pn+1 and so pn+1 = (f n+1 , ϕn+1 , c) is as required. The sequence p = (pn )n<ω satisfies α <  p  and  p  = supn<ω  f n  = supn<ω  Pn  which, by (e), is a proper limit. Hence we have  p  ∈ μo . Clearly we also have ϕ  P0 = ϕ0 , X ⊆ P0 , and so ϕ  X ⊆ ϕ0 , cn = c (n < ω). Now PTϕXc is defined as the set of all sequences which can be constructed as above. ω ⊆ CϕXc . In order to do so, we Next we show the existence of an ω–cub CϕXc ω define a suitable subset {pα | α < μ} ⊆ PTϕXc with CϕXc = { pα  | α < μ}. This will be achieved inductively by refining the above arguments. For α < μ, let   Yα := η ∈ ω> α | η strictly increasing .

By transfinite induction on α < μ we define  a sequence of traps pα = (pαn )n<ω and a family of partial traps pη (η ∈ α<μ Yα = Yμ ) such that the following holds: (i)  pα  ∈ μo (which is immediate from pα ∈ PTϕXc ); (ii)  pβ  <  pα  for all β < α; (iii)  pα  = supβ<α  pβ  for α ∈ μo ; (iv) α ≤  pα  (which is immediate from (ii)); (v) any pη (η ∈ Yα ) is a partial trap (f η , ϕ  Pη , c) of finite length l(η), where (pη  n )n≤l(η) can be extended to some p ∈ PTϕXc ;

347

9.2 The Black Boxes

(vi)  pη  <  pα  for all η ∈ Yα , and if l(η) = n + 1, then  pη(n)  <  Pη . For α = 0 we put Y0 = {∅}. Then it is easy to define p00 := (∅, ϕ  P0 , c) as above and to find a p0 ∈ PTϕXc with α <  p0 . Next consider α < μ and suppose pβ = (pβn )n<ω ∈ PTϕXc , pη (η ∈ Yβ ) are given for all β < α such that the above conditions hold. If α = β + 1 and η ∈ Yα , then pη is already defined for η ∈ Yβ . Otherwise η = σ ∧ β for some σ ∈ Yβ . Then we let pη extend pσ such that pη = (f η , ϕ  Pη , c) with f η from (9.2.5) and Pη such that  pβ  <  Pη  satisfying (a) to (e). This takes also care of (vi) for α. Note that α∗ := supη∈Yα {α,  pη } < μ because μ is regular and |Yα | < μ. Hence we can find a sequence of admissible traps pα ∈ PTϕXc as required in (i) – (vi). Now suppose that α is a limit ordinal. Then all pη s (η ∈ Yα ) are already defined, since l(η) is discrete. For α ∈ μo we choose any strictly increasing sequence σ = (αn )n<ω with αn < α and supn<ω αn = α. Note that σ  n ∈ Yα for all n < ω. We set pαn = pσ  n ; hence pα = (pαn )n<ω ∈ PTϕXc and  pσ(n−1)  <  Pσ  n  =  f (σ  n+1)  =  p(σ  n+1)  <  pσ(n+1)  for all 0 < n < ω by (v) and (vi). So (iii) also holds. If cf(α) > ω, then (iii) is not needed and we simply choose pα like in the case α = β + 1. ω The sequence of admissible traps pα is now established. Clearly the set CϕXc = { pα  | α < μ} ⊆ CϕXc of their norms is an unbounded ω–closed subset of μ, and so (9.2.4) is shown. 2 Without loss of generality, we redefine PTϕXc to be PTϕXc := {pα | α < μ}. Thus we deduce CϕXc ⊆ μo and CϕXc is an ω–cub.

(9.2.7)

We use these admissible traps for Constructing the genuine traps with coding set Γ First of all we define what we mean by a trap in our present context. Definition 9.2.25. A trap is a triple (f, ϕ, c) with the following properties: (i) f : ω → μ × λ is a stretched branch of the tree Tμ×λ .

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9 The Black Box and its relatives

 with Dom ϕ = P , and Im ϕ ⊆ P (ii) ϕ is a partial R–endomorphism of B (hence ϕ ∈ End P ).  is a κ–generated R–submodule. (iii) P ⊆ B (iv) [ P ] ⊆ T = Tμ×λ×κ is a subtree; (v) f × Tκ ⊆ [ P ]. (vi)  P  =  f  ∈ μo . (vii) c ∈ Γ.

 n Let T denote the family of all triples p = (f , ϕ , c ) with f = p p p p n<ω f  n n n and ϕp = n<ω ϕ such that (pn )n<ω = (f , ϕ , c)n<ω is an admissible trap in  of cardinality at most κ, ϕ through PTϕXc , where X runs through all subsets of B  and c through Γ which has cardinality at most λℵ0 . the R–endomorphisms of B, From this definition, Lemma 9.2.22 and the results just shown we have an immediate Corollary 9.2.26. (i) T is a family of traps p = (fp , ϕp , cp ) of cardinality λℵ0 . (ii) If (f, ϕ, c) is a trap, then  v  =  f  for all branches v ∈ Br(f × Tκ ). (iii) If p = q ∈ T, then fp = fq and thus Br(fp × Tκ ) ∩ Br(fq × Tκ ) = ∅.  a subset of cardinality at most κ, ϕ ∈ End B  and V ⊆ μo (iv) If c ∈ Γ, X ⊆ B a stationary subset of μ, then there is a trap p = (fp , ϕp , cp ) ∈ T such that p ‘catches’ X, ϕ, c, i.e. the following holds: (a) X ⊆ Pp := Dom ϕp . (b)  X  <  Pp . (c) ϕ  Pp = ϕp , cp = c. (d)  p  ∈ V .

Proof. (i) The members p = (f, ϕ, c) of T are unions of the admissible traps pα . By the inductive construction, the branch f is stretched with  f  =  P , where P = Dom ϕ is κ–generated and P ϕ ⊆ P holds because of (a) – (d) in the construction. So it is clear that p is a trap. An easy counting argument, similar to the proof of |PT| = λκ , shows that |T| = λκ which equals λℵ0 by assumption. Condition (ii) follows from Lemma 9.2.22 using the fact that branches of traps are stretched.

9.2 The Black Boxes

349

(iii) The members of T are distinct, in particular, their stretched branches are different by (9.2.6). Hence (iii) follows trivially. (iv) Let X, ϕ, c be given. Then CϕXc is an ω–closed unbounded subset of μ by (9.2.7). Hence CϕXc ∩ V = ∅ by Observation 9.2.23. The existence of p with (a) – (d) in (iv) is now immediate. 2 Similar to the Strong Black Box 9.2.2 we need an ‘enumerated’ version, which will be the General Black Box 9.2.27. The remaining conditions come from a suitable ordering of T. For the ordering of T we must work a little harder than in the strong case (see Corollary 9.2.7). We finally claim: The General Black Box 9.2.27. Let κ, λ, μ be cardinals such that μ is regular, κ+ ≤ μ ≤ λ, λκ = λℵ0 and let Γ be a set with |Γ| ≤ λℵ0 . Moreover, let R be an  as before satisfying |B| ≤ λ. S–ring and let B, B Then there are an ordinal λ∗ < (λℵ0 )+ with |λ∗ | = λℵ0 and a list pα = (fα , ϕα , cα ) (α < λ∗ ) of traps with the following properties for α, β < λ∗ : (i) If β ≤ α , then  pβ  ≤  pα . (ii) If β = α , then Br(fα × Tκ ) ∩ Br(fβ × Tκ ) = ∅. (iii) If β + κℵ0 ≤ α , then Br(fα × Tκ ) ∩ Br [ Pβ ] = ∅. (iv)  v  =  fα  ∈ μo for all branches v ∈ Br(fα × Tκ ).  is a subset of cardinality at most κ, c ∈ Γ, ϕ ∈ End B,  and if (v) If X ⊆ B o V ⊆ μ is a stationary subset of μ, then there is an ordinal α < λ∗ such that the trap pα ‘catches’ X, ϕ, c, that is, the following holds: (a) X ⊆ Pα := Dom ϕα . (b)  X  <  Pα . (c) ϕ  Pα = ϕα , cα = c. (d)  pα  =  Pα  =  fα  ∈ V .

Proof. By Corollary 9.2.26, it remains to show that T can be well–ordered such that Conditions (i) and (iii) are satisfied. Then, because of Corollary 9.2.26 (i), a well–ordered set T can be labelled by an ordinal λ∗ as posed in the theorem. We impose an ordering on T by defining a well–ordering on the sets Tα of all traps p ∈ T with  p  = α (α < μ), and by ranking the traps p according to their

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9 The Black Box and its relatives

norms, i.e.  q  <  p  for q ∈ Tβ , p ∈ Tα with β < α; in this way we ensure that Condition (i) holds. For p, q ∈ T with different norms, Condition (iii) follows almost immediately: assume  p <  q  and let P = Dom ϕp . On the one hand, if v ∈ Br [ P ], then v ⊆ τ ∈[P ] τ as graphs and thus  v  ≤  P  =  p . On the other hand, if v ∈ Br(fq × Tκ ), then  p  <  q  =  v  by Corollary 9.2.26. Therefore v∈ / Br [ P ] in the latter case, that is Br(fq × Tκ ) ∩ Br [ P ] = ∅. It remains to well–order the sets Tα (α < μ) in such a way that (iii) also holds for traps with the same norm, provided that there are at most κℵ0 other traps in between. Given any p = (f, ϕ, c) ∈ Tα with P = Dom ϕ, we have |[P ]| ≤ κ and so – counting countable subsets – the set Br [ P ] is of cardinality at most κℵ0 . By virtue of clause (ii), this means that the ‘immediate progeny’   D0 (p) := q ∈ Tα | Br(fq × Tκ ) ∩ Br [ P ] = ∅ is of size at most κℵ0 . By induction, the same holds for each set  Dn+1 (p) := D0 (q), q∈Dn (p)

and hence also for their union D(p) :=



Dn (p).

n<ω

Summarizing the above we now have, for each p ∈ Tα , an associated subset D(p) ⊆ Tα such that p ∈ D0 (p) ⊆ D(p) (cf. Definition 9.2.25 (v)), |D(p)| ≤ κℵ0 , and q ∈ D(p) =⇒ D0 (q) ⊆ D(p). For any p ∈ Tα , let P  := Dom ϕp . Then the following holds for any such p :    p ∈ D(p), p ∈ Tα \ D(p) =⇒ % & =⇒ p ∈ / D0 (p ) =⇒ Br(fp × Tκ ) ∩ Br P  = ∅. Now we put an arbitrary well–ordering ≤ on the set Δ = {D(p) | p ∈ Tα }. Moreover, for each D ∈ Δ, we define D∗ by  E. D∗ := D \ E
9.2 The Black Boxes

351

Every p belongs to a first D ∈ Δ, so to the corresponding D∗ . Hence the D∗ s (D ∈ Δ) form a partition of Tα in which each part has cardinality |D∗ | ≤ κℵ0 ; note that D∗ = ∅ is not excluded. Finally, we put well–orderings of type ≤ κℵ0 on each D∗ and require that p < p whenever p ∈ D∗ , p ∈ E ∗ and D < E for some D, E ∈ Δ. Thus these orderings canonically extend to a well–ordering on Tα . Then for p, p ∈ Tα and p ∈ D0 (p) follows that the first D ∈ Δ, which contains p and also contains p , thus either p < p or p ≤ p , but p, p both lie in the same D∗ , so there are fewer than κℵ0 elements of Tα between them. Hence p < p < p + κℵ0 or p ≤ p < p + κℵ0 , respectively. Thus p + κℵ0 ≤ p implies p ∈ / D0 (p), and by the implications displayed above the analogue of (iii) follows. The General Black Box is established. 2 The missing case ω ω The General Black Box 9.2.27 does not cover the case when the desired R–module  of countable rank; this is caused G should have a free base module B ⊆∗ G ⊆ B by the assumption κ+ ≤ μ, so ℵ1 ≤ μ. Note, we say that G has density character rk B because G/B is S–divisible (and S–torsion–free), and thus B is dense in G. However, the proof of Theorem 9.2.27 can be slightly modified, using a linear ordering induced by an ultrafilter quotient, which will cover the missing case. The proof is more technical, so we will only outline the substantial changes; for details we refer the interested reader to Shelah [365] and [171] (which uses [365]). The crucial idea for the General Black Box 9.2.27 is the fact that elements which enter the module G later (when G is under construction) are, whenever possible, algebraically independent of the earlier chosen elements. This is arranged by introducing a norm function with values in an uncountable cardinal. In the above proof this was done artificially when we attached certain branches from the tree Tμ to control the norm. The drawback is that we had to start with a base module B of rank at least ℵ1 . Thus we must now define the norm function differently: In order to allow a base module B of countable rank, we consider the countable tree T = ω> ω with infinite branch elements in Br(T ) = T = ω ω. We fix an ultrafilter D on ω and define a partial order on T by   g < h ⇐⇒ n < ω | g(n) < h(n) ∈ D. Using the canonical map T −→ T /D

(f → f ),

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9 The Black Box and its relatives

we obtain a total ordering (T , <) induced from (T /D, <). Moreover, there are a regular cardinal κ with ℵ0 < κ ≤ 2ℵ0 and a map ∗ : κ −→ T

(ν → ν ∗ )

such that the composite map κ → T /D (ν → ν = ν ∗ ) is an order monomorphism. Thus we find a copy of κ in T /D which we use to build a norm on elements of T . In particular, for g ∈ T , there is ν < κ with g/D < ν and (n − 1) divides ∗ ν (n) for all n < ω. We are now able to define a norm     : T −→ κ g →  g  = min{ν < κ | g/D < ν} which naturally extends to countable subsets X of ω: For n < ω let Xn := sup(X ∩ n), and put  X  :=  (Xn )n<ω  < κ. In the next step, the canonical submodules P are defined as before. Moreover, for a tree embedding f : T → T , a pair (f, ϕ) is a trap, if ϕ : P → P is an R–homomorphism and Im(f ) ⊆ [ P ] such that  [ P ]  =  v  ∈ κo for all v ∈ Br(T ) cofinal to ω. A proof similar to the old one, but more technical (see also [171, pp. 237 – 240]), provides the following variant of the General Black Box 9.2.27, which takes care of modules G with countable density character.  as above. If T = ω> ω is Theorem 9.2.28. Let R be an S–ring and let B, B  ℵ0 the tree over ω, then let B = τ ∈T Bτ be as in Section 1.1 with |B| ≤ 2 . ℵ Moreover, let Γ be a coding set of cardinality ≤ 2 0 . Then there are an ordinal λ∗ < (2ℵ0 )+ and a list pα = (fα , ϕα , cα ) (α ∈ λ∗ ) of traps with the following properties for α, β ∈ λ∗ : (i) If β ≤ α, then  pβ  ≤  pα . (ii) If β = α, then Br(Im fα ) ∩ Br(Im fβ ) = ∅. (iii) If β + 2ℵ0 ≤ α, then Br(Im fα ) ∩ Br[Pβ ] = ∅.  is a countable subset, c ∈ Γ and ϕ ∈ End B,  then there is α ∈ λ∗ (iv) If X ⊆ B such that the trap pα catches X, ϕ and c, that means the following holds: (a) X ⊆ Pα := Dom ϕα . (b)  X  <  Pα . (c) cα = c. (d) ϕ  Pα = vaα . (e)  pα  =  Pα  =  v  for all v ∈ Br(Im fα ).

9.3

The Shelah Elevator

353

9.3 The Shelah Elevator This section is based on Shelah’s method of constructing rigid families of (indecomposable) torsion–free abelian groups of arbitrary cardinality [360]. This method was generalized and applied to a wide range of other questions (see, for example, [171], [217], [219]).

The combinatorial part of the elevator As in the last sections, for any regular uncountable cardinal λ, we let λo = {α | cf(α) = ω}. This is a stationary subset of λ; it can be decomposed into λ pair–wise disjoint stationary subsets Ji (i < λ), in particular, |Ji | = λ (see Lemma 9.1.3). We choose bijections ιi : λ → Ji and define another bijection σ : λ × λ −→ λo by (i, j) → σ(i, j) = ιi (j). For any natural number n > 0, let   n↓ λ := τ ∈ n λ | τ (0) > τ (1) > · · · > τ (n − 1) , fix arbitrary bijections σn : n↓ λ → λ for n > 1, and put σ1 = idλ for n = 1. Given R–modules N, N  , M over a commutative ring R, m ∈ M and an R– homomorphism ϕ : N → N  , we write ϕ ⊗ m : N −→ N  ⊗ M

(x → xϕ ⊗ m)

for the homomorphism from N to N  ⊗ M ; the meaning of Hom(N, N  ) ⊗ m follows naturally from this. Dually, we use the notations m ⊗ ϕ ∈ m ⊗ Hom(N, N  ) in the obvious sense. Definition 9.3.1. Given R–modules M, M  and families S = {Ui | i ∈ I},

S = {Ui | i ∈ I}

of submodules of M and M  , respectively, we let   Hom(M, M  ; S, S ) := ϕ ∈ Hom(N, M  ) | Ui ϕ ⊆ Ui for all i ∈ I and End(M, S) := Hom(M, M ; S, S). If M = (M, Ui | i ∈ I) and similarly M = (M  , Ui | i ∈ I), then we also write HomR (M, M ) := Hom(M, M  ; S, S ) and EndR M := HomR (M, M ). If S consists of summands of M and N is another R–module, then S ⊗ N denotes the family of summands X ⊗ N (X ∈ S) of M ⊗ N .

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9 The Black Box and its relatives

Hence End(M, S) is a subalgebra of the R–algebra End M which is closed in the finite topology on End M . The converse, that any closed subalgebra of End M is of the form End(M, S) for a suitable S, is not obvious but follows from the Elevator 9.3.3 and its application Corollary 14.2.1; in fact |S| = 4 will suffice. The finite topology of all open subsets fin on End M is generated by the basis of 0 in End M consisting of all right annihilator ideals Ann F = {ϕ ∈ End M | F ϕ = 0} of End M with F a finite subset of M (see also Section 12.1). The proof of the following proposition is well–known and very easy. However, the result will be important later and we therefore include it. Proposition 9.3.2. Let M be an R–module and let fin denote its just defined finite topology. Then (End M, fin) is a Hausdorff, complete, topological R–algebra. The open sets fin are generated by a subbasis N = {Ann m | m ∈ M }. Moreover, End M/ Ann m ∼ = {mϕ | ϕ ∈ End M } ⊆ M .

Proof. The ideals Ann F (F ≤ M ) are intersections of all Ann m with m ∈ F . Thus N generates the finite topology. From ϕ ∈ Ann m for all m ∈ M follows ϕ = 0, and so fin is also Hausdorff. For ψ, ϕ ∈ End M and Ann m ∈ N, let ψϕ + Ann m be a neighbourhood of ψϕ. Moreover, Ann m is a right ideal and ψ(Ann mψ) ⊆ Ann m. Hence (ψ + Ann m)(ϕ + Ann mψ) ⊆ ψϕ + ψ(Ann mψ) + Ann m ⊆ ψϕ + Ann m, and so multiplication is continuous. The other algebra operations are obviously continuous, and thus (End M, fin) is a topological algebra. Now let ϕF ∈ End M (where F runs through all finite subsets of M ) be a Cauchy net in fin. Given m ∈ M , the Cauchy net satisfies ϕF − ϕF  ∈ Ann m for all finite subsets F, F  of M containing some fixed F0 . Thus mϕF is the same element of M for all large enough F, F  . Therefore we may define mϕ = mϕF for such F and for all m ∈ M . Obviously, this defines an endomorphism ϕ ∈ End M which is the limit of the given Cauchy net because ϕ − ϕF ∈ Ann m for these F s. Hence (End M, fin) is also complete. For the last assertion we only need to consider the map π : End M −→ M

(ϕ → mϕ)

and check its image and kernel, which is easily verified. We finish this section with presenting and proving Shelah’s Elevator.

2

9.3

The Shelah Elevator

355

Shelah’s Elevator 9.3.3. Let λ be an uncountable regular cardinal, let σ, σn be as above, and let H, G be free R–modules with bases {hi | i < λ} and {gn | n < ω}, respectively. Moreover, let S be the countable collection of the following summands of G ⊗ H: (i) W n = gn ⊗ H

(n < ω).

(ii) W1n = (g0 − gn ) ⊗ H (0 < n < ω).   (iii) W2 = i<λ j<λ R(g0 ⊗ hi − g1 ⊗ hσ(i,j) ).   (iv) W3 = i<λ j<λ R(g2 ⊗ hi − g3 ⊗ hσ(j,i) ).  (v) W4n = δ∈λo R(g4 ⊗ hδ − g4n+2 ⊗ hσ(η(n,δ),δ) ) (0 < n < ω) where, for each δ ∈ λo , a ladder η(δ) has been chosen, that is, η(δ) = (η(n, δ))n<ω is a strictly increasing sequence converging to δ.  (vi) W5n = i<τ ∈n↓ λ, i<λ R(g4n+1 ⊗ hσn (τ ) − g4n+3 ⊗ hσn+1 (τ ∧ i) ) (0 < n < ω). Then Hom(G ⊗ H, G ⊗ H ⊗ M ; S, S ⊗ M ) = idG⊗H ⊗M for every R–module M.

Proof. In order to show the non–trivial inclusion, let M be any R–module and consider   ϕ ∈ Hom G ⊗ H, G ⊗ H ⊗ M ; S, S ⊗ M . Then W n ϕ ⊆ W n ⊗ M and W1n ϕ ⊆ W1n ⊗ M renders homomorphisms ψn , ψ0n ∈ Hom(H, H ⊗ M ) such that (gn ⊗ m)ϕ = gn ⊗ mψn and ((g0 − gn ) ⊗ m)ϕ = (g0 − gn ) ⊗ mψ0n . Hence g0 ⊗ mψ0 − gn ⊗ mψn = (g0 ⊗ m)ϕ − (gn ⊗ m)ϕ = ((g0 − gn ) ⊗ m)ϕ = (g0 − gn ) ⊗ mψ0n = g0 ⊗ mψ0n − gn ⊗ mψ0n and so g0 ⊗ m(ψ0 − ψ0n ) + gn ⊗ m(ψ0n − ψn ) = 0 for all m ∈ M . Freeness of G yields ψ0 = ψ0n = ψn for all n > 0 and thus ϕ = idG ⊗ψ for ψ = ψ0 . (9.3.1)

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9 The Black Box and its relatives

Let 0 < n < ω, τ ∈ n↓ λ be arbitrary and assume hσn (τ ) ψ = 0, which can be  uniquely expressed as li=1 hi∗ ⊗ mi with i∗ < λ and mi ∈ M for i = 1, . . . , l. Since σn is bijective, there are also uniquely determined elements τi ∈ n↓ λ such that σn (τi ) = i∗ (1 ≤ i ≤ l). Hence hσn (τ ) ψ is uniquely represented as the sum hσn (τ ) ψ = hσn (τ1 ) ⊗ m1 + · · · + hσn (τl ) ⊗ ml . Recall that each τ = τ (0)∧ . . . ∧ τ (n − 1) is a sequence of length n. We now claim that the last entries of the τ s at (n − 1) satisfy τ (n − 1) ≤ τ1 (n − 1), . . . , τl (n − 1).

(9.3.2)

We show the claim (9.3.2) by transfinite induction on τ (n − 1) < λ, the last member of the sequence being τ . If τ (n − 1) = 0, then the assertion is trivially satisfied. So suppose that (9.3.2) has already been shown for all γ < τ (n − 1). If ν = νγ = τ ∧ γ ∈ n+1↓ λ, then, like above, we have hσn+1 (ν) ψ = hσn+1 (ν1 ) ⊗ m1 + · · · + hσn+1 (νk ) ⊗ mk for certain ν1 , . . . νk ∈ n+1↓ λ and for some non–zero m1 , . . . , mk ∈ M , provided hσn+1 (ν) ψ = 0. Condition (9.3.2) holds for ν by induction hypothesis and thus γ = ν(n) ≤ ν1 (n), . . . , νk (n). Now g4n+1 ⊗ hσn (τ ) − g4n+3 ⊗ hσn+1 (ν) lies in W5n and thus it is mapped by ϕ into W5n ⊗ M . Hence W5n ⊗ M ∩ W 4n+1 ⊗ M = 0 and hσn (τ ) ψ = 0 imply hσn+1 (ν) ψ = 0. Consequently, we can write (g4n+1 ⊗ hσn (τ ) − g4n+3 ⊗ hσn+1 (ν) )ϕ = g4n+1 ⊗ (hσn (τ1 ) ⊗ m1 + · · · + hσn (τl ) ⊗ ml ) −g4n+3 ⊗ (hσn+1 (ν1 ) ⊗ m1 + · · · + hσn+1 (νk ) ⊗ mk ). The displayed sum is an element in W5n ⊗ M . Consider τ1 : by the definition of W5n , the summand g4n+1 ⊗ hσn (τ1 ) ⊗ m1 must have a companion μ ∈ n↓ λ such that σn (τ1 ) = σn (μ) and σn+1 (μ∧ α) = σn+1 (νj ) for some α < λ and for a suitable 1 ≤ j ≤ k. It follows that τ1 ∧ α = νj because σn , σn+1 are bijective. Hence, each of the sequences τ1 , . . . , τl , is an initial segment of one of the sequences ν1 , . . . , νk . For ν we know from the displayed formula above that γ = ν(n) ≤ ν1 (n), . . . , νk (n) and hence γ < τ1 (n − 1), . . . , τl (n − 1) which holds for all γ < τ (n − 1). Therefore τ (n − 1) ≤ τ1 (n − 1), . . . , τl (n − 1), which completes the inductive proof of claim (9.3.2).

9.3

The Shelah Elevator

357

Next we define a cub C ⊆ λ such that, for any 0 = m1 , . . . , mn ∈ M , the following holds for i < j, j ∈ C: hi ψ = hi1 ⊗ m1 + · · · + hin ⊗ mn =⇒ i ≤ i1 , . . . , in < j.

(9.3.3)

The inequality i ≤ i1 , . . . , in is an immediate consequence of (9.3.2) for n = 1, using that σ1 is the identity on λ. We mentioned that i1 , . . . , in are uniquely determined by i, so we can define fi : λ → λ by f1 (i) = max{i + 1, i1 + 1, . . . , in + 1}, allowing n = 0 in case hi ψ = 0. Letting f2 (i) = sup{f1 (j) | j < i}, the function f2 is continuous and satisfies i ≤ f2 (i) ≤ f2 (j) for all i ≤ j. Thus C := {i ∈ λ | f2 (i) = i} defines a closed and unbounded subset of λ, since λ is uncountable and regular. Now assertion (9.3.3) is obvious. We proceed by showing that, for every δ ∈ C with cf(δ) = ω, there is some mδ ∈ M with hδ ψ = hδ ⊗ mδ . For hδ ψ = 0, this is trivially satisfied. So suppose ∃ 0 = m1 , . . . , mn ∈ M, ∃ i∗ < λ

(9.3.4)

such that hδ ψ = h1∗ ⊗ m1 + · · · + hn∗ ⊗ mn .

(9.3.5)

By (9.3.3) we may assume, w.l.o.g., that δ ≤ 1∗ < 2∗ < · · · < n∗ . Now g4 ⊗ hδ − g4k+2 ⊗ hσ(η(k,δ),δ) ) ∈ W4k and hence (g4 ⊗ hδ − g4k+2 ⊗ hσ(η(k,δ),δ) )ϕ = g4 ⊗ hδ ψ − g4k+2 ⊗ hσ(η(k,δ),δ) ψ ∈ W4k ⊗ M for all 0 < k < ω. Therefore, on the one hand, this element can be expressed as a sum of certain (g4 ⊗ hα − g4n+2 ⊗ hσ(η(k,α)α) ) ⊗ m with α ∈ λo , m ∈ M . On the other hand, (9.3.4) becomes g4 ⊗hδ ψ = g4 ⊗h1∗ ⊗m1 +· · ·+g4 ⊗hn∗ ⊗mn , the αs are just 1∗ , . . . , n∗ , and the coefficient for α = i∗ at g4 ⊗hi∗ −g4k+2 ⊗hσ(η(k,i∗ ),i∗ ) must be m = mi . Thus hσ(η(k,δ),δ) ψ = hσ(η(k,i1 ),i1 ) ⊗ m1 + · · · + hσ(η(k,in ),in ) ⊗ mn and so, in particular, cf(1∗ ) = · · · = cf(n∗ ) = ω. For k sufficiently large we also have η(k, 1∗ ) < · · · < η(k, n∗ ).

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Next we consider the element g0 ⊗ hη(k,δ) − g1 ⊗ hσ(η(k,δ),δ) ∈ W2 . Then (g0 ⊗ hη(k,δ) − g1 ⊗ hσ(η(k,δ),δ) )ϕ ∈ W2 ⊗ M . Using the same arguments as before and that the η(k, i∗ )s are distinct for large enough k, it follows that hη(k,δ) ψ = hη(k,i,1∗ ) ⊗ m1 + · · · + hη(k,n∗ ) ⊗ mn . By (9.3.3) it also follows that η(k, δ) ≤ η(k, 1∗ ) < · · · < η(k, n∗ ) < δ for all large k < ω. Moreover, recall from the above that 1∗ < · · · < n∗ < δ. This is only possible if 1∗ = δ and n = 1; thus hδ ψ = hδ ⊗ mδ follows. Now let i < λ be arbitrary. Since Ji ⊆ λo is stationary, we can find δ = σ(i, j) ∈ C ∩ Ji with cf(δ) = ω. Then (g0 ⊗ hi − g1 ⊗ hσ(i,j) )ϕ = g0 ⊗ hi ψ − g1 ⊗ hσ(i,j) ⊗ mδ ∈ W2 ⊗ M gives hi ψ = hi ⊗ mδ . Thus we have found, for each i < λ, elements mi ∈ M with hi ψ = hi ⊗ mi . Finally W2 ϕ ⊆ W2 ⊗ M yields mi = mσ(i,j) for all i < λ, whereas W3 ϕ ⊆ W3 ⊗ M implies mj = mσ(i,j) for all i < λ. Combining both statements, we deduce mi = mj = m for all i, j < λ, and ϕ = idG⊗H ⊗m follows, as desired. 2

Chapter 10

Independence results for cotorsion pairs

Most of the results on the existence of approximations in Chapters 2 – 8 rely on the completeness of the associated cotorsion pairs. The completeness has always been proved by showing that the cotorsion pairs involved are generated by a set. The proofs, using the deconstruction of cotorsion pairs, are rather involved. So we may wonder whether completeness and generation by a set do not actually hold for all cotorsion pairs, and whether they can be proved more directly. In this chapter we will show that there are indeed stronger results on completeness available, but only with the help of additional set theoretic assumptions. These results turn out to be independent of ZFC. In particular, there is no way of obtaining them by purely algebraic means of any kind. In fact, we will see that for some explicitly given cotorsion pairs C, the question whether C is generated by a set (or complete) is independent of ZFC + GCH. Rather than working with forcing extensions directly, we will use well–known combinatorial properties of various extensions of ZFC, notably the Generalized Weak Diamond Principle introduced in Chapter 9, and Shelah’s Uniformization Principle UP. Since Shelah’s solution of the Whitehead problem these combinatorial principles have been known to have a big impact on properties of extensions of modules. Section 10.1 of the present chapter is based on [353] and shows that the Generalized Weak Diamond Principle Ψ implies that many cotorsion pairs cogenerated by a set are also generated by a set, and hence complete. In Section 10.2, following [387], we will construct cotorsion pairs C cogenerated by a set, but not generated by any set, under Shelah’s Uniformization Principle. Combining this with the results of Section 10.1, we will thus provide examples of cotorsion pairs C such that the assertion “C is generated by a set” is independent of ZFC (+ GCH).

360

10.1

10 Independence results for cotorsion pairs

Completeness of cotorsion pairs under the Diamond Principle

In this section we will use the Generalized Weak Diamond as defined in Chapter 9 (see p. 301) to prove that many cotorsion pairs cogenerated by a set are also generated by a set, and hence complete. The lemma showing the role of Ψλ E goes back to [383]: Lemma 10.1.1. Assume Ψλ E . Let λ be a regular uncountable cardinal, and E a stationary subset of λ. Let R be a ring such that |R| ≤ λ, and N a module with |E(N )| ≤ λ. Let M be a λ–generated module with a λ–filtration (Mα | α < λ) such that Ext1R (Mα , N ) = 0 for all α < λ, and E = {α < λ | Ext1R (Mα+1 /Mα , N ) = 0}. Then Ext1R (M, N ) = 0.

Proof. First, we take an R–generating subset A ⊆ M of cardinality λ and a λ–filtration (Aα | α < λ) of A such that Mα = m∈Aα mR for all α < κ. This can be done by induction, since Mα is < λ–generated for each α < λ. Let (Bα | α < λ) be a λ–filtration of the Z–module B = E(N ). Denote by ν the inclusion of N into B, by π the projection of B onto B/N , and by να the inclusion of Mα into Mα+1 , for all α < λ. Take α ∈ E. Let Xα = HomR (Mα , N ) and Yα = Im HomR (να , N ). Since Ext1R (Mα+1 , N ) = 0 and Ext1R (Mα+1 /Mα , N ) = 0, there exists fα ∈ Xα \ Yα . Denote by oα the order of fα + Yα in the group Xα /Yα = Ext1R (Mα+1 /Mα , N ). Let α ∈ E. If oα = ℵ0 , we put pα = 2; if oα < ℵ0 , we let pα = oα . We equip the set of all mappings from Aα to Bα with an equivalence relation ∼α : u ∼α v, iff there are n ∈ Z and y ∈ Yα such that v = u + nfα  Aα + y  Aα . Note that the number n is unique (unique modulo pα ) provided that oα = ℵ0 (oα < ℵ0 ). For each α ∈ E, we take a partition Pα : map(Aα , Bα ) → pα such that Pα (u) = Pα (v), iff u ∼α v and the number n above is divisible by pα . Let ψ : E → ℵ0 be the mapping corresponding to the given setting by Ψλ E. In order to prove that Ext1R (M, N ) = 0, we shall construct g ∈ HomR (M, B/N ) \ Im HomR (M, π). By induction on α < λ, we define  gα ∈ HomR (Mα , B/N ) so that gα+1  Mα = gα for each α < λ, and gα = β<α gβ for all limit ordinals α < λ. Put g0 = 0. Assume gα is defined for an ordinal α < λ. We distinguish the following two cases: (I) α ∈ E and there exists f ∈ HomR (Mα+1 , B) such that Im(f να ) ⊆ Eα , Pα (f να  Aα ) = ψ(α), and gα = πf να . (II) = not (I).

10.1

Completeness of cotorsion pairs under the Diamond Principle

361

In case (I), take an f satisfying the conditions of (I). The injectivity of B yields the existence of hα ∈ HomR (Mα+1 , B) such that hα να = f να − fα . Put gα+1 = πhα . Then gα+1 να = πf να − πfα = gα . In case (II), the assumption of Ext1R (Mα , N ) = 0 yields the existence of hα ∈ HomR (Mα , B) such that gα = πhα . The injectivity of B gives an hα+1 ∈ HomR (Mα+1 , B) such that hα = hα+1 να . Put gα+1 = πhα+1 . Then gα+1  Mα = gα .  Finally, put g = α<λ gα . Then g ∈ HomR (M, B/N ). Suppose there is h ∈ HomR (M, B) such that g = πh . Note that the set {α < λ | Im(h  Mα ) ⊆ Bα } is closed and cofinal in λ. Since E is stationary, there is α ∈ E such that g  Mα = πhνα , Pα (hνα  Aα ) = ψ(α), and Im(hνα ) ⊆ Bα , where h = h  Mα+1 . Hence, case (I) occurs, and π(hα − h) = 0. Then yα = (hα − h)να ∈ Yα . Moreover, f να = hνα + fα + yα , whence ψ(α) = Pα (f να  Aα ) = Pα (hνα  Aα + fα  Aα + yα  Aα ) = Pα (hνα  Aα ), a contradiction. Thus g ∈ / Im HomR (M, π). 2 Denote by Ψ the assertion: “Ψλ E holds true for each regular uncountable cardinal λ and each stationary subset E ⊆ λ”. Lemma 10.1.1 implies the existence of many cotorsion pairs generated by a set (see Definition 3.2.6 for the notion of a κ–refinement): Theorem 10.1.2. Assume Ψ. Let N be a module such that ⊥ N is closed under pure submodules. (a) Let κ be a cardinal with κ ≥ |R| + |E(N )| + ℵ0 . Then for each module M ∈ ⊥ N there are an ordinal σ and a κ–refinement of M of length σ, (Mα | α < σ), such that Mα+1 /Mα ∈ ⊥ N for all α + 1 < σ. (b) The cotorsion pair cogenerated by N is generated by a set, and ⊥ N is a special precovering class.

Proof. (a) The existence of the κ–refinement R of M is proved by induction on the cardinality λ of M . There is nothing to prove for λ ≤ κ. Let λ be a regular cardinal > κ. We enumerate the elements of M , M = {mα | α < λ}, and let M0 = 0. Let α < λ. Since κ ≥ |R| + ℵ0 , Lemma 1.2.17 (a) gives a pure submodule P/Mα of M/Mα containing mα + Mα such that |P/Mα | ≤ κ. Since Mα is pure in M by inductive assumption, P is pure in M by Lemma  1.2.17 (b), and we let Mα+1 = P . If α is a limit ordinal, we let Mα = β<α Mβ which is a pure

362

10 Independence results for cotorsion pairs

submodule in M by Lemma 1.2.17 (c). Since λ is a regular cardinal, R has length λ, and |Mα | < λ for all α < λ, so R is a λ–filtration of M . Possibly taking a λ–subfiltration, we can w.l.o.g. assume that R is a λ–filtration with the following property: if α < β < λ are such that Ext1R (Mβ /Mα , N ) = 0, then also Ext1R (Mα+1 /Mα , N ) = 0. Since M ∈ ⊥ N and ⊥ N is closed under pure submodules, Ext1R (Mα , N ) = 0 holds for every α < λ, and Lemma 10.1.1 yields that the set   E = α < λ | Ext1R (Mα+1 /Mα , N ) = 0 is not stationary in λ. So there is a closed and unbounded subset C of λ such that E ∩ C = ∅. Taking the λ–subfiltration of R indexed by the elements of C,  /Mα ∈ ⊥ N for all we obtain a λ–filtration (Mα | α < λ) of M such that Mα+1 α < λ. By inductive assumption, we can refine this λ–filtration into a σ–filtration R which is a κ–refinement of M . If λ is singular > κ, we use Lemma 4.3.8 (or rather its version for modules with Q–filtrations consisting of pure submodules, cf. [49, §3]) for μ = κ and Q a representative set of all ≤ κ–presented elements of ⊥ N . We call a module M “free”, if M has a κ–refinement as in the claim of Theorem 10.1.2. For each regular cardinal κ < ρ < λ, consider the set Cρ of all pure submodules of M of cardinality < ρ. Since ⊥ N is closed under pure submodules, each element of Cρ is “free” by inductive assumption. Moreover, any subset X of M of cardinality < ρ is contained in an element of Cρ by Lemma 1.2.17 (a). By Lemma 1.2.17 (c), pure submodules of M are closed under unions of well–ordered chains. So M is ρ–Q–free for each regular cardinal κ < ρ < λ, and hence M is “free” by Lemma 4.3.8. (b) By part (a) and Lemma 3.2.8, the cotorsion pair C cogenerated by N is generated by a set, hence C is complete by Theorem 3.2.1 (b). In particular, ⊥ N is a special precovering class. 2 Remark 10.1.3. (a) Assuming Jensen’s Diamond Principle rather than just the Generalized Weak Diamond, we can obtain a stronger version of Lemma 10.1.1, where the assumptions of |R| ≤ λ and |E(N )| ≤ λ are relaxed just to |N | ≤ λ (see [181, Appendix]). The application to the deconstruction (and hence completeness) of the cotorsion pair cogenerated by N of course remains the same. (b) V = L provides for deconstruction, even in the cases when deconstruction is not available in ZFC (see Theorem 10.2.13 below). However, V = L is in general not sufficient to enable deconstruction to countable type. As an example, we consider again the class of all Whitehead modules, W1 . By [145], there exists (in ZFC) a non–cotorsion principal ideal domain R of cardinality 2ℵ1 such that for each ℵ1 –generated module M , M is Whitehead, iff all

10.1

Completeness of cotorsion pairs under the Diamond Principle

363

countably generated submodules of M are free. Moreover, since R is not cotorsion, all countably generated Whitehead modules are free (see [190]). Now let M be any ℵ1 –generated non–free module all of whose countably generated submodules are free (by [142, VII.1], we can even construct such an M with a prescribed Γ–invariant). Then M ∈ W1 , but M is not W1≤ω –filtered, and the class B = W1⊥ is not of countable type. (Notice the difference from non– cotorsion principal ideal domains R of cardinality ≤ ℵ1 : there, V = L already implies that all Whitehead modules are free by [49, §5].) Let us record a particular case of Theorem 10.1.2: Corollary 10.1.4. Assume V = L. Let R be a ring and C a cotorsion pair cogenerated by a set consisting of modules of injective dimension ≤ 1. Then C is generated by a set, hence C is complete. Corollary 10.1.4 can also be generalized in a different way from Theorem 10.1.2: Theorem 10.1.5. Assume Ψ. Let R be a ring, C = (A, B) be a hereditary cotorsion pair cogenerated by a set and such that B ⊆ I. Then C is generated by a set, hence C is complete.

Proof. By assumption there are an n < ω and a module N ∈ In such that A = ⊥1 N . Let μ = |R| + |E(N )| + ℵ0 . For 0 ≤ i ≤ n, let Ci = (Ai , Bi ) be a (hereditary) cotorsion pair with Ai = ⊥1 (B ∪ Ii ) = ⊥1 B ∩ ⊥1 Ii . Let Qi be a representative set of all ≤ μ–generated modules in Ai . By downward induction on i we will prove that every module M ∈ Ai is Qi – filtered. Let i = n. Since B ⊆ In , we have An = ⊥1 In , and the claim follows by Theorem 4.2.11 because R is right μ–noetherian. Let 0 ≤ i < n. We will proceed by induction on the minimal number λ of generators of the module M . There is nothing to prove for λ ≤ μ. Let λ > μ ⊆ π and M ∈ Ai . Consider an exact sequence 0 → K − → R(λ) − → M → 0. We may assume that the minimal number of generators of K is also λ. Note that K ∈ Ai , since Ci is hereditary. Moreover, K is a syzygy of a module from ⊥1 Ii , hence K ∈ ⊥1 Ii+1 . So we actually have K ∈ Ai+1 . By inductive premise, there is a Qi+1 –filtration K of the module K. Consider the family F for K arising from Theorem 4.2.6. Let G = {L ⊆ M | (∃AL ⊆ λ)(π(R(AL ) ) = L and K ∩ R(AL ) ∈ F)}. We claim that G ⊆ Ai . Indeed, let L be a module from G. Then for B ∈ Bi , we have the exact sequence 0 = Ext1R (M, B) → Ext1R (L, B) → Ext2R (M/L, B).

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10 Independence results for cotorsion pairs

However, Ext2R (M/L, B) ∼ = Ext1R (K/K ∩ R(AL ) , B) = 0, since Bi ⊆ Bi+1 and (A ) K/(K ∩ R L ) ∈ Ai+1 (by Theorem 4.2.6, because K ∩ R(AL ) ∈ F). It is obvious that 0 ∈ G, and that G is closed under well–ordered unions of chains. Now we show that for every regular κ ≤ λ with κ > μ and a subset X ⊆ M of cardinality < κ, there is a < κ–generated module L ∈ G containing X. Choose a subset A0 ⊆ λ of cardinality < κ such that X ⊆ π(R(A0 ) ). By Theorem 4.2.6, there is a < κ–generated module K0 ∈ F such that K ∩ R(A0 ) ⊆ K0 . Take A1 ⊇ A0 with K0 ⊆ R(A1 ) and |A1 | < κ. Iterating this process, we obtain a chain K0 ⊆ K1 ⊆ K2 ⊆ . . . of < κ–generated modules from F and a chain  A0 ⊆ A1 ⊆ A2 ⊆ . . . of subsets of λ of cardinality < κ. Define L = π(R( k<ω Ak ) ). Then L is the desired module from G. Let λ be regular. By the previous step we can select from G a λ–filtration M of M consisting of < λ–generated modules. Applying a modification of Theorem 4.3.2 (see Remark 4.3.3) for B = Ii , and Lemma 10.1.1 to M ∈ ⊥1 N , we obtain a subfiltration, N of M with consecutive factors from ⊥1 Ii , and then a subfiltration O of N whose consecutive factors are in A. However, by Lemma 3.1.2, these factors belong to Ai , and they are clearly < λ–generated. Hence, by inductive premise, they are Qi –filtered. It follows that M has the same property. If λ is singular, the properties of G proved above make it possible to apply Lemma 4.3.8 and conclude that M is Qi –filtered. Finally, for i = 0, we obtain that each module M ∈ A is A≤μ –filtered, so the claim follows as in Theorem 10.1.2 (b). 2

10.2

Uniformization and cotorsion pairs not generated by a set

One of the key tools of approximation theory is Theorem 3.2.1 saying that a cotorsion pair is complete provided it is generated by a set of modules. It is therefore important to know which cotorsion pairs have this property. Most cotorsion pairs considered so far do (in ZFC). Moreover, there are positive consistency results: by Corollary 10.1.4 above, it is consistent with ZFC + GCH that for any right hereditary ring R, any cotorsion pair cogenerated by a set is also generated by a set. In this section we present negative consistency results by constructing cotorsion pairs cogenerated by some set, but not generated by any set. This occurs under Shelah’s Uniformization Principle UP defined below. As a consequence, we infer that Corollary 10.1.4 is independent of ZFC + GCH.

10.2 Uniformization and cotorsion pairs not generated by a set

365

Let λ be an uncountable cardinal of cofinality ω and let E ⊆ {α < λ+ | cf(α) = ω}. We will need a slightly stronger notion of a ladder system than the one introduced in Chapter 9: a sequence (nα | α ∈ E) is a ladder system provided that for each α ∈ E, (nα (i) | i < ω) is a strictly increasing sequence of non–limit ordinals less than α such that supi<ω nα (i) = α. Denote by UPλ the assertion UPλ : “There exist a stationary set E ⊆ λ+ such that E ⊆ {α < λ+ | cf(α) = ω} and a ladder system (nα | α ∈ E) such that for each cardinal κ < λ and each sequence (hα | α ∈ E) of mappings from ω to κ there is a mapping f : λ+ → κ such that for each α ∈ E there exists jα < ω with f (nα (i)) = hα (i) for all i ≥ jα .” Denote by UP the assertion “UPλ holds for each uncountable cardinal λ of cofinality ω”. This is Shelah’s Uniformization Principle. It is known that UP is consistent with ZFC + GCH (cf. [143]). In order to apply UP, we need further notation: First, recall that a module M is a splitter, if Ext1R (M, M ) = 0. For example, if C = (A, B) is a cotorsion pair, then each K ∈ KerC = A ∩ B is a splitter; the elements of KerC are called the splitters for C. Moreover, K ∈ KerC is a strong splitter for C, if K (ω) ∈ KerC, that is, if (ω) ∈ B. K For a ring R and n < ω, let Cn = (Pn , Pn⊥ ). By Theorem 4.1.12, Cn is generated by a set, hence complete, for any ring R. If K ∈ KerC for a cotorsion pair C = (A, B), then clearly A ⊆ ⊥ K. The following example shows that the question of when ⊥ K = A is quite hard, even in particular cases: Example 10.2.1. (1) Let R = Z and K = Z. Then the (strong) splitters for C0 are the free groups. The elements of ⊥ K are the Whitehead groups. The question of whether P0 = ⊥ K is the famous Whitehead problem. Shelah proved that the answer to this problem is independent of ZFC + GCH. In fact, if K is any non–zero free group, then P0 = ⊥ K is independent of ZFC + GCH (see [143] and [383]). (2) Let R be an IC–domain, that is, let R be a Matlis valuation domain of global dimension 2 such that Ext1R (R/J, K) = 0 for each non–principal ideal J of R (where, as usual, Q denotes the quotient field of R and K = Q/R). Since R is a Matlis domain, K ∈ P1 . Moreover, C1 = (P1 , DI) = (⊥ Gen(δ), Gen(δ)),

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where DI is the class of all divisible modules and δ is the Fuchs tilting module (see Example 5.1.2). In particular, K is a strong splitter for C1 . Recall that a module M is a W2 –module, if Ext2R (M, R) = 0. As Ext2R (M, R) ∼ = Ext1R (M, K), W2 –modules coincide with the elements of ⊥ K. The shifted Whitehead problem asks whether P1 = ⊥ K. Bazzoni and Salce proved that the answer to the shifted Whitehead problem is independent of ZFC for any IC–domain R such that K is countable (cf. [44]). Our goal is to show that in many cases (including the ones mentioned in Example 10.2.1) UP implies that the cotorsion pair cogenerated by K is not generated by any set, where K is a strong splitter for a complete cotorsion pair C = (A, B). Then of course A = ⊥ K, which will in particular prove both the negative consistency result for the Whitehead problem and for the shifted Whitehead problem. Let C = (A, B) be a (complete) cotorsion pair generated by a set. By Theorem 4.2.11, there is a cardinal ρ such that each module in A is A≤ρ –filtered. The smallest such ρ is called the rank of C and denoted by ρC. For example, the cotorsion pair Cn has rank ρCn ≤ dim R for any ring R and any n < ω by Lemma 4.1.11. For a module K we will view K (ω) as the union of the increasing chain of submodules, (K (n) | n < ω). Let C be a cotorsion pair. A module K is called a local splitter for C, provided that K is a strong splitter for C, and there exists a non–split embedding ν : K (ω) → K (ω) such that for each n < ω, there exists a submodule Cn ⊆ K (ω) with ν(K (n) ) ⊕ Cn = K (ω) and ν(K (ω\n) ) ⊆ Cn . Example 10.2.2. Let R be a non–right perfect ring; so there is a strictly decreasing countable chain Ra0  Ra1 a0  . . .  Ran an−1 . . . a0  Ran+1 an . . . a0  . . . consisting of principal left ideals of R. Then the module K = R is a local splitter for C0 . This is witnessed by the non–split embedding ν : R(ω) → R(ω) defined by i < ω) is the canonical basis of the free module ν(1i ) = 1i − 1i+1 ai , where (1i | (ω) R , and by the modules Cn = n≤i 1i R. Note that F = Coker(ν) is a flat, but non–projective module (cf. [8, §28]). Example 10.2.3. Let R be an IC–domain (see Example 10.2.1 (2)). Let Q be the quotient field of R and K = Q/R. For each non–principal ideal J of R there is a non–split embedding ν  : K (ω) → K (ω) with Coker(ν  ) ∼ = Q/J. It is easy to check from the definition of ν  (or derive from Lemma 10.2.4 below)

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367

that for each n < ω, ν  (K (n) ) splits in K (ω) . By induction we can easily modify ν  into an embedding ν : K (ω) → K (ω) and construct submodules Cn (n < ω) of K (ω) so that Im(ν) = Im(ν  ) (so ν is non–split and Coker(ν) ∼ = Q/J) and ν(K (n) ) ⊕ Cn = K (ω) , ν(K (ω\n) ) ⊆ Cn for all n < ω. It follows that K is a local splitter for C1 . Lemma 10.2.4. Let R be a commutative non–perfect ring with a decreasing chain of principal ideals Ra0  Ra1 a0  . . .  Ran an−1 . . . a0  Ran+1 an . . . a0  . . . . Define ν ∈ EndR (R(ω) ) by ν(1i ) = 1i − 1i+1 ai , where (1i | i < ω) is the canonical basis of R(ω) . Let C = (A, B) be a cotorsion pair generated by a set and K be a strong splitter for C. Then K (ω) = K ⊗R R(ω) , and τ = idK ⊗R ν ∈ EndR (K (ω) ) is an embedding. If τ is non–split, then K is a local splitter for C.

Proof. We use the notation of Example 10.2.2. Since F is flat, τ is monic. Since F ∼ = Cn /ν(R(ω\n) ), we have τ (K (n) )⊕(K⊗R Cn ) = K (ω) and also τ (K (ω\n) ) ⊆ K ⊗R Cn for each n < ω. So, if τ is non–split, then K is a local splitter for C. 2 We fix our notation for the rest of this section: R denotes an arbitrary ring, C = (A, B) a cotorsion pair generated by a set and ρC the rank of C. Further, K denotes a local splitter for C, so K is a strong splitter for C, ν : K (ω) → K (ω) is a non–split embedding, and for each n < ω, Cn is a submodule of K (ω) such that ν(K (n) ) ⊕ Cn = K (ω) and ν(K (ω\n) ) ⊆ Cn . Lemma 10.2.5. Consider the exact sequence of modules ν

0 → K (ω) → K (ω) → N → 0. Then N ∈ / A.

Proof. Assume that N ∈ A. Since K is a strong splitter, we have K (ω) ∈ B, so the sequence splits, a contradiction. 2 Moreover, we fix a cardinal λ such that λ > |EndR (K)|, λ ≥ |K| + |R| + ρC and cf λ = ω. Let E ⊆ {α < λ+ | cf(α) = ω} and (nα | α ∈ E) be a ladder system. Let (Pα | α < λ+ ) be a sequence of modules defined as follows: Pα = K for all α ∈ λ+ \ E and Pα = K (ω) for α ∈ E. For α ∈ λ+ \ E, let μα denote the identity map of K onto Pα . For α ∈ E and i < ω denote by μα,i the i–th canonical

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10 Independence results for cotorsion pairs

 embedding of K into Pα and define να = ν ∈ EndR (Pα ). Put P = α<λ+ Pα . Clearly P ∈ A.  Let (Lα | α ∈ E) be a sequence of modules defined by Lα = i<ω Lα,i , where Lα,i = {μnα (i) (k) − να μα,i (k) | k ∈ K} ⊆ Pnα (i) ⊕ Pα . Then σα,i : K → Lα,i defined by σα,i (k) = μnα (i) (k) − να μα,i (k) is an isomorphism for all i < ω and α∈ E. Moreover, (Lα | α ∈ E) are R–independent submodules of P . Put L = α∈E Lα . Then L ∈ A. Finally put D = P/L. Note that |D| = λ+ . Lemma 10.2.6. D ∈ / A provided that E is stationary in λ+ .  Proof. For each α < λ+ , let Dα = ( β<α Pβ + L)/L. By the definition | α < λ+ ) is a λ+ –filtration of D. If α ∈ / E, of the Pα s and L above, (Dα  ∼ then Dα+1 /Dα = Pα /(Pα ∩ ( β<α Pβ + L)) = Pα /0 = K. If α ∈ E, then Dα+1 /Dα ∼ = N. = K (ω) /ν(K (ω) ) ∼ Assume D ∈ A. Since |D| = λ+ and |R| + ρC ≤ λ, Theorem 4.2.6 yields a λ+ –filtration (Aα | α < λ+ ) of D. Then the set C = {α < λ+ | Aα = Dα } is a cub in λ+ . If E is stationary, then there exist α ∈ C ∩ E and α ∈ C ∩ E with α <   α < λ+ . Put F = Aα /Aα = Dα /Dα . On the  one hand, F = Aα /Aα ∈ A by Lemma 3.1.2. On the other hand, Dα+1 ∩ (( α<β<α Pβ + L)/L) ⊆ Dα , so F ∼ = Dα+1 /Dα ⊕ (( α<β<α Pβ + L)/L)/Dα . Since Dα+1 /Dα ∼ = N is a direct summand in F , Lemma 10.2.5 gives F ∈ / A, a contradiction. 2 The key property of D implied by UPλ is as follows: Lemma 10.2.7. Assume UPλ holds. Take E ⊆ λ+ and a ladder system (nα | α ∈ E) as in UPλ , and define N and D as above. Then D ∈ ⊥ K.

Proof. Consider the presentation of D 0 → L → P → D → 0. Since P ∈ A, we have Ext1R (P, K) = 0. So it remains to verify that for any x ∈ HomR (L, K) there exists y ∈ HomR (P, K) such that y  L = x. Take x ∈ HomR (L, K). For each α ∈ E define hα : ω → EndR (K) by hα (i) = xσα,i for all k ∈ K and i < ω. Let f : λ+ → EndR (K) be as in UPλ (where κ = EndR (K)). Define y ∈ HomR (P, K) as follows:  (i) y  Pα = f (α)μ−1 α provided that α = nα (i) for some α ∈ E and jα ≤ i < ω.

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369

(ii) If α ∈ E, then Pα = Cjα ⊕ να (K (jα ) ) and να (K (ω\jα ) ) ⊆ Cjα . Define y  Cjα = 0, and for each i < jα yνα μα,i = f (nα (i)) − xσα,i provided that nα (i) = nα (i ) for some α ∈ E and i ≥ jα yνα μα,i = −xσα,i provided that nα (i) = nα (i ) for all α ∈ E and i ≥ jα . (iii) y  Pα = 0 otherwise. We prove that y  L = x. Let α ∈ E and i < ω. If i ≥ jα , then y  Cjα = 0, so yσα,i = yμnα (i) = f (nα (i)) = hα (i) = xσα,i . If i < jα and nα (i) = nα (i ) for some α ∈ E and i ≥ jα , then yσα,i = f (nα (i )) − (f (nα (i)) − xσα,i ) = xσα,i . If i < jα , but nα (i) = nα (i ) for all α ∈ E and i ≥ jα , then y  Pnα (i) = 0, so yσα,i = xσα,i . This proves that y and x coincide on all Lα,i (α ∈ E, i < ω), whence y  L = x. 2 Theorem 10.2.8. Assume UPλ holds. Then A  ⊥ K.

Proof. Since K ∈ KerC, we have A ⊆ ⊥ K. By Lemma 10.2.7 follows D ∈ ⊥ K. By Lemma 10.2.6 we conclude that D ∈ / A. 2

Corollary 10.2.9. (a) Let R be a non–right perfect ring and F be a non–zero free module. Assume UPλ holds for some λ > |EndR (F )| such that cf(λ) = ω. Then P0  ⊥ F . Moreover, D ∈ ⊥ F has projective dimension 1. (b) Let R be a Matlis valuation domain of global dimension 2 with the quotient field Q. Let K! = ! Q/R. Assume U Pλ holds for some λ such that λ > ! ! |EndR (K)|(= !R!) and cf λ = ω. Then P1  ⊥ K. Moreover, D ∈ ⊥ K has projective dimension 2.

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10 Independence results for cotorsion pairs

Proof. By Examples 10.2.2 and 10.2.3, Theorem 10.2.8 applies to the cotorsion pairs C0 and C1 , respectively. Moreover, D = P/L ∈ / Pi , while pairs P, L ∈ Pi , where i = 0 and i = 1, respectively. 2 Remark 10.2.10. If V = L and R is right hereditary, but not right perfect, then P0 = ⊥ F for any free module F of rank ≥ 2|R| (see [383, §3]). Similarly, if V = L and R is an IC–domain, then P1 = ⊥ K by [53, §4]. So with the additional assumption of R being right hereditary (R being an IC–domain), the assertion (a) (resp. (b)) of Corollary 10.2.9 is independent of ZFC + GCH. By Theorem 4.1.12, in either case V = L implies that the cotorsion pair cogenerated by K is generated by a set of modules. Lemma 10.2.11. Let C = (A, B) be a cotorsion pair generated by a set. Let K be a local splitter for C. Assume (⊥ K)⊥ = M ⊥ for a module M . (a) Assume A ⊆ P1 . Then (⊥ K)⊥ = B ⊥ for some B ∈ B ∩ ⊥ K. (b) Assume that R is a Prüfer domain with Q ∈ P1 and C = C1 . Then (⊥ K)⊥ = B ⊥ for a torsion module B ∈ B ∩ ⊥ K.

Proof. (a) Since C is complete, there is a special B–preenvelope of M , 0 → M → B → A → 0, with A ∈ A and B ∈ B. For a module N , consider the long exact sequence . . . → Ext1R (A, N ) → Ext1R (B, N ) → Ext1R (M, N ) → Ext2R (A, N ) = 0. Then B ⊥ ⊆ M ⊥ . Conversely, if N ∈ M ⊥ , then N ∈ (⊥ K)⊥ ⊆ A⊥ , so N ∈ A⊥ , and the long exact sequence gives N ∈ B ⊥ . Clearly B ∈ ⊥ (B ⊥ ) = ⊥ K. (b) Let B be as in part (a) for A = P1 and B = D. Then B = B  ⊕ Q(κ) for a cardinal κ and a divisible torsion module B  . By assumption, Ext1R (Q, K) ∼ = 2 ⊥ ⊥ ⊥ ⊥  ⊥ ⊥ ExtR (Q, R) = 0, so B = ( K) ⊆ Q , and (B ) = B . 2 Theorem 10.2.12. Assume UP and GCH. Let C = (A, B) be a cotorsion pair generated by a set. Let K be a local splitter for C and N = Coker(ν). Denote by K the cotorsion pair cogenerated by K. Assume that A ⊆ P1 . If R is a Matlis valuation domain and C = C1 , denote by T the class of all torsion divisible modules; otherwise put T = B. Anyway, assume that T is closed under unions of chains and T ∩ ⊥ K ∩ N ⊥ = 0. Then K is not generated by any set of modules.

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371

Proof. Assume that K is generated by a set, so (⊥ K)⊥ = B ⊥ for a module B. By Lemma 10.2.11, we can w.l.o.g. assume that B ∈ T ∩ ⊥ K. Let λ be a singular cardinal of cofinality ω such that λ > |EndR (K)|, λ ≥ |B| + |K| + |R| + ρC. Put κ = λ+ = 2λ . Then κλ = κ, so Theorem 3.2.1 (and Remark 3.2.2 (c)) gives an exact sequence 0 → B → A → A/B → 0, where A ∈ B ⊥ , and A/B is the union of a smooth chain (Aα /B | α < κ), and all the consecutive factors Aα+1 /Aα are isomorphic to B. By the assumption on the class T , Tα = A/Aα ∈ T ∩ ⊥ K for each α < κ. Take any α < κ. Then Ext1R (N, Tα ) = 0, since T ∩ ⊥ K ∩ N ⊥ = 0. So there / Im HomR (ν, Tα ). exists fα ∈ HomR (K (ω) , Tα ) such that fα ∈ Using UPλ , we define the module D = P/L as above. By Lemma 10.2.7, D ∈ ⊥ K. Since A ∈ B ⊥ and ⊥ (B ⊥ ) = ⊥ K, we infer that Ext1R (D, A) = 0. −1 Consider α ∈ E. Define ξα ∈ HomR (Lα , Tα ) by ξα = fα ( i<ω σα,i ) . Since Lα ∈ A and Aα ∈ B, we have Ext1R (Lα , Aα ) = 0. Applying HomR (Lα , −) to the exact sequence πα 0 → Aα → A → Tα → 0, we get the existence of ϕα ∈ HomR (L α , A) such that ξα = πα ϕα . 1 Define ϕ ∈ HomR (L, A) by ϕ = α∈E ϕα . Since ExtR (D, A) = 0, there exists ψ ∈ HomR (P, A) such that ψ  L = ϕ. Since |K| ≤ λ, the set   C = β < κ | ψ(μα (k)) ∈ Aβ ∀k ∈ K ∀α < β : α non–limit is a cub in κ. Since E is stationary, there exists γ ∈ E ∩ C. Put σ = ⊕i<ω σγ,i ∈ HomR (K (ω) , Lγ ). Then πγ ψσ = −πγ ψνγ . So fγ = ξγ σ = πγ ϕσ = πγ ψσ = −πγ ψνγ = ϑν, / Im HomR (ν, Tγ ). where ϑ ∈ HomK (ω) (Tγ ,). This contradicts fγ ∈ This proves that K is not generated by any set of modules.

2

We finish by an application of Theorem 10.2.12 to two cases of particular interest. Recall that W1 denotes the Whitehead cotorsion pair (cogenerated by R and generated by the class of all Whitehead modules). Similarly, for an IC–domain R, W2 denotes the cotorsion pair cogenerated by K (that is, generated by the class of all W2 –modules, see Definition 2.2.4 (b)). Theorem 10.2.13. Assume UP and GCH. (a) Let R be a countable Dedekind domain which is not a field. Then W1 is not generated by any set of modules.

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10 Independence results for cotorsion pairs

(b) Let R be an IC–domain. Then W2 is not generated by any set of modules.

Proof. (a) We apply Theorem 10.2.12 to C0 , where K = R and (ai | i < ω) is the set of all non–invertible non–zero elements of R. Then, on the one hand, N ∼ = Q, ⊥ so each module M ∈ N is cotorsion. On the other hand, R is not complete, so if M ∈ ⊥ R, then each countably generated submodule of M is projective (see e.g. [181, p.537]). If M ∈ N ⊥ ∩ ⊥ R, then M is torsion–free and cotorsion, hence pure–injective, but each countably generated submodule of M is projective. This yields M = 0 proving that N ⊥ ∩ ⊥ R = 0. (b) We apply Theorem 10.2.12 to the cotorsion pair C1 = (P1 , D), for K = Q/R and N = Q/J, where J is a non–principal ideal of R (see Example 10.2.3). Since the class of all divisible torsion modules is closed under unions of chains, it remains to prove that T ∩ ⊥ K ∩ N ⊥ = 0. Let M ∈ T ∩ ⊥ K. Then M is a coherent module (see [44, §2]). Assume that M ∈ N ⊥ . Then 0 = Ext1R (N, M ) → Ext1R (R/J, M ) → Ext2R (K, M ) = 0 is exact, so Ext1R (R/J, M ) = 0, and M = 0 because M is torsion, divisible and coherent (see [44, §3]). 2

Corollary 10.2.14. (a) Let R be a countable Dedekind domain which is not a field. Then the assertion “W1 is generated by a set of modules” is independent of ZFC + GCH. (b) Let R be an IC–domain. Then the assertion “W2 is generated by a set of modules” is independent of ZFC + GCH.

Proof. Clearly the group Z has injective dimension ≤ 1, and so does the module K (since the injective dimension of the IC–domain R is ≤ 2). So the claims follow by Theorem 10.2.13 and Corollary 10.1.4 (cf. Remark 10.2.10). 2

Remark 10.2.15. Also the completeness of a particular cotorsion pair can be independent of ZFC (+ GCH). Namely, Theorem 10.2.13 (a) can be improved in the case when R = Z: in [146], Eklof and Shelah constructed an extension of ZFC + GCH in which Q has no ⊥ Z–precover, so the class of all Whitehead groups is not precovering, and the cotorsion pair W1 is not complete.

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373

Open problems 1. Let C be a cotorsion pair in Mod–R cogenerated by a class of Enochs cotorsion modules. Is C complete? This is true when R is a Dedekind domain by Theorem 8.2.11. Notice that, if R is a right perfect ring, then the question asks whether all cotorsion pairs are complete. 2. Let C be a cotorsion pair in Mod–R cogenerated by a set containing at least one non–Enochs cotorsion module. Is the completeness of C independent of ZFC? This is true for the Whitehead cotorsion pair in case R = Z by [146] (see Remark 10.2.15). A related result was proved in [147]: if R is a Dedekind domain with a countable spectrum, then whether or not C is generated by a set is independent of ZFC (+ GCH). The assumption of C being cogenerated by a set is important here: there are many cotorsion pairs cogenerated by proper classes whose completeness is proved in ZFC. 3. Assume V = L. Let R be a ring. Is each cotorsion pair in Mod–R complete? Is the answer positive at least for the cotorsion pairs cogenerated by a set? For partially positive answers see Theorems 10.1.2 and 10.1.5.

Chapter 11

The lattice of cotorsion pairs

In this chapter we consider the lattice of cotorsion pairs (F, C) which are defined in Chapter 2 (Section 2.2) and investigated in Chapter 10. Recall that (F, C) is a cotorsion pair of R–modules, if F and C are classes of R–modules which are maximal with respect to Ext1R (F, C) = 0 for all F ∈ F and C ∈ C (see Definition 2.2.1). The cotorsion pairs are ordered correspondingly to the first component F, that is, (F, C) ≤ (F  , C  ) if F ⊆ F  . This convention agrees with Section 2.2 (and is dual to the ordering in Salce [350]). Moreover, the supremum and infimum of a family {(Fi , Ci ) | i ∈ I} of cotorsion pairs is given canonically (again, see Section 2.2): ) ) ( ( * '      ⊥   and . (Fi , Ci ) = ⊥ Ci , Ci (Fi , Ci ) = Fi , Fi i∈I

i∈I

i∈I

i∈I

i∈I

i∈I

Although the above definitions can be applied to arbitrary module categories, we shall mainly restrict our attention to cotorsion pairs of abelian groups here. More precisely, the main result of this chapter deals with the existence of many cotorsion pairs in the category of abelian groups; this follows by application of the Strong Black Box (Variant 3) 9.2.17. However, we first utilize the Strong Black Box to construct ultra–cotorsion–free R–modules which will help us to define distinct cotorsion pairs in Section 11.3.

11.1

Ultra–cotorsion–free modules and the Strong Black Box

In this section we will apply the Strong Black Box as given in Corollary 9.2.21 to prove the following theorem which is needed to investigate the lattice of cotorsion

11.1

Ultra–cotorsion–free modules and the Strong Black Box

375

pairs in the last part of this chapter. Recall the Definition 9.2.14 of ultra–cotorsion– free R–modules: an R–module G is ultra–cotorsion–free, if any submodule U ⊆ G with |G| = |U | and G/U S–cotorsion–free equals G. Also recall, that G is locally free, if any finite subset of G is contained in a free S–pure submodule of G (see Definition 1.1.12). Theorem 11.1.1. Let R be an S–ring, (Definition 1.1.1) and let |R| ≤ κ ≤ μ ≤ λ be infinite cardinals such that μκ = μ and λ = μ+ . Then there exists a locally free, ultra–cotorsion–free R–module G of cardinality λ. The existence of ultra–cotorsion–free modules was first shown for strong limit singular cardinals in [224] in order to answer a problem in Schultz [357] at a stage when the Black Boxes were not yet fully developed. For all relevant notions like the canonical summand, Signac–branch, or canonical Signac–pair we refer to Definition 9.2.15. Before we construct the desired module we show the following step lemma, for  from the Strong Black Box 9.2.17. which we also use the notations (of B and B)  let Step Lemma 11.1.2. Let M be an S–pure, S–cotorsion–free submodule of B,  b be an S–pure element of B ∩ M , and let π ∈ R. Moreover, let v = (vn )n<ω be a Signac–branch with vn ∈ M for all n < ω such that  b  <  v  (= supn<ω  vn ) and  [ m ] ∩ [ v ]  <  v  for all m ∈ M . Then qn vn + πb∗ M  = M, y = n<ω

is also S–cotorsion–free.

Proof. Let the assumptions be as above. First note that M = M + Ry k k<ω

with yk =

qn vn + π k b qk

n≥k

and πk =

qn rn for π = qn rn , qk n<ω

n≥k

i.e. π − qk π k ∈ R, respectively y − qk y k ∈ M , for all k < ω. Suppose there is a non–zero homomorphism  −→ M  . φ:R

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11 The lattice of cotorsion pairs

Then, by continuity, 1φ = 0 and so 1φ = m + ry k

(11.1.1)

for some m ∈ M, r ∈ R, k < ω. For r = 0 we deduce  [ πφ ] ⊆ [ 1φ ] = [ m ] for all π ∈ R and thus, by support arguments, Im φ ⊆ M , contradicting the S–cotorsion–freeness of M . Hence we now consider, in (11.1.1), the case r = 0. We may also assume that  let m = 0, since we could increase k. Now for any π ∈ R πφ = mπ + rπ y kπ for some mπ ∈ M, rπ ∈ R, k ≤ kπ < ω. Then, since πφ = π(1φ) by continuity, we have πm + πry k = mπ + rπ y kπ , respectively    qk qk  πm − mπ + πr y k − π y kπ = rπ − πr π y kπ =: g. q qk + ,-k . ∈M

From

  qk g = πm − mπ + πr y k − π y kπ qk

 and thus, on the one hand, it follows that g ∈ RM [ g ] ∩ [ v ] < v   q  by assumption. On the other hand, g = rπ − πr qkkπ y kπ and so  [ g ] ∩ [ v ]  =  v  (=  y  =  y kπ ) unless rπ = πr

qkπ qk

(i.e. g = 0). The latter implies πr

qk  qkπ q kπ ∈R ∩ πR R, = qk qk qk

 contradicting the S–cotorsion–freeness of R. and so πr ∈ R for any π ∈ R, Therefore there is no such homomorphism φ which means that M  is S–cotorsion– free, as required. 2 Now we are ready to construct the desired module.

11.1

Ultra–cotorsion–free modules and the Strong Black Box

377

Construction 11.1.3. Let (tβ = (Pβ , (vβ,n )n<ω , bβ , πβ ))β<λ be a family of traps as given by Corollary 9.2.21. We inductively define an increasing chain of S–pure, S–cotorsion–free submodules  (β < λ) Gβ ⊆ B such that Gβ ⊆ B +



γ . P

(11.1.2)

γ<β

 First we put G0 = B = α<λ Reα ; B is clearly an S–pure, S–cotorsion–free  satisfying (11.1.2). submodule of B Let β be a limit ordinal and suppose the Gγ s (γ < β) are given satisfying all the required conditions. We put  Gγ . Gβ = γ<β

 and Gβ satisfies (11.1.2). Moreover, Gβ is S–cotorsion– Then, obviously, Gβ ⊆∗ B  → Gβ , we obtain Im φ ⊆ Gγ for some free since, for a homomorphism φ : R γ  γ < β with 1φ ∈ G by (11.1.2) and by [ πφ ] ⊆ [ 1φ ] for all π ∈ R. Now suppose that Gβ is given and let tβ = (Pβ , (vβ,n )n<ω , bβ , πβ ) be the trap from the above family. If vβ,n ∈ Gβ for all n < ω, then we define yβ by qn vβ,n + πβ bβ yβ = n<ω

and put

0 / Gβ+1 = Gβ , yβ . ∗

From Step Lemma 11.1.2 we know that is also an S–pure, S–cotorsion– β+1  satisfies (11.1.2), since bβ ∈ Pβ , vβ,n ∈ free submodule of B. Moreover, G β . β (n < ω), and so yβ ∈ P P We can describe the above purification easily: Ryβk , Gβ+1 = Gβ + Gβ+1

k<ω

where

β for all k < ω yβ = yβ0 and yβ − qk yβk ∈ Gβ ∩ P

(cf. proof of Step Lemma 11.1.2).

378

11 The lattice of cotorsion pairs

If vβ,n ∈ / Gβ for some n < ω, then we do not extend Gβ , that is, we put β = G (yβ = 0).

Gβ+1

Finally let G=





Gβ = B +

β<λ

Ryβk .

β<λ k<ω

= (i.e. yβ = 0) happens ‘often’ since the prediction in Note that Corollary 9.2.21 can be applied to the set U = p(B) of all S–pure elements of B = G0 . Gβ

Gβ+1

It is immediate from the construction that |G| = λ and that G is an S–pure,  S–cotorsion–free submodule of B. Next we describe the elements of G. Lemma 11.1.4. Let G be as in Construction 11.1.3 and let g ∈ G \ B. Then there are a finite non–empty subset N of λ and kβ < ω (β ∈ N ) such that k Ryββ g ∈B+ β∈N

and [ g ] ∩ [ yβ ] is infinite ⇐⇒ β ∈ N. In particular, if  g  is a limit ordinal, then  g  =  yβ∗  =  tβ∗  for β∗ = max N.

Proof. Let the assumptions be as above. Since g ∈ / B, there is β0 < λ such that β +1 β 0 0 g∈G \ G . By construction, Gβ0 +1 = Gβ0 + Ryβk0 , k<ω

where yβn0 −

qk k y ∈ Gβ0 for all n ≤ k < ω. q n β0

Therefore g can be expressed as kβ

g = g1 + r0 yβ0 0 for some kβ0 < ω, r0 ∈ R, g1 ∈ Gβ0 . Now either g1 ∈ B , or there is β1 < β0 such that g1 ∈ Gβ1 +1 \ Gβ1 and, as before, we can write g1 as kβ

g1 = g2 + r1 yβ1 1

(kβ1 < ω, r1 ∈ R, g2 ∈ Gβ1 ).

11.1

Ultra–cotorsion–free modules and the Strong Black Box

379

We proceed like this and obtain ordinals β0 > β1 > . . . and elements gi ∈ Gβi +1 \ Gβi (g = g0 ) such that

kβ

gi = gi+1 + ri yβi i . Since the ordinals are well–ordered, there exists n < ω with gn ∈ B. Therefore g = gn +



kβ

ri yβi i ∈ B +

i


k

Ryββ

β∈N

with N = {βi | i < n}, where N = ∅, since g ∈ / B. Moreover, it easily follows from Corollary 9.2.21 (iii) that [ g ] ∩ [ yβ ] is infinite ⇐⇒ β ∈ N, and so the proof is finished.

2

Using the above lemma, we prove further properties of the module G. Recall that E ⊆ λo is a given stationary set (see Strong Black Box 9.2.17). Lemma 11.1.5. Let G be as in Construction 11.1.3 and define Gα (α < λ) by Gα := {g ∈ G |  g  < α}. Then the following conditions hold: (a) {Gα | α < λ} is a λ–filtration of G; (b) if α, β < λ are ordinals such that  tβ  = α , then Gα ⊆ Gβ ; (c) if α ∈ / E , then Gα+1 /Gα is free; and (d) if α ∈ E and Gα+1 /Gα = 0 , then Gα+1 /Gα contains a non–zero S– divisible submodule.

Proof. First we show (a). Let α < λ be arbitrary. Then we clearly have Gα ⊆ Gα+1 . Moreover, ! ! !! ℵ0 = (|R| · |α|)ℵ0 ≤ μℵ0 = μ < λ |Gα | ≤ !B α ! ≤ |Bα |

380

11 The lattice of cotorsion pairs

 (recall: Bα = δ<α Reδ ). It is easy to see that the increasing chain of the Gα s is smooth and that  G= Gα α<λ

holds. To see (b) let α, β < λ with  tβ  = α, and let g ∈ Gα . If g ∈ B, we have finished. Otherwise, by Lemma 11.1.4, k g∈B+ Ryγ g γ∈N

for some finite N ⊆ λ, kγ < ω such that [ g ] ∩ [ yγ ] is infinite ⇐⇒ γ ∈ N. This implies  tγ  =  yγ  ≤  g  < α =  tβ  for all γ ∈ N, and thus N ⊆ β by Corollary 9.2.21 (ii). Hence g ∈ Gβ , and so (b) is proved. Next we show (c). Let α < λ with α ∈ / E. If α is a limit ordinal, then, by Corollary 9.2.21 (i) and Lemma 11.1.4, there is no element of norm α in G, and so Gα+1 /Gα = 0 in this case. If α = δ + 1 (δ < λ), then  eδ  = α, and any element g ∈ G with  g  = α can be written as g = reδ + g  (r ∈ R, g  ∈ Gα ). Therefore Gα+1 /Gα = eδ + Gα  ∼ = R in this case. Thus Gα+1 /Gα is free for α∈ / E, as required. Finally, we show (d). To do so let α ∈ E with Gα = Gα+1 . Then there is an element g ∈ G with  g  = α. Hence it follows from Lemma 11.1.4 that there exists β < λ such that yβ = 0 and α =  yβ  =  tβ . This implies Gβ = Gβ+1 , and hence vβ,n ∈ Gβ with  vβ,n  < α (see Construction 11.1.3 and Definition 9.2.15 of a Signac–branch). Therefore vβ,n ∈ Gα for all n < ω. We also know that bβ ∈ Pβ ⊆ Gα . Thus we deduce yβ ≡ qn yβn mod Gα for all 2 n < ω, i.e. yβ + Gα is an S–divisible element of Gα+1 /Gα . Finally, we are ready to prove the main theorem of this section, that is, the existence of a locally free, ultra–cotorsion–free R–module G. Proof of Theorem 11.1.1. Let G be the R–module given in Construction 11.1.3.  is S–cotorsion–free. We already know that |G| = λ and that G ⊆∗ B

11.1

Ultra–cotorsion–free modules and the Strong Black Box

381

First we show that G is ultra–cotorsion–free. To do so let U  be a submodule of G and let ψ : G −→ G/U  be the canonical epimorphism. Suppose, for contradiction, that 0 = G/U  is S–  cotorsion–free and |U  | = λ = |G|. Since G/U  is S–reduced and G/B ⊆∗ B/B  is S–divisible, U cannot contain B. Therefore, by the S–torsion–freeness of G/U  (if and only if U  ⊆∗ G), there is an S–pure element b of B such that b∈ / U  = Ker ψ,  Let U = p(U  ) i.e. bψ = 0. We are going to show that πbψ ∈ G/U  for all π ∈ R.  be arbitrary. Then |U | = λ and so, by Corollary 9.2.21 (iv), there and let π ∈ R exists β < λ such that {vβ,n | n < ω} ⊆ U ⊆ U  ⊆ G, b = bβ and π = πβ . Let  Pβ  = α. By the definitions of a Signac–branch and of a trap we then obtain {vβ,n | n < ω} ⊆ Gα , where Gα ⊆ Gβ by Lemma 11.1.5. Therefore it follows from the Construction 11.1.3 that   0 = yβ = qn vβ,n + πb ∈ G n<ω

 and using the and so yβ ψ ∈ G/U  . Identifying ψ with its unique extension to B continuity of ψ, we deduce   qn vβ,n ψ + (πb)ψ = qn (vβ,n ψ) + π(bψ) = π(bψ) yβ ψ = + ,- . n<ω n<ω =0

 was arbitrary and bψ = 0, we deduce and thus π(bψ) ∈ G/U  . Since π ∈ R   0 = R(bψ) ⊆ G/U , contradicting the S–cotorsion–freeness of G/U  . Therefore such a submodule U  does not exist, i.e. for G/U  to be S–cotorsion–free we need G = U  or |U  | < λ, as required. It remains to show that G is locally free; i.e. we have to show that any finite subset of G is contained in a free S–pure  submodule of G. We inductively show thatG = α<λ Gα is locally free. Clearly G0 = 0 is locally free and so is Gα = δ<α Gδ , if α is a limit ordinal provided the Gδ s are locally free. Now suppose that Gα is locally free and consider Gα+1 . If α ∈ / E, then Gα+1 /Gα is free by Lemma 11.1.5 and thus Gα+1 is also locally free. So let α ∈ E and   Iα = β < λ |  Pβ  = α, Gβ = Gβ+1 .

382

11 The lattice of cotorsion pairs

Then Gα+1 = Gα +



Ryβk .

β∈Iα k<ω

Hence a finite subset of Gα+1 looks like X ∪ Y with X ⊆ Gα finite and Y = {yβk | k ≤ l, β ∈ I ∗ } for some finite I ∗ ⊆ Iα , l < ω. Without loss of generality we may assume that l is big enough such that  vβ,l  >  bβ   for all β, β  ∈ I ∗ ; & %  vβ,l  >  [ Pβ ] ∩ Pβ   for all β = β  ∈ I ∗ ;

(11.1.3)

 vβ,l  >  X  for all β ∈ I ∗ . Let δ = max{ vβ,l  | β ∈ I ∗ }. Then δ < α and X  := X ∪ {bβ | β ∈ I ∗ } ∪ {vβ,n | n < l, β ∈ I ∗ } is a finite subset of Gδ , and hence there exists an S–pure, free submodule X ∗ of Gδ (of finite rank) containing X  ; in particular, we have  X ∗  < δ.  We now put Y ∗ = β∈ I ∗ Ryβl . Then X ∪ Y ⊆ X ∗ + Y ∗ and, in fact, Y ∗ =  l ∗ β∈ I ∗ Ryβ by (11.1.3), i.e. Y is free of finite rank. ∗ ∗ Moreover, X ∩ Y = 0, since elements of X ∗ have a norm less than δ < α and non–zero elements of Y ∗ have norm α. Finally, we show that X ∗ ⊕ Y ∗ is an S–pure submodule of Gα+1 . To do so let g ∈ Gα+1 with sg ∈ X ∗ ⊕ Y ∗ for some s ∈ S. Since we could multiply with a suitable element of S, we may express g as g = g + rβ yβ β∈ I 

for some g  ∈ Gα , I  ⊆ Iα finite. This implies sg = sg  + srβ yβ = x + rβ yβl β∈ I 

β∈ I ∗

for some x ∈ X ∗ , rβ ∈ R, where the latter follows from sg ∈ X ∗ ⊕ Y ∗ . Simple support arguments show that I  ⊆ I ∗ , and so I  = I ∗ may be assumed. Moreover, we now have srβ ql yβl + srβ (yβ − ql yβl ) = x + rβ yβl , sg = sg  + + ,. β∈ I ∗ β∈ I ∗ β∈ I ∗ ∈ Gα

11.2

and hence

Rational cotorsion pairs

(srβ ql − rβ )yβl = x − sg  −

β∈ I ∗



383

srβ (yβ − ql yβl ) =: h ∈ Gα .

β∈ I ∗

Therefore, since the yβl s are linearly independent and  yβl  = α >  h  for all β ∈ I ∗ , we deduce srβ ql − rβ = 0, respectively rβ ∈ sR, for all β ∈ I ∗ . Thus  l ∗ β∈ I ∗ rβ yβ ∈ sY . Moreover,   rβ (yβ − ql yβl ) ∈ X ∗ ∩ sGα = sX ∗ , say x = sx (x ∈ X ∗ ). x = s g + β∈ I ∗

 Putting everything together we obtain sg = sx + s β∈ I ∗ rβ ql yβl , respectively  g = x + β∈ I ∗ rβ ql yβl by the S–torsion–freeness of G. Hence X ∗ ⊕Y ∗ is S–pure 2 in Gα+1 , which completes the proof. A trivial modification of the proof of the local freeness of G above together with Lemma 11.1.5 (d) is given by the following Proposition 11.1.6. Let G=



Gα

α<λ

be the λ–filtration of G as in Lemma 11.1.5 and let E ⊆ λ be the stationary subset given in the Strong Black Box 9.2.17. Then G/Gα is locally free, if and only if α∈ / E. As mentioned in the beginning, we shall make use of the module G constructed in this section later. In fact, we will need a family of such modules depending on different stationary sets E; thus we introduce the notation G = G(E) to refer to the module as obtained in Construction 11.1.3 satisfying the conclusion of Theorem 11.1.1 and also the other results.

11.2 Rational cotorsion pairs In the remaining part of this chapter, i.e. in this and the following section, we restrict our attention to the category of abelian groups. More precisely, we discuss the lattice structure of the class of all cotorsion pairs of abelian groups. The lattice structure is given by the definitions in Section 2.2, which have been recalled at the beginning of this chapter. Note, that most of the results presented below come from Göbel, Shelah, Wallutis [240].

384

11 The lattice of cotorsion pairs

The maximal element in the lattice considered here is (Mod−Z, D) , where D is the class of all divisible groups; it is, for example, generated by  the set of all cyclic groups of prime order or, equivalently, by the single group p∈Π Zp (Π = ‘all primes’). (Note that the notions ‘generated’ and ‘cogenerated’ are also dual to [350].) The minimal element is the cotorsion pair (P0 , Mod−Z) , where P0 is the class of all free groups; it is generated by Z. Another important and well– studied cotorsion pair is the classic one (F0 , EC) , where F0 denotes the class of all torsion–free groups and EC denotes the class of all cotorsion groups; it is generated by the rationals Q. Correspondingly, for a cotorsion pair (F, C) of abelian groups, we call the first component F the torsion–free class and the second component C the cotorsion class. Salce [350] noted that every cotorsion pair is generated by a class of groups which are torsion or torsion–free: Proposition 11.2.1. Every cotorsion pair is generated by torsion and torsion–free groups.

Proof. It is sufficient to prove that, if (F, C) is a cotorsion pair, then F ∈ F, if and only if tF ∈ F and F/tF ∈ F, where tF denotes the torsion subgroup of F . By the closure properties of F it hence remains to show that F ∈ F implies F/tF ∈ F. So let F ∈ F and let C ∈ C be arbitrary, i.e. Ext(F, C) = 0. Since all divisible groups are contained in C and any group is a direct sum of its divisible subgroup and a reduced group, we may assume that C is reduced. Moreover, we have that C is p–divisible for any prime p with Zp ∈ F because Ext(Zp , C) ∼ = C/pC; in particular, this is true for all primes p such that tF has a non–zero p–component tp F . Finally considering the exact sequence Hom(tF, C) −→ Ext(F/tF, C) −→ Ext(F, C) = 0, we deduce Ext(F/tF, C) = 0, respectively F/tF ∈ F, since Hom(tF, C) must be zero by the above. 2 By this proposition any cotorsion pair is the supremum of a cotorsion pair generated by torsion groups and a cotorsion pair generated by torsion–free groups. Now a cotorsion pair which is generated by torsion groups is always ≥ (F0 , EC), and we have already classified the sublattice of all cotorsion pairs in the interval between (F0 , EC) and (Mod–R, I0 ) in Theorem 8.2.11: these cotorsion pairs correspond one–to–one to subsets of the primes. So, we may restrict our attention

11.2

Rational cotorsion pairs

385

to the cotorsion pairs generated by torsion–free groups, i.e. to the cotorsion pairs ≤ (F0 , EC). Naturally we first consider torsion–free groups, which can be characterized by invariants, namely torsion–free rank–1 groups which we will often identify with the subgroups of the rationals. Corresponding to the latter we call a cotorsion pair generated by a rank–1 group a rational cotorsion pair. Rational cotorsion pairs were discussed in detail by Salce [350]. One of our first results on T ⊥ will be Proposition 11.2.4. It will be applied to prove that the sublattice of all rational cotorsion pairs can be identified with the well–known lattice of types. We construct examples to establish that the (obvious) lattice homomorphism is an isomorphism. Note that, for the lattice of types, it is known that there exist anti–chains of size 2ℵ0 , which equals the cardinality of the lattice (see [173]), and also ascending and descending chains of uncountable length, in fact, there are descending and ascending chains of cofinality at least ℵ1 (see [137]). Next we recall the definition of a cotorsion pair generated by a single group. For a (rank–1) group T let T ⊥ = {X | Ext(T, X) = 0}   (T ⊥ ) = Y | Ext(Y, X) = 0 for all X ∈ T ⊥ .   The pair ⊥ (T ⊥ ), T ⊥ is the cotorsion pair generated by T , where T ⊥ is the class torsion–free class. Thus of all T –cotorsion groups and ⊥ (T ⊥ ) the   ⊥ corresponding ⊥ ⊥ rational cotorsion pairs are of the form (T ), T for some rational (= rank–1) group T . As we did with the cotorsion pairs in general, we also order the rational cotorsion pairs according to the inclusion of the first component. Recall, a type τ is an equivalence class of sequences of the form (tp )p∈Π , where tp ∈ N ∪ {0} ∪ {∞}, and Π is the set of all primes in Z, and two such sequences are equivalent, if and only if they only differ in finitely many finite entries. Note, we shall slightly abuse the notation by representing a type with a sequence. Baer showed that rank–1 groups are uniquely determined by their types (up to isomorphism) and that there exists a rank–1 group of type τ for each possible type τ (see [173]). Now let T be the set of all types and let Crat be the set of all rational cotorsion pairs. We define  ⊥ Φ : (T , ≤) −→ (Crat , ≤) by τ → τ Φ = (T ⊥ ), T ⊥ ∈ Crat , and

⊥

where T is a rank–1 group of type τ . The aim of this section is to prove that the mapping Φ is an order–preserving isomorphism. First we show that Φ is order–preserving.

386

11 The lattice of cotorsion pairs

Lemma 11.2.2. Let R be rank–1 with t(T ) ≤ t(R). Then T ⊥ ⊇ R⊥  ⊥groups ⊥T, ⊥ ⊥ ⊥ ≤ (R ), R⊥ ). (and, equivalently, (T ), T

Proof. Since t(T ) ≤ t(R), there is a monomorphism ϕ : T −→ R (see [173, Vol II, Proposition 85.4]). Let G be an element of R⊥ , i.e. Ext(R, G) = 0. Now the short exact sequence ϕ

0 −→ T −→ R −→ R/T −→ 0 induces the exact sequence Ext(R/T, G) −→ Ext(R, G) −→ Ext(T, G) −→ 0, and hence Ext(T, G) = 0. Therefore G ∈ T ⊥ and thus R⊥ ⊆ T ⊥ .

2

It follows immediately from the above lemma that the mapping Φ is well– defined: Corollary 11.2.3. Let T, T  be rank–1 groups of the same type τ . Then the corresponding cotorsion classes T ⊥ and (T  )⊥ coincide. Note that, more generally, we have G ⊆ H implies G⊥ ⊇ H ⊥ , and G ∼ =H implies G⊥ = H ⊥ for any groups G, H. In order to show that Φ is an isomorphism we consider types τ = t(T ), ρ = t(R) with τ strictly less than ρ and we show that R⊥ is properly contained in T ⊥ , i.e. we construct groups G ∈ T ⊥ \ R⊥ . Throughout the remainder of this section let τ = (tp )p∈Π = t(T ) and ρ = (rp )p∈Π = t(R) with tp ≤ rp for all primes p. For τ to be strictly less than ρ, one of the following two conditions has to be satisfied: (1) There exists a prime q such that tq < ∞ and rq = ∞. (2) There is an infinite set P of primes such that tp < rp < ∞ for all p ∈ P . Before we can construct the required groups we need Salce’s ([350, Theorem 3.5]) characterization of T –cotorsion groups: Proposition 11.2.4. Let τ = t(T ) be as above. Then  Gτp ⇐⇒ G/Gτ is Q–cotorsion, G ∈ T ⊥ ⇐⇒ G/Gτ ∼ = p∈Π

11.2

Rational cotorsion pairs

where



Gτ =

387

ptp G, Gτp = G/ptp G

p∈Π

for tp < ∞, and

Gτp = Ext(Zp∞ , G)

for tp = ∞. Applying Proposition 11.2.4 to rank–1 groups gives the following: Corollary 11.2.5. Let X be a rational group of type t(X) = (xp )p∈Π and let τ = t(T ) be as above. Then X is an element of T ⊥ , if and only if xp = ∞ for almost all p with tp = 0 and whenever tp = ∞.

Proof. First recall that, for an abelian group G, Ext(Zpn , G) ∼ = G /pn G for any  Jp , where m is the rank of a p–basic subgroup of n ∈ N and Ext(Zp∞ , G) ∼ = m

G, and Jp is the additive group of the ring of p–adic integers. Now assume that X ∈ T ⊥ . Then  Ext(Zptp , X) X/Xτ ∼ = {p∈Π|tp =0}

by Proposition 11.2.4. However, X, and hence X /Xτ , is countable and thus Ext(Zptp , X) = 0 for ! ! ! ! ! almost all p with tp = 0 and whenever tp = ∞, since ! Mn !! ≥ 2ℵ0 , if |Mn | ≥ n<ω

2 for all n, and Jp ⊆ Ext(Zp∞ , X) unless Ext(Zp∞ , X) = 0. Therefore ptp X = X for almost all p with 0 < tp < ∞, and the rank of a p–basic subgroup of X is zero for all p with tp = ∞. In either case it follows that X is p–divisible, and 2 hence xp = ∞ for almost all p with tp = 0 and whenever tp = ∞. From the above it is clear that the primes with tp = 0 play a special role. In particular, it makes sense to divide the second case (2) into two subcases: (2a) There is an infinite set P of primes such that tp = 0 and 0 = rp < ∞ for all p ∈ P. (2b) There is an infinite set P of primes such that 0 < tp < rp < ∞ for all p ∈ P. We first consider case (1). Proposition 11.2.6. Suppose t(T ) = τ < ρ = t(R) such that (1) is satisfied. Then there exists a rank–1 group X which is an element of T ⊥ but not of R⊥ .

388

11 The lattice of cotorsion pairs

Proof. Suppose (1),i.e. there is a prime q such that tq < ∞ and rq = ∞. Let X = Z(q) = m n ∈ Q | (n, q) = 1 be the localization of the integers Z at the prime q, i.e. t(X) = (xp )p∈Π with xp = ∞ for all p = q and xq = 0. Then X ∈ T ⊥ \ R⊥ by Corollary 11.2.5. 2 Case (2a) is as easily tackled as the above: Proposition 11.2.7. Suppose t(T ) = τ < ρ = t(R) such that (2a) is satisfied. Then there exists a rank–1 group X which is an element of T ⊥ but not of R⊥ .

Proof. Suppose (2a), i.e. there is an infinite set P of primes p with tp = 0 and 0 = rp < ∞. We define X ⊆ Q by t(X) = (xp )p∈Π with xp = 1 for p ∈ P and xp = ∞ otherwise, i.e, 1 X=

1 |p∈P p

2

 ∪

1 | n < ω, p ∈ Π \ P pn

23 .

Then X is an element of T ⊥ \ R⊥ by Corollary 11.2.5.

2

It remains to consider case (2b). This is slightly more difficult, as we cannot expect to find a rank–1 group belonging to T ⊥ , but not to R⊥ by Corollary 11.2.5. In fact, we cannot even find a group of any finite rank belonging to T ⊥ , and not to R⊥ , as we shall see shortly. Beforehand we need: Lemma 11.2.8. Let T and X be rank–1 groups with Ext(T, X) = 0. Then Ext(T, X) has cardinality 2ℵ0 .

Proof. Let t(T ) = τ = (tp )p∈Π and t(X) = (xp )p∈Π as before. The short exact sequence 0 −→ Z −→ T −→ T /Z −→ 0 induces the exact sequence Hom(Z, X) −→ Ext(T /Z, X) −→ Ext(T, X) −→ Ext(Z, X) = 0. Now

T /Z ∼ =



Zptp ,

tp =0

and hence

Ext(T /Z, X) ∼ =

 tp =0

Ext(Zptp , X) =: E.

11.2

Rational cotorsion pairs

389

By assumption, Ext(T, X) = 0, and thus, by Corollary 11.2.5, there is either some prime q with xq < ∞ and tq = ∞, or there are infinitely many primes qn with xqn < ∞ and 0 = tqn < ∞ (n < ω). In the first case, Ext(Zq∞ , X) contains a copy of the q–adic integers Jq , and in the latter we have 4 t Ext(Zqtqn , X) ∼ = X qnqn X = 0 n

for all n < ω. Therefore, in either case, it follows that E has at least cardinality 2ℵ0 and, of course, the cardinality cannot be bigger. Finally, the mapping Ext(T /Z , X) −→ Ext(T, X) in the above sequence is an epimorphism with at most countable kernel, and thus the result follows. 2 Now we can proceed with: Proposition 11.2.9. Let t(T ) = τ < ρ = t(R) satisfying condition (2b), but neither (1) nor (2a), and let F denote the set of all finite rank torsion–free groups. Then T ⊥ ∩ F = R⊥ ∩ F.

Proof. Without loss of generality we may assume that tp = 0, if and only if rp = 0, tp = ∞, if and only if rp = ∞, and that the remaining set is P = {p ∈ Π | 0 < tp < rp < ∞}, which is infinite by assumption. Obviously, T ⊥ ∩ F ⊇ R⊥ ∩ F. So let G ∈ T ⊥ ∩ F be of rank n. We show G ∈ R⊥ by induction on n. For n = 1 this follows immediately from Corollary 11.2.5. So let n > 1 and consider the short exact sequence 0 −→ X −→ G −→ G /X −→ 0, where X is a pure subgroup of G of rank 1 and so G /X is torsion–free of rank n − 1. The above sequence induces the exact sequences Hom(T, G/X) −→ Ext(T, X) −→ Ext(T, G) −→ Ext(T, G/X) −→ 0 and Hom(R, G /X ) −→ Ext(R, X) −→ Ext(R, G) −→ Ext(R, G /X ) −→ 0.

390

11 The lattice of cotorsion pairs

Now Ext(T, G) = 0 by assumption, and thus also Ext(T, G /X ) = 0. Hence, by induction hypothesis, Ext(R, G /X ) = 0, since rk (G /X ) = n−1. Therefore the first of the two above sequences reduces to Hom(T, G /X ) −→ Ext(T, X) −→ 0 and so |Ext(T, X)| ≤ ℵ0 . By Lemma 11.2.8 this is only possible, if Ext(T, X) = 0. Therefore we also have Ext(R, X) = 0, since rk(X) = 1, and thus it follows from the second of the above sequences that Ext(R, G) = 0, i.e. G ∈ R⊥ . 2 Before we tackle the remaining case (2b) we recall some well–known facts (see also Theorem 1.2.19 and the subsequent results): Remark 11.2.10. (a) Algebraically compact groups are cotorsion. In particular, complete groups are cotorsion. (b) Let H be a reduced cotorsion group, and let U ⊆ H be a subgroup. Then U is cotorsion, if and only if H/U is reduced.   (c) The completion of Zpnp in the Z–adic topology is Zpnp for any set p∈P

p∈P

P of primes.

Proof. Part (c) is an easy exercise; (a), (b) can be found in Fuchs [173, Vol. I, pp. 232–233]. 2 Now we are ready for: Proposition 11.2.11. Suppose t(T ) = τ < ρ = t(R) such that (2b) is satisfied. Then there exists a group G which is an element of T ⊥ but not of R⊥ .

Proof. Suppose (2b), i.e. P of primes such that 0 <  there exists an infinite set tp < rp < ∞. Let H = Z(p) ⊆∗ Z(p) = H  , where Z(p) is the localization p∈P

p∈P

of the integers at the prime p. Note that H  and thus also its pure subgroup H is q–divisible for any prime q ∈ / P . We define G as a subset of H  by   G = (gp )p∈P ∈ H  | ∃ m, k ∈ N such that mgp ∈ Z and |mgp | ≤ kptp ∀p ∈ P . First we show that G is a pure subgroup of H  containing H. Let (gp )p∈P , (hp )p∈P ∈ G, i.e. there are m, n, k, l ∈ N such that mgp , nhp ∈ Z, |mgp | ≤ kptp , and |nhp | ≤ lptp for all p ∈ P . Then mn(gp + hp ) = n(mgp ) + m(nhp ) ∈ Z and |mn(gp + hp )| ≤ n|mgp | + m|nhp | ≤ nkptp + mlptp =

11.2

391

Rational cotorsion pairs

(nk + ml)ptp for all p ∈ P . Thus (gp )p∈P + (hp )p∈P ∈ G, i.e. G is a subgroup. As an immediate consequence from the definition we have that G is pure in H  . Now let (hp )p∈P be an element of H, i.e. hp = 0 for almost all p and hp = zp np ∈ Z(p) . Let N be the product of all np , and let K be the sum of all |hp | over all p with hp = 0. Then N hp ∈ Z and N |hp | ≤ N K(∈ Z) for all p ∈ P . Thus H ⊆ G. Next let  5 πt : H  −→ Ht = Z(p) ptp Z(p) p∈P

be the canonical epimorphism given by   (hp )p∈P πt = hp + ptp Z(p) p∈P .     ∼ Z(p) /ptp Z(p) ∼ Zptp Obviously, Hπt ∼ = = p∈P Zptp and thus Ht = p∈P

p∈P

is the completion of Hπt by Remark 11.2.10 (c). Therefore Ht is cotorsion by Remark 11.2.10 (a). Since an element of Ht can be represented by (gp )p∈P with gp ∈ Z and 0 ≤ gp < ptp , we also immediately have that Gπt = Ht . But Gπt ∼ = G/ p∈Π ptp G,  since G is a pure subgroup of H , and so Ker πt ∩ G =



   ptp Z(p) ∩ G = ptp H  ∩ G

p∈P

=

p∈P

   ptp H  ∩ G = ptp G = ptp G = Gτ .

 p∈P

p∈P

p∈Π

Thus we have shown that G /Gτ ∼ = Gπt = Ht is cotorsion. Therefore G is an element of T ⊥ by Proposition 11.2.4. Finally, we show that G is not an element of R⊥ . Following the same arguments as above, we have that Hr =



5 Z(p) prp Z(p)

p∈P

is the completion of

Hπr ∼ =



Zprp ,

p∈P

rp p G and that Gπr ∼ = G/Gρ and that Hr is cotorsion, where Gρ = p∈Π

πr : H  −→ Hr

392

11 The lattice of cotorsion pairs

is the corresponding epimorphism. Now Hπr ⊆ Gπr , and so Hr /Gπr is divisible as an epimorphic image of the divisible group Hr /Hπr . Therefore it is enough to show that Gπr = Hr to prove G∈ / R⊥ by Remark 11.2.10 (b). 1 1 We choose integers np (p ∈ P ) such that ptp + 2 − 1 ≤ np ≤ ptp + 2 . Suppose (np + prp Z(p) )p∈P ∈ Gπr . Then there is (gp )p∈P ∈ G such that np ≡ gp mod 1

prp Z(p) for all p ∈ P . Note that gp = np for almost all p, since ptp + 2 − 1 ≤ np ≤ kptp for all p ∈ P is impossible for a fixed k ∈ N. However, there are m, k ∈ N such that mgp ∈ Z and |mgp | ≤ kptp for all p ∈ P . So mnp ≡ mgp mod prp Z, and thus prp divides m(np − gp ) (in Z). Since tp < rp , we have that ptp +1 divides 1

1

|m(np − gp )| ≤ |mnp | + |mgp | ≤ mptp + 2 + mkptp = mptp (p 2 + k) 1

1

for all p ∈ P . However, for almost all p ∈ P, m, k < 12 p 2 and thus ptp (mp 2 + mk) < ptp +1 , contradicting m(np − gp ) = 0. Therefore (np )p∈P ∈ Hr \ Gπr , and this completes the proof. 2 As a consequence of the above results we can finally state: Theorem 11.2.12. The lattice of types (T , ≤) is isomorphic to the lattice of rational cotorsion pairs (Crat , ≤) via the mapping Φ

τ = t(T ) −→

⊥

 (T ⊥ ), T ⊥ .

With Theorem 11.2.12 we have fully described the lattice of all rational cotorsion pairs.

11.3

Embedding posets into the lattice of cotorsion pairs

We now turn our attention to the more general case of all cotorsion pairs. In this case we cannot find any obvious ‘candidate’ with which the lattice could be compared; note that there is a proper class of cotorsion pairs. However, knowing about the properties of the lattice of types or, equivalently, of the lattice of rational cotorsion pairs as mentioned above, it seems natural to ask if there exist ascending, descending and anti–chains of cotorsion pairs of arbitrary size. We shall see that the answer is affirmative in any of the cases. Note, the existence of descending chains of arbitrary length also follows from Göbel, Trlifaj (see [244]). In fact, here we show that any power set (P, ⊆) can be embedded into the lattice of all cotorsion pairs. Therefore any partial order can be embedded into the

11.3 Embedding posets into the lattice of cotorsion pairs

393

lattice of all cotorsion pairs. In fact, any poset can be embedded into the lattice of all singly generated cotorsion pairs, that is, cotorsion pairs generated by a single group. In order to prove the main theorem we will use the group (module) constructed in Section 11.1. We need to adjust the setting to our situation of abelian groups: we consider R = Z, S = N and κ = ℵ0 ; the choice of the cardinal λ follows later. Recall that the results of the first section imply the existence of an ℵ1 –free, ultra– cotorsion–free group G = G(E) of cardinality λ satisfying further properties (see Lemma 11.1.5); note, the notions of local freeness and ℵ1 –freeness coincide for abelian groups by Pontryagin’s criterion. Additionally, we shall here define groups GY and construct groups H X for subsets X, Y of an arbitrary, but fixed set I such that Ext(GY , H X ) = 0 ⇐⇒ Y ⊆ X. In this way we obtain an order–preserving and injective morphism from (P(I), ⊆) into the lattice of all cotorsion pairs by mapping the set Y onto the cotorsion pair generated by GY . It is known how to construct a group H such that Ext(G, H) = 0 for a given group G or even for a collection of groups. This method was introduced in [229] (using factor systems) and was further developed (replacing factor systems by homomorphisms in the spirit of Hom–Ext–sequences) in [149, 150] (see Theorem 3.2.1 and its proof) and [52]. We shall follow the arguments presented in [250], which use the vanishing of Ext from [149] to construct the groups H X (X ⊆ I). For this construction it is not important what the groups GY look like. However, the GY s need to satisfy certain properties to guarantee that Ext is non–zero in some cases. We find this amazing, as one would expect that it is obvious how to get non–vanishing Exts after all the hard work which had been done over decades in order to establish a method for making Ext vanish. But we also have some work to do to obtain non–zero Exts. The key to proving Ext(GY , H X ) = 0 for Y ⊆ X is the existence of a stationary set E such that H X is locally E–free, and GY is not; the definition of locally E–free is given in 11.3.4. We therefore have to choose the groups GY in such a way that they are locally E–free with respect to some stationary set E and not locally E  –free with respect to others. In fact, we have already constructed groups G = G(E) depending on a stationary set E and we shall define the groups GY depending on different stationary sets. To see the connection with a given stationary set E let us finally recall that the group G = G(E) has a λ–filtration  Gα G= α<λ

394

11 The lattice of cotorsion pairs

such that G/Gα is ℵ1 –free, if and only if α ∈ E (see Lemma 11.1.5). Using this fact we can show that G depending on E is locally E  –free for any stationary set E  disjoint from E but not locally E–free. Throughout this section let I be an arbitrary set, and let P = P(I) be the power set of I. Moreover, let λ ≥ |I| be a regular cardinal such that λ = μ+ for some cardinal μ with μℵ0 = μ. Note that such a cardinal always exists, e.g. take μ = |I|ℵ0 . The aim is to prove the Main Theorem 11.3.1. There is an embedding from (P, ⊆) into the lattice of all cotorsion pairs (C, ≤). We shall prove the main theorem in several steps. First we define an order– preserving mapping Φ : (P, ⊆) −→ (C, ≤) which will turn out to be injective. As before, we let λo = {α ∈ λ | cf(α) = ω}, which is stationary in λ and can bepartitioned into |I| disjoint stationary subsets (see Theorem 9.1.3); say λo = Ei . Let Gi = G(Ei ) be the ℵ1 –free group of cardinality λ as constructed i∈I

in 11.1.3, depending on the stationary set Ei (i ∈ I). Moreover, for each Y ⊆ I, we put

Gi . GY = i∈Y

We define, for Y ⊆ I,

⊥ Y Φ = (⊥ (G⊥ Y ), GY ) ∈ C.

Obviously, Φ is well–defined and, for Y  ⊆ Y ⊆ I, we have GY  ⊆ GY , and thus ⊥ G⊥ Y  ⊇ GY , i.e. Φ is order preserving. Note that G∅ = 0, and so ∅Φ = (F0 ,Mod– Z) is the minimal cotorsion pair; recall that F0 denotes the class of all free abelian groups. Also note that the image Y Φ is singly generated. In order to establish that Φ is injective we construct groups H X (∅ = X ⊆ I) such that Ext(GY , H X ) = 0, if and only if Y ⊆ X, i.e. if Y ⊆ X, then ⊥ H Y ∈ G⊥ Y \ GX . Construction 11.3.2. Let X be a fixed non–empty subset of I, and let ρ be a cardinal with ρλ = ρ. Moreover, let H be a set of cardinality ρ with a ρ–filtration  H= Hα such that |H0 | = λ and |Hα | = |α| · λ = |Hα+1 \ Hα | α<ρ

11.3 Embedding posets into the lattice of cotorsion pairs

395

for all α < ρ. We inductively define a group structure on H and call the obtained group H X . We fix free resolutions 0 → Ki → Fi → Gi → 0 of Gi with |Ki | = |Fi | = λ (i ∈ X), and we ‘enumerate’ all set mappings from all Ki s into H by  Ki H = {ϕα | α < ρ} i∈X

in such a way that each mapping appears ρ times. First let H0X = Z(λ) be a free group of rank λ. If α is a limit ordinal, and if the group structure HβX on Hβ is defined for all β < α such that HβX is a subgroup  X X , then let H X = Hβ have the induced group structure. of Hβ+1 α β<α

Now let the group structure HαX be given. If Imϕα ⊆ HαX , and if ϕα is a homomorphism, then let ϕ 6α = ϕα , and put ϕ 6α = 0 otherwise. In either case we X define Hα+1 to be the pushout ϕ  α

Ki −→ HαX ↓ ↓ X , Fi −→ Hα+1 ψα

6α and where 5Dom ϕα = Ki for some i ∈ X. Hence ψα is an extension of ϕ X X ∼ ∼ Hα+1 Hα = Fi /Ki = Gi . Finally let the structure on  HX = HαX α<ρ

be the induced one. Note that the cardinality of H X is obviously ρ for each non–empty set X ⊆ I. First we show that H X ∈ G⊥ Y for any set Y ⊆ X. Proposition 11.3.3. Let ∅ = X ⊆ I and let H X be as in Construction 11.3.2. X Then H X ∈ G⊥ Y , i.e. Ext(GY , H ) = 0, for any Y ⊆ X.  Proof. Since GY = Gi (Y ⊆ X), it is sufficient to show that i∈Y

Ext(Gi , H X ) = 0 for each i ∈ X. So let i ∈ X be fixed and consider the free resolution 0 → K i → Fi → Gi → 0 X of Gi as in Construction 11.3.2. Also let ϕ : Ki → H = HαX be a homoα<ρ

morphism. Since |Ki ϕ| ≤ |Ki | = λ < cf(ρ), there is an ordinal β < ρ such that

396

11 The lattice of cotorsion pairs

Imϕ ⊆ HβX . Moreover, by the enumeration of



Ki H

in Construction 11.3.2,

i∈X

6α , and thus there is an extension there is β ≤ α < ρ such that ϕ = ϕα = ϕ X ψα : Fi → H of ϕ. Therefore we have seen that every homomorphism from Ki into H X extends to a homomorphism from Fi into H X , and hence Ext(Gi , H X ) = 0 whenever 2 i ∈ X. This implies Ext(GY , H X ) = 0 for all Y ⊆ X, as required. It remains to show that Ext(Gi , H X ) = 0 whenever i ∈ / X. Although it seems to be the more likely case that Ext(A, B) = 0 for arbitrary groups A and B, there is some work to be done in order to prove this. The key to the proof is the following: Definition 11.3.4. Let λ be as above and let E be a stationary set in λ. We call a group A locally E–free, if, for any smooth ascending chain {Kα | α < λ} of subgroups Kα of A with |Kα | < λ for all α < λ, the set {δ ∈ E | Kδ+1 /Kδ is not ℵ1 –free} is not stationary in λ.  Note, it is very likely that α<λ Kα is a proper subgroup of A. First we investigate the groups Gi = G(Ei ) (i ∈ I) with respect to the just defined property. Proposition 11.3.5. Let i = j be elements of I, and let Gi = G(Ei ) be as above. Then Gi is locally Ej –free but not locally Ei –free.

Proof.  i By Lemma 11.1.5 and Proposition 11.1.6, there is a λ–filtration Gi = Gα of Gi such that α<λ

5 Gi Giα is ℵ1 –free, if and only if α ∈ / Ei . (∗) 5 i i Moreover, we know that Gα+1 Gα contains a divisible subgroup for any α ∈ Ei . Hence it follows immediately that Gi is not locally Ei –free, since {δ ∈ Ei | Giδ+1 /Giδ is not ℵ1 –free} = Ei

is stationary in λ. Thus it remains to show that Gi is locally Ej –free for any j = i. Let {Kα | α < λ} be a smooth  ascending chain of subgroups of Gi , all of cardinality less than λ, and let K = Kα . α<λ

11.3 Embedding posets into the lattice of cotorsion pairs

397

If |K| < λ, then there is an α0 < λ such that K = Kα = Kα0 for all α > α0 . Thus the set {δ ∈ Ej | Kδ+1 /Kδ is not ℵ1 –free} is bounded by α0 , and hence it is not stationary. Otherwise |K| = λ and {Kα | α < λ} is a λ–filtration of K. Also {K ∩ Giα | α < λ} is a λ–filtration of K, and thus there exists a cub C in λ such that Kα = K ∩ Giα for all α ∈ C. Let δ ∈ C ∩ Ej . Then Kδ = K ∩ Giδ , since δ ∈ C and Gi /Giδ is ℵ1 –free by (∗ ), since δ ∈ Ej which is disjoint from Ei . Therefore Kδ+1 /Kδ ⊆ K/Kδ = K/(K ∩ Giδ ) ∼ = (K + Giδ )/Giδ ⊆ Gi /Giδ is ℵ1 –free. Hence C is disjoint from {δ ∈ Ej | Kδ+1 /Kδ is not ℵ1 –free}, and so this set is not stationary. Thus we have shown that Gi is locally Ej –free. 2 Combining the above proposition with our main result on ultra–cotorsion–free modules (Theorem 11.1.1) we have the following immediate consequence: Corollary 11.3.6. Let A be an S–cotorsion–free locally Ei –free group for some i ∈ I. Then Hom(Gi , A) = 0.

Proof. Suppose, for contradiction, that there exists a non–zero homomorphism  Giα −→ A, ϕ : Gi = α<λ

and let Kα = Giα ϕ. Since Gi /ker ϕ ∼ = Im ϕ ⊆ A is S–cotorsion–free by assumption, it follows from Theorem 11.1.1 that the kernel of ϕ has to be ‘small’, i.e. |ker ϕ| < λ. Therefore there is α0 < λ such that ker ϕ ⊆ Giα for all α ≥ α0 as λ is regular. This implies 5  5 i   Kα+1 /Kα ∼ Gα /ker ϕ ∼ = Giα+1 Giα for all α ≥ α0 , = Giα+1 /ker ϕ and thus, by Lemma 11.1.5, {δ ∈ Ei | Kδ+1 /Kδ is not ℵ1 –free } = {δ ∈ Ei | δ ≥ α0 } is stationary, contradicting the local Ei –freeness of A.

2

We now proceed with investigating the relevant properties of the groups H X (∅ = X ⊆ I) constructed in 11.3.2. Since Gi = G(Ei ) is ℵ1 –free for all i ∈ I by Theorem 11.1.1, we immediately have:

398

11 The lattice of cotorsion pairs

Lemma 11.3.7. Let ∅ = X ⊆ I and let H X =

 α<ρ

HαX be as in Construc-

tion 11.3.2. Then H X and also H X /HαX (α < ρ) are ℵ1 –free. Next we consider the local Ei –freeness of H X . Proposition 11.3.8. Let ∅ = X ⊆ 5 I, i ∈ I \ X and let H X be as in Construction 11.3.2. Then H X and also H X H0X are locally Ei –free. 5 Proof. We shall show inductively that HαX and HαX H0X are locally Ei –free for 5 all α < ρ. Of course, H0X = Z(λ) and H0X H0X = 0 are locally Ei –free. In the following we restrict our attention to the HαX s, as there is no difference in the  X5 arguments when considering the Hα H0X s. First assume that HαX is locally Ei –free and consider a smooth ascending chain {Kγ | γ < λ} X with |K | < λ. Let K = of subgroups of Hα+1 γ



Kγ and let

γ<λ

  M = δ ∈ Ei | Kδ+1 /Kδ is not ℵ1 –free . Moreover, let   M1 = δ ∈ Ei | (Kδ+1 ∩ HαX )/(Kδ ∩ HαX )} is not ℵ1 –free and

  M2 = δ ∈ Ei | (Kδ+1 + HαX )/(Kδ + HαX ) is not ℵ1 –free .

By induction hypothesis, 5 X by  M1 is not 5stationary. Also,  M2 is not stationary X Hα ∼ Proposition 11.3.5, since Kγ + HαX HαX | γ < λ is a chain in Hα+1 = Gj for some j ∈ X. Thus there are cubs C1 and C2 such that Ml ∩ Cl = ∅ (l = 1, 2). Let C = C1 ∩ C2 , then C is also a cub. We prove that M ∩ C = ∅, and therefore M is not stationary. / M1 ∪ M2 , and thus Let δ ∈ C ∩ Ei . Then δ ∈   5    5  Kδ+1 ∩ HαX Kδ ∩ HαX and Kδ+1 + HαX Kδ + HαX are ℵ1 –free. Moreover, there is an epimorphism Kδ+1 /Kδ −→ (Kδ+1 + HαX )/(Kδ + HαX ) with kernel ((Kδ+1 ∩ HαX ) + Kδ )/Kδ ∼ = (Kδ+1 ∩ HαX )/(Kδ ∩ HαX ),

11.3 Embedding posets into the lattice of cotorsion pairs

399

and hence Kδ+1 /Kδ is ℵ1 –free as an extension of an ℵ1 –free group by an ℵ1 – X free group. Therefore δ ∈ / M , and so M ∩ C = ∅, respectively, Hα+1 is locally Ei –free. Now let α be a limit ordinal, and suppose that HβX is locally Ei –free for all  Kγ ⊆ HαX with |Kγ | < λ for all γ < λ. Moreover, β < α. Consider K = γ<λ

let M = {δ ∈ Ei | Kδ+1 /Kδ is not ℵ1 –free}. If K ⊆ HβX for some β < α, then M is not stationary by assumption. So assume otherwise, i.e. K ⊆ HβX for all β < α. Then the cofinality of α is less than or equal to λ. First we consider the case of cf(α) < λ. Let α = supν
are cubs Cν (ν < cf(α)) such that Cν ∩ Mν = ∅ , where  5     Kδ ∩ Hν is not ℵ1 –free . Mν = δ ∈ Ei | Kδ+1 ∩ Hν Now let C = Cν . Then C is also a cub by Theorem 9.1.1 (b). We show that ν
M ∩ C = ∅. Let δ ∈ C ∩ Ei . Then   5     Kδ+1 ∩ Hν Kδ ∩ Hν ∼ = Kδ+1 ∩ Hν + Kδ /Kδ is ℵ1 –free for each ν < cf(α), and so     Kδ+1 ∩ Hν /Kδ = Kδ+1 /Kδ = ν
 

  Kδ+1 ∩ Hν + Kδ /Kδ

ν
is also ℵ1 –free. Hence δ ∈ / M and so M ∩ C = ∅, as required. It remains to consider the case cf(α) = λ. Since |Kγ | < λ for each γ < λ, we may choose αγ < α such that Kγ ⊆ HαXγ and αγ = supδ<γ αδ whenever γ is a limit. Then α = supγ<λ αγ , since K ⊆ HβX for any β < α. Now let Hγ = HαXγ and let  5     Kδ ∩ Hγ is not ℵ1 –free (γ < λ). Mγ = δ ∈ Ei | Kδ+1 ∩ Hγ By assumption, there are cubs Cγ with Mγ ∩ Cγ = ∅. We define C to be the Cγ }. By Theodiagonal intersection C = Δ{Cγ | γ < λ} = {δ < λ | δ ∈ γ<δ

rem 9.1.1 (c), C is also a cub. As before, we show that M ∩ C = ∅ in order to establish that M is not stationary.

400

11 The lattice of cotorsion pairs

Consider an ordinal δ ∈ C ∩ Ei . Then δ is a limit ordinal and δ ∈ Cγ for all γ < δ. So  5      Kδ ∩ Hγ ∼ Kδ+1 ∩ Hγ = Kδ+1 ∩ Hγ + Kδ /Kδ is ℵ1 –free (γ < δ). It follows immediately that       Kδ+1 ∩ Hδ /Kδ = Kδ+1 ∩ Hγ + Kδ /Kδ



γ<δ

is also ℵ1 –free. Moreover, for all γ ≥ δ, we have   5     /Kδ Kδ+1 ∩ Hγ /Kδ ∼ Kδ+1 ∩ Hγ+1 =  5    5     Kδ+1 ∩ Hγ ∼ + Hγ Hγ ⊆ H X /Hγ Kδ+1 ∩ Hγ+1 = Kδ+1 ∩ Hγ+1 is ℵ1 –free. Therefore it follows, by transfinite induction, that     Kδ+1 ∩ Hγ /Kδ Kδ+1 /Kδ = δ≤γ<λ

is ℵ1 –free. Thus δ cannot be an element of M , which completes the proof.

2

Using the above result, we can finally show the last missing bit in order to establish the correctness of the main theorem. Proposition 11.3.9. Let ∅ = X ⊆ I, i ∈ I \ X, and let H X be as in Construc/ G⊥ tion 11.3.2. Then Ext(Gi , H X ) = 0, i.e. H X ∈ i .

Proof. Let 0 −→ Ki −→ Fi −→ Gi −→ 0 be a free resolution of Gi with |Ki | = |Fi | = λ. In order to show Ext(Gi , H X ) = 0 it is enough to find a homomorphism ϕ : Ki → H X which does not extend to a homomorphism ϕ  : Fi → H X . Let ϕ : Ki → H0X ⊆ H X be an isomorphism between the two free groups Ki , H0X of rank λ. Suppose, for contradiction, that there is ϕ  : Fi → H X5 with ∼  induces a homomorphism ϕ : Fi /Ki = Gi → H X H0X . ϕ   Ki =5ϕ. Then ϕ X X But H 5H0 is ℵ1 –free by Lemma 11.3.7; in particular, it is S–cotorsion–free. Also H X H0X is locally5Ei –free by Proposition 11.3.8. Hence Hom(Gi , H X H0X ) = 0 by Corollary 11.3.6. Therefore ϕ = 0, and so Fi ϕ  = H0 . But this implies Fi = Ki ⊕ker ϕ,  since for each f ∈ Fi there is k ∈ Ki with f ϕ  = kϕ  = kϕ and ker ϕ  ∩ Ki = ker ϕ = {0}. Then Gi ∼  = Fi /Ki ∼ = ker ϕ

11.3 Embedding posets into the lattice of cotorsion pairs

401

 of is free, contradicting that Gi is not free. Therefore there is no such extension ϕ ϕ, and thus Ext(Gi , H X ) = 0, as required. 2 Finally, the main result of this section is established:

Proof of Theorem 11.3.1 In the above results 11.3.2 – 11.3.9 we have shown that the mapping Φ : (P, ⊆) −→ (C, ≤) as defined at the beginning of this section is an order–preserving injection. Therefore we have proved the main theorem. 2 Theorem 11.3.1 serves as a striking example (see Example 2.2.2) to illustrate the complexity of the class of cotorsion pairs in the category of abelian groups. It is fairly obvious, and now no longer surprising, that a similar result will hold for many R–module categories. It is enough to assume that R is an S–ring to establish the existence of ultra–cotorsion–free modules (see Theorem 11.1.1) and to ensure that the above ‘freeness argument’ carries over from Z to the ring R. (This, however, is left to the reader.)

Chapter 12

Realizing algebras – by algebraically independent elements and by prediction principles

While the combinatorial principles Strong Black Box 9.2.2, the General Black Box 9.2.27 and Shelah’s Elevator 9.3.3, discussed in Chapter 9, can be applied to objects of arbitrarily large cardinality, providing objects of cardinality at least 2ℵ0 , it is also often desirable to produce similar objects of cardinality < 2ℵ0 , for instance countable ones. In this case we need ‘classical – more algebraic – arguments’ due to Corner [92]; a very particular case is older than that and comes from Baer [32]: it shows that any S–pure subgroup of the p–adic integers is indecomposable. Thus the existence of indecomposable torsion–free abelian groups of any rank ≤ 2ℵ0 follows at once. Here we will use an extension of Corner’s results given in Göbel, May [216].

12.1

Realizing algebras of size ≤ 2ℵ0

In this section we want to utilize the existence of sets of algebraically independent elements from Theorem 1.1.20. Again, let R be an S–ring (see Definition 1.1.1). Here we additionally assume that R is of cardinality |R| < 2ℵ0 . Besides realizing algebras of size ≤ 2ℵ0 , we also want to investigate modules whose non–trivial summands are proper infinite direct sums. Note that, although we realize the required algebra in this section, the property just mentioned will be shown later, namely in Chapter 15 (see Example 15.3.1). However, these special summands can only be recognized in the endomorphism ring with respect to some topology; a good candidate is the finite topology (End M, fin) introduced in Proposition 9.3.2; another one would be the ℵ1 –topology from [171]. However, we will restrict our attention to (End M, fin) which is a complete Hausdorff topological ring. This means that we want to realize complete algebras topologically as (End M, fin). The discrete case (when Ni below is 0) is the usual realization theorem. The next result extends [92, 93] (see also [216]). The notion of S–transcendence degree has been introduced in Definition 1.1.17; we also refer to the subsequent remarks.

12.1

Realizing algebras of size ≤ 2ℵ0

403

Theorem 12.1.1. Let R be an S–ring and let μ, λ be cardinals with ℵ0 ≤ μ < λ.  of R has S–transcendence degree at Also assume that the S–adic completion R least λ over R. Moreover, let (A, top) be an R–algebra which is a topological ring, where top denotes all open subsets of A. Then the following are equivalent: (a) (A, top) is Hausdorff, complete and admits a family of top–open right ideals {Ni ∈ top | i ∈ I} as neighbourhood base of 0, where |I| ≤ μ, |A/Ni | ≤ μ, and A/Ni (i ∈ I) is S–reduced and S–torsion–free. (b) (A, top) ∼ = (End G, fin) for some S–reduced and S–torsion–free R–module G of cardinality μ, where the isomorphism is algebraic and topological. (c) There exists a family {GX | X ⊆ λ} of S–reduced and S–torsion–free R– modules such that |X| ≤ |GX | ≤ μ + |X|, GX ⊆ GX  for X ⊆ X  ⊆ λ, and, for any X, X  ⊆ λ,   ∼ (A, top), if X ⊆ X (HomR (GX , GX  ), fin) = 0, if X ⊆ X  , where the isomorphisms are algebraic and topological. Note, we call such a system {GX | X ⊆ λ} as in (c) an A–rigid system of size λ or an A–rigid λ–family .

Proof. Clearly (c) implies (b). Also, it is easy to verify that (b) implies the first part of (a) (see Proposition 9.3.2). The last assertion of (a) is also immediate: for let I be the family of non–empty finite subsets of G and, for each i ∈ I, let Ni = AnnEnd(G) (i). By a standard argument, A/Ni can be embedded in the direct sum of |i| copies of G (see again Proposition 9.3.2). Thus it remains to prove one more implication; we assume (a) and show that (c) follows. If top is the discrete topology on A, then I could be a singleton. However, w.l.o.g. we assume I = {0, 1} and N0 = N1 = 0 in this case; this will be used in the last step of the  proof. Note, if top is not discrete then I is infinite. Now we put C = i∈I A/Ni . Then C is a right A–module which is S–reduced, S–torsion–free and |C| ≤ μ. Hence, by assumption, we may choose λ elements of  algebraically independent over C, and index them by the disjoint sets λ and C, R, that is, we denote these elements by wx , wc (x ∈ λ, c ∈ C).  is an R–module,  Since C we may define, for any X ⊆ λ,  ∩ S−1 HX with HX = C + cAwc + Cwx . GX = C c∈C

x∈X

 with |X| ≤ Therefore each such GX is an S–pure, S–dense A–submodule of C |GX | ≤ μ + |X|; note that |cA| ≤ μ, since cA ⊆ C. Moreover, if X ⊆ X  ⊆ λ

404

12

Realizing algebras

then GX ⊆ GX  . It also follows that GX a ⊆ GX ⊆ GX  for any a ∈ A, and, if ei = 1 + Ni (generating A/Ni ⊆ C) and ei a = 0 for all i ∈ I, then a ∈ i∈I Ni = 0 must be zero. Hence A is naturally embedded in HomR (GX , GX  ) as C is a faithful A–module. Equating coefficients of algebraically independent elements, we show first: If c ∈ C and cA is an S–pure submodule of C, then HX ∩ HX wc = cAwc .

(12.1.1)

We only have to show that HX ∩ HX wc ⊆ cAwc . So let x ∈ HX ∩ HX wc . Then we can write dwd ad + cx wx = c wc + dwd ad wc + cx wx wc sx = c + x∈X

d∈C

x∈X

d∈C

for suitable elements s ∈ S, c, c , c , cx , cx ∈ C and ad , ad ∈ A. Thus 0 = c +(cac −c )wc +



dad wd −

c =d∈C



dad (wd wc )+

d∈C



cx wx −

x∈X



cx (wx wc ).

x∈X

Equating coefficients of the algebraically independent elements we deduce cac = c and all other coefficients c , ad , cx , cx , cd , cd must be 0. Therefore sx = cac wc ∈  and so x ∈ cAwc because cAwc is S–pure in C.  Hence (12.1.1) follows. cAwc ∩ C Now let X, X  ⊆ λ and let ϕ : GX → GX  be an R–homomorphism. It will require some work to show that ϕ is the scalar multiplication by an element in A.  which we also We may extend ϕ, by continuity, to an R–endomorphism of C, call ϕ. Next we claim: If c ∈ C and cA is an S–pure submodule of C, then cAϕ ⊆ cA.

(12.1.2)

Given such an element c and any a ∈ A, choose s ∈ S such that s(ca)ϕ and s(cwc a)ϕ both lie in HX  . Then s(cwc a)ϕ = s(ca)ϕwc ∈ HX  ∩ HX  wc = cAwc , by (12.1.1), and so, by S–purity,  ∩ cA = scA, s(ca)ϕ ∈ sC i.e. (ca)ϕ ∈ cA by the S–torsion–freeness. In particular, we now have that, for each i ∈ I, there exists ai ∈ A such that ei ϕ = ei ai (recall that ei = 1 + Ni ). If i, j ∈ I are distinct and Nj ⊆ Ni , then

12.1

Realizing algebras of size ≤ 2ℵ0

405

(ei + ej )A is a direct summand of C. Thus, by (12.1.2), there exists a ∈ A such that ei ai + ej aj = (ei + ej )ϕ = (ei + ej )a = ei a + ej a and hence ei (ai − a ) + ej (aj − a ) = 0. Equating components in ei A and ej A leads at once to ai − aj ∈ Ni and aj − a ∈ Nj ⊆ Ni , whence ai − aj ∈ Ni . Thus the ai s form a Cauchy net, and, by completeness, we may suppose that they are all equal to some a ∈ A, i.e. ei ϕ = ei a for all i ∈ I. Finally, given i ∈ I and b ∈ A, by the continuity of the multiplication in A, we may choose j ∈ I \ {i} such that bNj ⊆ Ni . (In the discrete case we assumed I = 2, N0 = N1 and therefore the last assertion holds trivially.) Since (ei b + ej )A is a direct summand there exists a ∈ A with (ei b+ej )ϕ = (ei b+ej )a by (12.1.2). It follows that (ei b)ϕ + ej a = (ei b + ej )ϕ = (ei b + ej )a and thus (ei bϕ − ei ba ) + ej (a − a ) = 0. Equating coefficients once more gives a − a ∈ Nj , whence ba − ba ∈ bNj ⊆ Ni and so (ei b)ϕ = ei ba = ei ba. We conclude ϕ  C = a · idC and ϕ = a · idC by density. For X  X  we choose x ∈ X \ X  . Then Cwx ∩ HX  = 0 which is shown similar to (12.1.1). However, cwx ϕ = cawx ∈ Cwx ∩ HX  for every c ∈ C, and hence ca = 0. Consequently, a = 0 and thus ϕ = 0. The only remaining detail to prove is that, when X ⊆ X  , the identification of A with HomR (GX , GX  ) is topological. Since Ni = AnnA (ei ), continuity is immediate in one direction. For the other direction, it suffices to consider the homomorphisms that annihilate x + Ni for some x ∈ A, i ∈ I. By continuity, there exists j ∈ I such that xNj ⊆ Ni . Then Nj annihilates x + Ni , showing continuity in the other direction. 2 From Theorem 1.1.20 and Theorem 12.1.1 we obtain the following Corollary 12.1.2. Let |R| < 2ℵ0 and S countable. Every S–reduced, S–torsion– free R–algebra of cardinality < 2ℵ0 is the endomorphism algebra of an S–reduced, S–torsion–free R–module of the same cardinality. In the last proof we witnessed the miracle of equating coefficients of algebraically independent elements. The same miracle in Black Box proofs follows by strong disjointness of supports of elements (see, for example, the Strong Black Box 9.2.9). We also notice, that it is enough to know that the elements wx , . . . in

406

12

Realizing algebras

the proof above are quadratically independent, since only polynomials of degree at most 2 are involved. This is important because, if |R| is not strictly less than 2ℵ0 , then R might be close to its completion, and then it may not be possible to find enough algebraically independent elements while the existence of quadratically independent elements can still be established (see Goldsmith and Zanardo [252]). The same arguments as in the proof of Theorem 12.1.1 can be applied to R–algebras A of finite rank rk A = n. We say that A has finite rank n, if n is minimal such that there are a1 = 1, a2 , . . . , an generating an R–submodule U of A such that A/U is S–torsion. This notion is needed to work with R–modules which are only ‘S–torsion–free’; it agrees with the usual notion of finite rank for modules over domains. Let wi (i ≤ n) be algebraically independent elements as n  wi ai . The faithful (right) A–module G is defined by above and put w = i=1

 A ⊆ G = A, wA∗ ⊆ A and satisfies the following Corollary 12.1.3. Let R be an S–ring as above and let A be an S–reduced, S–  has transcendence degree at torsion–free R–algebra of finite rank rk A = n. If R  least n over A, then there is an A–module G ⊆ A of rank 2n such that End G ∼ = A. Remark: it is easy to replace G by any R–module of rank up to the transcen over A. The corollary extends [92] or [173, Vol. 2, p. 234] dence degree of R slightly. Proof. Clearly A ⊆ End G. Therefore let σ ∈ End G with 1σ = c and recall n n   wi ai . Hence wσ = wi (ai σ) and there are s ∈ S, bi , ci ∈ A such that w= i=1

i=1

s(wσ) = w(σs) = b0 + wc0 , ai σs = bi + wci

(i = 1, . . . , n).

We apply σs on the right to (the definition of) w. Substitution leads to an equation which we treat as before (using the miracle): n n n       wi ai c0 = wi bi + wj aj ci b0 + i=1

=

i=1

j=1

n

n n

i=1

wi bi +

wi wj aj ci .

i=1 j=1

Thus we again deduce b0 = 0, ai c0 = bi , aj ci + ai cj = 0 for i = j ≤ n, and ai ci = 0 for all i ≤ n. If i = 1, then c1 = 0 from the last equality, and so cj = 0 for all 1 < j ≤ n from the other equality for i = 1. Therefore the displayed

12.1

Realizing algebras of size ≤ 2ℵ0

407

equation reduces to wσs = wc0 , ai σs = bi = ai c0 . From a1 σ = 1σ = c = c0 now follows ai σs = ai c0 = ai cs  and so ai σ = ai c, since s ∈ S. Hence σ is scalar multiplication by c on U = i≤n ai R. However, G/A is S–divisible and A/U is torsion and thus Hom(G/U, G) = 0. The endomorphism σ − c induces a homomorphism G/U → G which must be trivial. Hence we derive σ = c ∈ A, as required. 2 The restrictions on the rank cannot be improved in general, as can be seen in case R = Z: for every integer n ≥ 2, there exists a ring A of rank n that is not isomorphic to the endomorphism ring of any torsion–free abelian group of rank < 2n (see Corner [93, p. 702, Proposition 3.1]). The problem of finding particular rings, where 2n can be replaced by n was raised by Hirsch and Zassenhaus [271] and answered in two papers, by Zassenhaus [398] for rings with free additive group of finite rank and, more generally, by Butler [77] for rings with locally free additive group of finite rank. In the latter, a torsion–free abelian group G is locally free, if Z(p) ⊗ G is a free Z(p) –module for each prime p (here Z(p) is the localization of Z at the prime p). (Warning: the same name also appears for modules with a different meaning; see Definition 1.1.12. The module–theoretic notion ‘locally free’ restricted to abelian groups is the same as ℵ1 –free.) Theorem 12.1.4. (Butler [77]) If A is a ring with locally free additive group A+ of finite rank n, then there is a torsion–free abelian group M of rank n such that A = End M . In a recent paper, Buckner and Dugas [75] obtained an easy proof of this theorem, if the additive group A+ is free (which is Zassenhaus’ Theorem) using the crucial ideas from Butler [77]. We include their arguments for this particular important case (and use another simplification due to the Essen undergraduate student Jasmin Matz). Proposition 12.1.5. For each non–constant polynomial f (x) with integer coefficients, there exist infinitely many primes p such that the congruence f (x) ≡ 0 mod p is solvable.

Proof. Suppose the congruence f (x) ≡ 0 mod pi is solvable for the distinct primes pi , i ≤ n; then there is an integer a such that f (a) = 0 and f (a) ≡ 0 mod p1 . . . pn . Choose an integer b ≡ a mod f (a)2 such that |f (b)| > |f (a)|. Then f (b) = f (a)(1 + cf (a)), where c is an integer and |1 + cf (a)| > 1 and coprime to the product p1 · · · pn . Hence the congruence is solvable for prime divisors 2 of 1 + cf (a), as well as for p1 , . . . , pn . We can proceed. Recall that endomorphisms act on the right:

408

12

Realizing algebras

Theorem 12.1.6. (Zassenhaus [398]) Let 1A ∈ A be a ring such that A+ is free abelian of finite rank. Then there exists a right A–module M (of the same rank) such that A ⊆ M ⊆ Q ⊗ A and End M = A. Moreover, the p–torsion part tp (M/A) of M/A is bounded for all primes p.

Proof. The setting for the construction of M : if X, Y are two torsion–free abelian groups, then we write QX := Q ⊗ X and identify Hom(X, Y ) with its canonical image in Hom(QX, QY ), so that a homomorphism ϕ : X −→ Y becomes the unique extension, a Q–linear map (also called) ϕ : QX −→ QY such that Xϕ ⊆ Y . If A is the given ring and a ∈ QA, then (as before) we write ar ∈ End QA+ for the scalar multiplication by a on the right on the vector space QA+ (al ∈ End QA+ for the scalar multiplication by a on the left) and emphasize this by writing (QA)r := {ar | a ∈ QA} ⊆ End QA+ . If End0 QA+ := {σ ∈ End QA+ | 1A σ = 0} denotes the annihilator ideal of End QA+ , then there is a ring split extension of Q–algebras End QA+ = End0 QA+ ⊕ (QA)r and we want to find a right A–submodule A+ ⊆ M ⊆ QA such that End M viewed as endomorphisms of QA equals Ar . This is to say that Ar = {σ ∈ End QA+ | M σ ⊆ M }. Essentially we have to show that the desired M forces End0 QA+ ∩ End M = 0. First we show the following claim from linear algebra. Claim 1: Let F be a free abelian group of finite rank, 0 = e ∈ F and ϕ ∈ End F . Let W = eZ[ϕ] be the ϕ–invariant subgroup of F generated by e and W∗ the purification of W in F . Then there exists (a least) k ∈ N such that kW∗ ⊆ W . Moreover, let c ∈ Z not be an eigenvalue of ϕ and n ∈ N. Then ne ∈ F (c − ϕ) implies that the determinant  (c − ϕ  W )  divides kn. Let χϕ  W (x) =  x − ϕ  W  be the characteristic polynomial of ϕ  W and mϕ  W (x) be the minimal of ϕ  W . Both are monic and in Z[x]. Let  polynomial m−1 i and, since W is ϕ–invariant, W = i a x mϕ  W (x) = xm + m−1 i i=0 i=0 Zeϕ follows. Hence χϕ  W (x) = mϕ  W (x) =: f (x). Since c is not an eigenvalue of ϕ, 0 is not an eigenvalue of c − ϕ and therefore the map c − ϕ must be injective and also bijective. Then the constant term q of the minimal polynomial of c − ϕ is not zero. Thus q(c − ϕ)−1 ∈ Z[c − ϕ], but any polynomial in c − ϕ is also a polynomial in ϕ, so that q(c − ϕ)−1 ∈ Z[ϕ] and eq(c − ϕ)−1 ∈ W . By hypothesis of the claim we have ne ∈ F (c − ϕ), hence neq(c − ϕ)−1 ∈ qF and neq(c − ϕ)−1 ∈ qF ∩ W ⊆ qF ∩ W∗ =  qW∗ by purity. This shows that i ne ∈ W∗ (c − ϕ) and thus kne ∈ W (c − ϕ) = ( m−1 i=0 Zeϕ )(c − ϕ).

12.1

Realizing algebras of size ≤ 2ℵ0

409

  m−1 zi eϕi (c−ϕ) and consider the polynomial h(x) = Suppose that kne = i=0 m−1 i i=0 zi x (c − x) − kn. Hence h(ϕ  W ) = 0 and f divides h. But f is monic and f , h have the same degree, so −zm−1 f (x) = h(x). It follows that −zm−1 f (c) = h(c) = −kn and f (c) =  (c − ϕ  W )  divides kn. Claim 2: Let 1A ∈ A be a torsion–free ring and F = {Pi = bi A | i < ω} be a countable family of principal right ideals such that bi is not a zero–divisor in A. Moreover, suppose that there are distinct primes pi , numbers di ∈ Z and natural numbers ri (i <ω) such that pri i di A ⊆ Pi , where pi and di are coprime. i Let M := A + i<ω p−r i Pi ⊆ QA. If y ∈ M and yr ∈ End M , then also y ∈ A. We now exploit that torsion–pi –components of abelian groups are fully invarii ant. Note that tpi (M/A) = (p−r i Pi +A)/A and tp (M/A) = 0 for all primes  p disv tinct from pi (i < ω). Since A/M is torsion, y = k with v ∈ A and k = i∈Iy psi i  for some finite index set Iy ⊆ ω. We will show by induction on i∈Iy si that y ∈ A. First assume that k = pi for some i < ω. Then 

i p−r i Pi

 v −ri i + A /A = (p−r Pi + A)/A. i Pi yr + A)/A ⊆ tpi (M/A) = (p pi −(ri +1)

This implies that pi

i Pi v ⊆ p−r i Pi + A and therefore

ri p−1 i di Pi v ⊆ di Pi + pi di A ⊆ Pi

by assumption. This shows that di Pi v ⊆ pi Pi and Pi v ⊆ pi Pi because di and pi are coprime. We infer that there is some a ∈ A such that bi v = pi bi a. Since bi is not a zero–divisor in A, we also get that v = pi a and thus y ∈ A. For the induction step consider the element yˆ = p−1 i ky for some i ∈ Iy . We have already shown that yˆ ∈ A since pi yˆ ∈ A. Hence the induction hypothesis applies and y ∈ A. The second claim is shown. The two claims will establish the theorem. We require, as mentioned above, that elements from the ideal End0 QA+ do not contribute to the endomorphism ring of the abelian group M we want to construct. Thus consider the family of all 0 = σ ∈ End A+ with 1A σ = 0, which we enumerate as {σi | i < ω}. (Recall that A has finite rank, and we may assume its rank is at least 2; thus this family has size ℵ0 .) We intend to show that this set is empty when restricting its members to endomorphisms of the submodule M under construction. By induction on i suppose that aj ∈ A, pj prime, cj ∈ Z and rj ∈ N for j < i are already chosen. For i < ω, there is some ai ∈ A such that ei := −ai σi = 0. If Wi = ei Z[(ai )r ], then ei ∈ Wi and there is ki ∈ N such that ki (Wi )∗ ⊆ Wi .

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Realizing algebras

By Claim 1 for ϕ = (ai )l and ei ∈ A for any ci ∈ Z not an eigenvalue of (ai )l and nei ∈ A(ci − (ai )l ) follows that  (ci − (ai )l  W )  = fi (ci ) divides nki . Here fi (x) is the minimal polynomial  x − (ai )l  Wi  of (ai )l  Wi . Since ci − (ai )l is a unit in the ring End(QA+ ) we can calculate the inverse of ci − (ai )l in terms of linear algebra. We get  ci − (ai )l  · 1A = (ci − ai ) · a, where a is the image of 1A under the adjugate of ci − (ai )l , hence an element from A. By Proposition 12.1.5 there are infinitely many primes q such that fi (x) mod q has a root in Z/qZ. Now pick such a prime q ∈ / {pj | 0 ≤ j ≤ i − 1} which does not divide ki . The set {c ∈ Z | fi (c) ≡ 0 mod q} is infinite, since any integer of the form ct = c + tq ∈ Z is a root of fi (x) mod q. Hence we may choose ci such that ci is not one (of the finitely many) eigenvalues of (ai )l and fi (ci ) ≡ 0 mod q. Then  ci − (ai )l  = q ri di with ri ≥ 1 and q does not divide di . By the above there is some a ∈ A such that (ci − ai )a = q ri di · 1A . It follows q ri di A ⊆ (ci − ai )A. If Pi := (ci − ai )A, then di (ci 1A − ai )σi = di ci (1σi ) − di ai σi = −di ai σi = di ei ∈ / Pi

(12.1.3)

by Claim 1, since q divides fi (ci ) =  (ci − ϕ  W )  but q does not divide di ki . Let pi := q and consider the right A–module −r pi i Pi ⊆ QA. M := A + i<ω

So clearly A ⊆ M ⊆ QA. Finally let ϕ ∈ End QA be acting on the right of QA+ such that M ϕ ⊆ M . Since A+ is finitely generated and M/A is torsion there is some m ∈ N such that Aϕm ⊆ A and 1ϕm ∈ A. Thus (1ϕm)r ∈ End M (because M is a right A–module) and σ := ϕm − (1ϕm)r ∈ End A+ . Note that / Pi 1σ = 0 and assume that σ = 0. Then σ = σi for some i < ω and di (ci −ai )σi ∈ by (12.1.3). Moreover, σi induces an endomorphism of M/A and it follows that −ri i (p−r i Pi )σi ⊆ pi Pi + A. From (ci − ai ) ∈ Pi we have that di (ci − ai )σi ∈ di Pi σi ⊆ di Pi + pri i di A ⊆ Pi + Pi = Pi . This contradiction shows that σ = 0 and therefore ϕ = (1ϕ)r and 1ϕ ∈ M . From Claim 2 follows 1ϕ ∈ A. Thus the theorem is shown. 2

12.2

ℵ1 –free modules of cardinality ℵ1

We begin this section with inviting the reader to go back to Proposition 9.3.2: in this context we would like to point out that in the last assertion of the proposition,

12.2 ℵ1 –free modules of cardinality ℵ1

411

if the submodule M is ℵ1 –free (locally free), then End M/ Ann m must be ℵ1 – free (locally free). In particular, if M is ℵ1 –free (or locally free and Pontryagin’s theorem holds) and if this quotient is countable, then End M/ Ann m must be free. This explains, why we must pose the following restrictions on a given algebra A to make it the endomorphism algebra of an ℵ1 –free (locally free) module. Throughout this section let A be an R–algebra (with 1) which is complete and Hausdorff in a topology admitting a basis of neighbourhoods of 0 consisting of a set N of right ideals N such that the quotients A/N are free as R–modules. We impose the cardinality restrictions |N| ≤ λ and κ =



+

sup |A/N |

N ∈N

≤ λ ≤ 2ℵ0 ,

(12.2.1)

where the plus sign denotes the cardinal successor κ ≥ ℵ1 which is regular. We are mainly interested in locally free R–modules of size ℵ1 . In [226] it was shown that any countable R–algebra with free R–module structure is the endomorphism algebra of a suitable locally free R–module of size ℵ1 . Here we want to strengthen this result and realize any suitable topological ring as such an endomorphism ring — algebraically and topologically; we will follow [105]. We consider the endomorphism ring End M of an R–module M together with the finite topology fin, as discussed in Proposition 9.3.2 and in the beginning of the previous Section 12.1. Thus our main result reads as follows. Theorem 12.2.1. Let R be an S–ring and let A be a topological R–algebra as described above. Then there exists a locally free R–module M of cardinality λ with End M ∼ = A, where the isomorphism is an algebraic and topological isomorphism. Theorem 12.2.1 was shown in Corner, Göbel [105] and the case, when A is discrete, follows from Göbel, Shelah [229]. Also other earlier results, like the existence of an ℵ1 –free group M of cardinality ℵ1 with Hom(M, Z) = 0, follow (see Eda [138] and Shelah [361], [140],[142]). Here the main difficulty is to work exclusively in ZFC. The first example of an ℵ1 –free module which is not free is the Baer–Specker module Rω , the direct product of countably many copies of the ring R, known for many years (cf. Baer [32] or [173, p.94]). Assuming CH, this module is an example of an R–module of cardinality ℵ1 = 2ℵ0 . However, it is surely (by the slenderness of R) a finite but not an infinite direct sum of summands = 0. Before we begin with constructing suitable modules M we would like to note, that it is fairly straight to replace the module M in the above theorem by a family

412

12

Realizing algebras

of modules MX with X ⊆ λ such that, for all X, Y ⊆ λ,  A, if X ⊆ Y ⊆ λ ∼ HomR (MX , MY ) = 0 otherwise.

The construction of modules In this subsection we first introduce all the needed notions and definitions for constructing the desired module, before we actually give the construction and prove some properties. For our assumptions on A, κ, λ we refer to (12.2.1). Let T be the binary tree, i.e. T = define the branching point of v, w by

ω> 2.

For (infinite) branches v, w of T we

br(v, w) = m, if v  m = w  m and v(m) = w(m). Note, that this definition makes only sense for finite branches, if one is not the restriction / extension of the other. Moreover, we say that a subtree T  of ω≥ 2 is perfect if, for any m < ω, there is at most one finite branch f ∈ T  of length m of the form f = v ∩ w for branches v, w ∈ T  , where v ∩ w = v  m with m = br(v, w). Next we want to deduce the existence of a perfect tree. In order to do so, we define the map σ : ω 2 −→ ω 2 (v → v σ ) by ⎧ ⎪ ⎨ v(n), σ v (m) = ⎪ ⎩ 0,

if m =

2n+1

+

n i=0

2n−i v(i)

(12.2.2)

otherwise.

n+1 ≤ m < 2n+2 , since v(i) ∈ {0, 1} and so Note, the first case have 2 n we n−i n inn−i v(i) ≤ i=0 2 = ni=0 2i < 2n+1 . i=0 2

Clearly v σ (0) = v σ (1) = 0 for any v ∈ ω 2. Now consider v, w ∈ ω 2 with 1 for some m. Then ≥ 2 and v(n) =  w(n) = 1, where v σ (m) = wσ (m) n = n−i m n n n+1 + n+1 + n−i w(i). From n−i v(i) = 2 v(i) = 2 2 m = 2 i=0 i=0 i=0 2 n n−i w(i) and v(i), w(i) ∈ {0, 1}, it follows that v  n = w  n and so we i=0 2 also have v  n + 1 = w  n + 1 by the above. Therefore, if v σ , wσ coincide, then so do v, w, that is, the above defined map σ is injective. For T σ := {v σ  m | m < ω, v ∈ ω 2}, we show the following:

12.2 ℵ1 –free modules of cardinality ℵ1

413

Proposition 12.2.2. Let σ be the map given by (12.2.2). Then (a) σ : ω 2 → ω 2 defines an injective tree embedding; (b) T σ is a perfect subtree of T with Br(T σ ) = Im σ.

Proof. Clearly T σ is a subtree of T with Br(T σ ) = Im σ. Moreover, we have already seen that the map σ is injective. Hence (a) holds, and for (b) it remains to show that T σ is perfect. Suppose, for pairs v, w and v  , w from ω 2, that their images under σ branch at the same level m, that is, m = br(v σ , wσ ) = br(v σ , wσ ). We may assume that v σ (m) = v σ (m) = 1 (thus wσ (m) = wσ (m) = 0). From the arguments preceding the proposition it follows that v  n + 1 = v   n + 1, where 2n+1 ≤ m < 2n+2 . Hence, by the definition of σ, we also have v σ  2n+1 = v σ  2n+1 and so the two branches v σ and v σ must coincide up to the branching point m, i.e. v σ  m = v σ  m (and so v σ  m+1 = v σ  m+1). Similarly, wσ  m = wσ  m and thus v σ ∩wσ = 2 v σ ∩ wσ , showing that T σ is perfect. The above proposition guarantees us the existence of a perfect tree. However, we shall need λ ≤ 2ℵ0 ‘disjoint’ perfect trees for the construction; their existence is what we deduce next. For an element v of ω≥ 2, we define the (1–)support of v to be [v]1 = {i < ω | v(i) = 1}, and for a set X ⊆ ω≥ 2, we call the set [X]br = {br(v, w) | v = w ∈ X} ⊆ ω of its branching points the (br–)support of X. Given a fixed sequence Nα (α < λ) of infinite, almost disjoint subsets of ω, we define Tα = {v σ  m | m < ω, v ∈ ω 2 with [v]1 ⊆ Nα infinite}

(α < λ).

Using Proposition 12.2.2, it is now easy to see that Tα (α < λ) is a sequence of perfect subtrees of T σ ⊆ T such that the Br(Tα )s are pair–wise disjoint and the [Tα ]br s (⊆ Nα ) are pair–wise almost disjoint; we fix this sequence for the remaining part of this section. Moreover, for each α < λ, let Vα ⊆ Br(Tα ) be a fixed subset of infinite branches of cardinality λ. Next, we let Γ := T × N × λ and define B∞ =



(τ, N, α)A

(τ,N,α)∈Γ

(12.2.3)

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Realizing algebras

to be the direct sum of cyclic A–modules (τ, N, α)A with annihilators AnnA (τ, N, α) = N, where we identify Γ with a subset of B∞ . The continuity of the multiplication in A implies that every element of B∞ is annihilated by some element of N. Moreover, it is clear, by our assumption on A, that B∞ is free of rank λ as an R–module. as a countable sum g =  Each element of the S–completion B∞ is expressible  (τ,N,α) g(τ,N,α) for suitable g(τ,N,α) ∈ (τ, N, α)A; we define the support of such an g by   [ g ] = (τ, N, α) ∈ Γ | g(τ,N,α) = 0 . The norm of g is the ordinal    g  = sup α + 1 | (τ, N, α) ∈ [ g ] for some τ ∈ T, N ∈ N . So, in particular,  (τ, N, α)  = α +1 for any (τ, N, α) ∈ Γ. The norm of a subset of B∞ is defined to be the supremum of the norms of its elements. We call an element b of B∞ basal, if it has the two properties that bA is a direct summand of the A–module B∞ and AnnA (b) ∈ N. We now fix a sequence bα (α < λ) running λ–times through the set of all basal elements in B∞ and put Nα = AnnA (bα )

(α < λ).

Next we construct a continuous increasing sequence Mα (α ≤ λ) of A– submodules of the S–completion B∞ . First we put M0 = {0}. Now assume inductively that, for some α < λ, Mα has been constructed so that  Mα  = α. Set

 gα =

bα , 0

if bα ∈ Mα otherwise

and, for v ∈ Vα , n < ω, let qi qi yv,n = (v  i, Nα , α) + gα v(i), qn qn i≥n

(12.2.4)

(12.2.5)

i≥n

where the qn s form a ’fast converging’ sequence as the tn s on page 15 (before Proposition 1.1.18), that is, q0 = 1, qn+1 = sn+1 qn2 . Note that the yv,n s do not

12.2 ℵ1 –free modules of cardinality ℵ1

need the additional index α, since the Vα s are pair–wise disjoint. Now we define Mα+1 = Mα + Fα ,

415

(12.2.6)

with Fα = T × {Nα } × {α}A + { yv,n | v ∈ Vα , n < ω }A .

(12.2.7)

Our choices of yv,n and gα ensure that Mα+1 is an S–pure submodule of B∞ and that  Mα+1  = α + 1. Note, that all the generators of Fα have annihilator Nα . Continuity takes care of the definition of Mα at a limit ordinal α. Finally, we define M to be equal to Mλ , that is,  M = Mλ = Mα = Fα . α<λ

α<λ

For the just constructed module M we show Proposition 12.2.3. |M | = λ and M/Mβ is locally free for each β < λ. For the notion of local freeness we refer the reader to Definition 1.1.12.

Proof. Clearly |M | = λ. For the second part of the assertion let β < λ be fixed and let M  be an arbitrary R–pure submodule of M/Mβ of finite rank, that is, M  = E∗S for some non–empty finite subset E ⊆ M/Mβ (see Definition 1.1.12 (i)). We have to show that E is contained in an S–pure free submodule of M/Mβ . In order to do so, we choose α < λ minimal with E ⊆ Mα /Mβ . First note that α > β must be a successor because E is a proper finite set, hence γ = α − 1 ≥ β exists. Also note that Mα /Mβ is a quotient of A–modules, hence it is itself an A–module. By induction, we may assume that Mγ /Mβ is locally free and thus it is enough to show that E ⊆ (U + Mβ )/Mβ ⊕ Mγ /Mβ ⊆∗S Mγ+1 /Mβ

(12.2.8)

for a free A–module (U + Mβ )/Mβ . First we want to find a suitable A–submodule U ⊆ Mα . For any m < ω, let be the set of elements in T of length ≤ m. By (12.2.5), (12.2.6) and (12.2.7), there are a finite set E  ⊆ Mγ+1 of representatives of the elements in E, a finite set F ⊆ Vγ , and a number m < ω such that

Tm

E  ⊆ U + Mγ , where U = T m × {Nγ } × {γ} ∪ {yv,m | v ∈ F }A .

416

12

Realizing algebras

Moreover, we may assume that [vm ] ∩ [wm ] = ∅ for all v = w ∈ F, where, for a branch v, we denote by vm = v  (ω \ m) the branch starting at m, and so [vm ] = {v  i | i ≥ m}. A simple support argument shows that the defining generators of U are A–independent modulo Mγ . Hence Mγ /Mβ ∩ (U + Mβ )/Mβ = 0 and thus (U + Mβ )/Mβ ∼ = (U + Mγ )/Mγ must be A–free. = ((U + Mγ )/Mβ ) / (Mγ /Mβ ) ∼ It is now easy to show that U is S–pure in M which then also implies the S–purity in (12.2.8): using support arguments, it follows immediately from the construction that Mγ+1 is S–pure in M . Hence, for h ∈ M \ Mγ+1 , we clearly have dh ∈ Mγ+1 , especially dh ∈ U , for any d ∈ S. So let h ∈ Mγ+1 . Applying the same arguments as above, we find a finitely generated A–submodule 

U  = T m × {Nγ } × {γ} ∪ {yv,m | v ∈ F  }A for some number m ≥ m and a finite set F ⊆ F  ⊆ Vγ with h ∈ U  . Again, we may also assume that m is chosen such that [vm ] ∩ [wm ] = ∅ for all v = w ∈ F  . One more support argument now shows that U is a summand of U  ; we leave it as an exercise to write down a complement of U in U  . Thus, if dh ∈ U for some d ∈ S, then h ∈ U follows from h ∈ U  , which shows that U is S–pure in Mγ+1 , respectively in M . Therefore (12.2.8) holds and so M/Mβ is locally free. 2 We finish this subsection with defining and considering the module B=



Bα , where Bα = T × {Nα } × {α}A for any α < λ.

α<λ

It is immediate from the construction that the right A–module M satisfies  B ⊆ M ⊆ B,  It is also clear that the annihilator of every element and that M is S–pure in B. of M is open in A. It thus follows that the natural map A → EndR (M ) is a topological embedding. Therefore, in order to prove that it is a topological isomorphism we only have to show that it is surjective.

12.2 ℵ1 –free modules of cardinality ℵ1

417

In fact, it will be enough to prove that each endomorphism ϕ ∈ EndR M has the property that bϕ ∈ bA for each basal element b ∈ B: assume that ϕ has this property, that is, bϕ = bab (b ∈ B basal), (12.2.9) for some ab ∈ A. For any x ∈ B, we may choose b = bx = (τ, Nα , α) ∈ Γ \ ([ x ] ∪ [ xϕ ]) satisfying AnnA (x) ⊇ N = AnnA (b). Easy calculations show that AnnA (b + x) = N and that the direct summand bA on the right–hand side of (12.2.3) may be replaced by (b + x)A. Therefore b + x is basal, and we have (b + x)ab+x = (b + x)ϕ = bϕ + xϕ = bab + xϕ, whence b(ab+x − ab ) = xϕ − xab+x . Our choice of b implies that the supports of the two sides of the above equation are disjoint and so both sides must vanish. Thus ab+x − ab ∈ AnnA (b) ⊆ AnnA (x) and so xϕ = xab+x = xab = xabx . It follows that, if, for the moment, we convert B into a directed set (B, ≤ ) by writing x ≤ x to mean AnnA (x) ⊇ AnnA (x ) for x, x ∈ B, then the net (abx )x∈B contains a cofinal Cauchy subnet, namely the subnet indexed by the basal elements. By the completeness of A, the whole net converges to an element a ∈ A, and so we have proved that xϕ = xa for all x ∈ B. Thus ϕ agrees with scalar multiplication by a on B and hence, by continuity, also on M . It therefore remains to show that bϕ ∈ bA for each basal element b ∈ B (see (12.2.9)). This is the goal of the next three subsections. The next one is devoted to the combinatorial part, a Pigeon–hole Lemma.

The Pigeon–hole Lemma In this subsection we apply the well–known pigeon–hole argument to a family of certain elements of M of size κ, that is, the ‘pigeons’ here are the elements of this family, respectively of its index set. Recall that κ is regular by assumption (see (12.2.1)). Pigeon–hole Lemma 12.2.4. Let β ≤ λ. Assume that there is a family of elements tv ∈ Mβ (v ∈ W ) of cardinality |W | = κ indexed by a subset W ⊆ Vα for some α < λ such that the following property is satisfied for some finitely generated A–submodule H of Mβ : If v, w ∈ W with br(v, w) = m, then tv − tw ∈ H + qm Mβ . (12.2.10) Then there exist a subset W  of W with |W  | = κ and a finite sequence β0 < · · · < βs < β of ordinals such that Fβi for all v ∈ W  . (12.2.11) tv ∈ i≤s

418

12

Realizing algebras

Proof. We prove the result by induction on β. The result is vacuous for β = 0. So we suppose β > 0 and assume that the corresponding result holds for all smaller ordinals. First we consider the case of β being a limit ordinal. If, for some γ < β, Mγ contains a subfamily tv (v ∈ Wγ ) of cardinality κ, then the result follows at once by our induction hypothesis, since we may suppose that γ is large enough for Mγ to also contain the finitely generated A–module H from (12.2.10). Hence a simple application of the modular law implies that Wγ satisfies the analogue of (12.2.10) for γ (in place of β). Assume now, for contradiction, that no such γ < β exists. Then one easily produces a continuous increasing sequence of ordinals βξ < β (ξ < κ) and branches vξ ∈ W (ξ < κ) such that tvξ ∈ Mβξ+1 \ Mβξ (ξ < κ). It is clear that we must have β = supξ<κ βξ . Moreover, there is no loss in generality if we also assume that H ⊆ Mβ0 . Now for each ξ < κ, the coset tvξ + Mβξ is a non–zero element of Mβξ+1 /Mβξ . By Proposition 12.2.3, this quotient is locally free and therefore S–reduced. So, for some mξ < ω, we have / qmξ Mβξ+1 /Mβξ or, in other words, tvξ + Mβξ ∈ tvξ ∈ / Mβξ + qmξ Mβξ+1 .

(12.2.12)

Since κ > ℵ0 , a pigeon–hole argument shows that there is a subset C ⊆ κ of size |C| = κ on which mξ is constant; say mξ = m (ξ ∈ C). Generally, if a set V of branches v ∈ ω 2 admits a bound n0 on the levels of its branching points br(v, w) (v, w ∈ V ), then, since v → v (n0 + 1) is visibly an injection into the finite set {v (n0 + 1) | v ∈ V }, the set V itself must clearly be finite. However, the set {vν | ν ∈ C} is infinite and so we may certainly choose indices ξ < η such that br(vξ , vη ) = n > m. Then, by our assumption (12.2.10) and by applying the modular law again, we have tvη − tvξ ∈ H + qn Mβη +1 and, since n > m and tvξ , H both lie in Mβη , it follows that tvη ∈ Mβη + qm Mβη +1 = Mβη + qmη Mβη +1 , contradicting (12.2.12). Therefore we can always find a γ < β such that Mγ contains a subfamily of tv s of size κ and so the limit case is finished.

12.2 ℵ1 –free modules of cardinality ℵ1

419

We now consider the case of β being a successor, that is, β = γ + 1 for some γ. Recall that, by induction hypothesis, the assertion of the lemma holds for all ordinals ≤ γ. Using support arguments and (12.2.4) – (12.2.7), we first show: Mγ ∩ Fγ = gγ A.

(12.2.13)

Choose any v ∈ Vγ and apply (12.2.5) for some n < ω with v(n) = 1. Then sn qn = qn+1 /qn and yv,n − sn qn yv,n+1 = yv,n − (qn+1 /qn )yv,n+1 = (v  n, Nγ , γ) + (qi /qn )(v  i, Nγ , γ) + gγ (qi /qn )v(i) i≥n+1



− (qn+1 /qn )

i≥n

(qi /qn+1 )(v  i, Nγ , γ) −

i≥n+1



 (qi /qn+1 )v(i)

i≥n+1

= (v  n, Nγ , γ) + gγ v(n) = (v  n, Nγ , γ) + gγ ∈ Fγ . However, on the one hand, (v  n, Nγ , γ) ∈ Fγ and gγ ∈ Mγ , thus gγ A ⊆ Fγ ∩ Mγ . On the other hand, h ∈ Fγ can be expressed as a sum h= yv,m av + gγ aγ + t (av , aγ ∈ A), v∈E

for some large enough m < ω, a finite subset E ⊆ Vγ , and t ∈ Bγ (recall: Bγ = T × {Nγ } × {γ}A ). We may assume that, for v = w ∈ E, [gγ ] ∩ [vm ] = ∅ = [vm ] ∩ [wm ]. If, additionally, h ∈ Mγ (so  h  ≤ γ) then, considering any (v  i, Nγ , γ) ∈ [vm ] ⊆ [yv,m ] (i ≥ m), leads to av ∈ Nγ = AnnA (yv,m ) (v ∈ E). Similarly, support arguments also imply t = 0. Hence h = gγ aγ and so Mγ ∩ Fγ ⊆ gγ A, that means (12.2.13) follows. It is now easy to finish the case β = γ + 1, respectively, the proof of the lemma. Let H ⊆ Mγ+1 , tv ∈ Mγ+1 (v ∈ W ) satisfy the hypothesis of the lemma. Using (12.2.6), we can write tv = t0v + t1v with t0v ∈ Mγ , t1v ∈ Fγ for all v ∈ W, so that tv − tw ∈ H + qm Mγ+1 whenever br(v, w) = m.

420

12

Realizing algebras

We can also find finitely generated A–submodules H 0 , H 1 of Mγ and Fγ , respectively, such that H ⊆ H 0 + H 1 . Thus, for v, w ∈ W with br(v, w) = m, we have (t0v − t0w ) + (t1v − t1w ) + h0 + h1 + qm g + qm f = 0 for suitable elements hi ∈ H i and g ∈ Mγ , f ∈ Fγ . Therefore, it follows from (12.2.13) that (t0v − t0w ) + h0 + qm g = (t1w − t1v ) − h1 − qm f ∈ Mg ∩ Fγ = gγ A. Hence (t0v − t0w ) ∈ H 0 , gγ A + qm Mγ , that is, the family {t0v | v ∈ W } and the finitely generated module H 0 , gγ A satisfy the assumption (12.2.10) for γ. Thus, by induction hypothesis, there are a suitable W  ⊆ W with |W  | = |W | = κ and  ordinals γ0 < · · · < γs < γ + 1 such that t0v ∈ i≤s Fγi for all v ∈ W  . If we additionally put γs+1 = γ then Fγi for all v ∈ W  tv = t0v + t1v ∈ i≤s+1

2

and this completes the induction.

Comparing branching points In this subsection we basically refine the set of ‘pigeons’ from the Pigeon–hole Lemma more and more. Note, however, that it will be convenient to also call the ‘new’ index sets W instead of W  , W  , . . .; the same goes for the involved ordinals and the considered family of ordinals. Proposition 12.2.5. Let m0 < ω. Assume that  we have a finite sequence β0 < · · · < βs < λ of ordinals and a family {tv ∈ i≤s Fβi | v ∈ W } of elements of M of cardinality |W | = κ indexed by a subset W ⊆ Vα for some α < λ. Then we can find natural numbers n < ω, m, j ∗ ≥ m0 , elements aj (j ≤ n) of A, a subsequence of ordinals, which we also call β0 < · · · < βs , and another family tv ∈ Fβi indexed by a ‘new’ W, i≤s

of cardinality κ with tv (v ∈ W ) being of the form yvj ,m aj , tv = j≤n

that is, the branches vj (j ≤ n) are the only parameters depending on v. Moreover, the new family of elements satisfies the following properties:

12.2 ℵ1 –free modules of cardinality ℵ1

(a)



421

[gβi ] ∩ [(vj )m ] = ∅ for v ∈ W, j ≤ n;

i≤s

(b) [(vj )m ] ∩ [(vj  )m ] = ∅ for v ∈ W, j = j  ≤ n; (c) the branches vj (v ∈ W, j ≤ n) are pair–wise distinct; (d) vj ∈ Vβi for some i = i(j) ≤ s which is independent of v; (e) vj  j ∗ is independent of v and the vj  j ∗ (j ≤ n) are pair–wise distinct; (f) [Tβi ]br ∩ [Tβi ]br ⊆ j ∗ for all i = i ≤ s and [Tβi ]br ∩ [Tα ]br ⊆ j ∗ for all βi = α; (g) Nj = Ann yvj ∈ N for all v ∈ W, j ≤ n; (h) aj + Nj ∈ A/Nj \ qj ∗ −1 (A/Nj ) for all j ≤ n.

Proof. First note, that the new family of tv s is obtained by passing toequipotent subsets of W (which we also call W ) and by adding a fixed element of i≤s Fβi to each tv . Within this proof, we will apply the pigeon–hole argument several times. A first application of the pigeon–hole argument leads us to the assumption (w.l.o.g.), that W in the hypothesis of the proposition satisfies br(v, w) > m0 for all distinct pairs v, w ∈ W.  By (12.2.5), any finite sum n yv,n an (for fixed v) can be reduced to one summand yv,m am modulo B for a large enough m. Hence, we also have that any tv (v ∈ W ) can be written as a sum of elements fvi of Fβi (i ≤ s), which are, by (12.2.7), of the form fvi = yvli ,mvi alvi + bvi + gβi avi (12.2.14) l≤kvi

with vli ∈ Vβi , bvi ∈ Bβi , avi , alvi ∈ A, kvi , mvi < ω depending on v. However, using (12.2.5) we may assume that, for a fixed v ∈ W , we have kvi = kv (i ≤ s) and mvi = mv ≥ m0 (i ≤ s) large enough such that  [gβi ] ∩ [(vli )mv ] = ∅ and [(vli )mv ] ∩ [(vl i )mv ] = ∅ i≤s

for l, l ≤ kv , i, i ≤ s, that is, (a) and (b) of the proposition are satisfied for any fixed v ∈ W . Next we ‘unify’ all possible parameters: applying the pigeon–hole argument to ω leads to a subfamily with common k = kv and m = mv for all v ∈ W , since

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Realizing algebras

|ω| = ℵ0 < κ = |W |. Using this argument on A/Nβi , we obtain the same elements of A, i.e. ai = avi , ali = alvi no longer depend on v; this is possible because |A/Nβi | < κ (see (12.2.1), also recall gβi = 0 or AnnA (gβi ) = Nβi by (12.2.4) and AnnA (f ) = Nβi for all ‘canonical’ generators f ∈ Fβi by (12.2.7)). Moreover, we can pass to an equipotent subfamily with common bi = bvi (v ∈ W ) by applying the pigeon–hole argument to Bβi , since |Bβi | = |Tβi | |A/Nβi | < κ. Therefore (12.2.14) now reduces to yvli ,m ali + bi + gβi ai fvi = l≤k

for all v ∈ W, i ≤ s. Note that the properties (a) and (b) are obviously satisfied for this family. Now we identify all pairs li (l ≤ k, i ≤ s) with natural numbers j ≤ n = sk and we put i(j) = i, if j corresponds to li for some l. Thus (d) is clearly satisfied and we have tv = yvj ,m aj + (bi + gβi ai ) (v ∈ W ). j≤n

i≤s

Again because of the pigeon–hole argument, we may assume that yvj ,m aj = 0 for all pairs j ≤ n. Since Ann yvj ,m = Ann vj = Nβi ∈ N (for i = i(j)) as in / Nβi and so we can choose m1 > m0 such that (12.2.5), we have aj ∈ aj + Nβi(j) ∈ A/Nβi(j) \ qm1 (A/Nβi(j) ) for all j. Hence (g) holds and (h) follows whenever we choose j ∗ > m1 . However, we need to choose j ∗ in such a way, that also (e) and (f) are satisfied. Property (f) is easily arranged by choosing j ∗ large enough, since the br–supports of the perfect trees are almost disjoint. In order to establish (e), we choose jv > m1 (v ∈ W ) such that all the vj  jv are distinct for any fixed v. Once more, by a pigeon–hole argument, we can pass to a subfamily such that j ∗ = jv and also such that vj  j ∗ does not depend on v; surely j ∗ > m1 and so (e) holds.  Finally, it only remains to ensure property (c). Subtracting the constant element (bi + gβi ai ) we obtain a family of new elements i≤s

tv =



yvj ,m aj .

j≤n

We apply the Δ–Lemma (see Jech [284, p. 225]) to the finite sets {vj | j ≤ n} (v ∈ W ). Therefore there exist a set Δ and an equipotent subset W such that {vj | j ≤ n} ∩ {wj | j ≤ n} = Δ for all distinct v, w ∈ W . Again, we subtract the constant

12.2 ℵ1 –free modules of cardinality ℵ1

423

 element d = x∈Δ yx,m ax from each tv (note: ax = aj for x = vj ). Hence all the vj s needed for expressing the new tv s are distinct and so (c) of the proposition follows, which finally provides the required new family. 2 Before we are ready to prove the main result of this section we need: Proposition 12.2.6. Let tv =



yvj ,m aj ∈

j≤n



Fβi (v ∈ W ⊆ Vα ) be a family

i≤s

of elements of M satisfying the assertion of Proposition 12.2.5. Moreover, let z ∈ M \ sz M for sz ∈ S, sz |qm0 with the property tv − tw ± qb z ∈ qb+1 M for all v, w ∈ W with b = br(v, w).

(12.2.15)

Then there is a map π : n+1 = {0, . . . , n} → s+1 such that, for all v = w ∈ W , (a) vj , wj ∈ Vβπ(j) for all j ≤ n; (b) br(vj , wj ) ≥ br(v, w) ≥ j ∗ for all j ≤ n, and the br(vj , wj )s are distinct; (c) there is some j ≤ n with br(vj , wj ) = br(v, w) and βπ(j) = α, gα = 0.

Proof. First note, we may assume (w.l.o.g.) that all relevant branching points are greater than m. Now by Proposition 12.2.5 (d), for each j ≤ n, there is a unique π(j) ≤ s such that vj , wj ∈ Vβπ(j) are branches of the perfect tree Tβπ(j) ; hence π and (a) are obvious. We first claim that the branching points kj = br(vj , wj ) (j ≤ n) are pair–wise distinct, which is part of (b). Suppose not, i.e. there are j = j  with kj = kj  . Then kj ≥ j ∗ follows from Proposition 12.2.5 (e) and thus we have π(j) = π(j  ) by property (f), that is, vj , wj , vj  , wj  come from the same tree Tβπ(j) . Moreover, property (e) also implies (wj  j ∗ =) vj  j ∗ = vj   j ∗ (= wj   j ∗ ) and hence (vj  kj =) vj ∩ wj = vj ∩ wj are distinct but of the same length kj = kj  , which is impossible for a perfect tree. Therefore, for establishing (b), it remains to prove that b = br(v, w) ≤ kj for all j ≤ n, which we will do simultaneously with showing (c). For this we need the hypothesis (12.2.15) in the form tv − tw ± qb z ∈ qb+1 M ⊆

qb+1 M. qm

From (12.2.5) follows yvj ,m − ywj ,m ±

qkj qk +1 gβπ(i) = j (yvj ,kj +1 − ywj ,kj +1 ). qm qm

(12.2.16)

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Now, if b = br(v, w) < k = min{kj | j ≤ n} then b + 1 ≤ kj for all j ≤ n and q thus we deduce yvj ,m − ywj ,m ∈ qb+1 M (j ≤ n). Hence m tv − tw =

qb+1 (yvj ,m − ywj ,m )aj ∈ M qm j≤n

q

q

s

q

b+1 b M follows. Thus z ∈ qbb+1 and so qb z ∈ qb+1 qm M = qm M ⊆ sz M , since m sz |qm0 and m0 < m < b, which contradicts the choice of z. Therefore the minimum k is at most b. Suppose k < b. We note that there is a unique j ≤ n with k = kj and k < kj  for all j  = j by our first claim in the proof. q M . Moreover, From k + 1 ≤ b and (12.2.15) follows tv − tw ∈ qk+1 M ⊆ qk+1 m by (12.2.5), qk+1 (yvj  ,m − ywj  ,m )aj  ∈ M qm 

j =j ≤n

and hence (yvj ,m − ywj ,m )aj ∈

qk+1 M. qm q

M and so we get Together with (12.2.16) it now follows that qqmk gβπ(j) aj ∈ qk+1 m qk+1 ∗ gβπ(j) aj ∈ qk M = sk+1 qk M . Recall k ≥ j −1 and Ann gβπ(j) = Nπ(j) , hence aj + Nπ(j) ∈ qj ∗ −1 (A/Nπ(j) ), contradicting Proposition 12.2.5 (h). Thus the first part of (c) follows, which immediately implies βπ(j) = α by Proposition 12.2.5 (f) (the second part) and so (b), (c) are shown. 2

Proof of the theorem The main theorem will follow from the Pigeon–hole Lemma (see page 417) and a corollary of the two results in the previous subsection (beginning on page 420). Corollary 12.2.7. Let tv =

 j≤n

yvj ,m aj ∈



Fβi (v ∈ W ⊆ Vα ) be a family of

i≤s

elements of M as in Proposition 12.2.5 and let z ∈ M \ sz M be as in Proposition 12.2.6 with the property (12.2.15) and such that Ann z ⊇ Ann gα . Then z ∈ gα A.

Proof. Let v = w ∈ W with b = br(v, w). By Proposition 12.2.5, we have (yvj ,m − ywj ,m )aj . Moreover, by Proposition 12.2.6 (c), there is a tv − tw = j≤n

distinguished j ≤ n with br(vj , wj ) = b, βπ(j) = α, and br(vj  , wj  ) > b for all j  = j. Thus qb+1 M tv − tw − (yvj ,m − ywj ,m )aj ∈ qm

12.2 ℵ1 –free modules of cardinality ℵ1

425

(compare (12.2.16) in the proof of Proposition 12.2.6). Hence, using the hypothesis (12.2.15), we now obtain ±qb z − (yvj ,m − ywj ,m )aj ∈

qb+1 M qm

and so, using (12.2.16) again, we have ±qb z −

qb qb+1 gα aj ∈ M. qm qm

Therefore ±qm z − gα aj ∈

qb+1 M ⊆ qb M, qb

and this is true for all b ∈ [W ]br . Since |W | = κ we have that [W ]br is unbounded and hence we may choose a sequence (v, w) of pairs v, w ∈ W with br(v, w) = b < ω converging to infinity. qb M = 0 and so ±qm z = gα aj . Note that 0 = gα = Thus ±qm z − gα aj ∈ b<ω

bα ∈ B∞ is basal with AnnA bα = Nα ⊆ AnnA z. Therefore aj ≡ qm a modulo Nα for some a ∈ A and so z = ±gα a ∈ gα A, as required. 2 We finally prove the main theorem of this section.

Proof of Theorem 12.2.1. By (12.2.9) it remains to show that b∗ ϕ ∈ b∗ A for all basal elements 0 = b∗ ∈ B. We fix a basal element b∗ ∈ B and suppose that / b∗ A. In particular, z = 0 and so there are m0 < ω and sz ∈ S with z = b∗ ϕ ∈ z ∈ / sz M and sz |qm0 . Now choose any α < λ such that b∗ = gα ∈ Mα and Ann gα = Nα ∈ N. We consider Fα as in (12.2.7) and the corresponding family tv = yv ϕ (v ∈ Vα ) of the images of the generators yv under ϕ. From (12.2.5) follows yv −yw ±qb gα ∈ qb+1 M for b = br(v, w) and hence tv − tw ± qb z ∈ qb+1 M for any v, w ∈ Vα with b = br(v, w). Applying the Pigeon–hole Lemma 12.2.4, we obtain a subfamily {tv | v ∈ W ⊆ Vα } satisfying the hypotheses of Proposition 12.2.5 and Proposition 12.2.6. Passing to the ‘new’ subfamily provided by Proposition 12.2.5, we can apply Proposition 12.2.6 and deduce z ∈ gα A from Corollary 12.2.7 – a contradiction. Thus it 2 follows b∗ ϕ ∈ b∗ A, as desired.

426

12.3

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Realizing all cotorsion–free algebras

Let R again be a fixed S–ring. In this section we want to expand Theorem 12.1.1 and remove the cardinal restriction λ ≤ 2ℵ0 . This appeared first in [120], however our main source will be [102]. We will also include [124] and for sake of clarity we will avoid the torsion case and the mixed case in [102] and restrict ourselves to the torsion–free, discrete case. A recent result from [104] is included to simplify applications of the main result (see Section 12.4). Our main cases of examples will be cotorsion– free modules, almost cotorsion–free modules as well as separable, torsion–free modules. We will use the following Definition 12.3.1. Let R be an S–ring. (i) A module G is S–separable (torsion–free), if G is the S–pure submodule of a product Rκ for some cardinal κ (see Section 1.4). (ii) A module G is almost S–cotorsion–free or ℵ0 –S–cotorsion–free (for S), if any homomorphism ϕ : F → G from the completion of a free module F = ⊕e<ω Re of countable rank into G is of finite S–rank. Recall that F is the completion of F in the S–topology. (iii) ϕ : F → G is of finite S–rank, if Gϕ ⊆ S−1 Eϕ for some finitely generated submodule E of F. Cotorsion–free modules over countable principal ideal domains can be characterized in many ways (see [120, 191, 246]). Proposition 12.3.2. The following conditions for an R–module G over a countable principal ideal domain (not a field) R are equivalent:  G) = 0, where R  is the completion of (a) G is cotorsion–free, that is HomR (R, R in the R–topology; (Hence S = R \ {0}.) (b) any cotorsion submodule of G is {0}; (c) G has no summands isomorphic to the quotient field Q(R) of R, or to R/pR (pR) (the p–adic completion of the p–localization R(pR) of R); or to R (d) The additive group of EndR G is cotorsion–free. (pR) ⊕ R (pR) Remark 12.3.3. The algebras Q(R) ⊕ Q(R), R/pR ⊕ R/pR and R are not isomorphic to endomorphism algebras EndR G of any R–modules G, as

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Realizing all cotorsion–free algebras

427

(pR) follows easily by counting summands. Thus we will exclude Q(R), R/pR, R as summands and by Proposition 12.3.2 this is the same as to say that the algebra A (as an R–module R A) is cotorsion–free. A similar argument as in the proof of Proposition 12.3.2 applies more generally for Dedekind domains R which are  of Dedekind domains R can be deS–rings: by Theorem 1.5.5 S–completions R scribed. Hence, similar to the above, it is easy to single out those particular co G) = {0} describing torsion modules of an R–module G responsible for Hom(R, that G is not S–cotorsion–free (see Definition 1.1.23). We leave it to the reader to extend Proposition 12.3.2 accordingly to a characterization of S–cotorsion–free modules over Dedekind domains.  G) = Proof of Proposition 12.3.2. (a) ⇐⇒ (b): if (b) holds, then HomR (R,  {0} because R is cotorsion and the class of cotorsion modules is closed under epimorphic images.  C) = 0 for Conversely we may assume that G is reduced because Hom(R, torsion modules or for divisible modules C = 0. Then it is easy to see that any  torsion–free, reduced cotorsion R–module C is an R–module and thus C is an  epimorphic image of a free R–module, so (b) follows from (a). (c) ⇐⇒ (b): the implication (b) ⇒ (c) is obvious, thus we will assume (c). The structure of cotorsion modules over principal ideal domains is known (see Fuchs [173, Vol. 1, Section 55]). Any non–trivial cotorsion module has a summand D (pR) , respectively. If D ∼ isomorphic to either Q(R), R/pR or R = Q(R), then D is injective, hence a summand of G. Thus we may assume that G is reduced. If G has a torsion submodule tG = 0, then tG ⊆∗ G and we can choose a basic element a such that aR ∼ = R/pR which is pure in tG, hence in G. However R/pR is pure–injective, hence aR must be a summand of G and G is torsion–free by (c). (pR) and f : R (pR) −→ D −→ G is this isomorphism, then we can If D ∼ = R (pR) −→ G is a choose k < ω maximal such that pk | f (1). Thus g = p−k f : R (pR) is pure–injective, pure embedding and Im g is a pure submodule of G. Also R hence Im g splits, which contradicts (c). Thus (b) and (c) are equivalent. (b) ⇒ (d): submodules of products of cotorsion–free modules are cotorsion– free. Then EndR G is cotorsion–free because EndR G −→ GG (σ → (gσ)g∈G ) is an embedding. (d) ⇒ (c): suppose that D is a summand of G and D is isomorphic to one of (pR) . Then EndR D ∼ the modules Q(R), R/pR and R = D and as a summand of G the module D is also isomorphic to a summand of EndR G. Hence EndR G is not cotorsion–free by ((c) ⇒ (a)), a contradiction. 2 The next part of Section 12.3 is devoted to an application of the Strong Black Box 9.2.2 (or Corollary 9.2.7). We obtain a result which shows that any S–

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cotorsion–free algebra is the endomorphism algebra of a class of modules M (which is not a set.) However, due to application of the Strong Black Box we cannot care much about the size of M . But as a bonus, this proof is much shorter than the one using the General Black Box 9.2.27: it is important to us to make this difference very clear. Thus we will include proofs based on very similar algebraic arguments in each case. Also note that the results of this section will include cotorsion–free modules as a particular case. The choice, using the General Black Box 9.2.27 or the Strong Black Box 9.2.2 or both is left to the reader.

The main realization theorem and the Strong Black Box In this subsection we shall apply the Strong Black Box as given in Corollary 9.2.7 to prove the following theorem. Theorem 12.3.4. Let R be an S–ring (see Definition 1.1.1) and let A be an S– cotorsion–free R–algebra such that |A| ≤ λ and for BA = ε<ρ Raε we have A . BA ⊆∗ A ⊆ B A denotes the S–adic completion of F := BA in the Hausdorff S– Again, B topology. Moreover, let |R| ≤ κ ≤ μ ≤ λ be infinite cardinals such that μκ = μ and λ = μ+ . Then there exists an S–cotorsion–free R–module G of cardinality λ such that EndR G = A. Before we can construct the desired module we need the following lemma which basically tells us how to obtain the module “step by step”.  Step Lemma 12.3.5. Let P = (α,ε) ∈ I ∗ Reα,ε for some I ∗ ⊆ λ × ρ and let M be an A–module as well as an S–cotorsion–free R–module with  P ⊆∗ M ⊆∗ B. Also suppose that there is a set I = {(αn , εn ) | n < ω} ⊆ [ P ] = I ∗ such that α0 < α1 < . . . < αn < . . . and Iλ ∩ [ g ]λ is finite for all g ∈ M (Iλ = [ I ]λ ). Moreover, let φ : P −→ M be such an R–homomorphism which is not multiplication by an element of A. Then there exists an element y of P such that  yφ ∈ / M  := M, yA∗ ⊆ B

12.3

429

Realizing all cotorsion–free algebras

. The module M  is where φ is identified with its unique extension φ : P −→ M again an A–module as well as an S–cotorsion–free R–module with M ⊆∗ M  ⊆∗  The element y can be chosen to be either B.  b ∈ P and for x = y = x or y = x + πb for suitable π ∈ R, qn eαn ,εn . 

n<ω

Proof. Let the assumptions be as above. Either x = n<ω qn eαn ,εn satisfies xφ ∈ / M, Ax∗ or not. In the latter case there are k < ω and a ∈ A such that qk xφ − xa ∈ M.

(12.3.1)

Since φ ∈ / A also qk φ ∈ / A, by purity arguments, and thus there is an S–pure element b of P such that qk bφ = b(qk φ) = ba. Hence, by the S–cotorsion–freeness  such that of M , there is π ∈ R πb(qk φ − a) ∈ /M

(12.3.2)

 Let z = x + πb and suppose where, again, we may assume that π is S–pure in R. zφ ∈ M, zA∗ . Then ql zφ − za ∈ M for some l ≥ k, a ∈ A; let ql = sqk . Therefore, using (12.3.1), we obtain that (ql zφ − za ) − s(qk xφ − xa)= ql xφ + ql πbφ − xa − ba π − ql xφ + sxa = x(sa − a ) + πb(ql φ − a ) is an element of M . Now [ x ]λ = Iλ , [ bql (φ − a ) ]λ ∩ Iλ is finite and [ x(sa − a ) + πb(ql φ − a ) ]λ ∩ Iλ is also finite. Hence sa − a = 0 and thus it follows from the above that πb(sqk φ −  This implies πb(qk ϕ − a) ∈ M contradicting sa) = sπb(qk φ − a) ∈ M ⊆∗ B. (12.3.2). Therefore either y = x or y = z satisfies yφ ∈ / M, yA∗ =: M  . Clearly M  is also an A–module.  it is S– It remains to show that M  is S–cotorsion–free. Since M  ⊆∗ B,   M ) = {0} we first describe the torsion–free and S–reduced. To prove Hom(R, purification of M + yA. For any k < ω we put qn qn eαn ,εn , π k = rn xk = qk qk n≥k

n≥k



where π = n<ω qn rn and z k = xk + π k b. Then, with y k = xk or y k = z k according to y = x or y = z, we deduce y k − sk+1 y k+1 ∈ P ⊆ M

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12

where sk+1 =

qk+1 qk

Realizing algebras

and M = M +



y k A.

k<ω

 −→ M  . Then, by Now suppose that there is a non–zero homomorphism φ : R continuity, 1φ = 0 and so there are m ∈ M, a ∈ A and k < ω such that 1φ = m + y k a. Clearly a = 0, since otherwise [ πφ ] ⊆ [ 1φ ] = [ m ] would imply that πφ ∈ M  i.e. 0 = φ : R  −→ M contradicting the S–cotorsion–freeness of M . for all π ∈ R, Moreover, we may assume that m = 0, since we could increase k. Now let πφ = π(1φ) = πm + πy k a = mπ + y kπ aπ ∈ M  with kπ ≥ k. Then

  qk qk πm − mπ = y kπ aπ − π y k − π y kπ a + π y kπ qk qk

and so

   qk qk  πm − mπ + π y k − π y kπ a = y kπ aπ − πa π . q qk + ,-k . ∈M + ,.  ∈ RM

The intersection of Iλ with the λ–support of the left hand side is finite while the q λ–support of the right hand side intersected with Iλ is infinite unless aπ = πa qkkπ q  contradicting the S–cotorsion– (note: qkkπ ∈ S). Thus πa ∈ A for all π ∈ R freeness of A. 2 We are now ready to construct the desired module. Construction 12.3.6. Let (φβ )β<λ be a family of canonical homomorphisms as given by Corollary 9.2.7. For any β < λ let

eγ A; Pβ = γ ∈ [φβ ]λ

then Dom φβ ⊆ Pβ . γ and pure R–submodules Gβ of B  such We inductively define elements yγ ∈ P that, for all γ < β < λ,

12.3

Realizing all cotorsion–free algebras

431

(i)  yγ  =  Pγ  (=  φγ ), (ii) Gβ = B  , yγ A (γ < β)∗ , and (iii) Gβ is S–cotorsion–free.  β Recall that B ⊆ B  = α<λ eα A. Also note that the G s are then clearly A– modules.  ; obviously B  is S–cotorsion–free, since A is,  = B Let G0 = B  ⊆∗ B by assumption, and it also satisfies the conditions (i) and (ii), since there are no relevant yγ s. γ all the required conNext let β be a limit ordinal and suppose  thatγG satisfies β ditions for any γ < β. We put G = γ<β G . Then Gβ certainly satisfies (i) and (ii). Moreover, Gβ is clearly S–torsion–free and S–reduced and so it remains  −→ Gβ be a homomorphism and let  Gβ ) = {0}. So let φ : R to show Hom(R,  we δ < β such that 1φ ∈ Gδ = B  , yγ A (γ < δ)∗ . Then, for each r ∈ R, have [ rφ ] ⊆ [ 1φ ] and hence  [ rφ ]λ ∩ [ yγ ]λ  <  yγ  for all γ ≥ δ (see Corol respectively, Im φ ⊆ Gδ and lary 9.2.7 (iii)). Therefore rφ ∈ Gδ for all r ∈ R, β thus φ ≡ 0, i.e. G is S–cotorsion–free. It remains to consider the successor case. Assume Gβ is given satisfying all the conditions. We consider φβ . Since  φβ  ∈ E ⊆ λo and Dom φβ is canonical there are (αn , 0) ∈ [ φβ ] (n < ω), such that α0 < α1 < . . . < αn < . . . and  φβ  = sup αn . n<ω

We put I = {(αn , 0) | n < ω}. Then Iλ ∩ [ g ]λ is finite for all g ∈ Gβ by (12.3.6) and condition (iii) in Corollary 9.2.7. We differentiate two cases.  φβ satisfies Im φβ ⊆ Gβ and φβ ∈ / A, then we apply the If φβ : Dom φβ → Im Step Lemma 12.3.5 to I as above (εn = 0 for all n), P = Dom φβ ⊆∗ B ⊆∗ Gβ  We deduce the existence of an element and M = Gβ ⊆∗ B. β  y = yβ ∈ Dom φβ ⊆ P and of an A–module Gβ+1 = Gβ , yβ A∗ = Gβ +



yβk A

k<ω

 such that which is an S–cotorsion–free, S–pure submodule of B yβ φβ ∈ / Gβ+1 ,

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Realizing algebras

where yβk =

qn eα ,0 qk n

n≥k

or yβk =

qn eα ,0 + π k b qk n

 b ∈ B). (yβ = yβ0 , π ∈ R,

n≥k

Note,  yβ  =  yβk  =  φβ  =  Pβ . β . Moreover, Gβ+1 satisfies (12.3.6), since yβ ∈ P β If Im φβ  G or φβ ∈ A, then we put yβk =

qn eα ,ε , yβ = yβ0 qk n n

n≥k

and Gβ+1 = Gβ , yβ A∗ = Gβ +



yβk A.

k<ω

Then we apply the Step Lemma 12.3.5. And also in this case, Gβ+1 satisfies the required conditions. Finally, we define G by  Gβ = B  + yβk A. G= β<λ

β<λ k<ω

This finishes the construction. It is an immediate consequence from the construction that G is an A–module of  Next we cardinality λ which is also an S–cotorsion–free, S–pure submodule of B. describe the elements of G. Lemma 12.3.7. Let G be as in Construction 12.3.6 and let g ∈ G \ B  . Then there are a finite non–empty subset N of λ and k < ω such that yβk A g ∈ B + β∈N

and [ g ]λ ∩ [ yβ ]λ is infinite ⇐⇒ β ∈ N. In particular, if  g  is a limit ordinal then  g  =  yβ∗  =  φβ∗  for β∗ = max N.

12.3

Proof. Let

Realizing all cotorsion–free algebras

g ∈ G = B +



433

yβn A.

β<λ n<ω

N

b

of λ, ∈ B  , k < ω, aβ,n ∈ A (β ∈ N  , n ≤ k) Then there are a finite subset such that g = b + yβn aβ,n . β∈N  n≤k

Since yβn −

qk k qn yβ

∈ B ⊆ B  this expression reduces to g =b+



yβk aβ

β∈N 

for some aβ ∈ A (β ∈ N  ), b ∈ B  . Putting N = {β ∈ N  | a%β = &0} (N = ∅ for g ∈ / B  ) the conclusion of the lemma follows, since [ yβ ]λ ∩ yβ  λ is finite 2 for β = β  by Corollary 9.2.7 (iii). Using the above lemma, we prove further properties of G. Lemma 12.3.8. Let G be as in Construction 12.3.6 and define Gα (α < λ) by Gα := {g ∈ G |  g  < α,  g A < α}. Then: β ⊆ Gβ+1 for all β < λ; (a) G ∩ P (b) {Gα | α < λ} is a λ–filtration of G; and (c) if β < λ, α < λ are ordinals such that  φβ  = α, then Gα ⊆ Gβ . Note, we used the upper index (β < λ) for the chain obtained by construction while we use the lower index (α < λ) for the new λ–filtration taking care of the norm. β for some β < λ. Since G0 = B  ⊆ Proof. First we show (a). Let g ∈ G ∩ P β+1  we assume g ∈ G \ B . Then, by Lemma 12.3.7, G g ∈ B + yγk A γ∈N

for some finite N ⊆ λ, k < ω such that [ g ]λ ∩ [ yγ ]λ is infinite for γ ∈ N . β we also have Since g ∈ P  7 8  β . [ g ]λ ⊆ [ Pβ ]λ = P λ

If  g  <  Pβ  then N ⊆ β by Corollary 9.2.7 (ii) and thus g ∈ Gβ ⊆ Gβ+1 .

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Otherwise, if  g  =  Pβ  (∈ λo ) then  g  =  yγ∗  =  φγ∗  for γ∗ = max N and [ g ]λ ∩ [ yγ∗ ]λ ⊆ [ φβ ]λ ∩ [ φγ∗ ]λ is infinite. Hence β = γ∗ by condition (iii) of Corollary 9.2.7 and so g ∈ Gβ+1 as required. Condition (b) is obvious. To see (c) let α, β < λ with  φβ  = α and let g ∈ Gα . If g ∈ B  we are finished. Otherwise, by Lemma 12.3.7, we have   g ∈ B + yγk A N ⊆ λ finite, k < ω γ∈N

with [ g ]λ ∩ [ yγ ]λ is infinite for γ ∈ N . This implies  φγ  =  yγ  ≤  g  < α =  φβ  for all γ ∈ N and thus N ⊆ β by Corollary 9.2.7 (ii), i.e. g ∈ Gβ , which finishes the proof. 2 Finally, we are ready to prove the main theorem of this subsection, i.e. the realization theorem.

Proof of Theorem 12.3.4. Let G be the A–module as constructed in Construction 12.3.6. We already know that G is an S–cotorsion–free R–module of cardinality λ. It remains to show EndR G = A. Obviously, A ⊆ EndR G (compare Section 1.1). Conversely, suppose there exists ψ ∈ EndR G \ A. Let ψ  = ψ  B, then ψ  ∈ / A, since ψ is uniquely determined by ψ  (B ⊆∗ G ⊆∗  Let B). I = {(αn , εn ) | n < ω} ⊆ λ × ρ such that α0 < α1 < . . . < αn < . . . and Iλ ∩ [ g ]λ is finite for all g ∈ G. Note, the existence of I can be easily arranged, e.g. let E λo , α ∈ λo \ E, εn ∈ ρ (n < ω) arbitrary and (αn )n<ω any ladder on α.  such that By the Step Lemma 12.3.5 there exists an element y of B yψ ∈ / G, yA∗ = G .

12.3

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435

By the Strong Black Box (Corollary 9.2.7) the set   E  = α ∈ E | ∃ β < λ :  φβ  = α, φβ ⊆ ψ  ⊆ ψ, [ y ] ⊆ [ φβ ]  φβ . is stationary, since |[ y ]| ≤ ℵ0 ≤ κ. Note, [ y ] ⊆ [ φβ ] implies y ∈ Dom Moreover, let C = {α < λ | Gα ψ ⊆ Gα }. Then C is a cub, since {Gα | α < λ} is a λ–filtration of G by Lemma 12.3.8 (b). Now let α ∈ E  ∩ C (= ∅). Then Gα ψ ⊆ Gα and there exists an ordinal β < λ such that  φβ .  φβ  = α, φβ ⊆ ψ and y ∈ Dom The first property implies Gα ⊆ Gβ by Lemma 12.3.8 (c) and the latter properties / A. imply φβ ∈ Moreover, Dom φβ ⊆ B with  Dom φβ A ≤  Dom φβ  = α and hence Dom φβ , and so also (Dom φβ )ψ are contained in Gα ⊆ Gβ . Therefore φβ : Dom φβ −→ Gβ with φβ ∈ / A and thus, on the one hand, it follows from the Construction 12.3.6 / Gβ+1 . that yβ φβ ∈ On the other hand, it follows from Lemma 12.3.8 (a) that yβ φβ = yβ ψ ∈ β ⊆ Gβ+1 – a contradiction. G∩P So we have shown that no such ψ exists and this means EndR G = A as required. 2 We would like to mention that one can also show, using standard arguments, that G is an ℵ1 –free A–module (cf. Observation 12.3.39 and Corollary 12.3.42). We finish this section with pointing out that the constructions and proofs in this sectioncan be simplified for |A|≤ κ. In this case we may work directly with B = α<λ eα A and with P = α∈I eα A as canonical summand provided that I satisfies |I| ≤ κ, (I ∩ μ)gα = I ∩ α (α ∈ I) and  I  =  P  ∈ λo (cf. Definition 9.2.1). The definition of the equivalence relation on the set C of all canonical homomorphisms has to be adjusted: φ, φ are of the same type, if [ φ ] ∩ μ = [ φ ] ∩ μ and there is an order–isomorphism f : [ φ ] → [ φ ] such that (aeα )φf = (aeα )fφ = (aeαf )φ for all a ∈ A, α ∈ [ φ ] (see Definition 9.2.3). All other adjustments are obvious. Note, the simplifications we can achieve in this way are due to the fact that the support function maps into λ rather than into λ × ρ. In fact, for |A| ≤ κ, we only need to assume that A is S–cotorsion–free, i.e. no “ρ”, respectively “F ”, is needed here.

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The main realization theorem and the General Black Box In order to derive simultaneously the realization theorem for different classes of modules (cotorsion–free, almost cotorsion–free or separable modules, respectively), we need some unifying notion depending on these cases. The reader who is only interested in realizing S–cotorsion–free algebras and who does not care so much for the size of the underlying modules can return to the quicker proof based on applications of the Strong Black Box in Theorem 12.3.4 or restrict his attention to the condition concerning S–cotorsion–free modules. Again, let R be an S–ring, which is a commutative ring with S as before such that R is S–torsion–free and S–reduced. Definition 12.3.9. We will say that an S–torsion–free R–algebra A is representable, if the following holds. (i) In the cotorsion–free case we assume that R A is S–cotorsion–free (for S as before). (ii) In the almost cotorsion–free case we assume that R A is almost S–cotorsion– free (see also Definition 12.3.1). (iii) In the separable case we  assume that there is a free R–basis {eα : α < κ} ∩  of A such that A := A α<κ Reα is an A–submodule of A. The last perhaps somewhat mysterious condition ensures that we have a supply of S–separable A–modules for constructing the basic module needed for the application of the Black Box. Fortunately, this condition holds for any countable R–algebra A with R A a free R–module over any principal ideal domain as we will show at the end of this chapter in Corollary 12.4.2. The result, which is of independent interest, comes from a recent paper (Corner, Göbel [104]): Theorem 12.3.10. If A is a countably generated, free R–algebra over a principal ideal domain, then we can find a basis of R A such that all scalar multiplications by elements in A can be represented by matrices with entries from R which are row–and–column–finite. We begin the construction of a module of cardinality λℵ0 = λκ for any representable R–algebra A of cardinality ≤ λκ and |R| ≤ κ; choose μ < λℵ0 to be a regular uncountable cardinal for the norm. As in the definition of norms in Chapter 1 let T = Tμ × Tλ × Tκ = Tμ×λ×κ be a tree of cardinality λ, with Tμ as norm tree and Tκ as the nursery. Hence nodes of T are triples τ = (τμ , τλ , τκ )

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Realizing all cotorsion–free algebras

with τμ ∈ Tμ , τλ ∈ Tλ and τκ ∈ Tκ and we let

 B= τ A ⊆ B := B

437

(12.3.3)

τ ∈T

in the cotorsion–free case as well as in the almost cotorsion–free case. Clearly B  is an S–pure and dense A–submodule of the A–module B. In the separable case   we apply our hypothesis that A = A ∩ α<κ eα R is an A–module and consider  

∩ ∩ τ A ⊆ B := B τA = B Reτ,eα , B := τ ∈T

τ ∈T

τ ∈T,α<κ

where eτ,eα is the basis element of τ A corresponding to eα in A. Again B is an S–pure and dense A–submodule of the A–module B. Now we are ready to define an ideal of EndR G for any module G between B and B depending on the three classes of modules under consideration. Definition 12.3.11. Suppose G is an R–module over the S–ring R with B ⊆ G ⊆ B. Any endomorphism ϕ : G → G which has an extension ϕ  : B → G is called inessential (for S). The set Ines G = {ϕ ∈ EndR G, ϕ is inessential} is a two–sided ideal of EndR G. More generally, if also B ⊆ G ⊆ B, then let  ⊆ G}. Ines(G , G) = {ϕ ∈ HomR (G , G) : B ϕ Note that any homomorphism is continuous in the S–topology. Thus by density any endomorphism ϕ of G has at most one extension ϕ  and in case of S–cotorsion– free or almost S–cotorsion–free modules the extension ϕ  : B −→ B also exists by completeness, but the image may not be in G. Also note that Ines(G , G) does not depend on G , if we identify any homomorphism in HomR (G , G) with its unique  Hence we often write extension to B. Ines(G , G) = Ines G. Now we can state our Main Theorem 12.3.12. Let A be a representable R–algebra of cardinality ≤ λκ with |R| ≤ κ, where λ is any cardinal such that λκ = λℵ0 . For any subset X ⊂ λℵ0 we can find an R–module GX of cardinality λℵ0 such that B ⊆ GX ⊆ B as above

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and for any X, X  ⊆ λℵ0 the following holds.  A ⊕ Ines GX , if HomR (GX  , GX ) ∼ = Ines GX , if

X ⊆ X X  ⊆ X

Moreover, GX  ⊆ GX ⊆ G if X  ⊆ X and G = Gλℵ0 . Remark 12.3.13. The algebra A ⊕ Ines GX on the right is a split extension, thus Ines GX is a two–sided ideal and the quotient algebra A can be identified with a complement. The result can also be strengthened as in Theorem 12.1.1 such that the isomorphism between the topological algebras also becomes a homeomorphism (see [102]). For clarity we postpone the discussion of mixed or torsion ℵ modules (cf. pp. 457 f.). If we choose a family Xα (α ∈ 2λ 0 ) of pairwise incomparable subsets Xα ⊆ λℵ0 and apply the Main Theorem 12.3.12, then the following is immediate. Corollary 12.3.14. Under the hypothesis of Main Theorem 12.3.12 there is a rigid ℵ family of R–modules Gα (α ∈ 2λ 0 ) of cardinality λℵ0 such that Hom(Gα , Gβ ) = Aδαβ ⊕ Ines Gβ . The proof is divided into three parts, the construction of the modules GX , properties of GX , followed by establishing the displayed equalities of the Main Theorem 12.3.12. The setting. We choose the tree T = Tμ×λ×κ, the modules B ⊆ B from above and the coding set Γ = λℵ0 + 1 (the ordinal sum, hence λℵ0 ∈ Γ). Next we want to code the tree structure (its branches) into the module and will find for each branch v ∈ Br(T ) an associated branch element in B with support [v]. Because [v] is infinite we must apply topology. As before let S = {si | i < ω} be an enumeration of S  with s0 = 1. Moreover, let R be an S–ring (Definition 1.1.1) and recall that qn = i≤n si . Definition 12.3.15. (Divisibility sequence of branch elements) Given a branch v ∈ Br(T ), then we can write the linearly ordered set v = {vn = v  n ∈ T | n < ω} with l(vn ) = n for all n < ω and let vk =

σ∈v,l(σ)≥k

(ql(σ) /qk )σ.

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We will also write v = v 0 , if it is clear that v ∈ B and call {v k | k < ω} a divisibility sequence. The name divisibility sequence should suggest that v k+1 sk+1 ≡ v k

mod B for all k < ω,

(see also the next observation). The following observation shows that multiples of branch elements can be recognized.

Observation 12.3.16. (a) A sequence v k ∈ B (k < ω) of a branch element v is a divisibility chain, that is v k − (qk+1 /qk )v k+1 ∈ B for all k < ω. (b) If a ∈ A, then [va] ⊆ [v]. (c) Moreover, for any k < ω we have a = 0 ⇔ v k a = 0 ⇔ [v k a] is finite ⇔ v \ [v k a] = ∅.

Proof. The first statement by Definition 12.3.15.  follows immediately ql(σ) σ, then va = ql(σ) σa by continuity of multiIf a ∈ A and v = σ∈v

σ∈v

plication by a and [va] ⊆ [v] is obvious. The implications from left to right in (c) are trivial, hence it remains to show that the last implication gives a = 0. If v \ [v k a] = ∅, then choose any σ ∈ v \ [v k a]. Hence (ql(σ) /qk )a = 0 and a = 0 follows from S–torsion–freeness. 2

The construction of the module G We continue constructing the modules in Theorem 12.3.12. By the General Black Box 9.2.27 there is a transfinite sequence pα = (fα , ϕα , cα ) (α < λ∗ ) of traps satisfying the conclusion of the General Black Box 9.2.27. As a technical device let ∞ denote a fixed element not in B. Let ξ < λ∗ and assume that we have found an ascending chain of pure A–submodules Gα (α < ξ) of B and elements bβ (β + 1 < ξ) of B ∪ {∞} such that, for all α < ξ,

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(I0 ) G0 = B. / Gα for all β < α. (Iα ) bβ ∈ (Iξ ) Gξ =

 α<ξ

Gα , if ξ is a limit ordinal.

If ξ = α + 1 is a successor ordinal, we distinguish two cases: (1) Suppose that Pα := Dom ϕα is the S–pure submodule Xα ∗ ⊆ B, (Xα ⊆ T of cardinality ≤ κ). Then Pα is called a canonical submodule of B and  Pα = τ A ∩ B. τ ∈Xα

Also suppose it is possible to choose a branch vα from the tree fα × Tκ and elements gα0 , gα1 ∈ Pα and Gα+1 , bα in such a way that (Iα+1 ) and each of the following conditions is satisfied: (IIα+1 )

Gα+1 = Gα + gα A∗ , where gα = gα0 + gα1 ;

(IIIα+1 )

||gα0 || < ||gα1 ||, [gα1 ] = vα and [gα1 a] is infinite for all 0 = a ∈ A;

(IVα+1 )

either (the strong case) bα = gα ϕα or (the weak case) bα = ∞.

We use the strong case, whenever possible, and in this case we call α strong. Otherwise α will be called weak. (2) If (1) is not satisfied, we call α useless and take Gα+1 = Gα , gα = 0, bα = ∞, so (IIα+1 ) holds in this case as well. (We will however show, that this case never arises). In every case (Iξ ) is obviously satisfied. The recursion therefore proceeds for all ξ < λ∗ and gives rise to the S–pure A–submodule    Gα = B + gα A ∗ of B. G= α<λ∗

α<λ∗

The algebra A acts faithfully on the A–module G by scalar multiplication and we can view A ⊆ End G naturally. In the case of S–separable modules, being an S–pure submodule of a product of copies of R shows immediately that G is an S–separable R–module. It needs work to show that G is S–cotorsion–free in the first case and almost S–cotorsion–free in the second case.

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Properties of the module G For this we also define the submodules GX of G for any X ⊆ λℵ0 . For any subset X ⊆ λℵ0 we use the elements cα from the sequence of traps pα = (fα , ϕα , cα ) and the coding set Γ = λℵ0 + 1 to define X  = {α < λ∗ : cα ∈ X} and the module GX =



Gα ⊆ G.

α∈X 

Using that B/B is divisible, we have an immediate Observation 12.3.17. If x ∈ B there is a sequence xk ∈ B (k < ω) such that for all n ≤ k, xk − (qk+1 /qk )xk+1 ∈ B and [xk ] ⊆ [xn ]. Note, that we have seen a special case of Observation 12.3.17 in Definition 12.3.15 and Observation 12.3.16. Definition 12.3.18. For a subset U of T and an ordinal ν < μ we define the part of U to the right of ν to be νU

= {τ ∈ U : ν ≤  τ }.

Recognition Lemma 12.3.19. Let g ∈ GX \ B be as above. (a) There is a unique α ∈ X  such that g ∈ Gα+1 \ Gα . (b) With α as in (a), there exist a strictly decreasing sequence of ordinals α = α0 > · · · > αr in X  with  Pαi  =  Pα  for i ≤ r and ν <  Pα  such that  ν [g] = F ∪ ν [vαi ] (disjoint union), i≤r

where F is a finite set of elements from T each of norm greater than  Pα . (c) Moreover, for each β < λ∗ with  Pβ  =  Pα  there exist a ∈ A, k < ω such that for almost all σ ∈ vβ , g  σ = σ(ql(σ) /qk a).

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Proof. (a) is immediate because the Gα (α < λ∗ ) constitute a continuous ascending chain.  k gαi ai , where b ∈ B, (b), (c) We may clearly write g = b + x + i≤r

 x  <  Pα  =  gαk i ai 

(i ≤ r).

We can absorb into x all terms of norm less than  Pα ; and since cf( Pα ) = ω, there is no term in b of norm  Pα . Then by Observation 12.3.16 any large enough / {α0 , . . . , αr }. ν <  Pα  will do for (b), and (c) follows at once, with a = 0 , β ∈ 2 We have at once from the Recognition Lemma 12.3.19 the Corollary 12.3.20. For any g ∈ GX , we have that g ∈ B, if and only if [g] is finite. The following lemma plays a crucial role in the closing stages of the proof of our main theorem. This lemma is not needed, if we work with the Strong Black Box 9.2.2. Compare the proof in Section 12.3. Recall the construction of G! Lemma 12.3.21. Let α < λ∗ , and write gvk = xk + v k with  xk  <  v  =  Pα  (as in the construction of G) and let Gα+1 (v) = Gα +



gvk A

k<ω

for all v ∈ Br(fα × Tκ ). Then there exists v ∈ Br(fα × Tκ ) such that bβ ∈ / Gα+1 (v) (β < α).

Proof. Suppose that the conclusion is false. Then for each v ∈ Br(fα × Tκ ) there exists β = β(v) < α such that bβ ∈ Gα+1 (v). Clearly bβ = ∞; so by (IVβ ) in the construction we have bβ = gβ ϕβ ∈ P β , and there exists a = av ∈ A, k = k(v) < ω with bβ − gvk a ∈ Gα ;

(12.3.4)

since bβ ∈ / Gα by (Iα ) certainly gvk a = 0. But then it follows by the construction of gvk that ν [gvk a] = ν [v k a] for some large enough ν <  Pα . Thus ν [gvk a] is an infinite subset of v. But the branches vγ (γ < α) are all of norm at most

12.3

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Realizing all cotorsion–free algebras

 Pα  =  v  and they are all distinct from v, by condition (ii) of the General Black Box 9.2.27. Hence the Recognition Lemma 12.3.19 (b) applied to (12.3.4) implies that an infinite subset of v lies in [bβ ] ⊆ [Pβ ]; and as [Pβ ] is a subtree of T the whole of v is contained in [Pβ ]. Thus v ∈ Br(fα × Tκ ∩ [Pβ ]) which forces β < α < β + κℵ0 by (iii) of the General Black Box 9.2.27. Thus we have proved that for each v ∈ Br(fα × Tκ ) there exist an ordinal β(v) and av ∈ A such that β(v) < α < β(v) + κℵ0 and bβ(v) − gvk(v) av ∈ Gα .

(12.3.5)

Let β0 be the least ordinal with β0 < α < β0 + κℵ0 , so that β(v) assumes fewer than κℵ0 = | Br(fα × Tk )| values. By a pigeon–hole argument there are two distinct branches v, w ∈ Br(fα × Tk ) with β(v) = β(w) = β (say). Subtracting the corresponding equations (12.3.5) gives k(w) − gvk(v) ∈ Gα . gw k(w)

Arguing as before we conclude that an infinite subset of v must lie in ν [gw w; and this is impossible because v, w are distinct branches.

aw ] ⊆ 2

The next corollary is now immediate. Corollary 12.3.22. There are no useless ordinals. An ordinal α < λ∗ is strong or weak as gα ϕα lies outside or in G, respectively. Calculating End G We choose a constant branch f ∈ Br(Tμ×λ ) such that f  n does not depend on n. If we take any infinite branch w from the tree f × Tκ and the corresponding ele/ G by support (see the Recognition ment w ∈ B, then obviously [w] = w and w ∈ Lemma 12.3.19). The same holds for any wa (0 = a ∈ A) by continuity of multiplication. Thus the first half of our next proposition follows from the definition of inessential homomorphisms. The second half is nothing but unique extensions of  homomorphisms GX −→ GX  to B. Proposition 12.3.23. (a) A ∩ Ines G = {0}.  | GX ϕ ⊆ GX  } for any X, X  ⊆ λℵ0 . (b) Hom(GX , GX  ) = {ϕ ∈ End B For the moment we will need the following notation. Let Δ = {(a, s) ∈ A × S : s = 1 or a ∈ / sA}.

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 \ (A ⊕ Ines GX ). Then there exists a canonical Lemma 12.3.24. Let ϕ ∈ End B submodule P of B such that P (sϕ − a) ⊆ GX for all (a, s) ∈ Δ.

Proof. We first prove that there exists a canonical submodule P such that P (ϕ − a) ⊆ GX for all a ∈ A.

(12.3.6)

To see this, consider any canonical submodule P containing an element w ∈ P with constant branch [w] = w ∈ T as above. Suppose we are unlucky and that P does not satisfy (12.3.6). Then there exists an element a ∈ A with P (ϕ − a) ⊆ GX .

(12.3.7)

But ϕ − a ∈ / Ines GX , so there is an element x ∈ B such that x(ϕ − a) ∈ / GX .

(12.3.8) 

Take a canonical submodule P  such that P ⊆ P  and x ∈ P .  We claim that P (ϕ − a) ⊆ GX for all a ∈ A. For otherwise we can find a ∈ A such that 

P (ϕ − a ) ⊆ GX . Subtracting (12.3.7) we get P (a − a ) ⊆ GX , that is a − a = 0 from the constant branch w ∈ P . Therefore x(a − a ) = 0 and the last displayed expression forces x(ϕ − a) ∈ GX , contrary to (12.3.8). 2 Next we want to show a lemma, which needs more work.  \ (A ⊕ Ines G). Then there exists x ∈ B such Lemma 12.3.25. Let ϕ ∈ End B that xϕ ∈ / GX , xA∗ .

Proof. Choose a canonical submodule as in Lemma 12.3.24 and pick an ordinal η < λ such that max{ P ,  P ϕ } <  η .

(12.3.9)

This is possible because [P ] and [P ϕ] are of cardinality at most κ < cf(μ) = μ. Consider the constant branch w = w(η). If we are lucky, then w = x ∈ B

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445

will answer. Suppose we are unlucky and that wϕ ∈ GX , wA∗ . Then there are a ∈ A, s ∈ S and a least n < ω such that wsϕ − wn a ∈ GX .

(12.3.10)

/ Δ, then qn = 1, a ∈ qn A and there is a ∈ A such that a = qn a , Now if (a, qn ) ∈ whence wsϕ − wa ∈ GX and w = w0 , contrary to the minimality of n. We come now to the crux of the proof, viz. the fact that (a, qn ) ∈ Δ. Lemma 12.3.24 provides an element z ∈ P such that / GX . z(qn sϕ − a) ∈

(12.3.11)

We claim that w + z answers the lemma. For if not, then for some n ≥ n, a ∈ A we have   (w + z)sϕ − (wn + z n )a ∈ GX , that is, subtracting (12.3.10), 



(zsϕ − z n a ) + (wn a − wn a ) ∈ GX .

(12.3.12)



But wn − (qn /qn )wn ∈ B ⊆ GX , so (12.3.12) is equivalent to 



(zsϕ − z n a ) + wn ((qn /qn )a − a ) ∈ GX .

(12.3.13)

Here the first term has norm less than  η , and the support of the second term is contained in a constant branch w of norm  η . This forces that this support must be finite (see the Recognition Lemma 12.3.19). This is only possible if  (qn /qn )a = a . Then (12.3.13) reduces to zsϕ − z n (qn /qn )a ∈ GX . Multiplying by qn we obtain z(qn sϕ − a) ∈ GX , contrary to (12.3.11). The lemma is proved. 2

Proof of the Main Theorem 12.3.12 We are now in the position to prove that End GX is a split extension of the subalgebra A by the ideal Ines GX and more as indicated in the Main Theorem 12.3.12.

Proof of the Main Theorem 12.3.12. Recall from Proposition 12.3.23 that the ideal Ines GX and the submodule A ⊆ End GX , scalar multiplication by a ∈ A are disjoint. Slightly more general, again considering the action on a constant branch we have A ⊕ Ines(GX ) ⊆ Hom(GX  , GX ), if X  ⊆ X.

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It will therefore be enough to prove that, Hom(GX  , GX ) ⊆ A ⊕ Ines GX

(12.3.14)

Hom(GX  , GX ) ⊆ Ines GX ⇒ X  ⊆ X.

(12.3.15)

and

We prove first that (12.3.14) implies (12.3.15). Assume (12.3.14), and suppose that ϕ ∈ Hom(GX  , GX ) \ Ines(GX ). Then ϕ = a + θ for some θ ∈ Ines GX and 0 = a ∈ A. Consider any x ∈ X  . The General Black Box 9.2.27 provides an α < λ∗ with pα = (fα , ϕα , x), where Γ = λℵ0 + 1. Therefore α ∈ X  ; and α is of course strong or weak by Corollary 12.3.22. Since θ ∈ Ines GX there is an s ∈ S such that sBθ ⊆ GX . Then the element z = sgα lies in sP α ∩ GX ; so it is mapped into GX by both θ and ϕ. Therefore za ∈ GX , because a = ϕ − θ. But a = 0, so [za] contains an infinite subset of vα , where  za  =  vα . Therefore za ∈ GX ∩ (Gα+1 \ Gα ) by the Recognition Lemma 12.3.19. This forces α ∈ X  , whence x ∈ X; and we have deduced that X  ⊆ X, as required in (12.3.15). It remains to prove (12.3.14). Now suppose for contradiction that there exists a homomorphism ϕ ∈ Hom(GX  , GX ) \ A ⊕ Ines GX .  Applying Lemma 12.3.25 Surely we may view ϕ as an endomorphism of B. we get x ∈ B such that xϕ ∈ / G, xA∗ .

(12.3.16)

The General Black Box 9.2.27 now provides an α < λ∗ such that pα = (fα , ϕα , x), x, xϕ ∈ P α , max{ x ,  xϕ } <  Pα , and ϕα = ϕ  P α .

(12.3.17)

The theorem will therefore be proved once we have shown that α is strong, and for this we shall appeal to Lemma 12.3.21. Consider any v ∈ Br(fα × Tκ ). We claim that there exists ε = εv ∈ {0, 1} such that (v + εx)ϕ ∈ / Gα , v + εxA∗ . For if not, there exist s ∈ S and a0 , a1 ∈ A with (v + εx)sϕ − (v n + εxn )aε ∈ Gα (ε ∈ {0, 1}),

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447

and the usual subtraction gives (xsϕ − xn a1 ) + v n (a0 − a1 ) ∈ Gα .

(12.3.18)

By (12.3.17) the first bracket has norm less than  Pα  =  v , so the Recognition Lemma 12.3.19 forces v n (a0 − a1 ) to vanish, whence a0 = a1 . Bearing in mind (12.3.16), we proved that for each v ∈ Br(fα × Tκ ) there exists an gv = v + εv x satisfying the hypothesis of Lemma 12.3.21 and such that / Gα+1 (v). gv ϕα ∈ It follows from the Recognition Lemma 12.3.19 that for some v ∈ Br(fα × Tκ ) we / Gα+1 (v) (β < α). Therefore α is indeed strong and the theorem have also bβ ∈ is proved. 2

Cotorsion–free modules The characterization of cotorsion–free R–modules given by Proposition 12.3.2 will motivate to study S–cotorsion–free modules and their endomorphism algebras. We apply Definition 12.3.9 and begin with a Lemma 12.3.26. Let R be an S–ring and A be a representable R–algebra in the cotorsion–free case (hence R A is S–cotorsion–free). Then G in the Main Theorem 12.3.12 is S–cotorsion–free.

Proof. Suppose that the lemma does not hold, and consider a non–zero homo −→ G. Let g = 1ϕ. Then g ∈ G ⊆ B, so for all r ∈ R  we have morphism ϕ : R rϕ = r1ϕ = rg, by continuity. Clearly  [rg] ⊆ [g] (r ∈ R). Now any direct sum of S–cotorsion–free modules is visibly S–cotorsion–free: hence B is S–cotorsion–free. But if g ∈ B, then Corollary 12.3.20 and the last  B), and this is zero, we have a condisplayed inequality imply that ϕ ∈ Hom(R, tradiction. Therefore g ∈ / B, and by the Recognition Lemma 12.3.19 there is a unique α < λ∗ such that g ∈ Gα+1 \ Gα . Multiplying ϕ by an element in S we may assume that g − gα a ∈ Gα

(12.3.19)

 Then rg = rϕ ∈ G. But for some a ∈ A, and clearly gα a = 0. Fix r ∈ R. [rg] ⊆ [g] cannot contain infinitely many elements of any branch vβ with β > α,

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and the Recognition Lemma 12.3.19 forces rg ∈ Gα+1 . Therefore for some s ∈ S, a ∈ A we have s(rg) − gα a ∈ Gα

(12.3.20)

 Thus [gα (a − sra)] is From (12.3.19) and (12.3.20) we get gα (a − sra) ∈ RG. contained in a finite union of supports of elements in Gα , and by the Recognition Lemma 12.3.19 its intersection with vα is finite. Therefore for almost all σ ∈ vα the term gα (a − sra)  σ vanishes, in other words, canceling qlσ , we have  = sσA. σa = σsra ∈ σA ∩ sσ A 



Thus without loss of generality we may assume that a = sa , where a ∈ A, and  (12.3.20) reduces to rg − gα a ∈ Gα . We now obtain a non–zero homomorphism  −→ A by setting rψ = a . The contradiction proves the lemma. ψ:R 2 From Lemma 12.3.26 and our Main Theorem 12.3.12 we have an important Theorem 12.3.27. Let A be an R–algebra with R A S–cotorsion–free. Moreover, suppose that |A| ≤ κ and λℵ0 = λκ . Then there are S–cotorsion–free R– submodules GX of G = Gλℵ0 of cardinality λℵ0 for all X ⊆ λℵ0 such that the following holds.  A, if X ⊆ X  ∼ HomR (GX , GX  ) = 0, if X ⊆ X 

Proof. By Main Theorem 12.3.12 it remains to show that Ines G = 0, but this follows from Lemma 12.3.26 because G is S–cotorsion–free. Note that submodules of S–cotorsion–free modules are also S–cotorsion–free, whence all GX s which are submodules of G are S–cotorsion–free as claimed. 2

Almost cotorsion–free, separable, slender and ℵ1 –free modules Over a complete ring R cotorsion–freeness is obviously a too strong notion because {0} is the only cotorsion–free R–module. We are forced to weaken our condition, if we wish to say anything of interest in such a case. The main result in this section gives a condition under which the inessential endomorphisms are precisely the endomorphisms of finite rank. This is the case for a homomorphism ϕ, if rk(Im ϕ) is finite. For a more general notion see Definition 12.3.1. We will consider the case of almost S–cotorsion–free modules. As above, the main result depends on a

12.3

Realizing all cotorsion–free algebras

449

Lemma 12.3.28. An R–module G over the S–ring R is almost S–cotorsion–free,  −→ if and only if for every S–reduced R–module M every homomorphism ϕ : M G is of finite S–rank. The proof is an easy exercise using the Definition 12.3.1. Thus we define the following ideal of an endomorphism ring. Definition 12.3.29. If End G is the endomorphism algebra of an R–module G, then let Fin G = {ϕ ∈ End G : the S–rank rk(Gϕ) is finite }. Similarly we may define Fin(G, G ) for R–modules G, G . A closely related notion is that of slenderness, which we discussed in Chapter 1 (see Definition 1.3.5):  Rei be the direct product of copies of If I is an indexing set, then let FI∗ = i∈I

the R–module R. An R–module G is slender, if every homomorphism Fω∗ −→ G maps almost all the ei to 0. However in this definition of slender modules Fω∗ can 6ω of Fω∗ , where be replaced by an S–pure submodule F   6I = rn ei(n) : (rn ) any null sequence in R, (i(n))n<ω ∈ I ω , F n<ω

6ω ⊆ Fω∗ and there is a 6I of  Rei in F ∗ . Hence F which is the S–adic closure of F I i∈I

homomorphism 6ω Fω∗ −→ F

 n<ω

rn en −→



 rn qn en .

n<ω

So it is clear that slender modules can be defined in two different ways as above, 6ω ⊆ Fω∗ the following and in many other ways as illustrated in Chapter 1. Since F proposition is also clear. Proposition 12.3.30. Every slender module is almost S–cotorsion–free. The main result of this section is a consequence of the following theorem which is of some independent interest and which will be proved after a couple of lemmas. Theorem 12.3.31. Let A and G be as in the Main Theorem 12.3.12. (a) If R A is almost S–cotorsion–free, then so is G. (b) If R A is slender, then so is G.

450

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Realizing algebras

First we must refine the notion of support. Definition 12.3.32. For each g ∈ G (as in the theorem) we define the top [g]∗ of the support of g to be the set  {σ ∈ T |  σ  =  g  and g  σ = 0} ( g  discrete) ∗ [g] = {α < λ∗ :  vα  =  g , g  σ = 0 for almost all σ ∈ vα } ( g  a limit ). It follows from the Recognition Lemma 12.3.19 that [g]∗ is always finite, and that [g]∗ = ∅ ⇐⇒ g = 0. 6I −→ G be a homomorphism with G as in the theorem. Lemma 12.3.33. Let ϕ : F Then (a) { ei ϕ  | i ∈ I} is finite and  [ei ϕ]∗ is finite. (b) i∈I

Proof. Part (a) follows at once from (b) and the Definition 12.3.32. Suppose that (b) does not hold. Then there exists a sequence (in ) of distinct elements of I such that the set  [fk ϕ]∗ Xn = [fn ϕ]∗ \ k
is non–empty, where fk = eik . Passing to a subsequence we may assume that the  fn ϕ  are either (i) all discrete or (ii) all limits, and that the sequence ( fn ϕ ) is non–decreasing.

(12.3.21)

(i) When the ( fn ϕ ) are all discrete, choose σn ∈ Xn ; then Definition 12.3.32 and (12.3.21) imply that sup  σn  = sup  fn ϕ ,

(12.3.22)

fn ϕ  σn = 0,

(12.3.23)

fk ϕ  σn = 0

(k < n).

(12.3.24)

(ii) When the ( fn ϕ ) are all limits, choose α(n) ∈ Xn ; then it follows from the definition of Xn that the ordinals α(n) are all distinct. Therefore the branches vα(n) are distinct. We may choose elements σn ∈ vα(n) to satisfy (12.3.22),

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Realizing all cotorsion–free algebras

(12.3.23) and (12.3.24). Passing again to a subsequence we may impose the normalization that, if σn ∈ [vβ ] for infinitely many n < ω, then they all do so.

(12.3.25)

Lastly, we choose an increasing sequence of natural numbers m(n) such that for all n < ω follows ql(σn ) qn | qm(n) and qm(n) fn ϕ  σn ≡ 0 mod qm(n+1) σn A

(12.3.26)

by virtue of (12.3.23) this is certainly possible.  6I because the ik Now we consider the element f = qm(k) fk . This lies in F k<ω

are distinct, and we may clearly write qm(k) fk + qm(n+1) f n , f= k≤n

6I . Therefore by (12.3.24), mod qm(n+1) σn A where also f n ∈ F f ϕ  σn = qm(n) fn ϕ  σn + qm(n+1) f n ϕ  σn ≡ qm(n) fn ϕ  σn ; for f ϕ, f n ϕ ∈ G, and σn A is S–pure in its completion. Thus (12.3.26) gives (12.3.27) f ϕ  σn ≡ qm(n) fn ϕ  σn ≡ 0 mod qm(n+1) σn A.  Therefore [f ϕ] contains all the σn , and since f ϕ = qm(k) (fk ϕ) by continuity, k<ω

(12.3.22) implies that  f ϕ  = sup  σn . The Recognition Lemma 12.3.19 now tells us that almost all the σn lie in a finite union of sets [vβ ], and by a pigeon– hole argument our normalization (12.3.25) forces the σn to lie in a single [vβ ], say v = vβ . Applying again the Recognition Lemma 12.3.19 (c) we obtain k < ω and a ∈ A with the property that, for all large n < ω, f ϕ  σn = σn (ql(σn ) /qk )a. But it follows from (12.3.26) that f ϕ  σn ∈ ql(σn ) qn σn A. Composing this with the last displayed equality shows that for all large n < ω we have (ql(σn ) /qk )a ∈ qn A = 0. So a = 0 and the last displayed equality ql(σn ) qn A, hence a ∈ n<ω

reduces to f ϕ  σn = 0, contrary to (12.3.27). The proof of (b) is complete.

2

6I (or F I ) is an endomorphism δ which mulDefinition 12.3.34. A dilatation of F tiplies each basis element ei by an element si ∈ S : ei δ = si ei (i ∈ I). The dilatations form a semigroup.

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Realizing algebras

6ω −→ G is a homomorphism (G as in the theorem), then Lemma 12.3.35. If ϕ : F there exist a dilatations δ and finite subsets X ⊆ T, Y ⊆ λ∗ such that



6ω δϕ ⊆ τA ⊕ gβ A. F τ ∈X

β∈Y

ω in place of F 6ω . The same is true with F 6ω −→ G is the Proof. Let D(X, Y ) be the direct sum in the lemma and if ϕ : F given homomorphism, then by Lemma 12.3.33 (b) the set  [en ϕ]∗ is finite Z= n<ω

and by Lemma 12.3.33 (a) we may write μ = max  en ϕ . We surely may assume that ϕ = 0, thus μ > 0. It will be enough to find a dilatation δ and a homomorphism 6ω −→ D(X0 , Y0 ), ψ:F where X0 , Y0 are finite subsets of T, λ∗ with the property that  en (δϕ − ψ)  < μ for all n < ω: For then max  en (δϕ − ψ)  < μ and we may assume inductively that there 6ω δ  (δϕ − ψ) ⊆ exist a dilatation δ  and finite subsets X, Y of T, λ∗ such that F D(X, Y ), so that 6ω δ  (δϕ − ψ) + F 6ω ψ ⊆ D(X0 ∪ X, Y0 ∪ Y ). 6ω δ  δϕ ⊆ F F (i) Suppose first that μ is discrete. Put X0 = {σ ∈ Z :  σ  = μ}, Y0 = ∅, and 6ω define for each f ∈ F f ϕ  σ. fψ = σ∈X0

6ω −→ D(X0 , Y0 ) is a homomorphism, and by the definition of X0 , Then ψ : F en (ϕ − ψ) is the sum of all terms of norm not more than μ − 1 in en ϕ. Therefore  en (ϕ − ψ)  < μ (n < ω), and the proof is complete in this case, with δ = id. (ii) Suppose that μ is a limit. Put X0 = ∅, Y0 = {β ∈ Z :  vβ  = μ}. Composing ϕ with a suitable dilatation we may suppose that for all k < ω, gα∗ A, where gα∗ = qnα gα . ek ϕ ∈ B + α<λ∗

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453

By the Recognition Lemma 12.3.19 (c), for each β ∈ Y0 and each k < ω there exists ak (β) ∈ A with the property that, for almost all σ ∈ vβ , ek ϕ  σ = σql(σ) ak (β).

(12.3.28)

Since Y0 is finite, we may choose a strictly increasing sequence of natural numbers, m(n) such that 2 | qm(n+1) , and (12.3.28) holds, qm(n)

(12.3.29)

if β ∈ Y0 , σ ∈ vβ , and l(σ) ≥ m(k). Take δ to be the dilatation ek δ = qm(k) ek (k < ω).

(12.3.30)

 6ω , where (rk ) is a null sequence in R. Now consider f = rk ek ∈ F  6ω . The purity of σA in its completion By continuity f δ = rk qm(k) ek ∈ F implies that for β ∈ Y0 , σ ∈ vβ , and l(σ) ≥ m(k) we have mod qm(n+1) σA f δϕ  σ ≡



rk qm(k) ek ϕ  σ = ql(σ) σ

k≤n



rk qm(k) ak (β),

(12.3.31)

k≤n

where we have used (12.3.28). But f δϕ ∈ G, and  f δϕ  ≤  vβ  by our choice of μ and Y0 . Therefore the Recognition Lemma 12.3.19 provides an s ∈ S and an a(β) ∈ A such that, for almost all σ ∈ vβ , sf δϕ  σ = ql(σ) σa(β).

(12.3.32)

Taking σ ∈ vβ with l(σ) = m(n), we then find from (12.3.31), (12.3.32) and (12.3.29) that for large n, 2 qm(n) (a(β) − s rk qm(k) ak (β)) ∈ qm(n+1) A ⊆ qm(n) A, k≤n

whence a(β) − s



rk qm(k) ak (β) ∈ qm(n) A.

k≤n

Letting n go to infinity, we find from purity that a(β) ∈ sA, and a(β) can be replaced by an element in sA; so we may assume that s = 1. The last sum implies that gβ∗ a(β) − gβ∗ rk qm(k) ak (β) ∈ qm(n) gβ∗ A, k≤n

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Realizing algebras

and again on letting n −→ ∞ we get gβ∗ rk qm(k) ak (β) = gβ∗ a(β) ∈ gβ∗ A. k<ω

Thus we obtain a well–defined homomorphism    6ω −→ D(X0 , Y0 ) ψ:F rk ek → gβ∗ rk qm(k) ak (β) . β∈Y0

Then for each k < ω we have ek ψ = β ∈ Y0 and almost all σ ∈ vβ ,



β∈Y0

k<ω

gβ∗ qm(k) ak (β). Therefore for each

ek ψ  σ = σql(σ) qm(k) ak (β) = ek δϕ  σ. Hence ek (δϕ − ψ) is a sum of terms whose norms are bounded below μ; the proof is complete in case (ii). The proof for Fω is essentially the same up to the end of (i). And in (ii) one  one has (rk ek )ϕ = rk (ek ϕ) by continuity, and has only to note that for rk ∈ R σrk ak (β) ∈ σA by our argument using support and purity. So one may again argue as for F. 2

Proof of Theorem 12.3.31. (a) Consider any homomorphism ϕ : Fω −→ G. Then by Lemma 12.3.35 there exist a dilatation δ and finite subsets X, Y of T, λ∗ such that Fω δϕ ⊆ D(X, Y ). But D(X, Y ) is visibly almost S–cotorsion–free, so δϕ is of finite S–rank, and it follows from the definition of a dilatation that ϕ is also of finite S–rank. Hence G is almost S–cotorsion–free. (b) The proof is similar but even easier, using a homomorphism Fω −→ G. 2 Lemma 12.3.36. Let R be an S–ring either of cardinality less than 2ℵ0 or a complete discrete valuation ring. Let M, G be S–reduced, torsion–free R–modules  −→ G be a homomorphism of finite rank. Then M ϕ = M ϕ. and let ϕ : M

Proof. (i) Assume first that |R| < 2ℵ0 . It will be enough to prove that ϕ = 0.  is naturally Suppose for a contradiction that xϕ = 0 for some x ∈ M . Since M   an R–module, the mapping r → (rx)ϕ is a homomorphism ψ : R −→ G with 1ψ = 0. Thus Proposition 1.1.26 applies. We derive 2ℵ0 ≤ |M ϕ| contradicting finite rank. (ii) Assume that R is a complete discrete valuation ring. Since every S–pure submodule of finite rank in a torsion–free R–module is a direct summand, M ϕ is ϕ = M ϕ, free of finite rank, and is therefore complete. It follows at once that M as required. 2

12.3

455

Realizing all cotorsion–free algebras

Theorem 12.3.37. Let R be an S–ring and either of cardinality less than 2ℵ0 or a complete discrete valuation ring. If R A is almost S–cotorsion–free or slender, then G from Theorem 12.3.12 satisfies Ines G = Fin G.  with F Proof. If ϕ ∈ Ines G, choose a surjective homomorphism π : F −→ B  free. Then F πϕ ⊆ G. By Proposition 12.3.30 and Theorem 12.3.31 the module G is almost S–cotorsion–free. Therefore there exists a free summand F0 of finite  this means that Bϕ  is of rank of F such that F πϕ ⊆ S−1 (F0 πϕ). Since F π = B,  ⊆ Bϕ ⊆ G and the finite rank, so ϕ ∈ Fin G. Conversely, if ϕ ∈ Fin G, then Bϕ theorem is shown. 2 By Theorem 12.3.37 we are able to determine the endomorphism algebra in various cases for G from our Main Theorem 12.3.12. Corollary 12.3.38. Let R be an S–ring either of cardinality less than 2ℵ0 or a complete discrete valuation ring and let R A be almost S–cotorsion–free. If GX (X ⊆ λℵ0 ) are the almost S–cotorsion–free R–modules given by the Main Theorem 12.3.12, then we have  A ⊕ Fin GX , if X ⊆ X  ∼ HomR (GX  , GX ) = Fin GX , if X ⊆ X  Observation 12.3.39. Let A be an R–algebra and G be an R–module from Main Theorem 12.3.12 in the separable case (Definition 12.3.9 (iii)). (a) The R–module G is S–separable. (b) If R is noetherian and R A is free, then G is also an ℵ1 –free R–module.

Proof. Part (a) of the observation is obvious because G, by construction, is an S–pure submodule of a direct product Rκ , hence G is S–separable by definition. Part (b) follows from Theorem 1.3.8 and Definition 1.1.12. 2

Corollary 12.3.40. Let R be an S–cotorsion–free S–ring and A be a slender R– algebra in the separable case. Moreover, let |A| ≤ λ, |R| ≤ κ and λℵ0 = λκ . Then there are S–separable R–submodules GX of G = Gλℵ0 of cardinality λℵ0 for all X ⊆ λℵ0 such that the following holds. HomR (GX , GX  ) ∼ =



A ⊕ Fin(GX , GX  ), Fin(GX , GX  ),

if if

X ⊆ X X ⊆ X 

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Realizing algebras

In order to apply the last Corollary 12.3.40 for particular R–algebras we must verify a crucial condition on A (see Definition 12.3.9 (iii)). However, this condition, by Theorem 12.3.10 (or see Proposition 12.4.1), holds automatically if R is a principal ideal domain and R A is free of countable rank. Thus we derive another very useful Corollary 12.3.41. Let R be a principal ideal domain and an S–ring which is S– cotorsion–free. If A is an R–algebra with R A is free of countable rank and λ is an infinite cardinal, then there are S–separable R–submodules GX of G = Gλℵ0 of cardinality λℵ0 for all X ⊆ λℵ0 such that the following holds.  A ⊕ Fin(GX , GX  ), if X ⊆ X  ∼  HomR (GX , GX ) = Fin(GX , GX  ), if X ⊆ X  From Observation 12.3.39 follows immediately that S–separable modules given by our Main Theorem 12.3.12 for free algebras R are ℵ1 –free. The result is also true in the S–cotorsion–free case, but it needs some work to show this. The proof of the next corollary stops after the first step, hence is simpler, if we only want to show that the R–modules are locally free; compare Definition 1.1.12. Observation 12.3.39 (b) also follows from the next corollary. Corollary 12.3.42. Let A be a free R–algebra and suppose that R is an S–ring and S–cotorsion–free such that |A| ≤ λ, |R| ≤ κ and λℵ0 = λκ . Then there are ℵ1 –free R–submodules GX of G = Gλℵ0 of cardinality λℵ0 for all X ⊆ λℵ0 such that the following holds.  A, if X ⊆ X  ∼ HomR (GX , GX  ) = 0, if X ⊆ X 

Proof. Let G = Gλℵ0 be the R–module given by Main Theorem 12.3.12 in the cotorsion–free case, where R A is a free R–module. Suppose that R A is freely generated by a set F , hence B in the construction is freely generated by a set T × F (see (12.3.3)). It remains to show that G is an ℵ1 –free R–module. We will use the Recognition Lemma 12.3.19: Any countably generated submodule of G is a submodule of a module U =  F , F  , where F  is a countable subset of T × F and F  is a countable set of ∗ infinite branch elements vαi i (i∗ , i < ω). Inductively we can enlarge U and  replace F  , F by ascending chains of finite subsets Fk , Fk (k < ω) with F  ⊆ k<ω Fk , F  ⊆ k<ω Fk such that the following holds: The elements in Fk have finite, pairwise disjoint support which is also disjoint from [Fk ]. The support [f ] for each branch elements f of the finite set Fk satisfies [f ] ⊆ [Fk \ {f }]. Thus Uk = Fk , Fk  is a free R–module. Possibly enlarging U

12.3

457

Realizing all cotorsion–free algebras

(thus changing F  ) we can also arrange that each branch element f of the finite set  \F  has support not in [F  ]. Therefore U is a summand of the corresponding Fk+1 k k k  2 Uk+1 with free factor and clearly U ⊆ k<ω Uk is a free R–module.

Endomorphism rings of Butler groups In this subsection we will address Butler groups and give references to results in our context. Butler groups were introduced by Butler [76] and extended to infinite rank by Bican and Salce [53]. A torsion–free abelian group B is a (Butler) B2 – group, if B is the union of a smooth chain of pure subgroups Bα of B (α < λ), for some ordinal λ such that each Bα+1 = Bα + Lα for a finite rank Butler group Lα . Moreover, Lα is a finite rank Butler group, if Bext(B, T ) = 0 for all torsion groups T . Particular Butler groups of finite rank are well studied in the book [316] by Mader. Butler groups can be studied by their endomorphism rings as well. This class is complex enough to allow another realization theorem which can be performed with the aid of the General Black Box. We quote from [128]. λℵ0 > Theorem 12.3.43. Let A be a ring such that AZ is a B2 –group and let λ = n |A| be a cardinal. If AZ is p–reduced for three distinct primes p, i.e. p A = 0, n<ω

then there exists a B2 –group H of cardinality λ such that End H ∼ = A. If AZ is a free abelian group, then H can be chosen so as to have typeset {τ1 , τ2 , τ3 }, where τ1 , τ2 are incomparable and τ0 = τ1 ∧ τ2 .

A discussion of realization theorems for torsion and mixed modules The Main Theorem 12.3.12 can be extended to include modules related to (separable) abelian p–groups and mixed modules [102, p. 451]. For transparency this was excluded above, however the proof which covers these two classes of modules is very similar. In order to state the results we need a description of the ideals of the endomorphism ring that correspond to Ines G in these categories. Let R be an S–ring with 1 = 0 as before and in the torsion case assume that S ={pn | n ∈ ω} G[s] is the for some p ∈ R which is not a zero–divisor of R. Then tS G = s∈S

(S–)torsion–submodule of an R–module G, where G[s] = {g ∈ G| sg = 0} is the s–socle of G. Moreover, G is (S–)torsion, if tS G = G and G is mixed, if 0 = tS G = G. We have two well–known definitions (for (i) see [173, Vol I, p. 195]). Definition 12.3.44. Let G, G be two R–modules and σ ∈ HomR (G, G ).

458

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Realizing algebras

(i) Then σ is small, if for each m < ω there is k < ω such that (pk G)[pm ]σ = {0}. The set Small(G, G ) denotes the collection of all small homomorphisms σ ∈ HomR (G, G ). Clearly Small G := Small(G, G) is a two– sided ideal of EndR G. (ii) Moreover, σ is S–bounded, if there exists s ∈ S such that (sG)σ = 0. Similarly Bd(G, G ) denotes the set of all bounded homomorphisms and Bd G := Bd(G, G) is a two–sided ideal of EndR G. Pierce established the remarkable result that the additive group of the ring A = End G/ Small G of an abelian p–group G is torsion–free, p–reduced and complete in the p–adic topology (hence a Jp –module) (see [173, Vol I, Section 46]). Thus (necessarily) we assume that the given algebra A over R is S–torsion–free and S– reduced in either case and that A is Hausdorff and complete in the S–topology in the torsion–case. Under these assumptions we next adjust the basic module B (see the General Black Box 9.2.27) to cover torsion and mixed modules. Let

B= τ A, τ ∈T 2 A in where AnnA τ = pl(τ ) A, if we are in the torsion case and AnnA τ = ql(τ ) the mixed case (see also Definition 1.1.1). Thus B is S–torsion in either case and we have enough room to find torsion–free elements in the mixed case (using 2 Aq ql(τ l(τ ) A. The main theorem in either case formally is as before. )

Main Theorem 12.3.45. Let A be an R–algebra of cardinality ≤ λκ with |R| ≤ κ as above (for the torsion or mixed case), where λ is any cardinal such that λκ = λℵ0 . For any subset X ⊂ λℵ0 we can find an R–module GX of cardinality λℵ0 such that B ⊆ GX ⊆ B as above and for any X, X  ⊆ λℵ0 the following holds.  A ⊕ Ines GX , if HomR (GX  , GX ) ∼ = Ines GX , if

X ⊆ X X  ⊆ X

Moreover, GX  ⊆ GX ⊆ G if X  ⊆ X and G = Gλℵ0 . Similar to the proof in Lemma 12.3.26 (that Ines G = 0 in the cotorsion–free case) we can now show the following (see [102, p. 471, Lemma 8.2 and Lemma 8.3]). Lemma 12.3.46. Let G, G be as GX , GX  in the main theorem above (torsion or mixed, respectively). Then we have Ines(G, G ) = Small(G, G ) in the torsion case and Ines(G, G ) = Bd(G, G ) in the mixed case.

12.4

Algebras of row–and–column–finite matrices

459

Hence we have immediately the Corollary 12.3.47. Let B ⊆ GX ⊆ B and B ⊆ GX ⊆ B be as in Theorem 12.3.45. Then the following holds in the torsion case.  A ⊕ Small GX , if X  ⊆ X ∼ HomR (GX  , GX ) = Small GX , if X  ⊆ X Moreover, in the mixed case follows  A ⊕ Bd GX , if ∼ HomR (GX  , GX ) = Bd GX , if

X ⊆ X X  ⊆ X

and tS GX = B for all X ⊆ λℵ0 .

12.4 Algebras of row–and–column–finite matrices Let A be an algebra of countable rank over  an arbitrary principal ideal domain R such that R A is a free R–module, say A = Rfi . Then for each a ∈ A there is i∈ω

a unique row–finite matrix (αij (a)) over R such that fi a = αij (a)fj (i ∈ ω). j∈ω

The following proposition (needed to deal with endomorphism rings of separable abelian groups) comes from Corner, Göbel [104, Section 2]. It was used to show that B, in the construction of separable modules, is also an A–module (see page 437). Proposition 12.4.1. In this situation the basis (fi )i∈ω may always be chosen so that the matrices (αij (a)) are all column–finite as well as row–finite.

Proof. Fixing an arbitrary sequence (gi )i∈ω of generators of R A, we embark on a recursive construction. We start by setting r(0) = 0, A0 = {0} and F0 = A, so that A+ = A0 ⊕ F0 and rk A0 = r(0) ∈ ω. Now assume recursively that for some n ≥ 0 we have constructed a free basis (fi )i
(i < n + 1),

(∗n )

460 and write

12

Realizing algebras

Fn = Un ⊕ Fn .

Then RA

= An ⊕ Fn ,

where An is free of finite rank. Let πn be the projection of R A onto An with Ker πn = Fn . The mapping

  c → cπn , ((cfj )πn )j
carries R A into the free module (An )1+r(n) of finite rank; so its kernel Fn+1 , say, is a direct summand of finite corank in R A. Since Fn+1 ≤ Ker πn = Fn ( ≤ Fn ), we may write Fn = Vn ⊕ Fn+1 , where Vn is free of finite rank. We now put An+1 := An ⊕ Vn = An ⊕ Un ⊕ Vn ; then also Fn = Un ⊕ Vn ⊕ Fn+1 , so that (∗n+1 ) holds. We now choose a free basis (fi )r(n)≤i
(12.4.2)

and gi ∈ An+1

(i < n + 1).

(12.4.3)

An obvious recursion now leads to a linearly independent sequence of elements (fi )i∈ω , an ascending sequence of summands An of finite rank, and a descending sequence of corresponding direct complements Fn . The sequence (fi )i∈ω is  clearly a basis for An , which coincides with A by (12.4.3). For each n the construction implies that fi ∈ An (i < r(n)), while fi ∈ Fn (r(n) ≤ i ∈ ω); therefore (fi )r(n)≤i∈ω is a free basis of Fn . Finally consider any a ∈ A. For any large enough n we have a ∈ An ; and then for r(n + 1) ≤ i ∈ ω we have fi a ∈ Fn by (12.4.2), which implies that αij (a) = 0 (j < r(n)). Hence for any large enough n all the non–zero entries in the first r(n) columns of (αij (a)) lie in the

12.4

Algebras of row–and–column–finite matrices

461

first r(n + 1) rows. Therefore the matrices (αij (a)) are indeed all column–finite. 2 Choosing the basis (fi )i∈ω as in Proposition 12.4.1, we write   A= ξi fi | ξi ∈ R and ξi → 0 i∈ω



for the S–closure of A = i∈ω Rfi in the corresponding Baer–Specker module  Rf . As a pure submodule of the Baer–Specker module A is S–separable, i i∈ω the virtue of our careful choice of basis is that we now have, automatically, the crucial Corollary 12.4.2. A is canonically an A–module.  Proof. Consider any element g ∈ A, say g = i∈ω ξi fi and a ∈ A. With the previous notation, ga = ξi αij (a)fj . i,j∈ω

For each positive integer s ∈ S, we know that for almost all i ∈ ω, ξi lies in sR. Since the matrix (αij (a)) is column–finite, there exists m ∈ ω such that for each exceptional i and all j > m. It follows that for all j > m the αij (a) = 0 coefficient i ξi αij (a) of fj lies in sR. Hence ga ∈ A, as required. 2

Open problems 1. Extend Corollary 12.3.42 to R–modules G which are ℵn –free (n > 1). The question is answered, if we assume additional set theory (see Theorem 9.1.19 and Theorem 9.1.22). A crucial result would be the existence of ℵn –free abelian groups G with |G| ≥ ℵn (or better |G| = ℵn ) and EndZ G = Z. Hence additional combinatorial methods are needed! Results in [142] concerning counterexamples of ℵn –free groups to Kaplansky’s test problems might be a start. It seems to be very complicated to replace n by an infinite ordinal. In fact, a result by Magidor and Shelah [318] tells us that infinite ordinals n with a G as above can not be very large.

Chapter 13

E(R)–algebras

13.1

Classical E(R)–algebras

Definition 13.1.1. Let R be a commutative ring. If M is an R–module, then its endomorphism ring EndR M is an R–algebra acting on M on the right. If M and EndR M are isomorphic as R–modules, then M is called an E(R)–module. If A is an R–algebra, then μ : A → EndR A mapping a ∈ A to right multiplication ar ∈ EndR A by a is an algebra–monomorphism, and if μ is an isomorphism, then A is an E(R)–algebra. An R–algebra A is a generalized E(R)–algebra, if it is isomorphic to EndR A. We will see in Proposition 13.1.9 that E(R)–algebras are precisely the commutative generalized E(R)–algebras. Every generalized E(R)–algebra is an E(R)– module and every E(R)–module admits a generalized E(R)–algebra structure, which is unique up to isomorphism. Thus E(R)–modules and generalized E(R)– algebras are essentially the same. The above notions are generalizations of E– rings, i.e. E(Z)–algebras. They go back to Schultz (see [357]). Many details about E–rings are collected in Feigelstock [163, 164]. The existence of torsion– free E–rings is important for the theory of torsion–free abelian groups, which is not the scope of this monograph (see the papers by Arnold, Mader, Pierce, Reid and Vinsonhaler [317, 334, 22]). For example Niedzwecki and Reid [330] proved that a torsion–free abelian group G of finite rank is cyclically projective over its endomorphism ring, if and only if G = R ⊕ A, where R is an E–ring and A is an E(R)–module. Moreover, E–rings of finite rank are classified up to quasi– isomorphism in Pierce, Vinsonhaler [335], and E–rings are also used to answer problems in homotopy theory (see [223]).

Excursion: localizations and cellular complexes Casacuberta, Rodríguez and Tai [80] noticed the role of E–rings in homotopy theory. We will now outline these connections and then come back to the main topic of this section, the structure on (generalized) E(R)–algebras. The connection between these two areas is established by localizations. This notion comes from

13.1 Classical E(R)–algebras

463

Bousfield [58] and Dror Farjoun [158], where its fundamental facts are shown. It also appears in [345, 80] and other works. Localizations are closely related to the notion of cogenerators in Gabriel [183]. By definition, a cogenerator is an object A (in a given category) that detects equivalences in the sense that, if a given morphism ϕ : X → Y induces, via composition with ϕ, an equivalence hom(ϕ, A) : hom(Y, A) → hom(X, A), (on the collections hom of all morphisms), then ϕ is itself an equivalence (of objects), where the notion of ‘equivalence’ must, of course, be specified.

Definition 13.1.2. Given a morphism ϕ : X −→ Y in some category (groups, R–modules, spaces, etc.), then an object A is said to be ϕ–local, if the induced map on the hom–sets hom(ϕ, A) : hom(Y, A) → hom(X, A), given by composition with ϕ, is bijective (or an equivalence in the category in which these hom–sets are assumed to reside). In other words, any morphism ψ ∈ hom(X, A) factors uniquely through ϕ. We say that ϕ is perpendicular to A and write ϕ⊥A. The object Y is a ϕ–localization of X, if ϕ⊥Y . A ϕ–equivalence of objects is a morphism ψ : M −→ N such that hom(ψ, A) : hom(N, A) → hom(M, A) is a bijection for all ϕ–local objects A. We now restrict ourselves to groups X, Y, A (or to R–modules X, Y, A over commutative rings R). Given ϕ : X −→ Y with ϕ⊥A, then there is an idempotent localization functor Lϕ from the category of groups (modules) into the reflective subcategory ϕ⊥ of all A such that ϕ ⊥ A. (In shorthand we write is ϕ ⊥ ϕ⊥ .) In other words, for every group (R–module) M there is a ϕ–equivalence or localization map ηM : M −→ Lϕ M with Lϕ M being ϕ–local (furthermore ηM ⊥ ϕ⊥ and ηX = ϕ), see Dror Farjoun [158]. This map ηM : M −→ Lϕ M is natural in the sense that any morphism of groups (modules) ψ : M → N induces a commutative square of groups (modules).

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13 E(R)–algebras

ηM

M

Lϕ (ψ)

ψ



N

/ Lϕ M

ηN

 / Lϕ N.

For any group (module) M and ϕ : X −→ Y the localization map ηM : M −→ Lϕ M is the unique ϕ–localization (see [158, 80]). Furthermore Lϕ (ψ) is uniquely determined by the commutative diagram above (uniqueness property). One of the major problems in the localization of groups (modules) is to decide what algebraic properties of M can be transferred to its localization Lϕ M by a localization ηM : M → Lϕ M . On the one hand it can be shown, but it is not obvious, that localizations of finite non–commutative groups are not necessarily finite, in fact can be arbitrarily large (see Lipman [311, 312] and Göbel, Shelah [234]). On the other hand localizations of abelian groups are abelian (see below) and bounded in size, if M above is a fixed torsion abelian group. However, it is a long–standing problem, if nilpotency of a group M implies nilpotency of Lϕ M . Another main question for localization is, whether or not an arbitrary localization Lϕ P of a finite p–group P is always a quotient of P . Aschbacher [23] established the case when the finite p–groups are nilpotent of class 3; see also Dror Farjoun, Göbel, Segev [159] for other problems. What are the localizations of a commutative ring R, e.g., if R is the ring of integers? This question can be answered and leads directly to E(R)–algebras. We follow the arguments in [344]. Notation 13.1.3. Let L be a fixed localization functor of groups (coming from some homomorphism ϕ as explained above) and denote by ηM : M −→ LM the localization map. Moreover, if ψ : M −→ N is a homomorphism, then let L(ψ) : LM −→ LN be the induced map as in the second column in the last diagram. Proposition 13.1.4. If ψ : M −→ N is a central homomorphism of groups (i.e. Im(ψ) is in the center of N ), then L(ψ) : LM −→ LN is also central. In particular, if M is abelian, then also LM is abelian.

13.1 Classical E(R)–algebras

465

Proof. Consider the commutative diagram ηM

M

ψ

/ LM

L(ψ)



N

ηN

 / LN

and let the group operation for the moment be multiplication. If y ∈ LN , then let y ∗ : LM −→ LN (x → (xL(ψ))y ) be L(ψ) followed by conjugation with y. Since ψ is central, also the diagram ηM

M

y∗

ψ



N

/ LM

ηN

 / LN

commutes, if y = zηN for some z ∈ N . Thus xL(ψ) = (zηN )−1 · xL(ψ) · zηN for all x ∈ LM . Now fix x ∈ LM and consider the trivial diagram ηN

N

id

id



N

/ LN

ηN

 / LN

and the diagram ηN

N

id

θ



N

/ LN

ηN

 / LN

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13 E(R)–algebras

with yθ = (xL(ψ))−1 · y · xL(ψ), which commutes by the last lines. Then again by the uniqueness property of localizations follows (xL(ψ))−1 · y · xL(ψ) = y for all x ∈ LM, y ∈ LN . Hence L(ψ) is central as well. The second statement is now obvious using the identity on M .

2

Theorem 13.1.5. Let L be a localization functor on the category of groups and R be a commutative ring with 1 = 0. If M is an R–module, then LM is an R– module as well and there exists a unique R–module structure on LM such that ηM : M −→ LM is an R–homomorphism. In particular, if ψ : M −→ N is an R–homomorphism, then L(ψ) : LM −→ LN is also an R–homomorphism.

Proof. With Proposition 13.1.4, the group LM is abelian. Let σ : R −→ EndZ M (r → Lr ) be the ring homomorphism which sends r ∈ R to scalar multiplication by r on M . The localization functor L yields a ring homomorphism EndZ M −→ EndZ (LM ) and composition σ  : R −→ EndZ (LM ) induces an R–module structure on LM . Note that for r ∈ R the following diagram commutes: ηM

M

σ(r)

L(σ(r))



M

/ LM

ηM

 / LM.

Now L(σ(r)) = σ (r), and ηM : M −→ LM is an R–homomorphism by commutativity of the diagram. From the uniqueness property of localizations also follows that the R–module structure is uniquely determined by the R–homomorphism ηM : M −→ LM . Now the second statement follows easily using that L is a functor of categories. 2 Finally, we obtain the link to E(R)–algebras as Theorem 13.1.6. Let L be a localization functor on the category of groups. If R is a commutative ring with 1 = 0 and LR = 0, then ηR : R −→ LR is a homomorphism of algebras and LR is an E(R)–algebra with 1LR = 1R ηR .

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13.1 Classical E(R)–algebras

Proof. By Theorem 13.1.5 the map ηR can be viewed as an R–homomorphism and LR becomes an R–module. Since LR is L–local, the L–equivalent homomorphism ηR : R −→ LR induces a bijective map EndZ LR −→ HomZ (R, LR), which restricts to an injection EndR LR −→ HomR (R, LR)

(13.1.1)

(see Definition 13.1.2). Given an R–homomorphism ψ : R −→ LR, then it follows by Theorem 13.1.5 that the induced map ψ := L(ψ) : LR −→ LR is an R–module homomorphism with ηR ψ = ψ. Hence (13.1.1) is bijective. The evaluation map δ : HomR (R, LR) −→ LR (ϕ → 1ϕ) yields an isomorphism and we can use (13.1.1) to endow LR with an R–algebra structure, where multiplication is induced by composition in EndR LR. Thus LR is an E(R)–algebra. 2. Answering a problem in homotopy theory, we can now deduce easily from Theorem 13.4.1 that there is a proper class of distinct homotopical localizations of the circle S 1 (for details see [344, 80]). Dualising ϕ–localizations Reversing all arrows in the basic definitions of localizations, we arrive at a dual theory of localizations, the theory of cellular complexes which has further impact in homotopy theory of spaces recently introduced and studied in papers by Dror Farjoun, Göbel and Segev [159, 160]. Thus the starting point is the following Definition 13.1.7. Let M be a group (an R–module). A cellular cover of M is a group homomorphism (an R–module homomorphism) ϕ : G −→ M such that the induced map Hom(G, ϕ) : End(G, G) → Hom(G, M ), given by composition with ϕ, is bijective. The group (the R–module) G is called the cellular covering group (module) and the map ϕ the cellular covering map. The basic properties of this setting are natural duals of properties of localization and it relates to generators in Gabriel [183]. Despite the difficulties with localizations it follows directly that cellular covers of nilpotent groups are nilpotent again (see [159]) and cellular covers of divisible abelian groups can be completely classified (see [160]). The general theory of cellular covers, however, can also become very complicated: this can be treated using ‘fully rigid systems’ and other methods from this monograph (see Farjoun et al. [161]).

Classical E(R)–algebras, the continuation We rush to mention the most basic properties of generalized E(R)–algebras.

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13 E(R)–algebras

Proposition 13.1.8. Let R be a commutative ring with 1. (a) Subalgebras of quotient fields of domains R and p–pure subrings of the ring Jp of p–adic integers are E(R)–algebras or E–rings, respectively. (b) Any E(R)–algebra is commutative.

Proof. (a) The first part of (a) is an easy exercise. If σ ∈ EndZ A and A is a p– pure subring of Jp , then σ extends uniquely to σ ∈ EndZ Jp . An easy topological argument shows that Jp is an E–ring, hence EndZ Jp = Jp and σ is multiplication by a p–adic number π = 1σ ∈ A. This shows that A is an E–ring as well. (b) If a ∈ A, then multiplication by a on the left is an endomorphism al of the R–module A, and as A is an E(R)–algebra we also find b ∈ A such that al = br . Hence ax = xb for all x ∈ A, thus a = b for x = 1. It follows that ax = xa for all x ∈ A, and A is commutative. 2 The existence of generalized E–rings, which are not E–rings, (shown only recently in [237]) will be proved at the end of this section using the Diamond Principle discussed in Section 9.1. A similar result based entirely on the axioms of ZFC (using the Strong Black Box 9.2) will appear in [212]. Let A be any R–algebra. If we consider A as a right A–module and view the algebra A as the subalgebra A := {ar | a ∈ A} of EndR A, then EndR A = A ⊕ AnnEndR A (1) is a ring split extension. It is obvious that the right–hand side of the displayed equation is in EndR A. Conversely, any σ ∈ EndR A gives rise to 1σ =: a and ar ∈ A . Moreover, 1(σ − ar ) = 0 and therefore σ − ar ∈ AnnEndR A (1). Clearly A ∩ AnnEndR A (1) = 0, and the last equality holds. The next proposition is useful for showing that an R–algebra is an E(R)–algebra, where we consider A as a right A–module; we will apply it, showing Theorem 13.2.1. Proposition 13.1.9. The following conditions for an R–algebra A over a commutative ring R are equivalent: (a) A is an E(R)–algebra. (b) A is a commutative, generalized E(R)–algebra. (c) If ϕ ∈ EndR A and 1ϕ = 0, then ϕ = 0 is the zero–endomorphism, i.e. HomR (A/1R, A) = 0.

13.1 Classical E(R)–algebras

469

(d) The evaluation map δ : EndR A −→ A (ϕ → 1ϕ) is an isomorphism. (e) If ϕ ∈ EndR A, then also ϕ ∈ EndA A, i.e. every R–endomorphism is an A–endomorphism (EndR A = EndA A) or equivalently (xy)ϕ = x(yϕ) for all x, y ∈ A. (f) EndR A is commutative.

Proof. (a)−→ (c). If A is an E(R)–algebra, then EndR A = {ar | a ∈ A} , and if 1ϕ = 0 for some ϕ ∈ EndR A, then ϕ = ar for some a ∈ A and 1a = 0. Thus a = 0 and also ϕ = 0. (c)−→ (d). From ar ∈ EndR A for all a ∈ A follows that δ is surjective. If ϕδ = 0 for some ϕ ∈ EndR A, then ϕδ = 1ϕ = 0 and ϕ = 0 is the zero– endomorphism by (c). Hence δ is an isomorphism. (d)−→ (a). The map ρ : A −→ EndR A (a → ar ) given by scalar multiplication on the right always satisfies aρδ = ar δ = a for all a ∈ A, hence ρδ = idA . We see that δρ : EndR A −→ EndR A acts on ϕ ∈ EndR A by ϕ(δρ) = (1ϕ)r . The element ϕδ determines ϕ uniquely by (d), thus ϕ and (1ϕ)r must coincide as ϕδ = 1ϕ = (1ϕ)r δ, and δρ = idEndR A follows. Hence ρ and δ are inverse maps and ρ is an isomorphism. This shows (a). (a)−→ (b) is obvious (see Proposition 13.1.8). (b)−→ (f) is trivial. (f)−→ (e). We must show that EndR A = EndA A. The inclusion EndA A ⊆ EndR A is obvious. Conversely, if ϕ ∈ EndR A and a ∈ A, then ϕ commutes with ar by (f) which yields (xa)ϕ = x(ar ϕ) = (xϕ)ar = (xϕ)a for all x ∈ A, thus ϕ is an A–homomorphism and ϕ ∈ EndA A. (e)−→ (c). If ϕ ∈ EndR A, then ϕ ∈ EndA A by (e). Therefore from 1ϕ = 0 follows xϕ = (1x)ϕ = (1ϕ)x = 0x = 0 for all x ∈ A, hence ϕ = 0 and (c) holds. 2

470

13.2

13 E(R)–algebras

Constructing torsion–free, reduced E(R)–algebras of rank ≤ 2ℵ0

We assume again that R is an S–ring (see Definition 1.1.1). Recall that we define the rank (or S–rank) rk A of an R–module A as the minimal  ∼ Rb cardinal κ such that A/U is S–torsion for some U = κ R, hence let U = b∈B

for a suitable basis B ⊆ A; it follows that U∗ = A, where U∗ denotes the S– purification of U in A. Similarly we define the algebra–rank of an R–algebra A, which is the minimal cardinality κ of a subset B of A such that U = R[B] is a free polynomial algebra over R and A/U is S–torsion. These notions are only interesting in the ‘torsion–free’ case. The following result comes from a recent paper [220], the case of countable E–rings is due to Faticoni [157], see also Corollary 13.2.3. Theorem 13.2.1. Let R be an S–ring and A a commutative R–algebra which is  has S–reduced and S–torsion–free of rank κ as an R–module. Suppose that R transcendence degree at least λ ≥ κ over A. Then for any κ ≤ μ ≤ λ we can find some S–reduced and S–torsion–free E(R)–algebra E of algebra–rank μ as an A–algebra such that A is an S–dense and S–pure subalgebra of E. Remark 13.2.2. The cardinal κ may be finite. The last assertion is equivalent to saying that E/A is S–divisible and sE ∩ A = sA for all s ∈ S. The second assertion of Proposition 13.1.8 (b) is a very special case of Theorem 13.2.1 for A := R := Z and S := {pn | n < ω}. Moreover, observe that A does not need to be a faithful R–module; for example, let R := Z[x], A := Z and S := {2n |n < ω}. Then A is an R–module, where the scalar multiplication is defined by zf (x) := zf (0) ∈ Z (for z ∈ Z and f (x) ∈ Z[x]). Now R is an S–ring and A is S–reduced and S–torsion–free, thus Theorem 13.2.1 provides an induced E(R)–algebra E, but A is not faithful, as an R–module (for example Ax = 0).  Proof. By hypothesis A/U is S–torsion for some U = Rb with |B| = κ. We b∈B

 select a system of algebraically independent elements ub , v (b ∈ B, v ∈ V ) of R over A such that |B| + |V | = μ. Fix b0 ∈ B, put u := ub0 and define  and E := F∗ ⊆ A.  F := A[u2b 1A + ub, v 2 1A | b ∈ B, v ∈ V ] ⊆ A  We want to show that Here F∗ denotes the S–pure subalgebra generated by F in A.  E is the desired E(R)–algebra. The algebra A is S–reduced and S–torsion–free, hence E is S–reduced and S–torsion–free. Furthermore, A is S–dense and S–pure  hence S–dense and S–pure in E. in A,

13.2 Constructing torsion–free, reduced E(R)–algebras of rank ≤ 2ℵ0

471

E is of algebra–rank μ: note that u2b 1A + ub, v 2 1A (b ∈ B, v ∈ V ) are algebraically independent over A. (Consider first the transcendentals v 2 1A (v ∈ V ), then u2 1A + ub0 , and finally u2b 1A + ub (b = b0 ), see below.) We now claim that E ∩ uE = 0.

(13.2.1)

It suffices to show F ∩ uF = 0. Put xb := u2b 1A + ub, yv := v 2 1A and suppose 0 = f ∈ F . Then f = fk (xb1 , . . . , xbn , yv1 , . . . , yvm ) + g, where fk is a non– zero homogeneous form of degree k, and g is a polynomial of degree < k in 2 ) + h, where f is a non– the xs and ys. Thus f = fk (u2b1 , . . . , u2bn , v12 , . . . , vm k zero homogeneous form of degree 2k and h is a polynomial of degree < 2k in the variables ub1 , . . . , ubn , v1 , . . . , vm . Thus f = 0 implies that the highest degree non–zero homogeneous form of f is of even degree in the ub s and vs. In particular, if f  ∈ F , then f = uf  is impossible, thus F ∩ uF = 0, and the claim (13.2.1) holds. This is all we need to finish the proof: we want to apply Proposition 13.1.9 (c) to show that E is an E(R)–algebra. Hence let ϕ ∈ EndR E with 1ϕ = 0. We only need to show that ϕ = 0 is the zero–endomorphism of R E. First we note  −→ A,  because E is S–dense and S–pure in A.  that ϕ extends uniquely to ϕ : A    by continuity. Moreover, we see that ϕ is an R–endomorphism of the R–module A 2  Thus (using that ub , u ∈ R) (u2b 1A + ub)ϕ = (u2b 1A )ϕ + (ub)ϕ = u2b (1A ϕ) + u(bϕ) = u(bϕ) ∈ E ∩ uE. From (13.2.1) follows bϕ = 0 for all b ∈ B, hence U ϕ = 0. Therefore ϕ induces a homomorphism A/U −→ E, which must be 0 because A/U is S–torsion and E is S–torsion–free. It follows that Aϕ = 0 and ϕ induces a homomorphism E/A −→ E, which is also 0 because E/A is S–divisible and E is S–reduced. Therefore ϕ = 0, and E is an E(R)–algebra. 2 We derive an immediate corollary. Corollary 13.2.3. Let R be an S–ring with |R| < 2ℵ0 and A a commutative R– algebra of rank κ < 2ℵ0 as an R–module which is S–reduced and S–torsion–free. If κ ≤ μ ≤ 2ℵ0 , we can find an S–reduced and S–torsion–free E(R)–algebra E of algebra–rank μ as an A–algebra such that A is an S–dense and S–pure subalgebra of E.

Proof. Apply Theorem 13.2.1 and Theorem 1.1.20.

2

Remark: for R := Z, S := Z \ {0} and κ := ℵ0 the Corollary 13.2.3 is due to Faticoni [157], however he uses different generators for E which would lead to

472

13 E(R)–algebras

rk E ≥ 2κ + 1, if κ is finite. For the sake of completeness we include an immediate consequence of Theorem 13.2.1 and a recent result concerning algebra–decompositions of E(R)–algebras. Corollary 13.2.4. Let R be an S–ring as in Theorem 13.2.1 without non–trivial idempotents. (a) If A in Theorem 13.2.1 is a super–decomposable R–algebra, then E is a super–decomposable E(R)–algebra (of size ≤ 2ℵ0 ). For A choose Example 15.1.1. (b) If R+ is also torsion–free and reduced, then there are arbitrarily large super–decomposable E(R)–algebras. (c) If R has only trivial idempotents 0 and 1, and R+ is torsion–free and reduced, then there are indecomposable E(R)–algebras (of finite rank, any rank ≤ 2ℵ0 and of arbitrarily large rank).

Proof. (a) If A is a super–decomposable R–algebra of size (or rank) < 2ℵ0 , then E is a super–decomposable E(R)–algebra by Theorem 13.2.1. See also Fuchs, Lee [179] for a different proof of (a). (b) See [174] for super–decomposable E(R)–algebras of rank larger than 2ℵ0 . (c) holds for A = R in Theorem 13.2.1. The E(R)–algebras constructed in Theorem 13.4.1 are indecomposable and of arbitrarily large rank. 2 Also note that there are no κ–super–decomposable E(R)–algebras E for κ infinite, because in an infinite decomposition of E the 1 must be expressed by elements from finitely many summands, but then there are non–zero endomorphisms ϕ with 1ϕ = 0, which contradicts Proposition 13.1.9 (c).

13.3

E(R)–algebras and uniquely transitive modules

In this section let R be a principal ideal domain that is not a field. Then R is an S– ring for a multiplicatively closed set S which is cyclically generated by some fixed prime p of R, say S = p. Let A be an R–algebra which is S–torsion–free and  of A is naturally an R–module  S–reduced. Then the S–completion A over the S– ℵ 0  completion R of R (see Lemma 1.1.6). Furthermore, if |R| < 2 and |A| < 2ℵ0 , we can apply Theorem 1.1.20 as in the last section:  of algebraically independent elements over A of size 2ℵ0 There is a set U ⊆ R (see Section 1.1 for the basic notions).

13.3 E(R)–algebras and uniquely transitive modules

473

We will use this fact heavily to settle the following problem raised by Farjoun in a very transparent way. This problem is formulated in terms of modules (originally for abelian groups). However, similar to Theorem 13.2.1, we will construct particular E(R)–algebras with additional algebraic properties motivated by this problem. We will proceed as follows: First we will state Farjoun’s problem for modules. Then we will translate it into a question on E(R)–algebras. Finally, we will answer this stronger question. This approach will show that the ring–theoretic setting is more suitable for its solution. The first answer was given after lengthy calculations in a recent paper by Göbel, Shelah [235]. The translation to rings comes from [209, 269]. We would like to remind the reader of a similar phenomenon in connection with Nöbeling’s group BX of all bounded integer valued functions on a given set X. By Bergman’s observation that BX is also a commutative ring with many idempotents, his new proof that BX is freely generated by certain characteristic functions on X became transparent and short (see Fuchs [173, Vol. 2, p. 173]). Definition 13.3.1. Let M be an R–module and let AutR M be the group of R– automorphisms of M , i.e. the group of units of EndR M . Moreover, let pR M be the set of all pure elements in M and recall that y ∈ pR M , if and only if for any x ∈ M, r ∈ R with rx = y follows r ∈ R∗ , where R∗ is the group of units of R. Then we say that (i) M is transitive, if AutR M acts transitively on pR M , i.e. for any x, y ∈ pR M there exists some ϕ ∈ AutR M with xϕ = y, (ii) M is uniquely transitive (a UT–module over R), if AutR M acts sharply transitive on pR M , i.e. (i) holds and if xϕ = y and xϕ = y for some ϕ, ϕ ∈ AutR M , then ϕ = ϕ . Also recall that an R–module M is of type 0 or R–homogeneous, if M is a torsion–free R–module and if every non–trivial element of M is an R–multiple of an element in pR M . Farjoun now asked for an answer to the following question (for R := Z): Find abelian groups different from Z and of type 0 which are also UT–groups. It is easy to check that Z is a UT–group, because pZ Z = {1, −1} and Aut Z = {idZ , − idZ }. The question naturally generalizes to R–modules. The condition on the type ensures that pR M = ∅, otherwise the answer would be trivial, as for example any Q– vector space would be a UT–module over Z. At the expense of more complicated

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13 E(R)–algebras

calculations, being of type 0 (R–homogeneous) is replaced in [235, 209, 210, 269] by the stronger condition that M is ℵ1 –free. Note that R–homogeneous modules M with |R| < |M | satisfy |pR M | = |M |, and a UT–module of this kind will not be a trivial object. In this first subsection we will transform the question into a ring–theoretic problem which immediately shows a link with E(R)–algebras. The key is the following Definition 13.3.2. Let R be a principal ideal domain. We call a commutative R– algebra A pure–invertible, if R A is an R–homogeneous module with pR A = A∗ . In particular the pure elements of the R–module R A have inverses in the R– algebra A. Clearly Z is pure–invertible and we will construct many examples below. We will see that to answer Farjoun’s question (in terms of rings, see Observation 13.3.7) it is sufficient to solve the following problem: Find E(R)–algebras that are at the same time pure–invertible R–algebras. We note that the existence of many transitive but not sharply transitive, torsion– free groups (assuming V=L) can be found in Dugas, Shelah [133]. The UT– modules constructed in [235] are of size ≥ 2ℵ0 , ℵ1 –free and have endomorphism rings isomorphic to group rings over non–commutative free groups, hence are not E–rings. The UT–modules arriving via E(R)–algebras in [210, 269] are also ℵ1 – free and naturally have size ≥ 2ℵ0 . A simpler Weak Diamond construction (applying ZFC + CH) in [209] provides ℵ1 –free UT–modules via E(R)–algebras also of size ℵ1 . The construction presented here uses only the axioms of ZFC and depends heavily on Theorem 1.1.20 as mentioned above. It is taken from a recent publication by Göbel, Herden [211].

UT–modules over principal ideal domains First we collect some properties of UT–modules and state the main result of this section. Observe that R itself is always a UT–module. It is also the only free UT–module by the following proposition. Proposition 13.3.3. (a) Any R–homogeneous UT–module over a principal ideal domain R of characteristic χ(R) = 2 is indecomposable. (b) If R is a complete discrete valuation domain, then the only R–homogeneous UT–module is R.

13.3 E(R)–algebras and uniquely transitive modules

475

Proof. (a) Let M be an R–homogeneous UT–module and suppose, M = M1 ⊕ M2 is decomposable with M1 = 0 and M2 = 0. Then also M1 and M2 are R–homogeneous modules. Let idM1 and idM2 be the identities on M1 and M2 , respectively. We can now define two automorphisms φ1 := idM1 ⊕(−1) · idM2 ∈ Aut M and φ2 := idM1 ⊕ idM2 ∈ Aut M with xφ1 = xφ2 = x for all x ∈ pM1 . From χ(R) = 2 follows φ1 = φ2 , hence pM1 = ∅ gives a contradiction. (b) If the rank of the R–homogeneous R–module M is at least 2, then we can consider a summand Rm ⊕ Rm of M (see [291, Corollary 1, p. 53]). We find a non–trivial automorphism of M which fixes m + m . Thus the R–homogeneous R–module M must be a copy of R. 2 We can now formulate the main theorem of this section: Main Theorem 13.3.4. Let R be a principal ideal domain with S as above such  has transcendence degree λ ≥ |R|. Then for any |R| ≤ μ ≤ λ we can that R  R E is an R– find a (commutative) E(R)–algebra E such that R ⊆∗ E ⊆∗ R, homogeneous UT–module of cardinality μ and a principal ideal domain. By our first remarks every principal ideal domain R with |R| < 2ℵ0 has a  of transcendence degree 2ℵ0 over R. Thus we have the immediate completion R Corollary 13.3.5. Let R be a principal ideal domain with S as above and |R| < 2ℵ0 . Then for any |R| ≤ μ ≤ 2ℵ0 we can find some (commutative) E(R)–algebra  R E is an R–homogeneous UT–module of cardinality E such that R ⊆∗ E ⊆∗ R, μ and a principal ideal domain. Corollary 13.3.5 can also be used directly to find indecomposable ‘endoprimal’ abelian groups (see [213] for details).

Pure–invertible algebras Next we collect some useful algebraic properties of pure–invertible algebras. Lemma 13.3.6. Let A be a pure–invertible R–algebra. Then A is a principal ideal domain with the same ideal structure as R.

Proof. The cyclic R–module 1R ⊆ A is isomorphic to R, because R A is R– homogeneous; we identify R = 1R ⊆ A as a subring. Now we use that A is pure–invertible: If a, a ∈ A, there are e, e ∈ pR A = A∗ with a = re and a = r e for suitable elements r, r ∈ R. If aa = 0, then rr = 0, hence either a = 0 or a = 0 and A is a domain. From a = re also follows aA = reA = rA. Now

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13 E(R)–algebras

let J = ai A | i ∈ I be any ideal of A. By the above there are suitable elements ri ∈ R with J = ri A | i ∈ I. Let r := gcd{ri | i ∈ I} be the greatest common divisor in R. Hence (13.3.1) rR = ri R | i ∈ I.   ri If x ∈ J, then there are ti ∈ A with x = i∈I ri ti = r · i∈Ir ti ∈ rA. Thus J ⊆ rA. Conversely, if ra ∈ rA, then by (13.3.1) follows r = i∈I ri ti ∈ ri R | i ∈ I for some ti ∈ R ⊆ A. Thus ra ∈ J and also rA ⊆ J. Therefore J = rA holds, and A is a principal ideal domain, where all ideals of A are of the form rA for some r ∈ R. Finally, we claim: For all r, s ∈ R the following holds: rR = sR ⇐⇒ rA = sA.

(13.3.2)

If rR = sR, then rA = sA is obvious. Conversely let rA = sA. Thus there exists some e ∈ A∗ with r = es and we may assume r, s = 0. We can now write r = dr and s = ds with 0 = d ∈ R a greatest common divisor of r and s in R. Thus there are a, b ∈ R with ar + bs = 1 and from r = es follows r = es , 1 = ar + bs = (ae + b)s and r , s ∈ A∗ . We now have A∗ ∩ R = R∗ , as otherwise for any t ∈ A∗ ∩ R \ R∗ the R–algebra A would be a t–divisible R– module and not R–homogeneous. Thus r , s ∈ A∗ ∩ R = R∗ , and it follows that rR = sR. 2 Observation 13.3.7. If the R–algebra A is pure–invertible and an E(R)–algebra, then R A is a UT–module.

Proof. First we use that A is an E(R)–algebra: hence A is commutative and EndR A = A by Proposition 13.1.9, where we identify a ∈ A with ar , the induced scalar multiplication on A (on the right) by a. By this identification follows AutR A = A∗ . From the other hypothesis and Lemma 13.3.6 we also get that A is a principal ideal domain and pR A = A∗ . Now we only have to translate these ring–theoretic conditions into the language of UT–modules: on the one hand AutR A acts transitively on pR A, because the automorphism induced by multiplication with ab ∈ A∗ maps a onto b for any a, b ∈ pR A = A∗ . On the other hand, if two homomorphisms ϕ1 , ϕ2 ∈ EndR A exist with aϕ1 = aϕ2 for some a ∈ pR A, then ϕ1 , ϕ2 are induced by multiplication with some elements s1 , s2 ∈ A, and s1 = s2 follows from as1 = as2 . Thus AutR A also acts sharply transitive on pR A, and A is a UT–module. 2 Let S be a multiplicatively closed subset of regular elements of the pure– invertible R–algebra A. As A is a domain, regularity here requires only 0 ∈ / S. We recall the following definition from page 7.

13.3 E(R)–algebras and uniquely transitive modules

477

Definition 13.3.8. If S is a multiplicatively closed subset of the domain T , then (as usual) TS := S −1 T = { st | t ∈ T, s ∈ S} denotes the localization of T at S. Thus AS ⊆ Q(A) is a canonical subring of the quotient field Q(A) because A is a domain. We need many pure–invertible R–algebras for constructing UT– modules. This will be provided by the next lemma.

The inductive step for the construction of UT–modules As in [209], it will be crucial that the class of all pure–invertible R–algebras is closed under certain ring operations, in particular under the passage to localizations. In the next subsection we will construct UT–modules by transfinite induction as an ascending, continuous chain. The basic inductive step is the following Lemma 13.3.9. Let R be a principal ideal domain, and A a pure–invertible R–  Suppose that U = {u, ua | a ∈ A} ⊆ R  is an algebra such that R ⊆ A ⊆ R. algebraically independent set over A. If  F := A[ua + ua | a ∈ A] ⊆ R,  is the S–purification of F , then the following holds: and if G := F∗ ⊆ R ∗ . (a) pR G is a multiplicatively closed subset of R (b) If A := GpR G is the localization of the R–algebra G at pR G, then R ⊆∗  A ⊆∗ R. (c) A is a pure–invertible R–algebra. (d) u is algebraically independent over A .

Proof. The R–algebra A is a principal ideal domain by Lemma 13.3.6. As on page 471, we claim that  is algebraically independent over A. (13.3.3) V := {ua + ua| a ∈ A} ⊆ R Put va := ua + ua and suppose f (va1 , . . . , van ) = 0 for distinct elements va1 , . . . , van ∈ V and some polynomial 0 = f ∈ A[x1 , . . . , xn ].

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13 E(R)–algebras

Then f = fk + g, where fk = fk (x1 , . . . , xn ) is a non–zero homogeneous form of degree k and g is a polynomial of degree < k in the xi s. It follows that 0 = f (va1 , . . . , van ) = fk (va1 , . . . , van ) + g(va1 , . . . , van ) = fk (ua1 , . . . , uan ) + uh1 (u, ua1 , . . . , uan ) + h2 (u, ua1 , . . . , uan ), where h1 and h2 are polynomials of degree < k in the xi s. Thus fk (ua1 , . . . , uan ) = 0, and fk = 0 follows. The contradiction proves (13.3.3). By (13.3.3) we can view F as a polynomial ring over the principal ideal domain A. Its monomials build a basis of FA . Moreover, as A is a pure–invertible R– algebra, hence R–homogeneous as an R–module, it follows that the R–algebra F is R–homogeneous as an R–module.

(13.3.4)

Next we show that the same holds for G, i.e. the R–algebra G is R–homogeneous as an R–module.

(13.3.5)

 are First we note that G = F∗ is an R–algebra. Moreover, F ⊆ G ⊆ R  has this property. torsion–free and p–reduced R–modules, because R Now let q be a prime different (also modulo units) from p. We consider elements g, h ∈ G with g = qh. Then pm g = g  ∈ F, pn h = h ∈ F for some m, n ∈ ω by purity and spn + tq = 1 for some s, t ∈ R because R is a principal ideal domain. It follows that g = qh =⇒ pn g  = qpm h =⇒ g  = (spn + tq)g  = q(pm h + tg  ), thus q| g in G ⇐⇒ q| g  in F.

(13.3.6)

Hence (13.3.5) follows from (13.3.6), (13.3.4) and the observation that G is p–reduced. Next we show (a). ·  follows pR G ⊆ R ∗ . If f g = qh for some prime element  = R ∗ ∪ pR From R ∗ . Hence q must be different q ∈ R, f, g ∈ pR G and h ∈ G, then qh ∈ R  (also modulo units) from the prime p. Moreover, pm f = f  ∈ F, pm g = g  ∈ F, pn h = h ∈ F for some m, m , n ∈ ω. It follows that 

f g = qh =⇒ pn f  g  = qpm+m h .

13.3 E(R)–algebras and uniquely transitive modules

479

Thus q| pn f  g  holds in F = A[ua + ua| a ∈ A]. By Lemma 13.3.6 the element q of R is also a prime element of A and distinct from p. We conclude that 

pn f  g  = qpm+m h =⇒ q| f  or q| g  in F, contradicting f, g ∈ pR G. Thus pR G is a multiplicative closed subset of G. From (a) follows that the commutative R–algebra A := GpR G is well–defined.  because the elements of pR G are units of R.  Hence (b) obviMoreover, A ⊆ R ously holds. Next we want to show (c). Observe that for any prime element q ∈ R and any ge ∈ A with g ∈ G, e ∈ pR G follows that q|

g in A ⇐⇒ q| g in G. e

Thus A is R–homogeneous by (13.3.5). It remains to show pR A = (A )∗ which immediately follows from g g ∈ pR A ⇐⇒ g ∈ pR G ⇐⇒ ∈ (A )∗ , e e and A is pure–invertible. Finally, we prove (d) and suppose that 0 = g ∈ F [x] with g(u) = 0. Then g can be expressed as g = (hx + f )xn for some n ∈ ω, h ∈ F [x] and some constant term 0 = f (va1 , . . . , van ) ∈ F , where 0 = f ∈ A[x1 , . . . , xn ]. It follows that 0 = g(u) =(h(u) · u + f ) · un = h(u) · un+1 + f (va1 , . . . , van ) · un = h (u, ua1 , . . . , uan ) · un+1 + f (ua1 , . . . , uan ) · un for some polynomial h . Thus f = 0 is a contradiction and (d) follows.

2

The construction of UT–modules Next we describe the construction of the R–algebras E needed for Theorem 13.3.4. By assumption on R in Main Theorem 13.3.4 we can choose an algebraically independent set  U := {u, uαγ | α, γ < μ} ⊆ R over R, which is enumerated as indicated by the cardinal μ. We will construct the pure–invertible R–algebra E for Theorem 13.3.4 as a union of a continuous,  Set E0 := R. ascending chain Eα (α < μ) of pure–invertible R–subalgebras of R.

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13 E(R)–algebras

Then E0 is pure–invertible. Suppose that the chain Eβ (β < α) of pure–invertible  is already defined. subalgebras of R We now distinguish two cases. If α = β + 1 ∈ μ is a successor ordinal, then we obtain Eα from Eβ : Choose {ua | a ∈ Eβ } ⊆ Uα := {uαγ | γ < μ} from U above and define  Eα := (Gα )pR Gα with Gα := Eβ [ua + ua| a ∈ Eβ ]∗ ⊆ R. From Lemma 13.3.9 and the induction hypothesis that Eβ is a pure–invertible  follows that this is also the case for Eα . R–subalgebra of R  If α ≤ μ is a limit ordinal, then by continuity we must let Eα := β<α Eβ . Pure–invertible is a property of finite character, hence it is obvious that the induction hypothesis carries on to Eα also in this case. (We use that {Eβ : β < α} is a pure chain.) Thus of this chain is completed. We get an  the construction  which is pure–invertible. R–subalgebra E := α<μ Eα of R, If |R| < μ ≤ λ, then it is clear that the Eα s constitute a μ–filtration (because | E0 | = | E1 \ E0 | = | R| and | Eα | = | Eα+1 \ Eα | = | R|| α| for all 0 < α <  is algebraically μ). Moreover, the set U> α := {u, uδγ | γ, δ ∈ μ, α < δ} ⊆ R independent over Eα by Lemma 13.3.9 (d). Finally, we state some basic properties of the R–algebra E which are immediate from the properties (a), (b), (c) of the Eα –chain above. Proposition  13.3.10. If (Eα )α∈μ is the chain as above (for some |R| ≤ μ ≤ λ) and E = α∈μ Eα , then the following holds.  and |E| = μ. (a) E is a pure–invertible R–algebra with R ⊆∗ E ⊆∗ R (b) The distinguished element u is algebraically independent over E. (c) E is an E(R)–algebra.

Proof. We must only show (c), so it remains to prove that the evaluation map δ : EndR E → E (ϕ → 1ϕ) is a ring isomorphism. The argument follows the proof given for Theorem 13.2.1 on page 470. The ring homomorphism δ is obviously surjective, hence it is enough to show Ker δ = 0, so let ϕ ∈ EndR E with 1ϕ = 0. We must show that ϕ = 0 is the zero–endomorphism of R E.  −→ R,  because E is p–dense First we note that ϕ extends uniquely to ϕ : R  Moreover, we see that ϕ is an R–endomorphism   by continuand p–pure in R. of R ity. Now choose any element a ∈ E. Then there is some β < μ with a ∈ Eβ ⊆ E. Thus ua + ua ∈ Eβ+1 ⊆ E according to our construction, and (ua + ua)ϕ = ua ϕ + (ua)ϕ = ua (1ϕ) + u(aϕ) = u(aϕ) ∈ E ∩ uE.

13.4 E(R)–algebras and the Strong Black Box

481

From Proposition 13.3.10 (b) follows E ∩ uE = 0. Hence (ua + ua)ϕ = 0 and also aϕ = 0. As this is the case for all a ∈ E, we have shown ϕ = 0, and E is an E(R)–algebra. 2 Finally, we come to the

Proof of Theorem 13.3.4. By Proposition 13.3.10 it remains to show that E is a UT–module. However, E is also a pure–invertible R–algebra and an E(R)– algebra by Proposition 13.3.10 (a) and (c), hence Observation 13.3.7 applies, and Theorem 13.3.4 is shown. 2

13.4

E(R)–algebras and the Strong Black Box

In this subsection we shall apply the Strong Black Box as given in Corollary 9.2.13 to prove the following theorem. Here we will assume additionally that R+ is torsion–free, not just S–torsion–free. Theorem 13.4.1. Let the domain R be an S–cotorsion–free S–ring (see also Definition 1.1.1). Moreover, assume that R+ is torsion–free, and let λ and μ be infinite cardinals such that μℵ0 = μ, λ = μ+ and |R| ≤ λ. Then there exists an E(R)–algebra A of cardinality λ which is also an S–cotorsion–free R–module. The existence of large E–rings was shown first by Dugas, Mader and Vinsonhaler [132] using the classic Black Box from [102]. Applying the Strong Black Box we will simplify the arguments, as demonstrated in [250]. Before we construct the desired E(R)–algebra we need the following lemma on submodules of the S–adic completion of the polynomial ring B = R[ xα | α < λ ] introduced in Section 9.2. The lemma will only be used after we have applied the Strong Black Box 9.2.13 for B. It needs very little knowledge about the notions related to the R–modules  R B and R B (see Section 9.2). Step Lemma 13.4.2. Let the domain R be an S–cotorsion–free S–ring with R+ torsion–free and P = R[ xα | α ∈ I ∗ ] for some I ∗ ⊆ λ, let M be an R–subalgebra  which is an S–cotorsion–free R–module. Also suppose  with P ⊆∗ M ⊆∗ B of B that there is a set   I = α0 < α1 < . . . < αn < . . . (n ∈ ω) ⊆ I ∗ = [ P ]X such that I ∩ [ g ]X is finite for all g ∈ M . Moreover, let φ : P −→ M be an R–homomorphism which is not the multiplication by an element of M . Then there

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13 E(R)–algebras

 where φ is identified exists an element y of P such that yφ ∈ / M  := M [ y ]∗ ⊆ B,  . Clearly M  is again an R–subalgebra of B, with its unique extension φ : P → M   which is also an S–cotorsion–free R–module with M ⊆∗ M ⊆ ∗ B. The element y can be chosen to be either y := x or y := x + πb for x := n∈ω qn xαn with  b ∈ P (for the qn  s see Definition 1.1.1). some suitable π ∈ R,

Proof. Let the assumptions be as above. Either qn xαn x= n∈ω

satisfies xφ ∈ / M [ x ]∗ or not. In the latter case there are k, m ∈ ω, ri ∈ M (i ≤ m) such that qk xφ = ri xi .

(13.4.1)

i≤m

Note that since φ is not the multiplication by an element of M , also qk φ is not the multiplication by an element of M , because M is S–torsion–free. We differentiate between two cases. First assume m ≤ 1. By the above P (qk φ − r1 ) = 0, hence there exists an element b = 0 of P such that 0 = b(qk φ − r1 ) = qk bφ − br1 ∈ M.  with By the S–cotorsion–freeness of M there is 0 = π ∈ R π(qk bφ − br1 ) ∈ / M.

(13.4.2)

Let z := x + πb and suppose zφ ∈ M [ z ]∗ . Then there are m ∈ ω, k ≤ l ∈ ω, ti ∈ M (i ≤ m ) such that ql zφ =



ti z i .

i≤m

Let s :=

ql qk

∈ S. Using (13.4.1) we obtain that

ql πbφ = ql zφ − sqk xφ =



ti (x + πb)i − s(r0 + r1 x).

i≤m

Since [ πb ] = [ b ], [ ql πbφ ] = [ bφ ] and [ x ] = {xαn | n ∈ ω}, respectively [ x ]X = I, we deduce m = 1 and t1 = sr1 by the assumption on I (looking

483

13.4 E(R)–algebras and the Strong Black Box 

at the tm xm αn –component of the above equation for some large n ∈ ω). Therefore ql πbφ = t0 − sr0 + sr1 πb and so

 sπ(qk bφ − br1 ) = t0 − sr0 ∈ M ⊆∗ B,

respectively π(qk bφ − br1 ) ∈ M, contradicting (13.4.2). Now suppose m > 1 in (13.4.1). We may assume that rm = 0 and so 0 =  with mrm ∈ M , using that R+ is torsion–free. Thus there is 0 = π ∈ R / M. πmrm ∈

(13.4.3)

Let z := x + π (i.e. b := 1 ∈ R ⊆ P ⊆ M ) and suppose ti z i for some m ∈ ω, k ≤ l ∈ ω, ti ∈ M (i ≤ m ); ql zφ = i≤m

let s :=

ql qk .

Using (13.4.1) we obtain ql πφ = ql zφ − sqk xφ =



ti (x + π)i − s

i≤m



ri xi .

i≤m

Comparing the supports again, we deduce m = m > 1, tm = srm and tm−1 + tm πm = srm−1 , and so  srm πm = srm−1 − tm−1 ∈ M ⊆∗ B, respectively, πmrm ∈ M, contradicting (13.4.3). Therefore, in both cases, either y := x or y := z satisfies yφ ∈ / M  := M [ y ]∗ .   Using support arguments the S–cotorsion–freeness  ofqnM follows from  Mqn = M [ yk | k ∈ ω ], where  yk := xk (+ πk b), xk := n≥k qk xαn , πk := n≥k qk rn with π := π0 := n∈ω qn rn for suitable rn ∈ R (see proof of Lemma 12.3.5). 2

We are now ready to construct the desired E(R)–algebra. Construction 13.4.3. Let (φβ )β<λ be a family of canonical homomorphisms as given by Corollary 9.2.13 (for arbitrary stationary E ⊆ λo , κ := ℵ0 ). For any β < λ let Pβ := Dom φβ = R[ xα | α ∈ [ φβ ]X ].

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13 E(R)–algebras

 (β < λ) such that Aβ is an We inductively define R–subalgebras Aβ of B  and S–pure, S–cotorsion–free R–submodule of B γ | γ < β] ⊆ B  Aβ = B[yγ | γ < β]∗ ⊆ B[P

(13.4.4)

γ ⊆ B.  holds as R–algebras for suitable yγ ∈ P 0 0 Let A := B = R[ xα | α < λ ]; clearly A satisfies (13.4.4), and also B=



 Rm ⊆∗ B

m∈M

is S–cotorsion–free since R is (see Section 9.2). If β is a limit ordinal and Aγ is given for all γ < β satisfying the above properties, then we define  Aβ := Aγ . γ<β

It is easy to check that Aβ satisfies the required conditions (see the Construction 12.3.6). It remains to consider the successor case. Suppose Aβ is given, fulfilling all the conditions. We consider φβ . Since  φβ  ∈ λo there are ordinals α0 < α1 < . . . < αn < . . . (n ∈ ω) in [ φβ ]X such that  φβ  = supn∈ω αn . We put I := {αn | n ∈ ω}. Then I ∩ [ g ]X is finite for all g ∈ Aβ by (13.4.4) and condition (iii) in Corollary 9.2.13. We distinguish between two cases: β satisfies Im φβ ⊆ Aβ and is not the multiplication by an element If φβ : Pβ → P of Aβ , then we apply the Step Lemma 13.4.2 to I as above, P := Pβ , M := β  and φ := φβ . We thus deduce the existence of an element yβ := y ∈ P Aβ ⊆∗ B β+1  with and of an S–pure, S–cotorsion–free R–submodule A of B Aβ+1 := Aβ [ yβ ]∗ = Aβ [ yβk | k ∈ ω ], which is also an R–algebra, such that / Aβ+1 , yβ φβ ∈ where either yβk :=

qn xα qk n

n≥k

or

yβk :=

qn xα + πk b qk n

n≥k

13.4 E(R)–algebras and the Strong Black Box

485

  for suitable b ∈ Pβ , rn ∈ R). (with yβ := yβ0 , πk := n≥k qqnk rn ∈ R Note that  yβ  =  yβk  =  φβ  =  Pβ . Moreover, Aβ+1 satisfies property β . (13.4.4), since yβ ∈ P If either  Im φβ  Aβ or φβ is multiplication by an element of Aβ , then we put yβk := n≥k qqnk xαn , yβ := yβ0 and Aβ+1 := Aβ [ yβ ]∗ = Aβ [ yβk | k ∈ ω ]. Then, also in this case, Aβ+1 satisfies the required conditions. Finally, we define A by  A := Aβ = B[ yβk | β < λ, k ∈ ω ]. β<λ

It is an immediate consequence from the construction that A is an R–subalgebra  of cardinality λ which is also an S–pure, S–cotorsion–free R–submodule of of B  Next we describe the elements of A. B. Lemma 13.4.4. Let A be as in Construction 13.4.3 and let g ∈ A \ B. Then there are a finite non–empty subset N of λ and k ∈ ω such that g ∈ B[ yβk | β ∈ N ] and [ g ]X ∩ [ yβ ]X is infinite ⇐⇒ β ∈ N. In particular, if  g  is a limit ordinal, then  g  =  yβ∗  =  φβ∗  for β∗ = max N.

Proof. Let g ∈ A = B[ yβn | β < λ, n ∈ ω ]. Then there are a finite subset N of λ and k ∈ ω such that g ∈ B[ yβn | β ∈ N, n ≤ k ] = B[ yβk | β ∈ N ], where the equality holds since yβn − qqnk yβk ∈ B for all n ≤ k. & % / N , since [ yβ ]X ∩ yβ  X is finite for β = Now [ g ]X ∩ [ yβ ]X is finite for all β ∈ β  by Corollary 9.2.13 (iii). Moreover, choosing the set N minimal under inclusion [ g ]X ∩ [ yβ ]X is obviously infinite for all β ∈ N . Clearly N = ∅ for g ∈ / B. Hence the first part of the lemma is proved. The second part is an easy exercise. 2 Using the above lemma, we find further properties of A. Lemma 13.4.5. Let A be as in Construction 13.4.3 and define Aα (α < λ) by Aα := {g ∈ A |  g  < α}. Then:

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13 E(R)–algebras

β ⊆ Aβ+1 for all β < λ. (a) A ∩ P (b) Aα is an R–subalgebra of A for all α < λ. (c) {Aα | α < λ} is a λ–filtration of A. (d) If α, β < λ are ordinals such that  φβ  = α, then Aα ⊆ Aβ . Note, we used the upper index (β < λ) for the chain in the construction, while we employ the lower index (α < λ) for the λ–filtration taking care of the norm.

Proof. The proof of (a), (c) and (d) is similar to the one of Lemma 12.3.8, using Lemma 13.4.4 instead of Lemma 12.3.7. Condition (b) follows from Aα = Bα [ yγk |  φγ  < α, k ∈ ω ], where Bα = R[ xγ | γ < α ].

2

Finally, we are ready to prove the main theorem of this subsection, i.e. the existence of an E(R)–algebra.

Proof of Theorem 13.4.1. Let A be the R–algebra as in Construction 13.4.3. We already know that A is an S–cotorsion–free R–module of cardinality λ. It remains to show EndR (R A) = A (identifying a ∈ A with ar ∈ EndR (R A)). Clearly A ⊆ EndR (R A). Conversely, suppose there exists an R–module homomorphism ψ : A −→ A which is not the multiplication by an element of A. Let ψ  := ψB, then also ψ  ∈ / A, since ψ is uniquely determined by ψ   Let (B ⊆∗ A ⊆∗ B).   I = α0 < α1 < . . . < αn < . . . (n ∈ ω) ⊆ λ such that I ∩ [ g ]X is finite for all g ∈ A. Note, the existence of I can easily be arranged, e.g. let E λo , α ∈ λo \ E and (αn )n∈ω any ladder on α. By the Step Lemma 13.4.2 (I ∗ := λ, P := B, M := A, φ := ψ  ) there  such that yψ ∈ exists an element y of B / A[ y ]∗ . By the Strong Black Box (Corollary 9.2.13) the set   E  := α ∈ E | ∃ β < λ :  φβ  = α, φβ ⊆ ψ  ⊆ ψ, [ y ]X ⊆ [ φβ ]X β where is stationary since |[ y ]X | = ℵ0 . Note, [ y ]X ⊆ [ φβ ]X implies y ∈ P Pβ := Dom φβ . Moreover, let C := {α < λ | Aα ψ ⊆ Aα }. Then C is a cub, since {Aα | α < λ} is a λ–filtration of A by Lemma 13.4.5 (c). Now let α ∈ E  ∩C (= ∅). Then Aα ψ ⊆ Aα , and there exists an ordinal β < λ such that  φβ  = α, φβ ⊆ ψ and y ∈ Pβ . The first property implies Aα ⊆ Aβ

13.5 Discussing ℵ1 –free E(R)–algebras of cardinality ℵ1

487

by Lemma 13.4.5 (d) and the latter properties together with yψ ∈ / A[ y ]∗ imply φβ ∈ / A, especially φβ ∈ / Aβ . Moreover, Pβ ⊆ B with  Pβ  = α, hence Pβ and also Pβ ψ are contained in Aα ⊆ Aβ (using Aα ψ ⊆ Aα ). / Aβ , and, on the one hand, we derive Therefore φβ : Pβ −→ Aβ with φβ ∈ from the Construction 13.4.3 that yβ φβ ∈ / Aβ+1 . On the other hand, it follows β ⊆ Aβ+1 – a contradiction. from Lemma 13.4.5 (a) that yβ φβ = yβ ψ ∈ A ∩ P So we have shown that no such ψ exists, and this means EndR (R A) = A as required. 2

13.5 Discussing ℵ1 –free E(R)–algebras of size ℵ1 Suppose for simplicity that (for the moment) R is a countable principle ideal domain and an S–ring. In the last section we established the existence of arbitrarily large ℵ1 –free E(R)–algebras A for suitable rings R. (Using some further algebraic consideration we can easily replace “ S–cotorsion–free ” in Section 13.4 by “ ℵ1 –free ”!) This construction (in ZFC) is fairly simple, if the E(R)–algebra A has size at least 2ℵ0 ; the above construction showed only technical difficulties. If A has size ℵ1 and we also want A to be ℵ1 –free at the same time, then we are closer to freeness and it becomes more complicated to force endomorphisms to be the scalar multiplication by elements from A. Assuming the Diamond Principle we showed in [243] that there exist (strongly) λ–free E(R)–algebras of cardinality λ for any regular not weakly compact cardinal λ > ℵ0 . Since the existence of ℵ2 –free E(R)–algebras of cardinality ℵ2 is undecidable in ordinary set theory ZFC (see [226, Theorem 5.1] and [243]), it is hopeless to conjecture that there are λ–free E–rings of cardinality λ in ZFC for cardinals λ larger than ℵ1 . However from Göbel, Shelah, Strüngmann [239] follows the existence of ℵ1 –free E(R)–algebras in ZFC of cardinality λ for every regular cardinal ℵ1 ≤ λ ≤ 2ℵ0 . Thus there are ℵ1 –free E(R)–algebras of size ℵ1 in ZFC. The construction and the new combinatorial setting (predicting elements rather than partial homomorphisms) are similar to the construction of indecomposable ℵ1 –free R–modules of cardinality ℵ1 with prescribed endomorphism algebras in Section 12.2.

13.6 Mixed E(R)–modules over Dedekind domains In this section let R be a Dedekind domain and let Q denote the quotient field of R. If S is a set of prime ideals of R, then an R–module M is S–divisible, if

488

13 E(R)–algebras

M = P M for all P ∈ S. The module M is S–reduced, if it has no non–trivial S– divisible submodule. If N ⊆ M , then N is S–pure in M (denoted by N ⊆S M ), if IN = N ∩ IM for all ideals I of R which are products of prime ideals from S.  For example, T := P ∈S R/P nP is an S–pure submodule of Π :=



R/P nP ,

P ∈S

where the nP are arbitrary positive integers. Both T and Π are S–reduced and they are divisible by all prime ideals not in S. Notethat Π/T is divisible: if U ∈ S then Π/T is a factor of the U –divisible module P ∈S\{U } R/P nP and hence it is U –divisible. If U ∈ / S, then Π/T is U –divisible, since Π is already U –divisible. We will also write RP m := R/P m as for cyclic abelian groups. Now we can reformulate the structure theory of E–rings from Schultz [357, p. 63, Theorem 6] for Dedekind domains; we will follow the arguments in Braun, Göbel [65]. Theorem 13.6.1. Let A be a generalized E(R)–algebra. Then the following holds. (a) The primary components of the torsion part T of A are cyclic:

R/P nP , T =

(13.6.1)

P ∈S

where S is a set of prime ideals and the nP s are positive integers. After Schultz, the elements of S are called the relevant prime ideals. (b) Using the above notation, let D be the largest S–divisible  submodule of A. Then D is an ideal, A/D is an S–reduced subalgebra of P ∈S R/P nP and T =

P ∈S

R/P nP ⊆S A/D ⊆S



R/P nP =: Π.

(13.6.2)

P ∈S

The proof of this theorem is based on two lemmas, which we are going to establish first. We recall some well–known facts about modules over Dedekind domains used below. Torsion modules decompose into primary components. Each torsion module M has a basic submodule, i.e. a pure submodule B, which is a direct sum of cyclic modules, and M/B is divisible. Basic submodules of M are not unique, but the number of cyclic summands R/P k is unique for all P and k. In particular this number is the largest cardinal κ for which (R/P k )(κ) is a direct summand of M .

13.6

Mixed E(R)–modules over Dedekind domains

489

Divisible submodules and pure bounded submodules are direct summands in any module. Divisible modules are direct sums of RP ∞ s and copies of Q, where RP ∞ is the P –component of Q/R. The number of summands RP ∞ , respectively Q is independent of the decomposition. Namely, in every decomposition of a divisible module M , the number of summands Q is the largest cardinal κ for which Q(κ) is a direct summand of M . The same result holds for RP ∞ . Recall that JP ∼ = EndR RP ∞ is the algebra of P –adic integers over R. This gives an embedding of RP ∞ into Hom(JP , RP ∞ ) identifying any element of RP ∞ with its induced evaluation map on JP . As we already mentioned, JP ∼ = EndR JP . Recall that Hom(A, B) is torsion–free, if A is divisible and Hom(A, B) is divisible, if A is torsion–free and B is divisible. The proof of the next two lemmas is now obvious. Lemma 13.6.2. Let M be a torsion cyclic module or the quotient field Q. Then for every cardinal κ EndR M (κ) ∼ (13.6.3) = M (λ) for some λ. If κ > 1, then λ > κ. Lemma 13.6.3. Let A be an E(R)–module. Suppose that A = B ⊕ C is a direct sum of R–modules. Then EndR B ⊕ EndR C is a direct summand of A as an R–module. Moreover, (as R–modules) A∼ = EndR A = EndR B ⊕ Hom(B, C) ⊕ Hom(C, B) ⊕ EndR C.

(13.6.4)

The two parts of the theorem are now proved separately.

Proof of Theorem 13.6.1 (a). By the preceding remarks, a basic submodule of the P –component AP of A is a direct sum of cyclic P –modules ∞

(R/P k )(κ(k)) , k=1 (κ)

where κ(k) is the largest cardinal κ for which RP k is a direct summand of AP , and thus of A. First we show that κ(k) is at most 1. (λ) From Lemmas 13.6.2 and 13.6.3 we have a direct summand RP k of A, where λ > κ(k) if κ(k) > 1. Hence κ(k) ≤ 1. Now we prove that the basic submodule of AP is cyclic. Otherwise RP k ⊕ RP l would be a direct summand of A for some k < l. Then we have as the R–module decomposition of EndR (RP k ⊕ RP l ) EndR RP k ⊕ Hom(RP k , RP l ) ⊕ Hom(RP l , RP k ) ⊕ EndR RP l ,

(13.6.5)

490

13 E(R)–algebras

which is a direct summand of A isomorphic to RP3 k ⊕ RP l , contradicting the previous paragraph. Hence the basic submodule of the P –component is cyclic. It follows that AP is a direct sum of a cyclic P –module (a basic submodule) (κ) (κ) and a divisible one RP ∞ , where κ is maximal with RP ∞ being a direct summand of A. We show that κ = 0 and hence AP is cyclic. (κ) (κ) Observe that EndR RP ∞ is torsion–free since RP ∞ is divisible. Now by a double application of Lemma 13.6.3 follows (κ)

(κ)

Hom(EndR RP ∞ , RP ∞ ) & A. (κ)

(13.6.6)

(κ)

Furthermore, as EndR RP ∞ is torsion–free and RP ∞ is divisible, (κ)

(κ)

Hom(EndR RP ∞ , RP ∞ ) (λ)

is divisible. Its P –torsion part is RP ∞ , where λ is the dimension of its P –socle: (κ) (κ) (κ) (κ) (κ) Hom(EndR RP ∞ , RP ∞ )[P ] ∼ = Hom((EndR RP ∞ ) ⊗ R/P, RP ) ' EndR RP . (13.6.7) (κ) The dimension of (EndR RP ∞ ) ⊗ R/P as vector space over the field R/P is obviously at least κ (e.g. since the canonical projections onto the different coordinates are linearly independent), which justifies the last relation in (13.6.7). If (κ) κ > 1, then we have λ > κ since the dimension of EndR RP is larger than κ by Lemma 13.6.2. But λ > κ contradicts the maximality of κ. Thus κ ≤ 1. We also have to exclude the case κ = 1. So suppose that RP ∞ & A. Then JP ∼ = EndR RP ∞ & A. So RP ∞ ⊕ JP & A. By Lemma 13.6.3 the R–module A has summands

JP ⊕ JP ∼ = EndR RP ∞ ⊕ EndR JP & A.

(13.6.8)

Hence RP ∞ ⊕ JP2 & A and RP2 ∞ & Hom(JP , RP ∞ )2 = Hom(JP2 , RP ∞ ) & A, which is impossible. So every P –component is reduced and cyclic. 2 Now we are ready to prove the second part of the main structure theorem.

Proof of Theorem 13.6.1 (b). If P ∈ S, then the P –component R/P nP of A is cyclic and pure in A. Hence it is a direct summand: A = R/P nP ⊕ BP .

(13.6.9)

Thus we have as R–module decomposition A ∼ = EndR A = EndR R/P nP ⊕ Hom(R/P nP , BP ) ⊕ Hom(BP , R/P nP ) ⊕ EndR BP .

13.6

491

Mixed E(R)–modules over Dedekind domains

The first three summands are all P –modules. But EndR R/P nP ∼ = R/P nP , so this is already the whole P –component of A. In particular, we must have Hom(BP , R/P nP ) = 0, thus Hom(BP ⊗ R/P, R/P ) ∼ = Hom(BP , R/P ) = 0, BP ⊗ R/P = 0 and BP is P –divisible. It follows that BP is the largest P –divisible submodule of A and hence is an ideal of A. Also, R/P nP is an ideal of A being the P –torsion part of A, and furthermore R/P nP · BP = 0 holds. So A = R/P nP × BP as R–algebras, and this gives rise to an algebra epimorphism ϕP : A → R/P nP . Combining the algebra homomorphisms ϕP together for all P ∈ S, we obtain a homomorphism  (13.6.10) R/P nP . ϕ: A → P ∈S

Obviously, torsion part T of A is just the canonical embedding  ϕ restricted to the of T = P ∈S R/P nP into P ∈S R/P nP . Now D := P ∈S BP is obviously the kernel of ϕ and an ideal. Since BP is the largest P –divisible submodule of A, every S–divisible submodule of A must be contained in D. We prove that D is S–divisible, and so it is the largest S–divisible submodule of A. If P, P  ∈ S are two different prime ideals, then A = R/P nP ⊕ R/P  nP  ⊕ B

(13.6.11)

holds similarly to (13.6.9). Now, as already mentioned, BP and BP  are uniquely determined, hence equating (13.6.9) and (13.6.11) leads to BP = R/P  nP  ⊕ B P ⊕ B. Thus B = B ∩ B  is P –torsion–free and P –divisible, and BP  = R/P n P P and so also D = P  ∈S (BP ∩ BP  ) must be P –divisible. So far we have proved (13.6.2) of Theorem 13.6.1, except that A/D is S–  pure in the product P ∈S R/P nP = Π. Since T is S–pure in Π and Π/T is S–divisible, this is equivalent to saying that (A/D)/T is S–divisible. But already A/T is S–divisible, as can easily be seen: for P ∈ S, the factor A/T is P – divisible being a factor of the P –divisible module BP by (13.6.9). 2 The classification of non–reduced E–rings generalizes to E(R)–algebras: Theorem 13.6.4. The non–reduced generalized E(R)–algebras over a Dedekind domain R are exactly the R–modules of the form Q⊕T where T is a torsion cyclic R–module. In particular they are E(R)–algebras.

Proof. The methods of the proof are similar to the proof of Theorem 13.6.1. Let A be a non–reduced generalized E(R)–algebra. The torsion part of A must be reduced by Theorem 13.6.1 (a). It follows that the divisible part must be Q(κ) for

492

13 E(R)–algebras

some κ ≥ 1. By Lemma 13.6.2, if κ > 1, then Q(λ) is a direct summand of A for some λ > κ. This is impossible, and therefore the divisible part of A is Q. So A splits as A = Q ⊕ T for some reduced module T . We have the R–module decomposition A∼ = EndR A = EndR Q ⊕ Hom(Q, T ) ⊕ Hom(T, Q) ⊕ EndR T.

(13.6.12)

The first three summands are divisible and EndR Q ∼ = Q. Since the divisible part of A is only Q, the second and third summands must be zero. In particular, Hom(T ⊗ Q, Q) ∼ = Hom(T, Q) = 0 means T ⊗ Q = 0 and T is torsion, so T is the torsion part of A. Hence the torsion part T of A is a direct summand, where  T = P ∈S R/P nP holds. Observing that EndR (Q ⊕



R/P nP ) ∼ =Q⊕

P ∈S



R/P nP

P ∈S

the set S has to be finite, and T is thus a torsion cyclic R–module.

2

The construction of mixed E(R)–modules The following example of mixed E(R)–algebras is due to Schultz  [356]. Let S be a set of prime ideals of R. Let A be a subalgebra with 1 of P ∈S R/P nP such that 

R/P nP ⊆S A ⊆S R/P nP . (13.6.13) P ∈S

P ∈S

Then it is easy to check that A is an E(R)–algebra and S is the set of relevant prime ideals. E(R)–modules whose torsion part is a direct summand: An R–module is an E(R)–module whose torsion part is a direct summand, if and only if, for some non–zero ideal I of R, it is of the form R/I ⊕ N , where N is a torsion–free I–divisible E(R)–module. (N is I–divisible means N = IN .) This follows directly from Theorem 13.6.1 (see also Theorem 13.6.4). Thus this class of E(R)–modules is classified modulo torsion–free E(R)– modules. In particular, the torsion E(R)–modules are precisely the cyclic ones.

13.6

Mixed E(R)–modules over Dedekind domains

493

E(R)–modules whose torsion part is not a direct summand: Using the notations and assumptions of Theorem 13.6.1, we have a pull–back diagram: A −−−−→ A/T ⏐ ⏐ ⏐ ⏐ (13.6.14)   A/D −−−−→ A/(D + T ). This suggests that we can construct mixed E(R)–algebras as pull–backs (see Lemma 13.6.7). It is a fruitful method, as it gives examples of arbitrarily large E(R)–modules A for which D (and therefore also T ) is not a direct summand as stated in the next theorem; the result appears in [65]. In contrast, the examples above from [356] have S–divisible part 0 and their cardinality is bounded essentially by the size of the ring. Let RS (or S −1 R) denote the localization of R (a subring of Q) at the set S of prime ideals. Theorem 13.6.5. Let R be a Dedekind domain and let S be an infinite proper subset of its spectrum of prime ideals. Moreover, let P0 be a fixed prime ideal not belonging to S. Then, provided that RP0 is not a complete discrete valuation domain, for any family {nP : P ∈ S} of positive integers there exists an arbitrarily large E(R)–algebra A whose torsion submodule is given by (13.6.1) and whose S–divisible part D is not a direct summand. Remark 13.6.6. The torsion–free rank of the mixed E(R)–algebras in the theorem is determined by the possibilities of ranks of torsion–free E(R)–algebras; see Theorem 13.2.1 (for small ranks) and Construction 13.4.3 (for arbitrarily large ranks). This indicates that the classification of mixed E(R)–modules is a difficult task. The next lemma is our main tool for constructing mixed (generalized) E(R)– algebras out of torsion–free ones: Lemma 13.6.7. Let S be an infinite set of prime ideals and B an algebra such that 

R/P nP ⊆S B ⊆S R/P nP =: Π (13.6.15) T := P ∈S

P ∈S

with 1 ∈ B. Moreover, let C be an S–divisible torsion–free generalized E(R)– algebra. Fix an algebra–isomorphism φ : C −→ EndR C satisfying: (a) There is a fully invariant ideal D of C and C/D ∼ = B/T .

494

13 E(R)–algebras

(b) The induced map C ∼ = EndR C −→ EndR (C/D) maps every c ∈ C to the right multiplication by the residue–class c + D. Then the pull–back A below is a generalized E(R)–algebra. 0 −−−−→ ⏐ ⏐  0 −−−−→ T −−−−→   

D ⏐ ⏐ 

D ⏐ ⏐ 

A −−−−→ ⏐ ⏐ 

C ⏐ ⏐ 

−−−−→ 0 (13.6.16)

0 −−−−→ T −−−−→ B −−−−→ B/T ∼ = C/D −−−−→ 0 ⏐ ⏐ ⏐ ⏐   0

0.

Moreover, A is commutative, exactly if C is commutative.

Proof. Note that B is automatically an E(R)–algebra, i.e. there exists an isomorphism ψ : B → EndR B mapping b to the right multiplication br , and that T is fully invariant in B. By hypothesis every endomorphism of C induces an endomorphism on C/D; similarly for B and B/T . We can regard T and D as canonical R–submodules of A, they are even ideals. Since B ∼ = A/D is S–reduced and D is S–divisible (as an ideal of the S–divisible algebra C), D is the largest S–divisible submodule of A. Similarly, since T is torsion and C ∼ = A/T is torsion–free, T is the torsion part of A. Hence we may form another pull–back, say X, of EndR C and EndR B and we have a composite diagram as below, where the mapping C/D → EndR (C/D) is the canonically induced map, and the dotted map from A to X is the mapping coming from the universal property of pull–backs. C

M

A

X C/D

B

End C

End C/D

O

End B

Now a straightforward check using the pull–back definition shows that X may be identified with those endomorphisms of A which induce endomorphisms on both B and C. Moreover, our hypotheses ensure that the composite diagram above

13.6

Mixed E(R)–modules over Dedekind domains

495

is commutative and so, by the pull–back property, A is isomorphic to X and may be identified with this set of endomorphisms on A. As can be seen from the first diagram, A/D ∼ = B and A/T ∼ = C, and so the set of endomorphisms of A inducing endomorphisms on B, C is precisely the set of endomorphisms of A leaving D and T invariant. But then the subalgebra X is clearly the whole EndR A, since every endomorphism must leave the torsion part T and the largest S–divisible submodule D invariant. Thus A is a generalized E(R)–algebra. It is obvious from the pull–back diagram (13.6.16) that A is commutative exactly if C is commutative. 2

Proof of Theorem 13.6.5. We are going to apply Lemma 13.6.7 to obtain the required algebras. Choose any P0 –cotorsion–free, torsion–free, S–divisible R– algebra E of module–rank < 2ℵ0 . (Take e.g. E := RS .) Since E is P0 –cotorsion–free, there is a set X, larger than any given cardinal, and an E(R)–algebra C such that C is an E–algebra with  E[X] ⊆ C ⊆ E[X],

(13.6.17)

where E[X] denotes the polynomial R–algebra over E in the set of variables X  its P0 –adic completion. (Such a C can be constructed in the following and E[X] way: let us start with a polynomial algebra C0 := E[X0 ]. In the P0 –adic completion of C0 , we choose an appropriate algebraically independent set X1 over C0 . Let X := X0 ∪X1 and C be the P0 –purification of C0 [X1 ] = E[X] in the P0 –adic completion of C0 . Here we apply the results about the existence of torsion–free E(R)–algebras C, see Theorem 13.2.1 for small ranks and Theorem 13.4.1 for arbitrarily large ranks.)  Let D be the ideal of We consider elements of C as (infinite) sums in E[X]. C consisting of all sums with zero constant term, i.e. the kernel of the algebra  homomorphism which substitutes 0 for every variable. Thus E ⊆ C/D ⊆ E.  and hence C/D, Since Π/T is a divisible, torsion–free algebra, we can identify E, as a subalgebra B/T of Π/T for some S–pure subalgebra B with 1 of Π. Now let us apply Lemma 13.6.7 for the data B, T , C and D. The result is an E(R)–algebra A such that A/D ∼ = B and A/T ∼ = C. We prove that D is not a direct summand of A. If it were a direct summand, then the projection onto D would be a multiplication by some p ∈ A with p2 = p (since A is an E(R)– algebra). Since the image of multiplication by p is D, we have p = p · 1 ∈ D.  the element p would be a non–zero idempotent sum with zero So, by D ⊆ E[X], constant term. But no such sum exists. 2

496

13.7

13 E(R)–algebras

E(R)–modules with cotorsion

This section comes from recent work in Göbel, Goldsmith [208]. We continue to consider R–modules over Dedekind domains R and let Q be the quotient field of R. Suppose then that A is a generalized E(R)–algebra but that A is not cotorsion–free. It follows from an easy extension of the well–known classification of cotorsion–free groups (see Remark 12.3.3) that A is either (i) not reduced, (ii) not torsion–free, or (iii) it is torsion–free and reduced, but contains a submodule isomorphic to JP , the algebra of P –adic integers, for some prime ideal P . Non–reduced E(R)–modules and torsion E(R)–modules: Recall from Theorem 13.6.4 that any non–reduced E(R)–module over a Dedekind domain R is of the form Q⊕T where T is a torsion cyclic R–module. In particular they are E(R)–algebras. Thus generalized E(R)–algebras of type (i) are classified. Next we consider the particular case of (ii) when A is torsion. We record from Theorem 13.6.1 (a) that A is a cyclic R–module (and in particular an E(R)– algebra). Thus this case of (ii) is also settled. Mixed E(R)–modules: The remaining part of case (ii) can now be dealt with. So assume that A is a reduced generalized E(R)–algebra which is mixed and let T  denote the torsion submodule of A. It follows from Theorem 13.6.1 (a) that T = P ∈S R/P nP for some set of prime integers nP ; let V denote the corresponding direct  ideals S and n P product V := P ∈S R/P . By Theorem 13.6.1 (b) now follows that A is an extension of an ideal D by an S–pure subalgebra of V which contains T ; the ideal D consists of those elements in A which have infinite P –height for all P ∈ S. It is well–known (see e.g. [173, Vol. 1, Chapter 58]) that A can be embedded as a pure submodule in its cotorsion completion A• = Ext(Q/R, A) and that this latter splits as T • ⊕ F • where, as before, T is the torsion submodule of A, and F is the torsion–free quotient A/T (see e.g. [173, Vol. 1, Lemma 55.2]). Since F is torsion–free, we know from [173, Vol. 1, Theorem 52.3 ] that Ext(Q/R, F ) ∼ = Hom(Q/R, D/F ), where D is the injective hull of F . Moreover, Hom(Q/R, D/F ) ∼ =

 P

Hom(RP ∞ , D/F )

497

13.7 E(R)–modules with cotorsion

∼ =



Hom(RP ∞ ,



RP ∞ ),

mP

P

where mP is the P –rank of D/F . From [173, Vol. 1, Proposition 44.3] we conclude that   (mP ) , F• ∼ (13.7.1) R = Hom(Q/R, D/F ) ∼ = P P

 (mP ) denotes the P –adic complea torsion–free pure–injective module, where R P tion of the free R–module R(mP ) . We continue to investigate T • for the generalized E(R)–algebra A: Lemma 13.7.1. The cotorsion completion of T is isomorphic to a direct summand of the cotorsion completion of V , V • ∼ = T • ⊕ Y for some Y . Moreover, V • ∼ =V.

Proof. It is easy to see that V /T is torsion–free and so, if we consider the short exact sequence 0 → T → V → X → 0, where X := V /T , we get an induced sequence 0 → T • → V • → Ext(Q/R, X) → 0. ∼ Hom(Q/R, D/X), where D is the Since X is torsion–free, Ext(Q/R, X) = injective hull of X. But it follows easily, see (13.7.1), that this latter is isomorphic to a direct product of P –adic completions of free R–modules for various prime ideals P . In particular it is torsion–free and since T • is cotorsion, the extension 0 → T • → V • → Ext(Q/R, X) → 0 splits, i.e. T • is a direct summand of V • , as required.  Furthermore, since V = P ∈S R/P nP , we have V • = Ext(Q/R, V ) ∼ = ∼ =

 P ∈S



Ext(Q/R, R/P nP ) ∼ =

P ∈S

Ext(R/P

np

, R/P

np

)∼ =





Ext(RP ∞ , R/P np )

P ∈S

EndR (R/P np ) ∼ = V.

P ∈S

2 It is now possible to shed some light on the structure of the ideal D of elements of infinite P –height (P ∈ S).

498

13 E(R)–algebras

The generalized E(R)–algebra A is a pure submodule of A• = T • ⊕F • , which by Lemma 13.7.1 and the discussion immediately preceding it is a pure submodule of   &   %   (mP ) = V ⊕ (mP ) V ⊕ R R R(mP ) P . P P ⊕ P ∈S

P

P ∈S /

For any R–module X let rS X denote the set of elements in X which have infinite P –height for all P ∈ S. Then rS is a radical, and obviously      (mP ) ) ⊕ r rS A ⊆ rS (V ⊕ R R(mP ) P . S P P ∈S

P ∈S /

The former term, however, is clearly zero and so we have established a slight extension of Theorem 13.6.1. Proposition 13.7.2. If A is a reduced generalized E(R)–algebra which is a mixed algebra and S its set of relevant prime ideals (see Theorem 13.6.1), then A is an extension  of an S–divisible ideal  D of A by an S–pure subalgebra of the algebra P ∈S R/P nP containing P ∈S R/P nP . Moreover, D is contained in   (mP ) for suitable cardinals m , where the set S c consists of those c R P ∈S

P

P

prime ideals of R not in S. Torsion–free reduced E(R)–modules with cotorsion: The final case to be considered is, when the generalized E(R)–algebra A is torsion– free reduced and contains a submodule isomorphic to the algebra JP of P –adic integers. Since E(R)–algebras of arbitrary large rank with many additional properties have already been constructed (see Theorem 13.2.1 and Theorem 13.4.1), there is no possibility of obtaining a characterization in this case. We can, however, obtain a complete characterization ‘modulo cotorsion–free modules’: Suppose that A is a generalized E(R)–algebra which is torsion–free reduced but not cotorsion–free. Then the set   S := P |P is a prime ideal of R, A has a (pure) submodule isomorphic to JP . is non–empty; we refer to S as the set of relevant prime ideals. Observe that, if some generalized E(R)–algebra A has a submodule isomorphic to JP , then it also has some pure submodule isomorphic to JP , which splits as a pure–injective submodule. Lemma 13.7.3. A torsion–free reduced  generalized E(R)–algebra A does not contain a pure submodule of the form ℵ0 JP for any P ∈ S.

13.7 E(R)–modules with cotorsion

499

Proof. Suppose for a contradiction that A did contain such a pure submodule, say B. Then we have a short exact sequence  → D → 0, 0→A→A  is the completion of A in the natural R–topology and D := A/A.  where A Since  A is torsion–free reduced, the quotient A/A = D is torsion–free and divisible (possibly = 0), see e.g. [181, Theorem VIII.2.1]. Thus we get an induced sequence  A) → Hom(A, A) → Ext(D, A). 0 → Hom(A,  A). Hence A ∼ = Hom(A, A) contains a pure submodule isomorphic to Hom(A, Now it follows from the standard classification of pure–injective modules, which  has also carries over to this situation (see e.g. [181, Proposition XIII.4.5]), that A  = R(κ) P ⊕ Y for some Y with Hom(JP , Y ) = 0. Note that κ is the form A  (ℵ0 ) of B in A  is again pure  and, since the closure B  = R an invariant of A P  and thus a summand, κ ≥ ℵ0 . Since A has a summand isomorphic to JP and A  A) has a summand isomorphic to has a summand R(κ) P , it follows that Hom(A,   Hom(R(κ) P , JP ). But now Hom(R(κ) P , JP ) ∼ = Hom( κ JP , JP ) ∼ = κ JP and  (2κ ) (see [173, Vol. 1, Example 40.1]). Thus A has this latter is isomorphic to R P

  (2κ ) and hence, so also has A. (2κ )  Since R a pure submodule isomorphic to R P P  has a direct summand which is the P – is pure–injective, this would mean that A completion of a direct sum of strictly more than κ copies of R – a contradiction. 2 The P –cotorsion radical of an R–module G is defined by rP G :=



Im φ.

φ: JP →G

Lemma 13.7.4. For each relevant prime ideal P ∈ S exists a decomposition A = JP ⊕ XP , where XP is P –cotorsion–free and fully invariant.

Proof. Firstly we show that there cannot be two decompositions A = A1 ⊕ X1 = (n ) A2 ⊕ X2 with Ai ∼ = JP i , n1 < n2 and P –cotorsion–free Xi . Suppose such decompositions A = Ai ⊕ Xi (i = 1, 2) exist. Then taking P –cotorsion radicals we get rP A = A1 = A2 , and this is wrong. We now construct recursively a sequence B0 := 0, B1 , B2 , . . . of summands of A (i) with B0 & B1 & B2 & . . . and Bi ∼ = JP . If this process was not terminating, then  the union of this sequence would be a pure submodule of A isomorphic to ℵ0 JP . This, however, is impossible since, by the previous Lemma, A has no

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13 E(R)–algebras

 (n) pure submodule isomorphic to ℵ0 JP , and so A = JP ⊕ XP for some P – cotorsion–free module XP and finite integer n. However, as A ∼ = Hom(A, A), it is immediate that n = 1 (n > 0 by P ∈ S). If now Hom(XP , JP ) = 0, then XP is fully invariant. If not, there exists a non– zero mapping σ : XP → JP and then Hom(XP , JP ) contains σπ : XP → JP for every P –adic integer π and thus contains a copy of JP . So A has a summand with at least two copies of JP – a contradiction. Thus XP is fully invariant as claimed. 2

Theorem 13.7.5. If A is a torsion–free reduced generalized E(R)–algebra, then for each relevant prime ideal P ∈ S there exists a decomposition A = AP ⊕ XP with AP ∼ = JP , where AP and the complement XP are unique. We have AP = rP A and XP is the largest P –divisible submodule of A. Moreover, XP is itself a generalized P –cotorsion–free E(R)–algebra.

Proof. By Lemma 13.7.4 exists a decomposition A = AP ⊕ XP with AP ∼ = JP , where XP is always P –cotorsion–free and fully invariant. Thus AP = rP A and also XP are unique (see e.g. [173, Vol. 1, Corollary 9.7]). Since XP is P –cotorsion–free and fully invariant, Hom(AP , XP ) = Hom(XP , AP ) = 0, and AP ⊕ XP = A ∼ = Hom(A, A) ∼ = JP ⊕ Hom(XP , XP ). Thus, using the uniqueness of AP and XP , we get that XP ∼ = Hom(XP , XP ), and XP is a generalized E(R)–algebra as claimed. If XP is not P –divisible, then there exists some P –reduced element x ∈ XP . Now we have an obvious embedding σ : x∗ → AP ∼ = JP of the pure submodule x∗ generated by x, which lifts to a non–zero homomorphism σ : XP → AP by the pure–injectivity of AP – a contradiction to Hom(XP , AP ) = 0. Thus XP is P –divisible, and it is the largest P –divisible submodule of A, as A = AP ⊕ XP .2

Corollary 13.7.6. If A is a torsion–free reduced generalized E(R)–algebra with a finite set S of relevant prime ideals, then A is a split–extension of a cotorsion–  JP . free S–divisible ideal D, which is itself an E(R)–algebra, by the algebra P ∈S

Proof. The result follows immediately from Theorem 13.7.5 by finite repetition. 2 Now it is also easy to see that the following holds.

13.7 E(R)–modules with cotorsion

501

Corollary 13.7.7. If A is a torsion–free reduced module over a complete discrete valuation domain R, then A is a generalized E(R)–algebra, if and only if A ∼ = R.

Proof. In one direction the proof is immediate. Conversely, suppose that A is an E(R)–algebra. Then the set of relevant prime ideals has exactly one member. Moreover, the only cotorsion–free R–module is the trivial module 0. Now the result follows immediately from Corollary 13.7.6. 2 Lemma 13.7.8. Any torsion-free reduced generalized E(R)–algebra A contains  a submodule of the form JP . P ∈S



AP will have this form, if we can show that  AP & A holds for every finite subset this sum is direct. But this is obvious, as

Proof. Clearly the submodule

P ∈S

P ∈S 

S  ⊆ S by finite repetition of Theorem 13.7.5.

2

Theorem 13.7.9. If A is a torsion–free reduced generalized E(R)–algebra, S the set of prime ideals P for which A contains a submodule isomorphic to JP and D the largest S–divisible submodule of A, then A is an extension of D by an algebra B with 

JP ⊆S B ⊆S JP =: Π. T := P ∈S

P ∈S

Moreover, (a) D is a cotorsion–free ideal and the intersection of a family of generalized E(R)–algebras. (b) B contains the identity of Π and is hence an E(R)–algebra.

Proof. It follows from Theorem 13.7.5 that for P ∈ S each element a of A can be expressed uniquely as a = jP + xP , where jP ∈ AP ∼ = JP , xP ∈ XP , and so the mapping sending a to the vector (. . . , jP , . . . ) is a well–defined algebra– homomorphism φ of A into Π. The kernel of this mapping φ is precisely D := XP . Since each XP is P –cotorsion–free, the intersection D is S–cotorsion– P ∈S

free and  hence, as no other prime ideals are relevant, is cotorsion–free. Note that φ  AP acts as the identity, since an element of the form jP ∈ AP when P ∈S

expressed as jI + xI (P = I ∈ S) must have jI = 0, because jI together with jP is I–divisible. Thus we have 

 T = AP φ ⊆ B = A/ Ker φ ⊆ Π. P ∈S

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13 E(R)–algebras

Since φ is an algebra–homomorphism, D is an ideal which is clearly the intersection of a family of generalized E(R)–algebras, and B is a subalgebra with 1. of A, every S–divisible subSince XP is the largest P –divisible submodule module of A must be contained in D = XP . We prove that D is S–divisible P ∈S

and so it is the largest S–divisible submodule of A. If P, P  ∈ S are two different prime ideals, then A = AP ⊕ AP  ⊕ X holds similarly to Theorem 13.7.5. Now, as XP and XP  are uniquely determined, XP = AP  ⊕ X and XP  = AP ⊕ X. Thus X = XP ∩ XP  is P –cotorsion–free and P –divisible, and so also D = P  ∈S (XP ∩ XP  ) must be P –divisible. Next we show that B is S–pure in Π. Since T is S–purein Π and Π/T is S– divisible, this is equivalent to saying that B/T ∼ AP is S–divisible. = (A/D)/ P ∈S  AP is S–divisible as can easily be seen: for P ∈ S, the facBut already A/ P ∈S  AP is P –divisible being a factor of the P –divisible XP by Theorem tor A/ P ∈S

13.7.5. The final claim that B is an E(R)–algebra follows from the general fact proved in Proposition 13.7.10 below. 2 Proposition 13.7.10. If B is a subalgebra with 1 of the product

JP ⊆S B ⊆S

P ∈S



 P ∈S

JP such that

JP ,

P ∈S

then B is an E(R)–algebra. ∼ Proof. If φ ∈ EndR B, then φ  JP = bP∈ JP for every P ∈ S, as JP = JP . With b = (. . . , bP , . . .) ∈ EndR JP is a fully invariant submodule of P ∈S     JP now φ  JP = b holds. But JP is dense in JP , and since P ∈S

P ∈S

P ∈S

P ∈S

maps are continuous, φ acts as multiplication by b on B with 1φ = b ∈ B.

2

Theorem 13.7.9 and Proposition 13.7.10 have an easy but remarkable Corollary 13.7.11. If A is a torsion–free reduced generalized E(R)–algebra, then A contains for every prime ideal P a submodule isomorphic to JP , if and only if A is isomorphic to some subalgebra 

JP ⊆S B ⊆S JP P ∈S

P ∈S

13.7 E(R)–modules with cotorsion

503

with 1, where S is the spectrum of prime ideals of R.

Proof. Observe that here D has to be cotorsion–free and divisible, thus D = 0.2 If the ideal D =

P ∈S

XP actually splits, then we can deduce a good deal more:

Corollary 13.7.12. If the ideal D splits, then A is a split–extension of the S– divisible cotorsion–free generalized E(R)–algebra D by an E(R)–algebra B, where 

JP ⊆S B ⊆S JP . P ∈S

P ∈S

Proof. If A = D ⊕ B, then Hom(D, B) = 0, since D is S–divisible and B is S–reduced. Now let σ : B → D be a homomorphism. Then T ⊆ Ker σ by the cotorsion–freeness of D, and σ is induced by a homomorphism σ  : B/T → D. Thus σ = σ  = 0, as B/T is divisible and D is reduced, and Hom(B, D) = 0 holds. So we have D⊕B =A∼ = Hom(A, A) = Hom(D, D) ⊕ Hom(B, B) ∼ = Hom(D, D) ⊕ B, where Hom(D, D) is obviously the largest S–divisible submodule. Thus D ∼ = Hom(D, D) as required. 2

The construction of E(R)–algebras with cotorsion Our final result in this section is a partial converse to Theorem 13.7.9; we show the existence of many E(R)–algebras A with non–splitting ideal D. The proof is an interplay of cotorsion and cotorsion–free, very similar to the correspondence of torsion and torsion–free in the proof of Theorem 13.6.5. It is interesting to observe that also the results are very similar: Theorem 13.7.13. Let R be a Dedekind domain and let S be an infinite proper subset of its spectrum of prime ideals. Moreover, let P0 be a fixed prime ideal not belonging to S. We choose any cardinal λ ≥ |R| with λℵ0 = λ. Then, provided that RP0 is not a complete discrete valuation domain, there exists a generalized E(R)–algebra A of cardinality λ such that A is an extension of a cotorsion–free ideal D by an algebra B, where

 T := JP ⊆S B ⊆S JP := Π. (13.7.2) P ∈S

Moreover, A does not split over D.

P ∈S

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13 E(R)–algebras

First we need again a pull–back lemma and repeat the arguments from Lemma 13.6.7, including once more the two diagrams: Lemma 13.7.14. Let T :=



JP ⊆S B ⊆S

P ∈S



JP =: Π,

P ∈S

where B is an S–pure subalgebra with 1 of Π and suppose that C is an S–divisible cotorsion–free generalized E(R)–algebra with a fixed algebra–isomorphism φ : C → EndR C satisfying: (a) There is a fully invariant ideal D of C and C/D ∼ = B/T . (b) The induced map C ∼ = EndR C −→ EndR (C/D) maps every c ∈ C to the right multiplication by the residue–class c + D. Then the pull–back A below is a generalized E(R)–algebra. 0 −−−−→ ⏐ ⏐  0 −−−−→ T −−−−→   

D ⏐ ⏐ 

D ⏐ ⏐ 

A −−−−→ ⏐ ⏐ 

C ⏐ ⏐ 

−−−−→ 0

0 −−−−→ T −−−−→ B −−−−→ B/T ∼ = C/D −−−−→ 0 ⏐ ⏐ ⏐ ⏐   0

0.

Proof. Note that B is automatically an E(R)–algebra, i.e. there exists an isomorphism ψ : B → EndR B mapping b to the right multiplication bR , and that T is fully invariant in B. By hypothesis every endomorphism of C induces an endomorphism on C/D; similarly for B and B/T . Similar to Lemma 13.6.7 we may form another pull–back, say X, of EndR C and EndR B and we have a composite diagram as below, where the mapping C/D → EndR (C/D) is the canonically induced map, and the homomorphism from A to X is the mapping coming from the universal property of pull–backs. A straightforward check using the pull–back definition shows that X may be identified with those endomorphisms of A which induce endomorphisms on both B and C. Moreover, our hypotheses ensure that the composite diagram above is commutative and so, by the pull–back property, A is isomorphic to X and may be identified with this set of endomorphisms on A. As can be seen from the first diagram, A/D ∼ = B and A/T ∼ = C, and so the set of endomorphisms of A inducing

505

13.7 E(R)–modules with cotorsion

C

M

A

X C/D

B

End C

End C/D

O

End B

endomorphisms on B, C is precisely the set of endomorphisms of A leaving D and T invariant. Now B/T is torsion–free and isomorphic to C/D, thus we may conclude that D is pure in the S–divisible torsion–free module C. Since A/D ∼ = B, and B is S–reduced, we conclude that D is the maximal S–divisible submodule of A and hence is invariant under all endomorphisms of A. Moreover, A/T ∼ = C, and C is cotorsion–free. We claim that T must be fully invariant. To see this, recall from [201] the notion of the hyper–cotorsion radical of a module: a module M is said to be hyper–cotorsion, if every non–trivial epimorphic image contains a non–trivial cotorsion submodule, and the submodule hM is the hyper–cotorsion radical of M , if hM is hyper–cotorsion and the quotient M/hM is cotorsion–free. The pair (hyper–cotorsion, cotorsion–free) is a cotorsion–pair. Cotorsion–pairs were discussed in Section 11. Further details of this notion may be found in [201]. Hence we see that T is the hyper–cotorsion radical of A and, as a radical, is fully invariant in A. Hence every endomorphism of A leaves D and T invariant, and so A is identified with the full endomorphism algebra of A, i.e. A is a generalized E(R)– algebra. 2

Proof of Theorem 13.7.13. Since S is a proper infinite set of prime ideals, we may choose E := RP0 , where P0 ∈ / S. Then, since the localization at P0 is not a complete discrete valuation domain, E is P0 –cotorsion–free torsion–free S–divisible, and so we can apply the realization theorems for cotorsion–free E(R)–algebras (see Theorem 13.2.1 and Theorem 13.4.1). Thus we may obtain an E(R)–algebra  the P0 –adic completion C such that C is an E–algebra with E[X] ⊆ C ⊆ E[X], of the polynomial ring over the set X of cardinality λ with coefficients from the ring E. If D is taken as the ideal of C consisting of those (infinite) sums with zero  Since Π/T is divisible torsion–free, we can constant term, then E ⊆ C/D ⊆ E.  and hence C/D, as a subalgebra B/T of Π/T for some S–pure subidentify E, algebra B of Π with 1. With this choice of B, T, C and D apply Lemma 13.7.14.

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13 E(R)–algebras

This yields the desired generalized E(R)–algebra A. Finally observe that A does not split over D, for if it did, the corresponding projection onto D would be multiplication by a non–zero idempotent in A. However, the image of 1 under this idempotent must lie in D, and so there would exist a non–zero idempotent sum with zero constant term; this is clearly impossible, and so we conclude that A does not split over D, as required. 2

13.8

Generalized E(R)–algebras

In this and the following two sections we will show the existence of generalized E(R)–algebras A which are not E(R)–algebras for particular S–rings R. For this we will assume that S = p is cyclically generated (by p). In the proof of Step Lemma 13.10.14 we will also use the following additional condition on R which implies that R is S–cotorsion–free. Definition 13.8.1. An R–module M is ΣS–incomplete, if for any sequence 0 =  / M . If M = R R, mn ∈ M (n ∈ ω) there are an ∈ {0, 1} with n∈ω pn an mn ∈ we say that ring R is ΣS–incomplete. All S–rings of size < 2ℵ0 are ΣS–incomplete as shown in Theorem 1.1.20 (see the construction of the elements wα before Proposition 1.1.18). Thus it follows easily that any S–ring R whose additive group R+ is a direct sum of S–invariant subgroups of size < 2ℵ0 is ΣS–incomplete as well. So we deduce a Corollary 13.8.2. If the additive group R+ of an S–ring R is a direct sum of S– invariant subgroups of size < 2ℵ0 , then R is ΣS–incomplete. In particular, if the additive group R+ is free and S = p generates the ordinary p–adic topology (for some prime p ∈ Z ⊆ R), then R is ΣS–incomplete. We will construct non–commutative R–algebras A (over a ΣS–incomplete ring R) with EndR A ∼ = A. Hence these As are generalized E(R)–algebras but not E(R)–algebras (see Proposition 13.1.9). The existence of generalized E–rings answers a problem in Schultz [357] and Vinsonhaler [391]. If κ is a cardinal, then let κo := {α | cf(α) = ω, α < κ} again. Here we will prove the following Theorem 13.8.3. Let R be a ΣS–incomplete ring for some cyclically generated S. If κ > |R| is a regular uncountable cardinal and E ⊆ κo a non–reflecting subset with ♦κ E, then there is an S–cotorsion–free non–commutative R–algebra A of cardinality |A| = κ with EndR A ∼ = A. Moreover, any subset of cardinality < κ is contained in an R–monoid–algebra of cardinality < κ.

13.9 Model theory for generalized E(R)–algebras

507

Recall from Theorem 9.1.5 that the set theoretic assumptions hold for the (proper) class of regular uncountable, not weakly compact cardinals in Gödel’s universe. Theorem 13.8.3 will appear in Göbel, Shelah [237]. The parallel result in ZFC is based on similar ideas and applications of the Strong Black Box 9.2.2 and will appear in [212]: Theorem 13.8.4. Let R be a ΣS–incomplete S–ring for some cyclically generated S. For any cardinal κ = μ+ with |R| ≤ μℵ0 = μ there is an S–cotorsion–free non–commutative R–algebra A of cardinality |A| = κ with EndR A ∼ = A. The cardinal conditions in Theorem 13.8.4 is inherited from applications of the Strong Black Box 9.2.2. It seems particularly interesting to note that the R–monoid structure of A comes from Kleene’s (classic) λ–calculus taking into account that elements of a generalized E(R)–algebra A are at the same time endomorphism of A, thus the same phenomenon appears as known from computer science and studied intensively in logic in the thirties of the last century. The problems concerning the semantics of computer science were solved four decades later by Scott [358, 359]. We will describe the construction of the underlying monoid M explicitly. We will also elaborate the needed details coming from model theory. The model theoretic background appears for example in Rothmaler [346].

13.9 Model theory for generalized E(R)–algebras Discussion We begin by defining terms for a skeleton and will establish a connection with λ–calculus. By definition of generalized E(R)–algebras A, endomorphisms of R A must be considered as members of A. Hence they act on R A as endomorphisms while they are elements of R A at the same time. Thus we will introduce the classical definitions from λ–calculus over an infinite set X of free variables and an infinite set Y of bound variables to represent those maps. First note that we can restrict ourselves to unary, linear functions because endomorphisms are of this kind. (The general argument to reduce λ–calculus to unary functions was observed by Schönfinkel, see [32, p. 6].) What are the typical terms of our final objects, the bodies? If x1 and x2 are members of the generalized E(R)–algebras A and a, b ∈ R, then also polynomials like σn (x1 , x2 ) = axn1 + bx32 belong to the algebra A, so there are legitimate functions pn (y) = λy.σn (y, x2 ) on A mapping y to σn (y, x2 ), and A must be closed under such ‘generalized polynomials’. This observation will be described in Definition 13.10.2 and taken care of in Proposition 13.9.20 and in our Main Lemma 13.10.16. A first description of these

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13 E(R)–algebras

generalized polynomials will also be the starting point of our construction and we begin with its basic settings.

The general notion of terms Let τ be a vocabulary with no predicates; thus τ is a collection of function symbols with an arity–function τ −→ ω defining for every function symbol its number of places, where the 0–place functions consist of an individual constant 1 and an infinite set X of free variables. Then the collection of unspecified (τ, X)–terms (briefly called ‘terms’) is defined inductively as the closure of the atomic terms under these function symbols (only), that is (i) The atomic terms: these are the 0–place functions. (ii) The closure: if σ0 , . . . , σn−1 are terms and F is an n–place function symbol from τ , then F (σ0 , . . . , σn−1 ) is a term. We also define inductively the (usual) length l(σ) of a term σ: let l(σ) := 0, if σ is atomic and l(σ) := k + 1 with k := max{l(σi ) : i < n}, if σ derives from (ii). If σ is an unspecified (τ, X)–term, then we also define inductively the finite subset F V (σ) ⊂ X of free variables of σ: (a) If σ is an individual constant, then F V (σ) := ∅ and if σ ∈ X, then F V (σ) := {σ}.  (b) If σ = F (σ0 , . . . , σn−1 ) is defined as in (ii), then F V (σ) := i


(σ, x) := σ(σ0 , . . . , σn−1 ), Sub σ00 ,...,σn−1 n−1  replacing every occurrence of xi by σi . If we replace (if necessary) free variables of the σi , we can find a sequence x with (σ(σ0 , . . . , σn−1 ), x ) ∈ t(τ, X). This is a

13.9 Model theory for generalized E(R)–algebras

509

good place for two standard notations: let b = b0 , . . . , bn −1  be a finite sequence of elements without repetition from a set B. If n = n and if (σ, x) is as above, we say that b is a sequence from B (suitable) for x and write σ(b) = Sub xb (σ, x). A free variable x ∈ X is a dummy variable of (σ, x), if x ∈ Im(x) \ F V (σ), and we say that (σ, x) is X–reduced, if it has no dummy variables, i.e. F V (σ) = Im(x). Trivially, for any (σ, x) we get a natural X–reduced term by removing those entries of x that correspond to dummy variables. In this case x = x0 , . . . , xn−1  becomes x = xi0 , . . . , xit−1  for some 0 ≤ i0 < i1 < · · · < it−1 ≤ n−1 and we can use substitution to replace x by the more natural sequence xi ,...,xi



t−1     x = x0 , . . . , xt−1 : if σ  := Sub x00,...,xt−1  (σ, x ), then (σ, x ) = (σ , x ) (by an axiom below).

Model theory of skeletons via λ–calculus The vocabulary of a skeleton and its laws Let Y be an infinite set of so–called bound variables (used as variables for function symbols) and (as before) let y = y0 , . . . , yn−1  be a finite sequence of elements from Y without repetition. Also in this particular case of the vocabulary τ sk of a skeleton the collection τ sk will consists of an individual constant 1, of variables and of function symbols (only) defined inductively as τksk (k ∈ ω); moreover, let   sk := sk sk sk sk := τ sk . τ
sk sk The theory of skeletons, i.e. the axioms T
(i) (Step k = 0) If x ∈ X, then 1x = x1 = x and 1 · 1 = 1 belong to T0sk .

510

13 E(R)–algebras

(ii) (Step k = m + 1) Tksk comprises the following laws: (a) If (σ, x) ∈ t(τ sk , X), x0 ∈ F V (σ) and Fσ(y0 ,...,yn ) is a function symsk \ τ sk related to the term (σ, x), then bol in τ≤k
Fσ(y0 ,...,yn ) (x1 , . . . , xn ) = Fσ(y0 ,...,yn ) (x1 , . . . , xn ). (c) If π is an injective map {1, . . . , n} −→ ω \ {0} and σ  (x0 , x1 , . . . , xn ) := σ(x0 , xπ(1) , . . . , xπ(n) ), then Fσ(y0 ,...,yn ) (x1 , . . . , xn ) = Fσ (y0 ,...,yn ) (xπ(1) , . . . , xπ(n) ). sk for i = 1, 2 and T sk ( (σ , x) = (σ , x), then (d) If (σi , x) ∈ τ≤m 1 2 ≤m

Fσ1 (y0 ,...,yn ) (x1 , . . . , xn ) = Fσ2 (y0 ,...,yn ) (x1 , . . . , xn ). (e) If σ, σ  , σ  are terms but not function symbols, then products are associative, i.e. (σσ  )σ  = σ(σ  σ  ). Remark 13.9.1. (a) Recall that T ( (σ, x), means that (σ, x) follows by the axioms T . For convenience (as for free groups) we denote the empty product by 1. (b) Using the notion of λ–calculus for (ii) (a), the unary function Fσ(y0 ,...,yn ) (x1 , . . . , xn )

is

λy0 .σ(y0 , x1 , . . . , xn )

and it acts as xλy0 .σ(y0 , x1 , . . . , xn ) = σ(x, x1 , . . . , xn ). sk (k ≤ ω) are equations; thus we have an immediate imporThe axioms in T
13.9 Model theory for generalized E(R)–algebras

511

sk is a variety with vocabulary τ sk for (k ≤ Observation 13.9.2. The theory T
Proof. See Grätzer [254, p. 167] or Bergman [51, Chapter 8].

2

We immediately derive one of our central definitions. sk and τ sk = τ sk be as in Observation 13.9.2. Definition 13.9.3. Let T sk := T<ω <ω sk Any T –model (an algebra satisfying T sk ) is called a skeleton, and two skeletons are called isomorphic, if they are isomorphic as T sk –models (see e.g. [51, p. 262] or [346, p. 5]).

For applications it is useful to recall the following sk –skeleton. Definition 13.9.4. of (free) generators of a T
For any set B we will construct a skeleton B freely generated by B. For this we need Reduction of terms Freeness can easily be checked by the usual rewriting process (as in group theory). sk , X) its reduced form red(σ, x) := Thus we define for each term (σ, x) ∈ t(τ
The reduction of terms (i) If σ = x ∈ X, then σ r = x and if σ = 1, then σ r = 1.

(13.9.2)

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13 E(R)–algebras

(ii) If σ = σ  σ  and σ  , σ  are reduced, but σ  is not of the form Fσ0 (y0 ,...,yt ) (σ1 , . . . , σt ), then σ r = σ is reduced. (iii) Suppose that σ  and σi (i ≤ t) are reduced and σ = Fσ (y0 ,...,yt ) (σ1 , . . . , σt ), sk , X) the corresponding specified term. First with (σ  , x0 , . . . xt ) ∈ t(τ
|u|

) σi

: i ∈ u.

See below for a normalization. (iv) If σ = σ  σ  and σ  , σ  are reduced terms, but σ  is a unary function of the form Fσ0 (y0 ,...,yt ) (σ1 , . . . , σt ), then σ r = σ0 (σ  , σ1 , . . . , σt ). We are ready for the sk ) is reduced, if σ r = σ. Definition 13.9.5. An unspecified term σ ∈ t(τ
x = x1 , . . . , x|u|  and use substitution σ  = Subx1 ,...,x

|u| 

sk (σ r , x ), thus T
implies (σ r , x ) = (σ  , x ) by (13.9.1) (ii) (b) and (c). In (iv) we order the union of the free variables F V (σi ) (i ≤ t) after making them pairwise disjoint by free substitutions. Thus we have a definition and a consequence of the last considerations. Definition–Observation 13.9.6. Every term (σ, x) can be reduced to a (normalsk ( red(σ, x) = (σ, x). Let tr (τ sk , X) be ized) reduced term red(σ, x) with T
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sk , X) (so in particular function Thus we consider only elements from tr (τ
Definition 13.9.7. Using induction, we say when two reduced elements σ1 , σ2 ∈ . sk ) are essentially equal and will write σ = σ2 . tr (τ
Observation 13.9.8. sk , X). tr (τ
. (a) The relation = is an equivalence relation on the set

.  sk and σ = sk , X) for σi ∈ tr (τ
Proof. This is immediate by induction using (13.9.1).

2

Using normalization of x and b from Definition-Observation 13.9.6, we now can deduce a sk –model and (σ, x) ∈ t(τ sk , X) with red(σ, x) = Proposition 13.9.9. If M is a T
Note that reduction of terms is defined for each k ≤ ω, thus formally it depends sk , X) ⊆ (τ sk , X) for all h ≤ k ≤ ω. Next we show that the on k. Moreover, (τ
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sk ), then T sk ( σ = σ r . (c) If σ ∈ t(τ
The skeleton has the following important property. Corollary 13.9.11. For σ1 , σ2 ∈ τ sk the following are equivalent: (a) T sk ( σ1 = σ2 . . (b) σ1r = σ2r .

Proof. (b) −→ (a): from the Definition–Observation 13.9.6 follows T sk ( σ1 = σ1r , σ2 = σ2r and by Observation 13.9.8 is T sk ( σ1r = σ2r , thus T sk ( σ1 = σ2 . . (a) −→ (b): from (a) follows T sk ( σ1r = σ2r . Thus σ1r = σ2r by Definition 13.9.7 and Observation 13.9.8. 2 The skeleton freely generated by X Next we construct and discuss free skeletons based on reduced terms. We will use the infinite set X of free variables to construct a skeleton MX which is freely generated by a set which corresponds by a canonical bijection to X. . sk By Observation 13.9.8 (a) we have an equivalence relation = on τ sk := τ<ω sk with equivalence classes [σ] for any σ ∈ t(τ ). Let MX = {[σ] | σ ∈ t(τ sk )}. The equivalence classes [σ] of atoms σ are singletons by Definition 13.9.7 (i). If B = {[x] | x ∈ X}, then ι : X −→ B (x → [x]) is a bijection and we see that MX is a skeleton with basis B ⊆ MX . Moreover, [ ] is compatible with the application of function symbols: If F = Fσ ∈ τ sk with (σ, x) ∈ t(τ sk , X) and x = x0 , . . . , xn−1  is an . n–place function symbol and σi = σi (i < n), then .  F (σ0 , . . . , σn−1 )r = F (σ0 , . . . , σn−1 )r by Observation 13.9.8 (b), thus F ([σ0 ], . . . , [σn−1 ]) = [(F (σ0 , . . . , σn−1 )] (σi ∈ tr (τ sk )) is well–defined as follows from Observation 13.9.8 (b). We have the following

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Theorem 13.9.12. If X is an infinite set (of free variables) and MX is defined as above, then the following holds.   (a) MX = [σ] | σ ∈ tr (τ sk ) . (b) MX is a skeleton with n–place functions [F ] : (MX )n −→ MX ,



 [σ0 ], . . . , [σn−1 ] →

[F (σ0 , . . . , σn−1 )]

for each n–place function symbol F = Fσ(y0 ,...,yn−1 ) for (σ, x) ∈ tr (τ sk , X) with F V (σ) = {x0 , . . . , xn−1 }. (c) MX is freely generated by B = {[x], x ∈ X}, called the free skeleton over X. Using ι above we identify B and X.

Proof. The axioms (13.9.1) are satisfied, e.g. the crucial condition (ii) (a) follows by definition of [F ]. 2

Remark 13.9.13. In the construction of the free skeleton MX we also used an infinite set Y of bound variables. However, it follows by induction that another infinite set Y  of bound variables leads to an isomorphic copy of MX . Thus we do not mention Y in Theorem 13.9.12. sk –model M for k ≤ ω. Then B is a Lemma 13.9.14. Let B be a subset of the T
Proof. If B is a basis of M , then by Definition 13.9.4 the two conditions of the lemma hold (see Proposition 13.9.10 (a) and (c) for (a)). Conversely suppose that (a) and (b) hold. It is easy to extend inductively a bijection B −→ X to an isomorphism between M and the free skeleton MX as in Proposition 13.9.10. Thus B is a basis. 2

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Model theory of bodies The vocabulary of bodies and their laws Let R be a fixed S–ring with S ⊆ R cyclically generated by an element p as sk is the vocabulary of explained in the introduction of Section 13.8. Recall that τ
(13.9.3)

(i) Tksk ⊆ Tkbd . sk is an n–place function symbol, a ∈ R (i ≤ t) and (ii) Linearity: If F ∈ τ
F (x1 , . . . , xl−1 ,

t

ai xli , xl+1 , . . . , xn )

i=1

=

t i=1

ai F (x1 , . . . , xl−1 , xli , xl+1 , . . . , xn ).

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(iii) The usual module laws: Let a, b ∈ R and w, y, z ∈ M (M a Tkbd –model). (a) 0 + y = y, z + y = y + z, w + (y + z) = (w + y) + z. (b) 1y = y, a(by) = (ab)y, a(z + y) = az + ay, (a + b)y = ay + by and y + (−1)y = 0. bd is a variety with vocabulary τ bd for each Observation 13.9.15. The theory T
Proof. See Grätzer [254, p. 198, Theorem 3] or Bergman [51, Chapter 8].

2

bd and τ bd := τ bd be as in Observation 13.9.15. Definition 13.9.16. Let T bd := T<ω <ω bd Any T –model (an algebra satisfying T bd ) is called a body and two bodies are isomorphic if they are isomorphic as T bd –models (see e.g. [51, p.262] or [346]).

Observation 13.9.17. Any (generalized) E(A)–algebra is a body.

Proof. Generalized E(A)–algebras satisfy EndR A = A. Thus any function symbol Fσ of τ bd can be interpreted on A as a function and the axioms (13.9.1) and (13.9.3) hold. 2

But note that only free bodies derive from skeletons. bd , X) Linearity of unary body functions from t(τ
(b) al ∈ R for l < t,  bd ( σ = (c) and T
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Proof. (We will now suppress the index ‘< k’.) We let (σ, x) ∈ t(τ bd , X) and prove the lemma by induction on the length of σ. If σ is atomic, then (σ, x) is a monomial and there is nothing to show. If F is an m–place function symbol from τ bd and σ = F (σ0 , . . . , σm−1 ) with F V (σ l ) ⊆ F V (σ), then by induction hypothesis for σl there are polynomials σl = i
Proof. This is an easy induction on the length of σ. Case 1: if σ = 1 and σ = x0 , then the claim holds trivially. Case 2: if σ = F (σ0 , . . . , σm ), then the claim follows from the axioms (13.9.3) (ii) of T bd . Similarly, if σ = F+ (σ1 , σ2 ) and σ = Fa (σ1 ), then the linearity follows by definition of these functions and induction hypothesis. 2 Recall the notion from λ–calculus in Remark 13.9.1 (b). Proposition 13.9.20. The weak completeness of bodies. Let M be a body and (σ, x) a polynomial (a term in t(τ bd , X)) with x = x0 , . . . , xm  and d1 , . . . , dm ∈ M . Then there is (σ  , x1 , . . . , xn ) ∈ t(τ bd , X) and the following holds: (a) M is an R–module. (b) The unary function λ.zσ(z, d1 , . . . , dm ) : M −→ M (z → σ(z, d1 , . . . , dm )) is the R–endomorphism λy.yσ  (z, d1 , . . . , dm ) ∈ EndR (M ) (y → yσ  (d1 , . . . , dm )). Remark: we will show here that there is a function symbol (σ  , x) ∈ t(τ bd , X) such that σ(d, d1 , . . . , dm ) = dσ  (d1 , . . . , dm ) for all d ∈ M (see axioms (13.9.1) (ii) (a) of the skeletons).

Proof. By the axioms (13.9.3) of T bd , it is clear that M is an R–module. It

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remains to show (b). Let (σ, x ) ∈ t(τ bd , X) with x = x0 , . . . , xm  and x = x1 , . . . , xm . By Lemma 13.9.18, there are monomials (σl , x ) ∈ t(τ sk , X) and al ∈ R such that al σl . σ= l
From the construction of τ sk , we also have function symbols Fσl (y0 ,...,ym ) satisfying the axioms (13.9.1) and (13.9.3). Thus σl (x ) = x0 Fσl (y0 ,...,ym ) (x), and we put σ  (x) =



al Fσl (y0 ,...,ym ) (x).

l
For (b) it now remains to show σ(d, d1 , . . . , dm ) = dσ  (d1 , . . . , dm ) for all d ∈ M which will follow from σ(x0 , x) = x0 σ  (x). We use the three displayed formulas and calculate al σl (x0 , x) = al (x0 Fσl (y0 ,...,ym ) (x)) σ(x0 , x) = l
= x0



l
 al Fσl (y0 ,...,ym ) (x) = x0 σ  (x).

l
Hence (b) follows.

2

From skeleton to the bodies The monoid structure of skeletons Recall from Theorem 13.9.12 that the skeleton on an infinite set X of free variables is the set MX = {[σ] | σ ∈ tr (τ sk )} with n–place functions   [F ] : (MX )n −→ MX , [σ0 ], . . . , [σn−1 ] → [F (σ0 , . . . , σn−1 )] for each n–place function symbol F = Fσ(y0 ,...,yn−1 ) with (σ, x) ∈ tr (τ sk ) and F V (σ) = {x0 , . . . , xn−1 }. To ease notations, we will also write Roman letters for the members of MX , e.g. m = [σ] ∈ MX . The set MX has a distinguished element 1 = [1], and m1 = 1m = m holds for all m ∈ MX (thus MX is an applicative structure with 1). In order to turn MX into a monoid, we first represent MX as a submonoid of Mono(MX ) (the injective maps on MX ), say ι : MX −→ Mono(MX ):

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Let a = [σ] ∈ MX and σ  ∈ t(τ bd ). We will use induction: if a = [1], then = σ  , if a = [x], then [σ  ]a = σ  x and if a = [Fσ(y0 ,...,yn−1 ) x1 , . . . , xn−1 ] is a unary function as above, then [σ  ]a = σ(σ  , x1 , . . . , xn−1 ) (so aι = [λy.yσ]). Thus aι maps any m = [σ  ] ∈ MX to m(aι) = m(λy.yσ) = [mσ] ∈ MX , which can be represented by a reduced element using (13.9.2). If a = b ∈ MX , then 1(aι) = a = b = 1(bι) thus ι : MX −→ Mono(MX ) ⊆ MX is an embedding. We define multiplication of elements a, b ∈ MX as composition of functions (aι)(bι) = (ab)ι. This is to say that from a = [σ], b = [σ  ] we get the product as the equivalence class of λy.((yσ  )σ). We will write a · b = ab and will often suppress the map ι. From Mono(MX ) follows that also MX is a monoid. Also note that [x][x ] = [x ][x] for any free variables x, x ∈ X. We get an [σ  ]a

Observation 13.9.21. The free skeleton (MX , ·, 1) with the composition of functions as product is a non–commutative (associative) monoid with multiplication defined as above by the action on MX : if [σ], [σ  ] ∈ MX , then [σ  ] · [σ] = [λy.((yσ  )σ)]. Free bodies from skeletons Finally, we will associate with any skeleton M its (canonical) body BR M : let BR M be the R–monoid algebra RM of the monoid M . Moreover, any n–place function F : M n −→ M extends uniquely by linearity to F : BR M n −→ BR M . We deduce a Lemma 13.9.22. If R is a commutative ring as above and M a skeleton, then the R–monoid algebra BR M of the monoid M is a body. If the skeleton MX is freely freely generated by X as a body. Moreover, its generated by X, then also BR M is  R–module structure is R BR MX = m∈M Rm.

Proof. It is easy to see that BR M (with the linear n–place functions) is a body. We first claim that X, viewed as {[x] : x ∈ X} ⊆ BR MX is a basis. First apply Lemma 13.9.18 to the R–monoid BR MX : any (σ, x) ∈ t(τ bd , X) can be written  sk as a polynomial σ = l al σl with monomial (σl , x) ∈ t(τ , X). Moreover, any σl is viewed as an element of Mono(MX ), so axiom (13.9.1) (ii) (a) applies and σl becomes a product of elements from X. Thus X generates BR MX . The monomials of the skeleton M extend uniquely by linearity to polynomials of the 2 free R–module R BR MX = m∈M Rm from its basis M . We will also need the notion of an extension of bodies. Definition 13.9.23. Let B and B be two bodies, then B ≤ B (B extends B), if and only if B ⊆ B as R–algebras and if (σ, x) ∈ t(τ bd , X) and Fσ is a function symbol

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521

with corresponding unary, R–linear function F of B , then its natural restriction to B is the function for B corresponding to Fσ . Example 13.9.24. Let X ⊆ X  be sets of free variables and B, B be the free bodies generated by the free skeletons obtained from X and X  , respectively. Then B ≤ B . In this case we say that B is free over B.

13.10 Constructing proper generalized E(R)–algebras Technical tools for the construction The endomorphism ring EndR BR M of the R–module R BR M has natural elements as endomorphisms, the linear maps, our (generalized) polynomials interpreted by the specified terms σ ∈ t(τ bd , X) acting by scalar multiplication on BR M as shown in Proposition 13.9.20 (b). The closure under these polynomials is governed by the properties of E(R)–algebras. Thus we would spoil our aim to construct generalized E(R)–algebras, if we ‘lost these R–linear maps’ on the way. Definition 13.10.1. Let (σ, x) ∈ t(τ bd , X) with x = x0 , . . . , xn . If B is a body, d = d1 , . . . , dn  with d1 , . . . , dn ∈ B, then we call σd (y) := λy.σ(y, d) the (generalized) polynomial over B with coefficients d. Similarly, for (σ, x) ∈ t(τ sk , X) we call σd (y) := λy.σ(y, d) the (generalized) monomial over B with coefficients d. Note that σd (y) is a sum of products of elements di and y. Here we must achieve (full) completeness of the final body, thus showing that any endomorphism is represented. By a prediction principle we will kill all endomorphisms that are not represented by t(τ bd , X), thus the resulting structure will be complete: any R–endomorphism of an extended body B will be represented by a generalized polynomial q(y) belonging to B, hence B will be complete or, equivalently, a generalized E(R)–algebra. The fact that BR M is not just an R–linear closure (or A–linear closure for some algebra A), makes this final task to get rid of undesired endomorphisms harder than in the case of realizing algebras as endomorphism algebras (where the closures are not that floppy). Definition 13.10.2. Let B be a body and G := R B. Then ϕ ∈ EndR G is called represented ( by q(y) ), if there is a generalized polynomial q(y) with coefficients in B, such that gϕ = q(g) for all g ∈ G. If all elements from EndR G are represented, then B is a generalized E(R)– algebra.

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As for other algebraic structures, we have the following Lemma 13.10.3. Let R be a principal ideal domain and B := BR M be a body, then B ⊆ B is a basis of B, if B generates B and if one of the following equivalent conditions holds. (a) If B := BR MX is the body generated by the free skeleton MX and X → B is a bijection, then this map extends to a T bd –isomorphism B → B. (b) B is independent in B, i.e. if (σ1 , x), (σ2 , x) ∈ tr (τ bd , X) and the sequence b from B is suitable for x such that σ1 (b) = σ2 (b), then T bd ( σ1 = σ2 . (c) For all bodies H and maps ϕ : B → H there is an extension ϕ : B → H as T bd –homomorphism.

Proof. This is well–known for varieties (see Grätzer [254, p. 198, Theorem 3] or Bergman [51, Chapter 8]), so it follows from Observation 13.9.15. 2 Freeness Proposition 13.10.4. Let R be an S–ring as above, X ⊆ X  be sets of variables and BR MX ⊆ BR MX  be the corresponding free bodies. If u ∈ B := BR MX and v ∈ X  \ X, then w := u + v ∈ B := BR MX  is free over B, i.e. there is a basis X  of B with w ∈ X  ⊇ X.

Proof. We will use Lemma 13.10.3 (c) to show that the set X  := (X  \ {v}) ∪ {w} is a basis of B . First note that X  also generates B , thus B = BR MX  . Given ϕ : X  → H for a body H, we must extend this map to ϕ : B → H. Let ϕ := ϕ (X  \ {v}) and note that the set X  \ {v} = X  \ {w} is independent. Thus if B0 := BR MX  \{v} , then ϕ extends to ϕ : B0 → H by freeness, and from u ∈ B0 follows the existence of uϕ ∈ H. We now define ϕ: if ϕ  B0 := ϕ , then ϕ (X  \ {v}) = ϕ = ϕ (X  \ {v}). Thus it remains to extend ϕ to ϕ : B → H in such a way that wϕ = wϕ. If wϕ =: h ∈ H, then we must have h = wϕ = (u + v)ϕ = uϕ + vϕ. Hence put vϕ := h − uϕ = h − uϕ . Now ϕ : B → H exists, because X  is free, ϕ ⊆ ϕ and wϕ = (u + v)ϕ = uϕ + h − uϕ = h = wϕ, thus ϕ ⊆ ϕ holds as required. 2 The following corollaries (used several times for exchanging basis elements) are immediate consequences of the last proposition. Corollary 13.10.5. Let X be a basis of the body B, v ∈ X, B be the subbody of B generated by X \ {v}, and w ∈ B , then X  := (X \ {v}) ∪ {v + w} is another basis of B.

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Corollary 13.10.6. If X ⊆ X  and BR MX ⊆ BR MX  , then any basis of BR MX extends to a basis of BR MX  .

Proof. If X  is a basis of BR MX , then (X  \ X) ∪ X  is a basis of BR MX  . The easy argument is left as an exercise. 2 The last corollaries have another implication. Corollary 13.10.7. Suppose that Bα (α ≤ δ) is an ascending, continuous chain of bodies such that Bα+1 is free over Bα for all α < δ. Then Bδ is free over B0 , and if B0 is free, then Bδ is free as well. The proof of the next lemma is also obvious. It follows by application of the distributive law in T bd and collection of summands with r. Lemma 13.10.8. Let q(y) be a generalized polynomial and r ∈ R. Then there is a polynomial q  (y, y  ) such that q(y1 + ry2 ) = q(y1 ) + rq  (y1 , y2 ).  Proof. By Lemma 13.9.18 we can write σ = l 1. Also let, without loss of generality, n be maximal for the chosen monomial m1 (y). By the distributive law the monomials of the polynomial q(g + v1 + v2 ) include those monomials induced by m1 (y) replacing all entries of the variable y by

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arbitrary choices of v1 and v2 . Let m1 be one of these monomials. If there are more such monomials mi (i ≤ k) like m1 obtained by substitutions into monomi als mi (y) of q(y) with ki=1 mi = 0, then replacing all v1 s and v2 s by ys gives k  i=1 mi (y) = 0 contradicting the minimality of the above sum. Thus m1 represents a true monomial (not canceled by others) of q(g + v1 + v2 ), and as n > 1 we may also assume that v1 and v2 both appear in m1 . This monomial does not exist on the right–hand side of the equation in the lemma – a contradiction. Thus (a) holds. (b) First in the given equation substitute v1 := y, v2 := 0 and v1 := 0, v2 := y, respectively. Thus q(g + y) = q1 (y) + c and q(g + y) = q2 (y) + c , where c := q2 (0), c := q1 (0) ∈ B0 . Subtraction now yields 0 = q1 (y) − q2 (y) + (c − c ), 2 thus q1 (y) − q2 (y) = c − c does not depend on y, as required.

Preparing the Step Lemmas for the construction In order to establish the Step Lemmas, we next prepare some preliminary results.  Let Xω := n∈ω Xn be a strictly increasing sequence of infinite sets Xn of free variables and fix a sequence v0 ∈ X0 , vk ∈ Xk \ Xk−1 (0 < k < ω) of elements. Moreover, let Mα := MXα be the skeleton and Bα := BR Mα be the body generated by Xα for α ≤ ω, respectively. Note that by our identification Bα is an R–algebra and, restricted to the module structure, Gα := R Bα is an R–module, which is free by Lemma 13.9.22. Recall that S = p generates our S–topology on S–reduced R–modules for some p ∈ R (with n∈ω pn R = 0). Thus the S– topology is Hausdorff on Gα , and Gα is naturally an S–pure R–submodule of its ω . If α and pick particular elements wn ∈ G α ; we write Gα ⊆∗ G S–completion G lk ∈ N (k < ω) is increasing and ak ∈ {0, 1}, then we define ω plk −ln ak vk ∈ G (13.10.1) wn := wn (vk , lk , ak ) := k≥n

and easily check that wn − pln+1 −ln wn+1 = an vn ∈ Gn for all n ∈ ω.

(13.10.2)

Proposition 13.10.10. Let ak ∈ {0, 1} and lk ∈ N be as above. If Xω+1 := (Xω \ {vn | an = 1, n > 0}) ∪ W with W := {wn |an = 1, n > 0} and Mω+1 := MXω+1 , Bω+1 := BR Mω+1 , Gω+1 := R Bω+1 , then the following holds: ω . (a) Gω ⊆∗ Gω+1 ⊆∗ G (b) Gω+1 /Gω is p–divisible, thus an S−1 R–module.

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(c) Xω+1 is a basis of the (free) skeleton Mω+1 . freely generated by the skeleton Mω+1 , thus Bω+1 = (d) The R–algebra Bω+1 is RMω+1 and Gω+1 = m∈Mω+1 Rm. (e) Bω+1 is free over Bn (as body). ω . From wn , wn+1 ∈ ω and Gω+1 ⊆ G Proof. Claim (a): Clearly Gω ⊆∗ G Xω+1 , an = 1 and (13.10.2) follows vn ∈ Gω+1 . Hence vn ∈ Gω+1 for all n ∈ ω ω /Gω , and Gω+1 ⊆∗ G ω and Gω ⊆ Gω+1 follows at once. Thus Gω+1 /Gω ⊆ G follows, if Gω+1 /Gω is p–divisible. This is our next Claim (b): by definition of the body Bω+1 , any element g ∈ Gω+1 is the sum of monomials in Xω+1 . If wn , wn+1 are involved in such a monomial, then we apply (13.10.2) and get wn = pln+1 −ln wn+1 + an vn , which is wn ≡ pln+1 −ln wn+1 mod Gω . Let m be the largest index of wn s which contributes to g. By the above we can remove all wi of smaller index i < m and also write wm ≡ pwm+1 mod Gω . Thus g + Gω is divisible by p, and Gω+1 /Gω is an S−1 R–module. Claim (c): it is enough to show that Xω+1 is free, because Xω+1 generates Mω+1 by definition of the skeleton. First we claim that for an = 1 X  := (Xω \ {vn }) ∪ {wn } is free (n ∈ ω). If wn is a finite sum, then X  is free by Corollary 13.10.5. Otherwise we apply the characterization of a basis by Lemma 13.10.3 (b). Let (σ1 , x), (σ2 , x) ∈ tr (τ bd , X) with x = x1 , . . . , xk  be such that σ1 (y) = σ2 (y) for some yi ∈ X  , y = y1 , . . . , yk  and suppose that y1 = wn (there is nothing to show, if wn does not appear among the yi s, because X  \ {wn } is free; without loss of generality we can relabel the yi s such that y1 = wn ). Now we consider the above ω and note that the support [yi ] ⊆ Xω of the elements equation as an element in G yi (i > 1) is finite, while wn as an infinite sum has infinite support{vk | k ≥ n, ak = 0}. Thus we project σi (y) onto a free variable v from [wn ] \ 1
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y1 , . . . , ym ∈ Wn := {wi | ai = 1, i ≤ n} and ym +1 , . . . , ym ∈ Xω+1 \ W . Hence   y1 , . . . , ym ∈ Xω \ {v1 , . . . , vn } ∪ Wn , which is free by the last claim (13.10.3). Thus Xω+1 is free and (c) follows. (d) is a consequence of (c) and Lemma 13.9.22. (e) We note that (by (c)) the body Bω+1 is freely generated by Xω+1 and also (using (13.10.2)) by the (free) set X  := (Xω+1 \ {wi | ai = 1, i ≤ n}) ∪ {vi : ai = 1, i ≤ n}. However, Xn ⊆ X  which generates Bn , hence (e) also follows. 2 Throughout the remaining part of this section we use the notations from Proposition 13.10.10. Moreover, we assume the following, where we consider Bω as R–algebra: Let ϕ ∈ EndR Gω \ Bω with Gn ϕ ⊆ Gn for all n ∈ ω.

(13.10.4)

Lemma 13.10.11. Let ϕ be as in (13.10.4). If w0 ϕ ∈ Gω+1 , then the following holds: (a) There exist m ∈ ω and a generalized polynomial q0 (y) over Bω such that w0 ϕ = q0 (wm ). (b) There exists an n∗ > m such that q0 is a polynomial over Bn∗ .

Proof. (a) If w0 ϕ ∈ Gω+1 , then there exists some m ∈ ω such that w0 ϕ ∈ BR MXω ∪{w0 ,...,wm } . Using wi ≡ pwi+1 mod Gω it follows that w0 ϕ ∈ BR MXω ∪{wm } , and there is a generalized polynomial q0 (y) over Bω such that w0 ϕ = q(wm ). (b) The coefficients of q0 are in some Bn∗ for some n∗ > m. 2

The three Step Lemmas for constructing generalized E(R)–algebras We will use the notations from Proposition 13.10.10 and (13.10.4) and begin with our first Step Lemma, which will stop ϕ from becoming an endomorphism of our final module.

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13.10 Constructing proper generalized E(R)–algebras

Step Lemma 13.10.12. Let G := Gω and ϕ ∈ EndR G be an endomorphism as in (13.10.4) such that / BR MXk ∪{gk } , for all k ∈ ω there exists some gk ∈ Xk+1 \ Xk with gk ϕ ∈ and let Gω+1 be defined with wn := wn (gk , lk , 1) as in (13.10.1) for suitable elements lk ∈ ω. Then ϕ does not extend to an endomorphism in EndR Gω+1 .

Proof. We inductively define an ascending sequence lk ∈ ω: set l−1 := 0. If C ⊆ Gω is a submodule, then the S–closure of C is defined by  C := (pn Gω + C). n∈ω

It is then closure of C in the S–adic topology, which is Hausdorff on Gω (i.e. n∈ω p Gω = 0). In particular, C is S–closed (C = C), if C is a summand of Gω (e.g. C = 0 is S–closed). / BR MXk ∪{gk } and BR MXk ∪{gk } is a summand By hypothesis we have gk ϕ ∈ of Gω , so it is closed in the S–topology. There is an l ∈ ω such that gk ϕ ∈ / l BR MXk ∪{gk } +p Gω . If lk−1 is given, we may choose lk := l such that lk > 3lk−1 . Hence lk ∈ ω and Gω+1 are well–defined, and Proposition 13.10.10 holds; in particular, Gω ⊆∗ Gω+1 , thus plk Gω+1 ∩ Gω = plk Gω , and we have found a sequence / BR MXk ∪{gk } + plk Gω+1 . lk ∈ ω with lk+1 > 3lk and gk ϕ ∈

(13.10.5)

Suppose now for contradiction that ϕ ∈ EndR Gω extends to an endomorphism of Gω+1 ; this extension is unique, and we call it also ϕ ∈ EndR Gω+1 . In particular w0 ϕ ∈ Gω+1 ; by Lemma 13.10.11 (a), (b) there is a generalized polynomial q0 (y) with coefficients in Bn∗ for some n∗ ∈ ω and with w0 ϕ = q0 (wm ) for some m ∈ ω. We choose n > max{n∗ , m} and use (13.10.1) to compute w0 : w0 =

n

plk −l0 gk + pln+1 −l0 wn+1 .

k=0

Application of ϕ gives w0 ϕ ≡

n−1

plk −l0 (gk ϕ) + pln −l0 (gn ϕ) mod pln+1 −l0 Gω+1 .

k=0

If k < n, then gk ∈ Gn and gk ϕ ∈ Gn by the choice of ϕ. The last equality becomes w0 ϕ ≡ pln −l0 (gn ϕ) mod (pln+1 −l0 Gω+1 + Gn ), hence q0 (wm ) ≡ pln −l0 (gn ϕ)

mod (pln+1 −l0 Gω+1 + Gn ).

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Finally, we determine pln −l0 (gn ϕ) in terms of BR MXn ∪{gn } : From n > m and the definition of wm in (13.10.1) we get wm =

n−1

plk −lm gk + pln −lm gn + pln+1 −lm wn+1 ,

k=m

thus q0 (wm ) ≡ q0

 n−1

plk −lm gk + pln −lm gn



mod pln+1 −lm Gω+1 ,

k=m

and pln −l0 (gn ϕ) ≡ q0 (wm ) ≡ q0

 n−1

plk −lm gk + pln −lm gn



mod (pln+1 −lm Gω+1 + Gn ).

k=m

Now we use (again) that gk ∈ Gn for all k < n and that q0 (y) is a polynomial with coefficients in Bn . The last equation can be reduced to pln −l0 (gn ϕ) ∈ BR MXn ∪{gn } + pln+1 −lm Gω+1 , hence gn ϕ ∈ BR MXn ∪{gn } + pln+1 −lm −ln Gω+1 . Note that ln+1 > 3ln by the choice of the ln s, hence ln+1 − lm − ln > ln , thus we get a formula gn ϕ ∈ BR MXn ∪{gn } + pln Gω+1 that contradicts (13.10.5), and Step Lemma 13.10.12 follows.

2

Step Lemma 13.10.13. Let ϕ ∈ EndR G be an endomorphism as in (13.10.4). Moreover, suppose there are elements uk , gk ∈ Xk+1 \ Xk (for each k ∈ ω) with gk ϕ = qk1 (gk ) and uk ϕ = qk2 (uk ), where qk1 , qk2 are generalized polynomials over B0 such that qk1 (y) − qk2 (y) ∈ / B0 , i.e. y appears in the difference. If Gω+1 is defined with wn := wn (gk + uk , lk , 1) as in (13.10.1) for suitable elements lk ∈ ω, then ϕ does not extend to an endomorphism in EndR Gω+1 .

Proof. Let ck := gk +uk . The set X  := (Xω \{uk | k < ω})∪{ck | k < ω} is now a basis of Bω by Corollary 13.10.5, thus Proposition 13.10.10 applies and Gω+1 is well–defined. By definition of wn and (13.10.2) we have pln+1 −ln wn+1 +cn = wn , and as in the proof of Step Lemma 13.10.12 we get plk −l0 (ck ϕ) = q0 ( plk −lm ck ) w0 ϕ = q0 (wm ) =⇒ k≥0

k≥m

13.10 Constructing proper generalized E(R)–algebras

529

for n∗ , m, q0 (y) as in Step Lemma 13.10.12. Furthermore, ck ϕ = (gk + uk )ϕ = gk ϕ + uk ϕ = qk1 (gk ) + qk2 (uk ). Thus



  l −lm   plk −l0 qk1 (gk ) + qk2 (uk ) = q0 pk (gk + uk ) ,

k≥0

k≥m

where qk1 (gk ) ∈ R BX0 ∪{gk } and qk2 (uk ) ∈ R BX0 ∪{uk } , and arguments similar to Lemma 13.10.9 apply: in every monomial of q0 (y) the variable y appears at most once as there are no mixed monomials on the left–hand side, and the same holds for qk1 (y), qk2 (y). Furthermore, the variable y does not appear in qk1 (y) − qk2 (y), 2 which contradicts our assumption on the qki s. The next lemma is the only place where we will use that R is ΣS–incomplete in order to find a sequence ak ∈ {0, 1} (see Definition 13.8.1). Recall that this condition follows by Corollary 13.8.2, if R+ is a direct sum of S–invariant subgroups of size < 2ℵ0 . Hence it will be sufficient, if R+ is free and S defines the usual p–adic topology on R (for some prime p ∈ Z ⊆ R). Step Lemma 13.10.14. Let R be a ΣS–incomplete S–ring, let ϕ ∈ EndR G be an endomorphism as in (13.10.4) and let q(y) be a generalized polynomial over B0 such that gϕ − q(g) ∈ G0 for all g ∈ G. Moreover, suppose that for all k ∈ ω there are elements gk ∈ Xk+1 \ Xk such that gk ϕ − q(gk ) = 0. If Gω+1 is defined with wn := wn (gk , lk , ak ) as in (13.10.1) for suitable elements lk ∈ ω, ak ∈ {0, 1}, then ϕ does not extend to an endomorphism in EndR Gω+1 .

Proof. Choose gk ∈ Xk+1 as in the lemma, and put hk := gk ϕ − q(gk ) = 0. By assumption on ϕ and q(y) it follows that hk ∈ G0 . Let wn := wn (gk , lk , ak ) be defined as in (13.10.1) for a suitable sequence of elements ak ∈ {0, 1} and lk := k for all k ∈ ω. We again define Gω+1 as in Proposition 13.10.10, using the new choice of elements wn . Note that G0 is a free R–module. By the assumption that R is ΣS–incomplete there is now a sequence ak ∈ {0, 1}   k  pk ak hk ∈ / G0 . However, with k∈ω k∈ω p ak hk ∈ G0 by the choice of k / Gω+1 by the definition of Gω+1 . Recall w0 := h k∈ω p ak hk ∈ k , hence k k∈ω p ak gk and suppose that w0 ϕ ∈ Gω+1 . We compute w0 ϕ =

 k∈ω

 pk ak gk ϕ = pk ak (gk ϕ) k∈ω

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13 E(R)–algebras

and

pk ak hk =

k∈ω

k∈ω

=



k∈ω



pk ak (gk ϕ − q(gk )) pk ak (gk ϕ) −



pk ak q(gk ) = w0 ϕ −

k∈ω



pk ak q(gk ).

k∈ω



From k∈ω pk ak hk ∈ / Gω+1 follows k∈ω pk ak q(gk ) ∈ / Gω+1 . However, by the definition of bodies, the map taking g → q(g) for any g ∈ Gω+1 is an enk , and also w = domorphism of G ω+1 0 k∈ω p ak gk ∈ Gω+1 , hence q(w0 ) =  k / Gω+1 , and ϕ does not k∈ω p ak q(gk ) ∈ Gω+1 is a contradiction. Thus w0 ϕ ∈ extend to an endomorphism of Gω+1 . 2

The final stage: construction of generalized E(R)–algebras

 Lemma 13.10.15. Let κ be a regular, uncountable cardinal and B := α∈κ Bα a κ–filtration of bodies. Also let Gα := R Bα and G := R B. Then the following holds for any ϕ ∈ EndR G: (a) If there is g ∈ G such that gϕ ∈ / (Bα ){g} , then there is also h ∈ G free over / (Bα ){h} . Bα such that hϕ ∈ (b) If there are g ∈ G and a generalized polynomial q(y) over Bα such that / gϕ−q(g) ∈ / (Bα ){g} , then there is also h ∈ G free over Bα with hϕ−q(h) ∈ (Bα ){h} .

Proof. If g ∈ G satisfies the requirements in (a) or (b), respectively, then choose any element h ∈ G which is free over Bα . If h satisfies the conclusion of the lemma, then let h := h , and the proof is finished. Otherwise let h := h + g which is also free over Bα by Proposition 13.10.4. In this case h ϕ ∈ (Bα ){h } or h ϕ − q(h ) ∈ (Bα ){h } , respectively. It follows that hϕ ∈ / (Bα ){h} or hϕ − q(h) ∈ / 2 (Bα ){h} , respectively. The next lemma is based on results of the last section concerning the Step Lemmas and Lemma 13.10.15. We will construct first the κ–filtration of Bα s for application using ♦κ E for some non–reflecting subset E ⊆ κo . Construction of a κ–filtration of free bodies: The  body B (and the R–module G) will be constructed as a κ–filtration B := Bα of bodies. We choose this κ–filtration such that α∈κ

|Bα | = |α| + |R| = |Bα+1 \ Bα |

13.10 Constructing proper generalized E(R)–algebras

531

for all α ∈ κ, and let {gα | α ∈ E} be the family of Jensen–functions given by ♦κ E. Fix for each α ∈ E a strictly increasing sequence αk ∈ α \ E with sup αk = α. k∈ω

This is possible, because E consists of limit ordinals cofinal to ω only, and we can pick αk as a successor ordinal. We will use the same Greek letter for a converging sequence and its limit, so the elements of the sequence only differ by their suffix. As E is non–reflecting, we may also choose a strictly increasing, continuous sequence αν , ν ∈ cf(α) with sup αν = α and αν ∈ α \ E, ν∈cf(α)

if cf(α) > ω. This is crucial, because the bodies Bα of the (continuous) κ– filtration B must be free in order to proceed by a transfinite construction. This case does not occur for κ = ℵ1 . Using Lemma 13.10.12, Lemma 13.10.13 and Lemma 13.10.14 inductively, we define the body structure on Bα . We begin with B0 := 0, and by the continuity of the ascending chain the construction can be reduced to an inductive step passing from Bα to Bα+1 . We will carry on our induction hypotheses of the filtration at each step. In particular the following three conditions must hold: (i) Bα is a free body. (ii) If β ∈ α \ E, then Bα is a free body over Bβ . (iii) Suppose α ∈ E and that there exists some strictly increasing sequence αk (k ∈ ω) with supk∈ω αk = α such that (13.10.4) and the hypothesis of one of the three Step Lemmas (which we apply one after another: Lemma 13.10.12, Lemma 13.10.13 or Lemma 13.10.14) hold for ϕ := gα and Gk := Gαk (k ∈ ω). Then we identify Bα+1 with Bω+1 from the Step Lemmas (so gα does not extend to an endomorphism of Bα+1 ). We have one more choice for the inductive step from α to α + 1: if the hypotheses of condition (iii) are violated, for instance, if α ∈ / E, we choose Bα+1 := (Bα ){vα } adding any new free variable vα to the body. Next we must check that conditions (i) to (iii) extend to α+1. If the hypotheses of condition (iii) are violated, this is obvious. In the other case the Step Lemmas are designed to guarantee: Condition (i) is the freeness of Bω+1 in Proposition 13.10.10. Condition (ii) needs that Bα+1 is a free body over Bβ . However, Bβ ⊆ Bαk for a large enough

532

13 E(R)–algebras

k ∈ ω. Hence (ii) follows from the freeness of Bαk over Bβ (inductively) and the freeness of Bα+1 over Bαk (by Proposition 13.10.10 and Corollary 13.10.7). In the limitcase we have two possibilities: if cf(α) = ω, then supk∈ω αk = α, hence Bα = Bαk and Bα is a free body with the help of (i) and (ii) by induction k∈ω

(see Corollary 13.10.7). If cf(α) > ω, then by our set theoretic assumption (E is non–reflecting) and we have a limit  supν∈cf(α) αν = α of ordinals αν not in Bαν by (i) and (ii) is again a free body E. The union of the chain Bα = ν∈cf(α)  (see Corollary 13.10.7). Thus we proceed and obtain B = Bα , which is a α∈κ

κ–filtration of free bodies. It remains to show the Main Lemma  13.10.16. Assume ♦κ E. Let κ be a regular, uncountable cardinal and B = α∈κ Bα be the κ–filtration of the bodies just constructed. Also let Gα := R Bα and G := R B. Then every ϕ ∈ EndR G is represented in B.

Proof. Suppose for contradiction that ϕ is not represented in B. Let E ⊆ κo be given from ♦κ E, let {gα : α ∈ E} be the family of Jensen–functions and define a stationary subset Eϕ := {α ∈ E| ϕ  Gα = gα }. Note that C := {α ∈ κ| Gα ϕ ⊆ Gα } and C  := {α ∈ κ| α is limit in C} are cubs, thus Eϕ := Eϕ ∩ C  is also stationary, and for every α ∈ Eϕ holds (13.10.4) for some suitable strictly increasing sequence αk (k ∈ ω) with supk∈ω αk = α as in (iii) above. As a consequence we see that there exists some α ∈ Eϕ satisfying one of the following conditions: / R (Bβ ){g} . (i) For every β < α there is g ∈ Gα such that gϕ ∈ (ii) There is β < α such that gϕ ∈ R (Bβ ){g} for all g ∈ Gα (not case (i)), but for every β < α and every generalized polynomial q(y) over Bβ represented / Gβ . by an endomorphism of G there is g ∈ Gα with gϕ − q(g) ∈ (iii) There is β < α such that gϕ ∈ R (Bβ ){g} and there is a generalized polynomial q(y) over Bβ with gϕ − q(g) ∈ Gβ for all g ∈ Gα (neither (i) nor (ii) holds), but ϕ is not represented by B. By Lemma 13.10.15 we may assume that the elements g in (i), (ii) and (iii), respectively, are free over Bβ . If (i) holds, then there exists some proper ascending sequence αk (k ∈ ω) with supk∈ω αk = α and property (13.10.4) and elements gk ∈ Gαk+1 such that gk is free over Bαk and / R (Bαk ){g} for all k ∈ ω. gk ϕ ∈ We identify Gαk with Gk in Step Lemma 13.10.12 and note that condition (iii) of the construction applies. By construction of Gα+1 (as a copy of Gω+1

13.10 Constructing proper generalized E(R)–algebras

533

from Lemma 13.10.12), the endomorphism ϕ  Gα does not extend to EndR Gα+1 . However, ϕ ∈ EndR G, thus Gα+1 ϕ ⊆ Gγ for some γ < κ. Finally note that Gα+1 is the S–adic closure of Gα in Gγ , because Gα is S–dense in Gα+1 and Gα+1 is a summand of Gγ , hence S–closed in Gγ . We derive the contradiction that indeed ϕ  Gα+1 ∈ EndR Gα+1 . Hence case (i) is discarded. Now we turn to case (ii). Suppose that (ii) holds (in particular condition (i) is not satisfied). In this case there exist an ascending sequence αk (k ∈ ω) with supk∈ω αk = α and property (13.10.4) as above, elements gk , uk ∈ Gαk+1 (also free over Bαk ) and generalized polynomials qk1 (y), qk2 (y) over Bα0 such that gk ϕ = qk1 (gk ) and uk ϕ = qk2 (uk ). Moreover, we can force the polynomials qk1 (y) − qk2 (y) to be non–constant over Bα0 . Step Lemma 13.10.13 and condition (iii) of the construction apply, and we get a contradiction as in case (i). Thus also case (ii) is discarded. Finally, suppose for contradiction that (iii) holds (hence (i) and (ii) are not satisfied). In this case there exist an ascending sequence αk (k ∈ ω) with supk∈ω αk = α and property (13.10.4) as above and elements gk ∈ Gαk+1 (also free over Bαk ) such that gk ϕ − q(gk ) = 0. Moreover, we can force gϕ − q(g) ∈ Gα0 for all g ∈ Gα . We now apply Step Lemma 13.10.14 and condition (iii) of the construction; the argument from case (i) gives a final contradiction. Thus the Main Lemma holds. 2

Proof of Main Theorem 13.8.3. Let B be the body over the S–ring R constructed at the beginning of this section using E as in Theorem 13.8.3; moreover, let G := R B and A be the algebra structure of B. Thus |A| = κ and by the construction and Proposition 13.10.10 any subset of size < κ is contained in an R–monoid–algebra of cardinality < κ; the algebra A is the union of a κ–filtration of free bodies Bα . By Lemma 13.10.16 every element ϕ ∈ EndR A is represented by a generalized polynomial q(y) with coefficients from B (see Definition 13.10.2). Thus gϕ = q(g) for all g ∈ A, and ϕ = q(y) ∈ A. It follows that A = EndR A is the R–endomorphism algebra of A. Finally recall that there is Bα ⊆ A which is an R–monoid–algebra over a non– commutative monoid from Observation 13.9.21. Thus A cannot be commutative either, and Theorem 13.8.3 is shown. 2

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13 E(R)–algebras

Open problems 1. Consider the new notion of a cellular cover in Section 13.1. Try to transfer results from localizations to cellular covers or prove problems known for localizations first for cellular covers. It turns out that cellular covers can be treated more easily; see examples [159, 160, 161]. 2. Show the existence of large E(R)–algebras which are also absolute for suitable rings (e.g. choose R = Z). See Chapter 14 for the notion of absolute structures and for applications of Theorem 14.4.1.

Chapter 14

Modules with distinguished submodules

In many R–module categories we do not find the set S to make R an S–ring as in Definition 1.1.1. Thus the Black Box techniques do not apply. The most striking counterexamples for this drawback are fields. In this case, but also in many other situations, we can circumvent this problem and still realize rings as endomorphism rings by using the Shelah ‘elevator’, see Section 9.3. Let κ be any cardinal, possibly κ < ω and R be any commutative ring. In view of Definition 9.3.1, we will use the following traditional abbreviation. Definition 14.0.1. Rκ − Mod denotes the category of R–modules M with κ distinguished submodules M = (M, M α : α < κ). If κ = n < ω, we will also write M = (M, M 0 , . . . , M n−1 ). Call M as above Rκ –module. Also recall that the homomorphisms between two Rκ –modules M, M must respect their distinguished submodules M k , M k , thus we define   HomR (M, M ) = HomR M, M  ; M k , M κ | k < κ   = σ ∈ HomR (M, M  ) : M k σ ⊆ M k for all k < κ . If X is any R–module, then M ⊗ X = (M ⊗ X, M k ⊗ X : k < κ) is again an Rκ –module, defined by the tensor product over R.

14.1 The five–submodule theorem, an easy application of the elevator We recall from Section 9.3 Shelah’s elevator theorem. Theorem 14.1.1. Let λ be an uncountable, regular cardinal, σ, σn as above, H, G be free R–modules with basis {hi | i < λ} and {gn | n < ω}, respectively. Moreover, let S be the countable collection of the following summands of G ⊗ H: (i) W n = gn ⊗ H

(n < ω).

536

14

Modules with distinguished submodules

(ii) W1n = (g0 − gn ) ⊗ H (0 < n < ω).   (iii) W2 = i<λ j<λ R(g0 ⊗ hi − g1 ⊗ hσ(i,j) ).   (iv) W3 = i<λ j<λ R(g2 ⊗ hi − g3 ⊗ hσ(j,i) ).    (0 < n < ω) (v) W4n = δ<λo R g4 ⊗ hδ − g4n+2 ⊗ hσ(η(n,δ),δ) where, for each δ < λo , a ladder η(δ) has been chosen, that is, η(δ) = (η(n, δ))n<ω is a strictly increasing sequence converging to δ.   (0 < (vi) W5n = n↓ λ, i<λ R g4n+1 ⊗ hσn (τ ) − g4n+3 ⊗ hσn+1 (τ ∧ i) ) i<τ ∈  n<ω . Then Hom(G ⊗ H, G ⊗ H ⊗ M ; S, S ⊗ M ) = idG⊗H ⊗M for every R–module M. We first reformulate the elevator theorem in terms of Rω –submodules M = (M, M i | i < ω), that is of R–modules M over any commutative ring R = 0 with a countable family of distinguished submodules M i . Corollary 14.1.2. Let R = 0 be a commutative ring, λ be any regular, uncountable cardinal and G = n<ω Rgn and H = i<λ Rhi be free R–modules of rank ℵ0 and λ, respectively. If F = G ⊗ H, then there are submodules F n ⊆ F (n < ω) such that the following holds for the Rω –modules F = (F, F n | n < ω). (a) F n has a complement in F , and both F n and its complement have rank λ for each n < ω. These submodules are freely generated by elements of the form gn ⊗ hi and gn ⊗ hi − gm ⊗ hj . (b) If Y is an R–module, then HomR (F, F ⊗ Y ) = 1F ⊗ Y . The last corollary has an immediate consequence. Lemma 14.1.3. If X is an R–module and HomR (F, F ⊗ Y ) = 1F ⊗ Y (is as above), then also HomR (F ⊗ X, F ⊗ Y ) = 1F ⊗ HomR (X, Y ).

Proof. We must only show that any homomorphism ϕ : F ⊗ X −→ F ⊗ Y is of the form 1F ⊗ ψ for some ψ ∈ HomR (X, Y ). There is a homomorphism X −→ HomR (F, F ⊗ Y ) (x → (− ⊗ x)ϕ) because the tensor product is bilinear. The dual map η : Y −→ HomR (F, F ⊗ Y ) (y → 1F ⊗ y)

14.1 The five–submodule theorem, an easy application of the elevator

537

is onto by assumption and clearly injective, hence η −1 exists. If we put η −1

ψ : X −−−−→ HomR (F, F ⊗ Y ) −−−−→ Y, then (f ⊗ x)(1F ⊗ ψ) = f ⊗ (xψ) = (f ⊗ x)ϕ, hence ϕ = 1F ⊗ ψ.

2

Our next lemma is almost the same as Lemma 14.2.10. Lemma 14.1.4. Let ρ ≤ λ be cardinals and A be an R–algebra which is generated by at most ρ elements (as an algebra) and let X, Y be (right)∗ A–modules. If H = i<λ hi R is a free R–module,  then there  are submodules X ⊆ H ⊗ X and Y ∗ ⊆ H ⊗ Y R–isomorphic to ρ X and ρ Y , respectively, such that   HomA (X, Y ) = ϕ ∈ HomR (X, Y ), X ∗ (1H ⊗ ϕ) ⊆ Y ∗ .

Proof. Let {ai | i < ρ} be a set generating A as R–algebra. Choose two disjoint subsets of the free generators of H of size ρ and label these elements as hi , hi (i < ρ). If i < ρ, then Xi := {hi ⊗ x + hi ⊗ xai | x ∈ X} is a submodule of H ⊗ X ∗ isomorphic to X, hence X = i<ρ Xi∗ is a canonical summand of H ⊗ X which is isomorphic to ρ X. Similarly we define Y ∗ as a submodule of H ⊗ Y . If ψ ∈ HomA (X, Y ), then it is immediate that X ∗ (1αβ ⊗ ψ) ⊆ Y ∗ . It remains to show that any ϕ ∈ HomR (X, Y ) with X ∗ (1H ⊗ ϕ) ⊆ Y ∗ is an A– homomorphism. For x ∈ X we can find y ∈ Y such that (hi ⊗ x + hi ⊗ xai )(1H ⊗ ϕ) = hi ⊗ (xϕ) + hi ⊗ (xai )ϕ = hi ⊗ y + hi ⊗ yai . Thus xϕ = y, (xai )ϕ = yai = (xϕ)ai for all i < ρ, and ϕ is an A–homomorphism. 2 The next lemma is the key for passing from Rω –modules to R5 –modules. Let R = 0 be a commutative ring, and let E = Re1 ⊕ Re2 ⊕ Re3 and G = ⊕n<ω Rgn be free R–modules with generators e1 , e2 , e3 and gn (n < ω), respectively. In the tensor product E ⊗R G choose four distinguished submodules: U 1 = Re1 ⊗G,

U 2 = Re2 ⊗G,

U 3 = Re3 ⊗G and U 4 = R(e1 +e2 +e3 )⊗G.

Furthermore, let ρ, ν : ω −→ ω be two order preserving maps such that (i) Im ν ⊆ Im ρ, (ii) Im ρ \ Im ν is infinite, (iii) jump condition: [ν(k), ν(k) + k] ∩ Im ρ = {ν(k)} for all k < ω,

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where [ν(k), ν(k) + k] denotes the interval {n < ω : ν(k) ≤ n ≤ ν(k) + k}. It is easy to construct the functions ρ and ν by induction; the pair ρ, ν remains fixed in what follows. We also let U0 =



R(e1 ⊗ gi + e2 ⊗ gi+1 ) ⊕

i<ω



R(e3 ⊗ gν(i) ) ⊆ E ⊗ G

i<ω

and U 0 =



R(e1 ⊗ gi + e2 ⊗ gi+1 ) ⊕

i<ω



R(e3 ⊗ gρ(i) ) ⊆ E ⊗ G.

i<ω

From (i) we obtain U 0 ⊆ U 0 , while (ii) implies that U 0 /U 0 is a free R–module of infinite rank. We start with the following lemma; its proof is a modification of the proof of Proposition 3.1 in [197]. Lemma 14.1.5. Let R = 0 be a commutative ring, and let X, Y be R–modules. Then the R5 –modules   j = U j ⊗ X (j = 0, 1, 2, 3, 4) , CX = CX = E ⊗ G ⊗ X, CX   C Y = CY = E ⊗ G ⊗ Y, CY0 = U 0 ⊗ Y, CYj = U j ⊗ Y (j = 1, 2, 3, 4) satisfy HomR (CX , C Y ) = 1E⊗G ⊗ HomR (X, Y ). Here U 0 /U 0 is free of infinite rank.

Proof. It is enough to show that the left–hand side is contained in the right–hand side. So assume φ ∈ HomR (CX , C Y ). From the invariance of the subspaces U j (j = 1, 2, 3, 4) we conclude that there exists φj ∈ HomR (G ⊗ X, G ⊗ Y ) (j = 1, 2, 3, 4) such that (ej ⊗ g ⊗ x)φ = ej ⊗ (g ⊗ x)φj (j = 1, 2, 3) and [(e1 + e2 + e3 ) ⊗ g ⊗ x]φ = (e1 + e2 + e3 ) ⊗ (g ⊗ x)φ4 for all g ∈ G, x ∈ X. Comparison yields φ1 = φ2 = φ3 = φ4 = φ , thus φ = 1E ⊗ φ

with φ ∈ HomR (G ⊗ X, G ⊗ Y ).

We note that U 0 ∩ (U 1 ⊕ U 2 ) = U 0 ∩ (U 1 ⊕ U 2 ) =

j<ω

R(e1 ⊗ gj + e2 ⊗ gj+1 ),

14.1 The five–submodule theorem, an easy application of the elevator

539

and therefore, for x ∈ X, we have (e1 ⊗gi +e2 ⊗gi+1 )⊗x ∈ (U 0 ∩(U 1 ⊕U 2 ))⊗X and

R(e1 ⊗ gj + e2 ⊗ gj+1 ) ⊗ Y. e1 ⊗ (gi ⊗ x)φ + e2 ⊗ (gi+1 ⊗ x)φ ∈ j<ω



We also note that G ⊗ X = i<ω Rgi ⊗ X and G ⊗ Y = all i, j < ω, there exist φij ∈ HomR (X, Y ) such that (gi ⊗ x)φ = (gj ⊗ xφij ).

 i<ω

Rgi ⊗ Y . For

j<ω

Hence there are yij ∈ Y (i, j < ω) such that e1 ⊗ (gj ⊗ xφij ) + e2 ⊗ (gj ⊗ xφi+1,j ) j<ω

j<ω 

= e1 ⊗ (gi ⊗ x)φ + e2 ⊗ (gi+1 ⊗ x)φ = (e1 ⊗ gj + e2 ⊗ gj+1 ) ⊗ yij j<ω

=



(e1 ⊗ gj ) ⊗ yij +

j<ω



(e2 ⊗ gj+1 ) ⊗ yij .

j<ω

We get

(e1 ⊗ gj ) ⊗ (xφij − yij ) +

j<ω

(e2 ⊗ gj ) ⊗ (xφi+1,j − yi,j−1 ) j≥1

+ (e2 ⊗ g0 ) ⊗ xφi+1,0 = 0, whence equating coefficients we obtain xφij = yij , xφi+1,j = yi,j−1 (j > 0) and also xφi0 = 0 (i ≥ 1). It follows that xφij = xφi+1,j+1 , whence gj ⊗ xφij = gj ⊗ xφ0,j−i . There is a k < ω such that (gi ⊗ x)φ = gj ⊗ xφij = gi+j ⊗ xφ0j . (14.1.1) j<ω

j≤k

We now keep the element x ∈ X fixed and note that by definition of φ that the set {j < ω | xφ0j = 0} must be finite. Thus k in the formula (14.1.1) is the largest element of this set and does not depend on i. Finally, we observe that



U 0 ∩ U 3 = R(e3 ⊗ gρ(i) ) and U 0 ∩ U 3 = R(e3 ⊗ gν(i) ), i<ω

i<ω

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thus e3 ⊗ gν(k) ⊗ x ∈ U 0 ∩ U 3 ⊗ X and (e3 ⊗ gν(k) ⊗ x)φ ∈



R(e3 ⊗ gρ(i) ) ⊗ Y.

i<ω

If we pick k as in (14.1.1), then we can find elements yj ∈ Y such that e3 ⊗



(gρ(j) ⊗ yj ) = (e3 ⊗ gν(k) ⊗ x)φ = e3 ⊗ (gν(j) ⊗ x)φ

j<ω

= e3 ⊗

(gν(k)+j ⊗ xφ0j ). j≤k

  Again equating coefficients, we have j<ω gρ(j) ⊗ yj = j≤k gν(k)+j ⊗ xφ0j , and by the jump condition (iii) it follows that xφ0i = 0 for 1 ≤ i ≤ k. Hence also xφ0i = 0 for all x ∈ X. We get φ0i = 0 for all 0 = i < ω, and if we put φ00 = ψ, 2 then (gi ⊗ x)φ = gi ⊗ xψ and φ = 1E ⊗ 1G ⊗ ψ, as desired. Using the R5 –modules investigated in Lemma 14.1.5, we now construct the R5 –modules that appear in Main Theorem 14.1.6 of this section. If X is an A– module of the R–algebra A and α ⊆ λ is a subset of λ, then we will define the R5 –module CαX using the modules in Lemma 14.1.5. Recall that from above j CH⊗X = E ⊗ G ⊗ (H ⊗ X) and CH⊗X = U j ⊗ (H ⊗ X) for j = 1, 2, 3, 4. If we put j 0α , CH⊗X | (j = 1, 2, 3, 4)), CαX = (CH⊗X , CH⊗X 0α : then it remains to define CH⊗X 0 0 Since U /U is free of countable rank, we make room and write U 0 = U 0 ⊕ C with

C= (Ran ⊗ G ) ⊕ Rb1 ⊕ Rb2 ⊆ E ⊗ G and G a copy of G. n<ω

Fix an R–isomorphism from G to G , we get an induced isomorphism σ : F = G ⊗ H −→ G ⊗ H, which carries the distinguished submodules F n ⊆ F from Corollary 14.1.2 to F n = F n σ ⊆ F  = G ⊗ H, so Corollary 14.1.2 applies to F .  Let Hα = i<α Rhi and define submodules of (E ⊗ G) ⊗ (H ⊗ X): 0 0 We have CH⊗X ⊕ C ⊗ H ⊗ X = CH⊗X and select 0α 0 = CH⊗X ⊕ CH⊗X

n<ω

0 Ran ⊗ F n ⊗ X ⊕ Rb1 ⊗ X ∗ ⊕ Rb2 ⊗ Hα ⊗ X ⊆ CH⊗X

14.2

The four–submodule theorem, a harder case

541

0α is and note that X ∗ ⊆ H ⊗ X with X ∗ from Lemma 14.1.4. Hence CH⊗X 0 0α 0 sandwiched CH⊗X ⊆ CH⊗X ⊆ CH⊗X . If Y is another A–module and also β ⊆ λ, then ϕ ∈ Hom(CαX , CβY )

is an R–homomorphism E ⊗G⊗H ⊗G⊗X −→ E ⊗G⊗H ⊗G⊗Y which maps j j 0 0 CH⊗X ϕ ⊆ CH⊗Y for j = 1, 2, 3, 4 and CH⊗X ϕ ⊆ CH⊗Y . By Lemma 14.1.5 there is ψ ∈ HomR (H ⊗ X, H ⊗ Y ) such that ϕ = 1E⊗G ⊗ ψ. From G ⊆ E ⊗ G follows (G ⊗ H ⊗ X)ϕ = G ⊗ H ⊗ X(1E⊗G ⊗ ψ) = G ⊗ (H ⊗ X)ψ ⊆ G ⊗ H ⊗ Y. 0α and therefore If n < ω, f ∈ F n and x ∈ X, then an ⊗ f ⊗ x ∈ CH⊗X 0β . an ⊗ (f ⊗ x)ψ = (an ⊗ f ⊗ x)ϕ ∈ CH⊗Y

It follows that (F n ⊗ X)ϕ ⊆ F n ⊗ Y for all n < ω. By the remark above Corollary 14.1.2 applies to ϕ = ϕ  G ⊗H ⊗X, hence ϕ = 1G ⊗H ⊗ψ  for some ψ  ∈ HomR (X, Y ). If g ∈ G , then g ⊗ (h ⊗ x)ψ = (g ⊗ h ⊗ x)ϕ = g ⊗ h ⊗ (xψ  ) for all h ∈ H, x ∈ X, hence (h ⊗ x)ψ = h ⊗ (xψ  ), so ϕ = 1E⊗G⊗H ⊗ ψ  . Also b1 ∈ G and therefore Rb1 ⊗ X ∗ ⊆ Rb1 ⊗ H ⊗ X forces X ∗ ψ ⊆ Y ∗ and by Lemma 14.1.4 follows ψ  ∈ HomA (X, Y ). If α ⊆ β it is immediate that ϕ = 1E⊗G⊗H ⊗ψ  is as required. If α ⊆ β, there is i ∈ β \α and (b2 ⊗hi ⊗x)ϕ = b2 ⊗ hi ⊗ (xψ  ) ∈ Rb2 ⊗ Hβ ⊗ Y holds only if xψ  = 0 for all x ∈ X, thus ϕ = 0 in this case. Also recall that CR is a free R–module of rank λ, hence the following theorem holds. Main Theorem 14.1.6. If λ is an uncountable, regular cardinal A is an R–algebra over a commutative ring R which is generated by at most λ elements, and X, Y are A–modules, then we can find R5 –modules CαX (α ⊆ λ) on a free R–module F of rank λ such that for α, β ⊆ λ the following holds.  1F ⊗ HomA (X, Y ), if α ⊆ β β HomR (CαX , CY ) = 0, if α ⊆ β.

14.2 The four–submodule theorem, a harder case R4 –modules were first investigated in a paper by Kronecker [303], who showed that the category of Q4 –modules is not of finite representation type. This famous result can be considered as the beginning of the evolution of representation theory for algebras. Our Main Theorem 14.2.12 of this section can be seen as the infinite version of Kronecker’s result.

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A main case of our results in this section strengthens the main theorem of the last section. This is the following Corollary 14.2.1. Let R = 0 be a commutative ring, λ any infinite cardinal, and A an R–algebra which is generated by no morethan λ elements. Embed A in EndR M by scalar multiplication, where M = λ A. Then there exists an R4 – module M = (M, M 0 , M 1 , M 2 , M 3 ) with A = End M, where End M = {ϕ ∈ EndR M : M k ϕ ⊆ M k for k < 4}. The proof of this result collects ideas from Shelah [360] (the Shelah Elevator), its extension to modules with distinguished submodules [170] and Ringel [340] for the existence of non-splitting R4 –modules of rank 2. We will follow the presentation in [217]. Families of Rκ –Modules Let κ be any cardinal, λ be an infinite cardinal and Rκ −Mod be defined as above. This is a category with direct sums and enough projectives and injectives. We must define some types of families of Rκ –modules. We define the standard λ–family of R–modules to be a family {Fα | α ⊆ λ} of free R–modules of rank λ such that Fα ⊆ Fβ , if α ⊆ β, and such that there exists an exact sequence 0 −→ F∅ −→ Fλ −→



Qi −→ 0 all Qi (i < λ) free R–modules of rank λ,

i<λ

and which restricts for every α ⊆ λ to an exact sequence 0 −→ F∅ −→ Fα −→



Qi −→ 0.

i<α

It is easy to see that a standard λ–family exists and is unique up to isomorphism. If α ⊆ β ⊆ λ, we denote the inclusion Fα ⊆ Fβ by 1αβ . Observe that 1αβ is a split injection since Fβ /Fα is free. Let F be an Rκ –module. We say that F is λ–free if F k has a complement in F for every k < κ such that the F k s and their complements are free R–modules of rank λ. All the tensor products will be tensor products of R–modules, and we shall write ⊗ instead of ⊗R . If M is an R–module, we let F ⊗ M denote the Rκ –module constructed on F ⊗ M with (F ⊗ M )k = F k ⊗ M for every k < κ. Definition 14.2.2. A rigid λ–family of Rκ –modules {Fα : α ⊆ λ} is a family of Rκ –modules constructed on the standard λ–family {Fα | α ⊆ λ} such that:

14.2

The four–submodule theorem, a harder case

543

(i) the exact sequences of the standard λ–family are the exact sequences

0 −→ F∅ −→ Fα −→ Qi −→ 0 in Rκ − Mod; i<α

(ii) Fα and Qi are λ–free (α ⊆ λ, i < λ) and (iii) if M and N are R–modules and α ∪ β ⊆ λ, then Hom(Fα ⊗ M, Fβ ⊗ N ) = 1αβ ⊗ HomR (M, N ) if α ⊆ β and is 0 if α ⊆ β. In order to construct rigid λ–families, it will be useful to have the notion of a weaker type of family. Let us call {Xα : α ⊆ λ} a weak λ–family of Rκ –modules if: (iv) each Xα is a λ–free Rκ –module, and for each α ⊆ β ⊆ λ there is a split inclusion 1αβ : Xα −→ Xβ , which is an Rκ –homomorphism; (v) whenever α = α0 ∪ α1 is a partition of α, then there exists an exact sequence 0 −→ X∅ −→ Xα0 ⊕ Xα1 −→ Xα −→ 0; (vi) if N is an R–module and α ∪ β ⊆ λ, then Hom(Xα , Xβ ⊗ N ) = 1 ⊗ N if α = β, and is zero, if α \ β is infinite. It is immediate that a rigid λ–family is also a weak λ–family, since condition (i) implies that, if α = α0 ∪ α2 is a partition of α, then Fα is the pushout of Fα0 and Fα1 by F∅ . What is not so clear is that the existence of a weak λ–family implies the existence of a rigid λ–family. Before showing this, we prove the following lemma: Lemma 14.2.3. Condition (iii) in Definition 14.2.2 may be replaced by the following condition: If N is an R–module and α ⊆ β ⊆ λ, then Hom(Fα , Fβ ⊗ N ) = 1αβ ⊗ N .

Proof. We must show that condition (iii) follows from the stated condition, assuming (i) and (ii). First assume that α ⊆ β and that ϕ : Fα ⊗ M −→ Fβ ⊗ N . On the one hand, the bilinearity of the tensor product gives a homomorphism M −→ Hom(Fα , Fβ ⊗ N ) by m −→ (− ⊗ m)ϕ.

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On the other hand, the map N −→ Hom(Fα , Fβ ⊗ N ) given by n −→ 1αβ ⊗ n is onto by hypothesis. But 1αβ is a split injection, hence this map is also injective. The composite of its inverse with the homomorphism above yields a homomorphism ψ : M −→ N such that f 1αβ ⊗ mψ = f (1αβ ⊗ mψ) = (f ⊗ m)ϕ for f ∈ Fα . Thus ϕ = 1αβ ⊗ ψ ∈ 1αβ ⊗ HomR (M, N ), as is required in (iii). Now assume that α ⊆ β and put γ = α ∪ β. By properties of the standard λ–family, there exists a free complement for Fβ in Fγ which contains a basis element f coming from Fα . Let ϕ : Fα ⊗ M −→ Fβ ⊗ N . The composition of ϕ with 1βγ ⊗ 1N : Fβ ⊗ N −→ Fγ ⊗ N must have the form 1αγ ⊗ ψ for some ψ ∈ HomR (M, N ) by the case proved above. Let m ∈ M . Applying the composition to f ⊗ m, we get f ⊗ mψ ∈ Fβ ⊗ N . Thus the choice of F forces mψ = 0, and we conclude ϕ = 0. Thus (iii) is shown. 2

Proposition 14.2.4. If there exists a weak λ–family of Rκ –modules, then there exists a rigid λ–family of Rκ –modules.

Proof. Let {Xα | α ⊆ λ} be a weak λ–family of Rκ –modules. Partition λ into infinite subsets αj for j ∈ λ ∪ {∞}, and define Qj = Xαj . Moreover, for each i < λ, partition αi into two infinite subsets αi0 and αi1 , and put Qi0 = Xαi0 and Qi1 = Xαi1 . For each i < λ, choose an exact sequence 0 −→ X∅ −→ Qi0 ⊕ Qi1 −→ Qi −→ 0

(∗)

according to Definition 14.2.2 (v). Note that each Qi is λ–free, and if N is an R–module, then Hom(Qi , Qi ⊗ N ) = 1 ⊗ N and Hom(Qi , Qj ⊗ N ) = 0 for i, j ∈ λ ∪ {∞}, i = j. The exact sequence (∗) determines a class in Ext1 (Qi , X∅ ) (i < λ), in the category Rκ − Mod. We shall write Ext instead of Ext1 . We have homomorphisms      −−−→ Ext −−−→ Ext i<λ Ext(Qi , X∅ ) − i<λ Qi , X∅ − i<λ Qi , Q∞ , where the left map is a natural isomorphism and the right map is induced by the natural  Q∞ . The class of (∗) for every i determines a class in map X∅ −→ Ext i<λ Qi , Q∞ , hence we may choose an exact sequence 0 −−−−→ Q∞ −−−−→ Fλ −−−−→



i<λ Qi

−−−−→ 0

representing this class. We now define Fα for α ⊆ λ to be the submodule of Fλ such that the above exact sequence restricts to the following exact sequence. 0 −−−−→ Q∞ −−−−→ Fα −−−−→



i<α Qi

−−−−→ 0.

14.2

545

The four–submodule theorem, a harder case

We denote F{i} by Fi . Note that F∅ = Q∞ . It will be useful to know that we have a commutative diagram with exact rows 0 −−−−→ X∅ −−−−→ Qi0 ⊕ Qi1 −−−−→ ⏐ ⏐ ⏐ ⏐   0 −−−−→ Q∞ −−−−→

Qi −−−−→ 0   

−−−−→ Qi −−−−→ 0.

Fi

To see this, consider the commutative square Ext



i<λ Qi , X∅

⏐ ⏐ 

Ext(Qi , X∅ )



−−−−→ Ext

−−−−→



i<λ Qi , Q∞



⏐ ⏐ 

Ext (Qi , Q∞ ) .

  From the way Fλ was obtained, there is a class in Ext i<λ Qi , X∅ that maps across to the class defining Fλ and down to the class of (∗). Thus the class of (∗) maps to the class defining Fi in Ext (Qi , Q∞ ). This yields the desired diagram. Clearly {Fα | α ⊆ λ} is the standard λ–family of R–modules, and Definition 14.2.2 (i) is satisfied. Condition (ii) of Definition 14.2.2 follows from Definition 14.2.2 (i), since Q∞ and Qi are λ–free. Thus we need only to verify the condition in Lemma 14.2.3. Let N be an R–module, α ⊆ β ⊆ λ, and φ : Fα −→ Fβ ⊗ N . Then φ induces a commutative diagram with exact rows 0 −−−−→

Q∞ ⏐ ⏐ 1⊗n

−−−−→



−−−−→

Fα ⏐ ⏐ φ

0 −−−−→ Q∞ ⊗ N −−−−→ Fβ ⊗ N −−−−→



i<α Qi

⏐ ⏐ φ

j<β (Qj

−−−−→ 0

⊗ N ) −−−−→ 0

by first noting that the composition through φ resulting in Q∞ −→

(Qj ⊗ N ) j<β

must be zero, hence φ restricts to a map Q∞ −→ Q∞ ⊗ N . This restriction must be of the form 1 ⊗ n for some  n ∈ N . Consequently, φ induces the map φ, which must be of the form φ = i<α (1 ⊗ ni ) for ni ∈ N . Fix i < α. Replacing φ by φ − (1 ⊗ ni ), and restricting to the exact sequence 0 −−−−→ Q∞ −−−−→ Fi −−−−→ Qi −−−−→ 0,

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we obtain 0 −−−−→

X∅ ⏐ ⏐ 

0 −−−−→

Q∞ ⏐ ⏐ n−ni 

−−−−→ Qi0 ⊕ Qi1 −−−−→ ⏐ ⏐ 

Qi   

−−−−→ 0

−−−−→

Qi ⏐ ⏐ 0

−−−−→ 0

Fi ⏐ ⏐ 

0 −−−−→ Q∞ ⊗ N −−−−→ Fβ ⊗ N

−−−−→

−−−−→



j<β (Qj

⊗ N ) −−−−→ 0.

The zero map on the right forces the middle vertical composition to factor through Qi0 ⊕ Qi1 −→ Q∞ ⊗ N, thus it is also zero. Since X∅ −→ Q∞ splits, and X∅ is a non–trivial free module, it follows that n = ni . It is apparent that Fi is generated by the images of Q∞ and Qi0 ⊕ Qi1 , and that these map to zero in Fβ ⊗ N . Since this is true for every i < α, and since Fα = i<α Fi , we conclude that φ = 1 ⊗ n. 2

Construction of rigid λ–families In this subsection we shall construct rigid λ–families of Rκ –modules for three different types of (κ, λ). First we shall construct an ω–family of R4 –modules. To begin, we construct a family of R[x]–modules. Fix non–constant, monic polynomials pi ∈ R[x] (i < ω), such that given i = j, there exist qi , qj ∈ R[x] with pi qi + pj qj = 1. For example, one may take p0 = x and pi+1 = p0 . . . pi + 1. Let S be the multiplicative subset of R[x] generated by the {pi }. For each subset α ⊆ ω, define Lα = {f /pi0 . . . pik } ⊆ S−1 R[x] which is S–torsion– free of rank 1. Lemma 14.2.5. Let α∪β ⊆ ω, let N be an R–module, and let φ : Lα −→ Lβ ⊗N be an R[x]–homomorphism. If α = β, then ξφ = ξ(1φ) for every ξ ∈ Lα , where 1φ ∈ R[x] ⊗ N . If α \ β is infinite, then φ = 0.

Proof. By partial fractions one sees that pi Lβ has a free complement in Lβ as an R–module. It follows that Lβ ⊗ N is S–torsion–free. Since Lα has rank 1, we may conclude that ξφ = ξ(1φ) for every ξ ∈ Lα . We may think of R[x] ⊗ N as N [x], and of S−1 R[x] ⊗ N as the module localization S−1 N [x]. Thus 1φ = n(x)/s, where n(x) ∈ N [x] and s = pj0 . . . pjk for distinct j0 , . . . , jk < β. If α = β, then the ‘type’ of Lα does not permit denominators in 1φ, hence 1φ ∈ N [x] = R[x] ⊗ N . If α \ β is infinite, then n(x) = 0, hence φ = 0 in this case. 2

14.2

The four–submodule theorem, a harder case

547

Next we define two R–submodules of Lα that will be used  to definejan R4 – module Xα . Let di be the degree of pi , and put Ui = 0≤j
Proof. By Proposition 14.2.4, it suffices to find a weak ω–family of R4 –modules. If α is infinite, it is clear that Xα is an ω–free R4 –module. If α ⊆ β ⊆ ω, then the natural inclusion 1αβ : Xα ⊆ Xβ is a split homomorphism which is also an R4 –morphism. If α = α0 ∪ α1 , then the natural inclusions Xαi −→ Xα and Xα0 ∩α1 −→ Xαi (i = 0, 1) induce the sum map Xα0 ⊕ Xα1 −→ Xα , and the skew diagonal map Xα0 ∩α1 −→ Xα0 ⊕ Xα1 (x → (x, −x)). Since Uα = Uα0 + Uα1 and Uα0 ∩α1 = Uα0 ∩ Uα1 , it is clear that the sum map is onto, and the kernel is the image of the skew diagonal map. In fact, the same is true for each layer of submodules, thus we have the exact sequence of R4 –modules 0 −−−−→ Xα0 ∩α1 −−−−→ Xα0 ⊕ Xα1 −−−−→ Xα −−−−→ 0. Since Xα is ω–free if α is infinite, in order to produce a weak ω–family of R4 – modules, we fix an infinite subset α0 of ω which has an infinite complement. We will then take the modules {Xα | α0 ⊆ α ⊆ ω}, but reindexed by the subsets of ω \ α0 in an obvious fashion. The only remaining thing to be verified is condition (vi) of Definition 14.2.2. Let φ : Xα −→ Xβ ⊗ N . Since Xαi φ ⊆ Xβi ⊗ N for 0 ≤ i ≤ 2, we may regard φ as given by an R–homomorphism φ : Wα −→ Wβ ⊗ N , where the effect on Xα0 is the same as the restriction to Uα ⊆ Wα . Since Xβ3 ⊗ N = graph(x ⊗ 1N ), the inclusion Xα3 φ ⊆ Xβ3 ⊗ N implies that (xu)φ = x(uφ) for u ∈ Uα . Now consider the R[x]–homomorphism R[x] −→ Lβ ⊗ N which maps 1 to 1φ. This map and φ agree on R · 1, hence, by the pushout property, they extend to an R–  for every y ∈ Lα , since homomorphism φ : Lα −→ Lβ ⊗ N . But (xy)φ = x(y φ)  this is true on both Uα and R[x]. Therefore φ is an R[x]–homomorphism, and Lemma 14.2.5 applies. If α \ β is infinite, then φ = 0, hence φ = 0. If α = β, then 1φ ∈ R[x]⊗N . But 1φ = 1φ ∈ (R[x]⊗N )∩(Wα ⊗N ). From the decomposition

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Wα + R[x] = Uα ⊕ R · 1 ⊕ x · R[x], we see that 1φ ∈ R · 1 ⊗ N , thus φ ∈ 1 ⊗ N . 2 To construct rigid λ–families of Rκ –modules for κ infinite, we only need one appropriate module. Lemma 14.2.7. Let κ be infinite. If there exists a λ–free Rκ –module F such that Hom(F, F ⊗ N ) = 1 ⊗ N for every R–module N , then there exists a rigid λ–family of Rκ –modules.

Proof. By Proposition 14.2.4, it suffices to produce a weak λ–family of Rκ – modules. Let {fi , fi , fi | i < λ} be a free basis for F. We may assume that the submodules F k are indexed by k < κ, k > 0. For each α ⊆ λ, define Fα = F , Fαk = F k (0 < k < κ), and Fα0 = fi , fj | i < α, j < λ. We have therefore defined a family {Fα | α ⊆ λ} of Rκ –modules. Each Fα is clearly λ–free, and the inclusion Fα ⊆ Fβ (α ⊆ β) is trivially a split R–homomorphism. If α = α0 ∪ α1 is a partition of α, we have an exact sequence 0 −→ F∅ −→ Fα0 ⊕ Fα1 −→ Fα −→ 0 given by the skew diagonal and sum maps. It is immediate that this induces an exact sequence 0 −→ F∅k −→ Fαk0 ⊕ Fαk1 −→ Fαk −→ 0 for every k < κ. Thus we have an exact sequence 0 −→ F∅ −→ Fα0 ⊕ Fα1 −→ Fα −→ 0 as required in Definition 14.2.2. Our hypothesis on F implies that Hom(Fα , Fβ ⊗ N ) ⊆ 1 ⊗ N for α ∪ β ⊆ λ. The submodules Fα0 and Fβ0 and the fact that the fi are part of a free basis for F guarantee that condition (vi) of Definition 14.2.2 is satisfied. Thus we have a 2 weak λ–family of Rκ –modules. To deal with regular λ, we shall adapt a construction that appears in Section 9.3. Proposition 14.2.8. If λ is a regular cardinal, then there exists a rigid λ–family of Rω –modules.

14.2

The four–submodule theorem, a harder case

549

Proof. An R4 –module F is an Rω –module by taking F k = F 0 for 4 ≤ k < ω, thus Proposition 14.2.6 allows us to assume that λ is uncountable. We shall utilize the construction of Section 9.3 to obtain an Rω –module F that satisfies the hypotheses of Lemma 14.2.7. We shall write down F taken from Shelah’s Elevator 9.3.3 (= Theorem 14.1.1) F 0,n = W n , F 1,n = W1n , F 2 = W2 , F 3 = W3 , F 4,n = W4n , Fn5 = W5n (0 < n < 5). Thus Hom(F, F ⊗ N ) = 1 ⊗ N from the Elevator–Theorem 14.1.1. We note that each submodule is clearly free of rank λ and has free complements of rank λ. 2 For singular cardinals, we take κ equal to the cofinality. Proposition 14.2.9. If λ is a singular cardinal and κ = cf(λ), then there exists a rigid λ–family of Rκ –modules.

Proof. Choose an increasing sequence of regular cardinals {λk | k < κ} whose supremum is λ. By Proposition 14.2.8, choose a λk –free Rω –module Fk such that Hom(Fk , Fk ⊗ N ) = N ⊗ 1 for every R–module N . Put F = k<κ Fk , and choose a basis element fk ∈ Fk which lies outside of Fk0 (k < κ). We now make F into an Rκ –module by giving κ submodules.  For every (k,  k1 ) ∈ κ × κ with k,k 0 1 k < k1 , take the submodule F = Fk ⊕ k1 ≤k2 <κ Fk2 . For every n < ω,  n n take the submodule F = k<κ Fk . Finally take the submodule   F  = f0 − fk | k < κ ⊕ Fk0 . k<κ

It is easy to see that F is λ–free. To verify that F satisfies the hypothesis of Lemma 14.2.7, let N be an R– module and φ ∈ Hom(F, F ⊗ N ). Since Fk = k1 >k F k,k1 , it follows that Fk φ ⊆ Fk ⊗ N . But F n φ ⊆ F n ⊗  N also, hence Fkn φ ⊆ F n  ⊗ N for every k < κ and n < ω. Consequently, φ = k<κ (1k ⊗ nk ) relative to k<κ F k . But F  φ ⊆ F  × N , hence f0 ⊗ n0 − fk ⊗ nk = (f0 − fk )φ ∈ F  ⊗ N . By the way the fk s were chosen, we conclude that f0 ⊗ n0 − fk ⊗ nk ∈ (f0 − fk ) ⊗ N , and 2 hence n0 = nk for every k < κ. Thus φ = 1 ⊗ n. Proof of the theorem We wish to extend the scope of rigid λ–families to deal with an arbitrary R–algebra if λ is sufficiently large. Let A be an R–algebra and let {Fα | α ⊆ λ} be the standard λ–family of R–modules with exact sequences

Qi −→ 0 (α ⊆ λ). 0 −→ F∅ −→ Fα −→ i<α

550

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Modules with distinguished submodules

If M is a right A–module, put Mα = Fα ⊗ M for α ⊆ λ, which is a family of right A–modules. We define an A–rigid λ–family of Rκ –modules to be Rκ – module structures on Mα and Qi for every right A–module M , α ⊆ λ and i < λ, such that k (i) Mαk is a direct  summand of Mα such that Mα and its complement are R– isomorphic to λ M (k < κ), and Qi is λ–free (i < λ);

(ii) the exact sequence for Fα tensored with M yields an exact sequence

0 −→ M∅ −→ Mα −→ (Qi ⊗ M ) −→ 0 (α ⊆ λ); i<α

(iii) and, if M and N are A–modules and α ∪ β ⊆ λ, then Hom(Mα , Nβ ) = 1αβ ⊗ HomA (M, N ) if α ⊆ β, and is zero if α  β. It is clear that a rigid λ–family of Rκ –modules is an R–rigid λ–family. The proof of the theorem will be accomplished by a ‘transitivity’ result. First we will note in a lemma that one additional submodule suffices to restrict R–homomorphisms to A–homomorphisms. This submodule will be included before the reduction to R4 –modules. In preparation for the lemma, let A be an R–algebra which can be generated by a family {ai | i < λ} of no more than λ elements. Let {Gα | α < λ} be a rigid λ–family of Rκ –modules, and choose a free basis {gi , gi | i < λ} for G∅ . If M is a right A–module, put Mi = {gi ⊗ m + gi ⊗ mai | m ∈ M } (i < λ). the Mi s Then Mi is a submodule of Gα ⊗ M thatis R–isomorphic to M . In fact,  ∗  ∗ ∼ are independent, and we define M = i<λ Mi . Note that M = λ M and   ∗ that i<λ (gi ⊗ M ) is a complement for M in G0 ⊗ M . We also recall the very useful Lemma 14.1.4 in the new setting: Lemma 14.2.10. With the notation and assumptions as above, let M and N be A–modules, α ⊆ β and φ ∈ Hom(Gα ⊗ M, Gβ ⊗ N ). Then φ ∈ HomA (M, N ) ⊗ 1αβ , if and only if M ∗ φ ⊆ N ∗ . We now prove the key reduction step. Proposition 14.2.11. Let λ ≥ κ ≥ ρ, with κ infinite. If there exists a rigid λ– family of Rκ –modules and a rigid κ–family of Rρ –modules, then there exists an A–rigid λ–family of Rρ –modules for every R–algebra A which can be generated by no more than λ elements.

14.2

The four–submodule theorem, a harder case

551

Proof. Let {Gα | α ⊆ λ} be a rigid λ–family of Rκ –modules. We adopt the notation used in Lemma 14.2.10. We may assume that the submodules of Gα giving the Rκ –module structure are indexed by k with 2 ≤ k ≤ κ. For convenience of notation, we shall introduce two new submodules, which will not be, however, part of the Rκ –module structure. We have an exact sequence 0 −→ G∅ −→ Gλ −→



Yi −→ 0,

i<λ

where  the Yi are λ–free. We define new submodules of Gα and Yi by G1α = Gα , 0 Gα = i<λ R(gi +gi ), Yi1 = Yi and Yi0 = 0 (α ⊆ λ, i < λ). It is straightforward to check that the exact sequence above restricts to an exact sequence 0 −→ Gk∅ −→ Gkα −→



Yik −→ 0

i<α

for every α ⊆ λ and k < κ. 6 α | α ⊆ λ} of Rρ –modules for every A– We shall construct a natural family {M module M , such that the family meets all the requirements of an A–rigid λ–family 6α is isomorphic to except being of the form Fα ⊗ M . We shall then show that M 6 α } can be used to put the required structure on {Fα ⊗ M }. Fα ⊗ M , and that {M First we define submodules of Gα ⊗ M by (Gα ⊗ M )k = Gkα ⊗ M for 1 ≤ k ≤ κ and (Gα ⊗ M )0 = M ∗ . We note that (Gα ⊗ M )k is a summand of Gα ⊗ M for every k < κ. At this point and elsewhere, there will be no problem regarding one tensor product embedded in another, since inclusions will be induced by split R–homomorphisms. Choose a rigid κ–family {Hγ | γ ⊆ κ} of Rρ –modules and k 6α =  write Hk for H{k} . We now define M k<κ Hk ⊗ (Gα ⊗ M ) (α ⊆ λ), where we may regard everything as embedded in Hκ ⊗ Gλ ⊗ M . 6αr , we must decompose Hκ . We have an exact 6α and M In order to analyze M sequence 0 −→ H∅ −→ Hκ −→



Zk −→ 0,

k<κ

with each Zκ κ–free, and which restricts to exact sequences 0 −→ H∅ −→ Hκ −→ Zk −→ 0 and 0 −→ H∅r −→ Hκr −→ Zkr −→ 0 (k < κ, r < ρ). Let r < ρ. Since Zk is κ–free, we may choose a free submodule Ckr of Hκ of rank κ, and a summand Ckr of Ckr such that Ckr and its complement

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Modules with distinguished submodules

r are free of rank κ, and such that Hκ = H∅ ⊕ Ckr and Hkr = H∅r ⊕ C k (k < κ). r (Choose Ck first.) Moreover, of {Zk } implies that k<κ Hk =     the independence r r r H∅ ⊕ k<κ Ckr and k<κ Hk = H∅ ⊕ k<κ Ck . Recalling that (Gα ⊗ M )1 = G1α ⊗ M = Gα ⊗ M , we obtain decompositions 

 6α = (H∅ ⊗ Gα ⊗ M ) ⊕ M Ckr ⊗ (Gα ⊗ M )k , k<κ

and

6r = (H r ⊗ Gα ⊗ M ) ⊕ M α ∅



 Ckr ⊗ (Gα ⊗ M )k .

k<κ

6 α | α ⊆ λ} has appropriate properties. We know We now show that the family {M r r that H0 and Ck and their complements in H∅ and Ckr are free of rank κ. Since  6αr is a Gα ⊗ M and (Gα ⊗ M )k are isomorphic to λ M , we conclude that M r 6α such that both M 6 and its complement are isomorphic to direct summand of M α  M . To exhibit the exact sequence λ

6 ∅ −→ M 6 α −→ 0 −→ M (Qi ⊗ M ) −→ 0 i<α

we first note that the exact sequence 0 −→ G∅ −→ Gα −→



Yi −→ 0

i<α

yields 0 −→ G∅ ⊗ M −→ Gα ⊗ M −→

(Yi ⊗ M ) −→ 0. i<α

The exact sequence 0 −→ Gk∅ −→ Gkα −→



Yik −→ 0 (k < κ)

i<α

yields 0 −→ (G∅ ⊗ M )k −→ (Gα ⊗ M )k −→

(Yik ⊗ M ) −→ 0, i<α

where a direct argument is used for k = 0. Combining these exact sequences with the decompositions as above, we obtain the desired exact sequence of Rρ – modules, where  

Ckr ⊗ Yik Qi = (H∅ ⊗ Yi ) ⊕ k<κ

Qri = (H∅r ⊗ Yi ) ⊕



k<κ

Ckr ⊗ Yik

 (i < α, r < ρ).

14.2

The four–submodule theorem, a harder case

553

 It is easily seen that Qi is λ–free, and in fact Qi = k<κ Hk ⊗ Yik . To investigate the mapping property, let M and N be right A–modules, α∪β ⊆  β . Let k < κ and put kˆ = κ \ {k}. Observe that 6 α −→ N λ and φ : M  β ⊆ (Hκ ⊗ (Gβ ⊗ N )k ) + (Hˆ ⊗ Gβ ⊗ N ). N k Since (Gβ ⊗ N )k and Hkˆ are summands of Gβ ⊗ N and Hκ respectively, we conclude that (Hκ ⊗ (Gβ ⊗ N )k ) ∩ (Hkˆ ⊗ Gβ ⊗ N ) = Nβk ⊗ Hkˆ .  β to Hˆ ⊗ (Gβ ⊗ N )/(Gβ ⊗ N )k Thus we have a natural homomorphism from N k whose kernel is contained in Hkˆ ⊗ (Gβ ⊗ N )k . Composing this map with the restriction φk of φ to Hk ⊗ (Gα ⊗ M )k , we obtain a homomorphism   Hk ⊗ (Gα ⊗ M )k −→ Hkˆ ⊗ (Gβ ⊗ N )/(Gβ ⊗ N )k , which must be zero by the rigidity of the {Hγ }s. Therefore (Hk ⊗ (Gα ⊗ M )k )φk ⊆ Hkˆ ⊗ (Gβ ⊗ N )k . Again, by rigidity, we have φk = 1kκ ⊗ ψ k for ψ k : (Gα ⊗ M )k −→ (Gβ ⊗ N )k . Since (Gα ⊗ M )1 = Gα ⊗ M , ψ k is the restriction of ψ 1 . Thus ψ 1 is a map Gα ⊗ M −→ Gβ ⊗ N such that (Gα ⊗ M )k ψ 1 ⊆ (Gβ ⊗ N )k for every k < κ. By Lemma 14.2.10, we conclude that if α  β, then ψ 1 = 0 and hence φ = 0. If α ⊆ β, then there exists ψ ∈ HomA (M, N ) such that ψ 1 = 1αβ ⊗ ψ. Consequently, φ is the restriction of 1κ ⊗ 1αβ ⊗ ψ. Conversely, if ψ ∈ HomA (M, N ) and α ⊆ β, then 1κ ⊗ 1αβ ⊗ ψ restricts to a homomorphism of Rρ –modules. Finally, we must define a standard λ–family {Fα } and put an Rρ –module structure on Fα ⊗ M . If we take A = R, all ai = 1, and M = R we in the preceding, k obtain a family of R–modules {Fα | α ⊆ λ}, where Fα = k<κ Hk ⊗ Gα . Each Fα is a free R–module of rank λ, and we have exact sequences 0 −→ F∅ −→ Fα −→



Qi −→ 0,

i<α

where each Qi is free of rank λ. Thus we have the standard λ–family. The earlier 6 α , say for r = 0, tells us that decomposition for M Fα = (H∅ ⊗ Gα ) ⊕



k<κ

 (Ck0 ⊗ Gkα ) ,

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14

Modules with distinguished submodules

where the choice of ai = 1 gives (Gα ⊗ R)0 = R∗ = G0α . Thus we have 

 Fα ⊗ M = (H∅ ⊗ Gα ⊗ M ) ⊕ (Ck0 ⊗ Gkα ⊗ M ) . k<κ

Since Gkα ⊗ M = (Gα ⊗ M )k for k = 0, we have an obvious identification 6 α except that G0 ⊗ M of every part of Fα ⊗ M with a corresponding part of M α   ) ⊗ M and is not the same as (Gα ⊗ M )0 . We have G0α ⊗ M = (g + g i i i<λ (Gα ⊗ M )0 = M ∗ = i<λ Mi . We define an R–isomorphism (gi + gi ) ⊗ M −→ Mi by (gi + gi ) ⊗ m −→ gi ⊗ m + gi ⊗ mai . Combining this with the other identifications, we obtain an isomorphism 6 α for every α ⊆ λ. θα : Fα ⊗ M −→ M 6 α to put an Rρ – We use this isomorphism and the Rρ –module structure on M module structure on Mα = Fα ⊗ M . We must verify that these identifications do not alter the exact sequence and mapping properties. The exact sequence property is unaltered, since θα restricted to F∅ ⊗ M is clearly θ∅ . For the mapping property, if ψ ∈ HomA (M, N ) and α ⊆ β, then one easily checks that θα (1κ ⊗ 1αβ ⊗ ψ) = θβ by evaluating at 6 α is preserved for Mα . 2 c ⊗ (gi + gi ) ⊗ m. Thus the mapping property for M The theorem will now follow quickly. Theorem 14.2.12. Let λ be an infinite cardinal and let A be an R–algebra that can be generated by no more than λ elements. Then there exists an A–rigid λ– family of R4 –modules.

Proof. First suppose that λ is a regular cardinal. Then Proposition 14.2.8, Proposition 14.2.6 and Proposition 14.2.11 yield the conclusion. Now suppose λ is a singular cardinal. Then κ = cf(λ) is regular and κ < λ, hence Proposition 14.2.9 and Proposition 14.2.11 plus what we have just proved for regular cardinals yield the conclusion. 2 Proof of Corollary 14.2.1. Choose M = N = A and apply Theorem 14.2.12 for 2 an A–rigid λ–family of R4 –modules. If M is an A–module, then Mα ∼ =  We close with two remarks on the theorem. λ λ M for every α ⊆ λ. By choosing 2 incomparable subsets of λ, we obtain 2 λ

14.3 A discussion of representations of posets

555

 distinct R4 –module structures on the single R–module λ M , say Mi (i ∈ 2λ ), such that  A ⊗ 1, if i = j Hom(Mi , Mj ) = 0, if i = j. The second remark concerns the possibility that the submodules Mαi and Mαj give an inner direct sum decomposition Mα = Mαi ⊕ Mαj for some i, j < 4. Suppose that the weak λ–family of Definition 14.2.2 has the property Xα = Xαi ⊕ Xαj for some i and j. In the proof of the proposition, the Qs will then have the property, and consequently so will the Fα . But then Hλ in the proof of Proposition 14.2.11 will also have the property, and it is easy to see that Mα will as well. Reviewing the construction of a weak ω–family of R4 –modules just before Proposition 14.2.6, we see that the decomposition Mα = Mαi ⊕ Mα1 will thus hold for i = 0, 2 or 3. Therefore End(Mα ) can be characterized as all endomorphisms in EndR Mα that commute with three certain projections. It is an interesting question whether this can be reduced to two projections.

14.3 A discussion of representations of posets Note that all indecomposable R4 –modules of finite–dimension over fields R were determined by Brenner [68]. If we view 4 = {0, 1, 2, 3} in Theorem 14.2.12 as an anti–chain of length 4, then it is obvious that the results of the last section should be extended to posets I replacing 4. This is a classical topic in the representation theory of algebras and we will shortly discuss some results related to the above. An RI –module is a family M = (Mi , i ∈ I) of distinguished submodules Mi of an R–module M such that Mi ⊆ Mj whenever i ≤ j in I. We immediately pass to a natural extension of RI –modules: Let R = 0 be any commutative ring and I = (I, <) be a finite poset. Following Simson [374] we consider the category CIR whose objects (–I–spaces over R) are systems M = (Mi , σij ) (i, j ∈ I), where Mi is an R–module, and σij for all i ≤ j ∈ I is an R–homomorphism Mi −→ Mj subject to two conditions. •  For this let max(I) denote the set of maximal elements or peaks of I. If Mi = p≥i Mp denotes the direct sum taken over all p ∈ max(I) with p ≥ i, then we have a canonical R–homomorphism σi• = p≥i σip : Mi → Mi• and require: (1) Compatibility condition: For all i ≤ j ≤ k ∈ I, we have σij σjk = σik and σii = idMi is the identity map on Mi . (2) Weak injectivity condition: σi• : Mi → Mi• is injective for all i ∈ I.

556

14

Modules with distinguished submodules

Particular examples of peak–I–spaces over R can be derived from direct sums

Mp M• = p∈max(I)

and submodules Mi ⊆ Mi• , where σij is induced by the canonical projection Mi• → Mj• . Finally RI –homomorphisms ϕ : M → M between objects M = (Mi , σij )  ) are I–tuples ϕ = (ϕ ) and M = (Mi , σij i i∈I of R–homomorphisms ϕi : Mi → Mi (i ∈ I)  s, i.e. compatible with the given σij , σij ϕi

Mi −−−−→ Mi ⏐ ⏐  ⏐ σij ⏐ σij   ϕj

Mj −−−−→ Mj . If I has precisely one maximal element p, we may ignore the maximal element and consider J = I \ {p}, which can be regarded as a subposet of I. Then σi• in (2) becomes σip and all the maps σij (i < j) are likewise injective. Hence CIR ∗ of R–modules with distinguished submodules may be viewed as the category CJR partially ordered by (J, <), which was introduced by Gabriel [24, 25, 26]. The anti–chain (J, <) = {0, 1, 2, 3} was the main target of the last section. This category was investigated over fields R in [36, 68, 67, 112, 113, 184, 185, 186, 189, 273, 292, 293, 297, 298, 326, 327, 329, 340, 341, 343, 372, 373, 374, 328] (see also Simson [371] for further references). Kleiner [297, 298] character∗ of finite representation type. This led to the well-known ized all posets J with CJR Kleiner-list of 5 minimal counterexamples, the partially ordered sets (1,1,1,1) , (2,2,2), (1,3,3,), (N, 4) and (1,2,5). (See also P4 − P8 of Simson’s list below and note that the smallest elements in the diagrams are the top modules, which are not mentioned above.) ∗ is not of finite representation type, if and only if at least Then the category CJR one of these orderings can be (peak–embedded) into J. If this is the case, then the result parallel to Theorem 14.2.12 holds for R– ∗ (see [56]). modules in the category CJR However, we can say much more about the structure of modules in CIR , if I is not of finite representation type. Similar to Kleiner’s result, Simson [374]

14.3 A discussion of representations of posets

557

characterized all minimal posets for CIR not being of finite representation type (see the list of minimal posets below). Using this result we can show that in this case Theorem 14.2.12 extends even further. Working through 114 counterexamples yields the following (see [219]): Theorem 14.3.1. For any finite poset I, the following are equivalent: (1) If R is any commutative ring, λ is an infinite cardinal such that an R-algebra A is generated by λ elements, then we can find a λ–family of peak–I–spaces MX (X ⊆ λ) in CIR such that for all X, Y ⊆ λ  A, if X ⊆ Y HomR (MX , MY ) = 0, if X  Y (2) The category of peak–I–spaces CIR over any field R is not of finite prinjective type. (3) There is a finite poset of the list of Simson’s 114 critical posets presented next, which can be peak–embedded into I. Hence Theorem 14.2.12 and the parallel result using Kleiner’s list of partially ordered sets ([56]) are particular cases of Theorem 14.3.1 (see also [241, 242] for related results concerning the Kronecker module). The dimension vectors attached to the orderings give additional information but can be neglected in the present context, (see [241, p. 216, Theorem 1.7] for the explanation of prinjective type and equivalent conditions). S IMSON ’ S MINIMAL POSETS P1 , . . . , P110

OF INFINITE PRINJECTIVE TYPE

Part 1. The infinite series   , P   , P2,n P1,n, P2,n+1 , P2,n+1 3,n, P3,n, P3,n, n ≥ 0 1 •

1 •

1 •

1 •

1

1

1

1

1

↓)↓)↓)· · · )↓)↓     

P1,n :

P2,n+1 :

 P2,n+1

1 •

:

2 •

2 •

(n + 2 star vertices), 2 •

2 •

*↓) * ) · · · * ) *↓)        

1

1

2

1 •

1 •

2 •

+

2

,-

2

n+1 2 •

2

.

1 •

1 •

)↓* ) · · · * )↓*     2 2 2 2 + ,. n+1

1

1

558

 P2,n

14 2 •

:

Modules with distinguished submodules

2 •

2 •

2

2

2 •

1 •

1 •

*↓) *↓*· · · *↓ * )↓*       1

1

2

+

,-

2

.

n

1•

↓

P3,n :

1•

2

2

•

2

2

2

+ ↓

1•

•1

2

•

1

.

n

2

1

↓

2

•

•

•1

*↓*↓*↓· · · * ) ↓)       1

2

2

2

+

2

,-

↓

2

1•

•

1

.

n

1•

:

•

2

,-

1•

 P3,n

2

•

*↓*↓*↓· · · *↓* ↓)        1

 P3,n :

2

•

2

2

•

•

1 •

1

↓) *↓*↓· · · * )↓*      1

2

2

+

2

2

,-

•

.

.

n

 We complete the infinite series P2,n+1 for n = −1 and P3,n for n = 0 by the following four–peak and three–peak posets respectively. 1• 1 ↓ ↑ 1•

2•

 : *↓) P3,0   

P2,0 : *↓)    1

1

1

1

1

1

.

Part 2. One–peak enlargements P4 − P8 of Kleiner’s posets. •1

P4 = K1∗ :

1 •

1 •

)↓* ∗ 2

1 •

1 •

↓ P5 = K2∗ :

•1

•1

↓ ↓ •1 •1 •1 )↓* ∗ 3

559

14.3 A discussion of representations of posets

•1

•1

↓

↓

•1

•1

↓ ↓ • •1 •1 )↓* ∗

2

P6 = K3∗ :

4

•1

↓

•1

•1

↓

↓ •1 ↓ •2 •1 ↓ ↓ 3 • •2 •1 P8 = K5∗ : )↓* ∗ 6 .

•1

1•

↓

•2

•1

↓) ↓ ↓ •1 •1 2• )↓* P7 = K4∗ : ∗ 5

Part 3. Two–peak posets P9 − P31 . •1

1 •

↓

2 •

•

↓

3

•1

↓

1

•1

1

↓)↓ * ∗ + 1

↓

1 •

•

9

1

•

•

2

2 10

•1

↓ 1

3

•

↓

2

•

•

↓* )↓ * + ∗ 2

2

1

•

1•

↓

1•

2 •

1 •

•1

1•

•

•

3

11

3

•

1

1•

•

1

1•

•

1

↓ ↓ 2

↓

3

4

1•

↓

1•

•

4 •

2 •

↓ )↓

2•

5

3 •

1 •

↓

2 •

1•

↓

↓*↓ ↓ )↓ ∗ • 2• • 2 1 )↓ *1

•

↓* )↓ *1 ∗ + 3

1 •

17

+ 5

1• 18

2 3*↓ •

•

+ 6

4

↓

•

1• 19

•

16

•

1

•

1

•

1

↓

1• 2

↓* )↓ * + ∗ 4

•

↓*↓ 3 ↓ ∗ • • • 3 1 )↓ *2

15

1

12

2

•

1• 1

1

↓*2 )↓ + ∗

1

↓*↓ ∗ +

14

1

↓ ↓ ↓

•

↓

•1

+ 4

13

•

•1

↓

3

*↓ 2 ↓ ∗ • • •1 1 1 )↓ *

•1

4

•

↓

•

1

↓* )↓ * + ∗

•1

2

•1

•

↓*1 )↓ + ∗

•

2*↓

1

↓

↓*3)↓ * + ∗

3

4

2 •

20

560

14

1•

1•

1•

2•

↓

2 •

↓

↓

↓

1•

2•

4

3

Modules with distinguished submodules

↓*↓ * ∗ +

1•

*↓ 2 ∗ 1• 1• • 1 ↓ ↓ )↓

•1

1•

2•

+ 5

21

1•

↓

∗

↓

3

↓

↓

1• 1•

2

22

1•

•1

1•

•

•

1•

↓*3 )↓ + ∗ 5

•

4

2

•

2

↓

↓

•2

1•

+ 6

4 25

2

↓

1•

)↓ *

•

↓

•2

↓

•

1•

•

↓

3

1

)↓*↓ ∗ +

•2

↓ •1 ↓

1

•

•

1 *↓

*↓

1•

1•

↓

1•

•

1

↓

1•

)↓ *1

•

↓

•

↓*↓ ∗ +

26

5

↓

•

↓

5

1•

2

↓

3

•

•

•

↓* )↓ *2 ∗ +

23

4

1 •

6

24

3 •

↓*↓ ∗ •1 2 ↓

•2

↓

3

•1

2

3

2

•

•2

)↓ * + 6

27

28

1•

↓

1•

1•

1•

1•

1•

1•

1•

1• 1•

↓

↓

↓

↓

1•

↓

1•

↓

↓

•

1

1•

•

•

3

1•

5

2

2 *↓

)↓*↓ ∗ +

↓ ↓

↓

29

3 •

↓

4 •

1•

)↓*↓ ∗ + 6

2 •

↓)↓ ∗ +

2 30

3

4 31

.

Part 4. Posets P32 , . . . , P110 with 3 ≤ | max Pj | ≤ 5. 3 •

1

•

↓ ∗ 2

•

1

) * ∗ •1 2* ∗ 32 2

1 • 1 •

↓)1

2•

•

2

3

2 •

2

1 • 2

↓

•

1•

1

3

)↓ ) ↓ *↓ ∗ ∗ ∗ 36

1

↓) 1 • • ∗1 • ↓*1 ) * ∗ ∗ 33 1

2

↓

•

2

•1

2

•

•

↓)↓ *↓ ∗ ∗ ∗ 34

1

3

2 •

↓

1

•

2

1

1

2

•2

•

3

1

↓*↓*↓ ∗ ∗ ∗

↓

•

2 37

2

•2

•

↓) * ) *↓ ∗ ∗ ∗ ∗ 38

1

3

3

•

•

3

1

↓)↓*↓ ∗ ∗ ∗ 35

•

2

∗

2, 2

1

561

14.3 A discussion of representations of posets

2• 2 •

2 • 1 •

3

↓) • ∗

2

•

•

)↓)↓*1 ∗ ∗ 2

1

•

1 39

4

3 •

2 •

3 •

1 •

)↓ ) ↓ * ↓* ∗ ∗ ∗ 40 2

4

*↓ ∗ •

1

2

3 •

↓

↓)↓ •1 ∗ ↓*↓ 2 ∗ ∗ 43

1•

3

2•

↓

1•

•

1

*1 ) •

•

2

1

↓)↓*↓ ∗ ∗ ∗

1

1 •

1 47

↓

3

44

2

1•

1

3

↓)1

1•

•

2

2

1 •

1•

2

1

)↓)↓)↓ ∗ ∗ ∗ 48

2

2

•

•

1

1

4 •

2 •

↓ * ↓ ) *↓ ↓) *↓) *↓ ∗ ∗ ∗ ∗ 51 ∗ ∗ ∗ ∗ ∗ 52

3

2

1

•

2

•

3

)

2

•1

↓

3•

•1

•

2 •

↓)↓)↓* ∗ ∗ ∗ 55 1

3

↓ •

3 •

1 •

•1

*↓

•2

↓* )↓ * ↓ ∗ ∗ ∗ 59 2

*↓

∗ 1

4

•1

1

3

4

2

•1

2

3

1

1 •

↓)↓ ∗2 ↓)↓) ∗ ∗ ∗ 46 2

45

•

1•

1

3

2

1

1 •

•

↓

1 •

2

•

•

↓*↓)↓ ∗ ∗ ∗

49

3

2

2 •

4

↓

4

1•

1 50

3

2 • 1

3

•

•

↓)1

3•

•

4

4

1 •

56

3 •

4

2

2

4

2 •

57

3

1 •

•

•1

↓*↓*2 ) ↓ ∗ ∗ ∗

•

2

←−• 1 ↓ ↓ •

2

↓* )↓)↓* ∗ ∗ ∗ 54

•

•1

3

42

2

←−• 1 ↓ ↓

4

3

•

*↓ ↓)1 2 ∗ •1 • • • 1 ↓*3 )↓* ∗ ∗ 53

•

↓*2 ) ↓ ∗ ∗

2

2

2

•

↓*↓ ∗ ∗

1

1•

2

•

↓

∗

)1 • ↓

•

4

2

↓)↓)↓ ∗ ∗ ∗

1

↓

3

↓)↓) ∗ ∗ ∗

•

1•

4

1•

•

1•

1•

1•

2

2 •

1 •

1•

1•

↓)↓) ∗ ∗ ∗

2

•

↓)

↓

2

1•

2

↓

1•

1 •

↓) ∗ ↓ 1

1•

↓) • ∗ 1 ) ↓ *1 1 ∗ 41

1• 1 •

2 •

4

↓*↓

•

1•

3

5

↓)↓)↓* ∗ ∗ ∗

1

•2

58

•1

↓

2

•

•1

↓)1 1 *↓ ∗ • • •2 1 )↓ * ↓ ∗ ∗ 60 4

2

2

2 •

•

2 •

4 •

2 •

*↓ 5 3 ↓ *↓*↓ 3 ↓ ∗ 1• • • • ∗ ∗ • • • 1 ↓ * )↓ * 2 1 3 1 )↓*2 ∗ ∗ 61 ∗ 62 4

6

6

562

14

1•

Modules with distinguished submodules

1•

↓

1•

•

↓*↓ ∗ • 3

1

1

4

•

•

) ↓ *↓ ∗ ∗ 63 6

2

↓ 5 3 4 • • • 1• ↓ * ) ↓ *↓ ∗ ∗ ∗ 64 4

6

2

6

1•

3•

3

5

↓ 4 ↓ 2 • 3• • 1• ↓ * ) ↓ )↓ ∗ ∗ ∗ 66

↓*↓ ∗ •1 2 ↓ 3 4 • • 1• ) ↓ *↓ ∗ ∗ 65

1•

3

3 •

•

↓

4

4

2

1•

↓

2

1•

•

1

3 •

•

3 •

*↓

•3

↓*↓ ↓ 2 2 4 ↓ 2 • • • • •3 • ∗ • 2 1 ) *3 )↓ ↓)↓)↓)↓ ∗ ∗ 67 ∗ ∗ ∗ ∗ 68 5

4

•2

∗

•2

*↓

•2

↓ ↓ •1 •2 ↓*↓ ∗ ∗ 71 5

3

5

•1

•3

↓ ↓ 2 •1 •3 • ↓* ↓ * ∗ ∗ 69

1

4

5

4

3 •

4 •

↓

4

↓

↓)

1•

•

1•

•2

↓*↓ ∗ ∗ 5

∗ 2

70

3

•2

↓

•1

1

1

∗

∗ 1

3

*↓

4

•1

•

↓ ↓) •1 •2 ∗ ↓*↓ 2 ∗ ∗ 72 5

3

1 •

1

↓ * ↓ ↓) ∗ •1 •2 ∗ 2 ↓*↓ 2 ∗ ∗ 73 5

3

4 •

•

↓

4

↓) •2 ∗ ↓*↓*↓ 2 ∗ ∗ ∗ 74

3

5

1•

•

3

1• •2

↓

4•

•2

2 4 ↓) ↓ • • •2 2• ∗2 *↓ * ↓ * ↓ ↓)↓)↓) ∗ ∗ ∗ ∗ 75 ∗ ∗ ∗ ∗ 76 2

4

3

5

•

1

•

3

1

3

5

3

79

*↓ •1 ∗ •1 ↓ 2 *↓ 1 • • •3 1) ↓ * ↓ ∗ ∗ 80

↓

•2

1•

•2

1•

↓

1•

↓ 4 ↓ • •2 ↓*↓*↓ ∗ ∗ ∗ 77

1•

3

5

3

↓ •5 ↓*↓ 2 • 4∗ 1• ↓)↓ ∗ ∗ 3

4

78

1•

↓

1•

•4

↓*↓ •1 3∗ ↓ 2 • 1• ↓)↓ ∗ ∗ 3

4

•2

5

2

2

•

4 •

2 •

•1

*↓

•3

↓)↓)↓*↓ ∗ ∗ ∗ ∗ 81 1

3

5

2

1•

↓

1•

•1

4 •

2 •

*↓

•3

↓* )↓ * ↓ ∗ ∗ ∗

3

5

2

82

563

14.3 A discussion of representations of posets

2 •

*↓ •1 ↓ •1 ↓ 4 •1 • 3• )↓*↓ ∗ ∗ 84

1∗ 1 •

3 •

1 •

↓*↓ 2 *↓ ∗ • • •3 2 1) ↓ * ↓ ∗ ∗ 83 5

1

3 •

•

5 •

2

3 •

4 •

↓*↓* )↓*↓ ∗ ∗ ∗ ∗ 87 2

4

6

2

6

1•

2

1

•

3

5 •

3

•3

↓*↓ •1 2∗ ↓ •1 ↓ 2 • 1• ↓)↓ ∗ ∗ 88

3 •

2

•

↓*↓*↓ 2 ∗ ∗ •1 • 2 4 ↓)↓ ∗ ∗ 85

↓ • ↓*↓* )↓*2 ∗ ∗ ∗ 86 1

3

2

4

•

4

5

•

∗ 1

•1

•

6

2

2 •

•

*↓

3

•

•

↓*↓ 2 ∗ 1• • 4 ↓)↓ ∗ ∗ 89 3

4 •

* ↓* ↓ ∗ ∗ •1 1 3 ↓ 2 • 1• ↓)↓ ∗ ∗ 90

5

4

3

4

4

2

•

↓) • 1 ∗1 ↓

1•

↓

1•

•1

4

•2

↓

↓ •1 ↓ 2 • •1 ↓*↓ ∗ ∗

1•

↓

1• 91

3

6

2

•

•

↓*↓)↓ ∗ ∗ ∗ 5

3

4

92

2 •

1•

↓

4

1•

6

•

2

•

•

↓*↓*↓)↓ ∗ ∗ ∗ ∗ 95 3

5

1

•

3

3 •

4

•2

1•

↓*↓ ↓ ∗ •1 •2 ↓*↓ 2 ∗ ∗ 99 5

4

1•

↓

3

6

↓

2

3 •

↓ 2 *↓ • •1 1• ↓*↓*↓ ∗ ∗ ∗ 100 4

5

2

*↓ 5 3 4 ∗ •1 • • • 1 ↓* )↓*↓ ∗ ∗ ∗ 96

•2

*↓ ∗ •1 ↓ 6 2 1 • • 1• ↓*↓)↓ ∗ ∗ ∗ 93

3

2

3

4

1•

•3

↓*↓ 6 2 ∗ •1 • • 2 ↓*↓)↓ ∗ ∗ ∗ 94 5

3

4

4 •

•

↓) *↓ 3 4 ∗ ∗ • • • 1 3 1 )↓*↓ ∗ ∗ 97 6

2 •

2

2

4 •

1

3

3

2

6 •

2 •

3

4

↓)↓) *↓)↓ ∗ ∗ ∗ ∗ ∗ 98 5

3 •

*↓ 2 * ↓ ∗ •1 • •1 ↓*↓*↓ 1 ∗ ∗ ∗ 101 4

•

1 •

3 •

2 •

•3

*↓

•1

↓*↓*↓*↓ ∗ ∗ ∗ ∗ 102 2

4

3

2

564

14

Modules with distinguished submodules

1•

↓

2

1•

2 •

•

↓ 1 *↓) • •2 ∗ )↓*↓ 1 ∗ ∗ 103

1•

4

2 • 2

2

4

1

1 •

−→ ↓ ↓

•

2•

•

1

3

3

1

*↓)

•

1

•

•2

•1

2

3

14.4

3

3

•

•

1 *↓) • • 2 ∗1

↓* )↓*↓ ∗ ∗ ∗

2

4

2

105

1•

↓

1•

2

1

•2

•

3

2

• −→• ↓ ↓

↓*↓ * ↓ ∗ ∗ ∗

3

1 106

•1

,↓

•1

•

2

2

•

1

2

1 •

1 •

*↓)1 2•

•

2

3

2 •

↓)↓)↓*↓ ∗ ∗ ∗ ∗ 109 1

1

•1

↓)↓)↓* ∗ ∗ ∗ 1

2

2

↓ ↓ •2 •1 ↓*↓*↓ ∗ ∗ ∗ 108 1

1

↓)↓ ) ↓) ∗ ∗ ∗ ∗ 107

•

2 •

*↓ 1 *↓) ∗ • • • 2 ∗1 1 1 )↓*↓ ∗ ∗ 104

↓

•1

110

.

Absolutely indecomposable modules

A module M (over any ring R) is said to be absolutely indecomposable, if it is indecomposable in every generic extension of the universe, and again equivalently, if M is not equivalent in L∞ω to any R–module that has a non–trivial direct decomposition. The only restriction on the extended model of set theory is that it has the same ordinals. We assume that it is a model of ZFC and that any set of the original model has the same members in the extended universe. (This is the case when the extended universe is obtained by forcing.) Recently the existence of large absolutely indecomposable R–modules (of size below the first Erd˝os cardinal κ(ω)) for commutative rings R with an infinite set of primes was shown in [238]. This section and [238] replace the problematic arguments [144],[142, Chapter XV, page 491, ()]. According to Shelah, it is very likely that larger abelian groups will always decompose in a suitable extension of the given universe. As in [238, 176] we will realize first R as an endomorphism algebra of Rω –modules and then pass to R–modules. So this section also provides a new (more robust) construction of indecomposable abelian groups. In Fuchs, Göbel [176] this result was extended to R5 –modules for commutative rings R, thus (by the trick of mixing primes; see page 577, Case A) to R–modules over a ring R with at least four primes. Moreover, in [176] (in the fashion of the last section) any faithful

14.4 Absolutely indecomposable modules

565

R–algebra of size < κ(ω) will be realized as an endomorphism algebra which remains the same in every generic extension of the given universe (in which the construction took place). Our strategy will now be to realize R as the endomorphism algebra of a large Rω –module, then to extend this object to a fully rigid system and pass to R–modules with enough primes. We will follow the exposition in Göbel, Shelah [238], blended with some ideas from [176], an earlier paper that had to wait for publication. It is also remarkable that we will encode valuated trees into the modules and will apply the existence of absolutely rigid families of valuated trees from Shelah [363] that originate from work by Milner and Nash– Williams on well–quasi–orderings of trees, today also used in operations research (especially, stochastic optimization), see for example [325]. There is a whole industry transporting symmetry properties from one category to another: for example consider a tree or a graph (with extra properties if needed) together with its group of automorphisms. Then encode the tree or the graph into an object of your favored category in such a way that the branches (or vertices) of the tree (of the graph) are recognized in the new structure. If the new category is the class of abelian groups, we argue by (infinite) divisibility and in case of groups and fields we use of course infinite chains of roots (with legal primes) etc. It happens that the automorphism group we start with becomes (modulo inessential maps: inner automorphisms in case of groups and Frobenius automorphisms in case of fields) the automorphism group of an object of the new category. The reader who wants to see more on this strategy may want to consider papers by Heineken [267], [64] (in case of groups), Corner, Göbel [102] in case of modules (with group rings as the first category) or Fried and Kollar [172], [129] in case of fields and [130] for automorphism groups of geometric lattices. Here we also argue with symmetry properties of trees, but they are of a different kind.

Rigid families of valuated trees and the first ω–Erd˝os cardinal We first describe the results on valuated trees. Then we want to encode the valuated trees into modules with distinguished submodules. Let κ(ω) denote the first ω-Erd˝os cardinal. This is defined as the smallest cardinal κ such that κ → (ω)<ω , i.e. for every function f from the finite subsets of κ to 2 there exist an infinite subset X ⊂ κ and a function g : ω → 2 such that f (Y ) = g(|Y |) for all finite subsets Y of X. The cardinal κ(ω) is known to be strongly inaccessible; see Jech [284, p. 392]. Thus κ(ω) is a large cardinal. We should also emphasize that κ(ω) may not exist in every universe and if κ(ω) exists, then it will also exist in Gödel’s model V=L of set theory (see [363]).

566

14

Modules with distinguished submodules

If λ < κ(ω), then let Tλ = ω> λ = {f : n −→ λ : with n < ω and n = Dom f } (as before) be the tree of all finite sequences f (of length or level lg(f ) = n) in λ. Since n = {0, . . . n − 1} as ordinal, we also write f = f (0)∧ f (1)∧ . . . ∧ f (n − 1). If n = 0, we write f = ⊥ for the empty function, the root of Tλ . By restriction we have g = f  m for any m ≤ n and obtain all initial segments of f . We will write g
f . Thus g ≤ f ⇐⇒ g ⊆ f as graphs ⇐⇒ g
f. A subtree T of Tλ is a subset which is closed under initial segments, and a homomorphism between two subtrees of Tλ is a map that preserves levels and initial segments. Note that a homomorphism does not need to be injective or preserve . The tree T is valuated, if with the tree we have a valuation map v : T −→ ω. In this case ω is just a set (with the trivial ordering) and thus v is not in conflict with valuations on trees used elsewhere. In the following a tree will always come with a valuation and Hom(T1 , T2 ) denotes the valuated homomorphisms between subtrees T1 and T2 , i.e. if vi is the valuation of Ti (i = 1, 2) and ϕ is such a valuated homomorphism, then v2 (ηϕ) = v1 (η) for all η ∈ T1 . Shelah [363] showed the existence of an absolutely rigid family of 2λ valuated subtrees of Tλ . Theorem 14.4.1. If λ < κ(ω) is an infinite cardinal and Tλ = ω> λ, there is a family Tα (α ∈ 2λ ) of valuated subtrees of Tλ (of size λ) such that in any generic extension of the universe the following holds for α, β ∈ 2λ . Hom(Tα , Tβ ) = ∅ =⇒ α = β.

Proof. The result is a consequence of the Main Theorem 5.3 in [363, p. 208]. The family of rigid trees is constructed in [363, p. 214, Theorem 5.7] and the proof that the trees are rigid, follows from Theorem 5.8 using Theorem 2.11 in [363]. In Shelah’s notation κ(ω) is the first beautiful cardinal > ℵ0 . 2 In a forthcoming paper we will apply Shelah’s very useful theorem on rigid valuated trees to other algebraic structures (e.g. fields). In [222] we will also include a more direct proof (in the style of the Black Box from Chapter 9) of the above Theorem 14.4.1 . It should be emphasized that this property of rigid families of valuated trees fails whenever λ ≥ κ(ω), as the well–ordering number of these trees is λ ≥ κ(ω). Using a related (but negative) result from [362], Eklof, Shelah [144], observe that their arguments apply to any family of structures, so we can apply their results to R–modules. Consequently, we can just quote without proof:

14.4 Absolutely indecomposable modules

567

Theorem 14.4.2. (Eklof, Shelah [144]) Let λ be a cardinal ≥ κ(ω) and R = 0 any ring of cardinality < κ(ω). (a) If {Mν | ν < λ} is a family of non–zero left R–modules, then there are distinct ordinals μ, ν < λ, such that in some generic extension V [G] of the universe V , there is an injective homomorphism φ : Mμ → Mν . (b) If M is an R–module of cardinality λ, then there exists a generic extension V [G] of the universe V , such that M has an endomorphism that is not a multiplication by an element of R. Thus, if we wish to construct modules with absolute endomorphism rings, then we must, and therefore in what follows we will, restrict the size of modules under consideration to cardinals below κ(ω). Corollary 14.4.6 will exhibit the cardinal κ(ω) as the precise border line.

The main construction Let R = 0 be a commutative ring. As we shall write endomorphisms as before on the right, it will be convenient to view R–modules as left R–modules. Next we define a free R–module F of rank λ over a suitable indexing set (obviously) used to encode valuated trees Tα from Theorem 14.4.1 into the structure when turning the free R–module F into an Rω –module F with ω distinguished submodules. We enumerate a subfamily of λ valuated trees from Theorem 14.4.1 by the indexing set, the tree I = ω> (ω> λ). Thus Tη

with valuation map vη : Tη −→ ω (η ∈ I)

without repetition. Next inductively define subsets Sn ⊆ n (ω> λ) such that the following holds. (0) S0 = {⊥}. (1) If Sn is defined, then Sn+1 = {η ∧ ν | η ∈ Sn , ⊥ = ν ∈ Tη }.  Let S = Sn and also let η ∧ ⊥ = η for ⊥ ∈ Tη ⊆ Tλ . n<ω

Put Snk = {η ∧ ν ∈ S : lg η = n, lg ν = k} ⊆ Sn+1 . Here ν = ∧ ν0 . . . ∧ νk−1 with νi < λ is a sequence of ordinals and η = η 0 ∧ . . . ∧ η n−1 with η i ∈ Tη0 ∧ ...∧ ηi−1 a sequence of branches from special trees. Moreover, write Tηk = {ν ∈ Tη : lg ν = k} ⊆ Tη and Tηk = {ν ∈ Tη : vη (ν) = k} (η < I). Now we define the free R–modules:

568 1. F =

14



Reη .

η∈S

2. Fnk =





η∈Sn ν∈Tηk

3. F nk =





η∈Sn

R(eη∧ ν  k−1 − eη∧ ν ).



(

η∈Sn

4. Fnk =

Modules with distinguished submodules

ν∈Tηk

(



ν∈Tηk

Reη∧ ν ).

Reη∧ ν ).

5. F0 = R(eη − eη )| η, η  ∈ S and F1 = Re⊥ .  We note that F0 = R(e⊥ − eη ) and F = F0 ⊕ F1 . ⊥ = η ∈ S

Next we define Rω –modules. These are R–modules with ω distinguished submodules. We enumerate the distinguished submodules by a suitable countable indexing set ·

·

·

W = {0, 1} ∪L1 ∪L2 ∪L3 of disjoints sets with Li a copy of ω ×ω (i = 1, 2, 3). Then an Rω –module X is an R–module X with a family of submodules Xi (i ∈ W ). We will also say that X is a free Rω –module, if X, Xi , X/Xi (i ∈ W ) are free R–modules. In particular,   s F = F, F0 , F1 , Fnk , F pq , Fm : (n, k) ∈ L1 , (p, q) ∈ L2 , (s, m) ∈ L3 (14.4.1) is a free Rω –module. We will eventually ease the notation and identify the distinguished submodules of F with Fi (i ∈ W ) and write F = (F, Fi | i ∈ W ). If X, Y are Rω –modules, then ϕ as in the first part of this section is an Rω – homomorphism (ϕ ∈ HomR (X, Y)), if ϕ ∈ HomR (X, Y ) and Xi ϕ ⊆ Yi for all i ∈ W , where Y = (Y, Yi | i ∈ W ). We also write HomR (X, X) = EndR X. We want to show the following Theorem 14.4.3. Let R be a commutative ring with 1 = 0 and |R|, λ < κ(ω). A free R–module F of rank λ can be made into a free Rω –module F = (F, Fi : i ∈ W ) such that EndR F = R holds in any generic extension of the given universe. Note that the size of R and the rank λ can be arbitrary < κ(ω); in particular R = Z/2Z. If λ is finite, then we can choose directly a suitable finite family of Fi s with the required endomorphism ring. Take a basis {ei (i < λ)} of F , set Fi := Rei (i < λ) and Fλ+i := R(e0 + ei ) (0 < i < λ), thus (changing names) the implication of the next lemma holds, and the argument in the proof of Theorem 14.4.3 on page 570 applies. Otherwise λ is infinite and we can apply Theorem 14.4.1. So we choose F = (F, Fi | i ∈ W ) as in the formula (14.4.1) depending on the trees from Theorem 14.4.1. Then clearly it remains to show EndR F = R. We first prove the following crucial

569

14.4 Absolutely indecomposable modules

Lemma 14.4.4. Let ϕ ∈ EndR F with F as in (14.4.1) and F =



Reη . If

η∈S

η ∈ S, then

eη ϕ ∈ Reη .

Proof. Let η ∈ S be fixed and recall that Tηk = Tη ∩ k λ. We consider its successors η ∧ ν in S with ⊥ = ν ∈ Tη and let lg η = n, lg ν = k. Thus η ∧ ν ∈ Snk and ν ∈ Tηk . If ϕ ∈ EndR F, then we claim eη∧ ν ϕ = rνl eρνl ∧ σνl  (14.4.2) l
with ρνl ∈ Sn , σνl ∈ Tρνl k = k λ ∩ Tρνl and 0 = rνl ∈ R. If eη∧ ν ϕ = 0, we choose lν = 0 and have the empty sum which is 0. By definition of F nk follows eη∧ ν ∈ F nk , thus eη∧ ν ϕ ∈ F nk showing that eη∧ ν ϕ is of the desired form (14.4.2). We will now use F to derive further restrictions of the expressions in (14.4.2). If ν1 ∈ Tηk+1 , then ν0 = ν1  k ∈ Tηk and eη∧ ν0  − eη∧ ν1  ∈ Fn k+1 , hence w := (eη∧ ν0  − eη∧ ν1  )ϕ ∈ Fn k+1 as well. Using (14.4.2) and the definition of Fn k+1 , we get that w is of the form   rν0 l eρν l ∧ σν0 l  − rν1 l eρν l ∧ σν1 l  = swi eρwi ∧ νwi  k − eρwi ∧ νwi  l
0

1

l
i
with ρwi ∈ Sn , νwi ∈ Tρwi k+1 and 0 = swi ∈ R. Now we collect terms of length k and k + 1, respectively and it follows that rν0 l eρν l ∧ σν0 l  = swi eρwi ∧ νwi  k length k: l
length k + 1:



l
0

rν1 l eρν

1l

i
=



swi eρwi ∧ νwi  .

i
We will apply the last two displayed equations and suppose for contradiction that  / Reη . Hence eη ϕ = rl eηl and there is η 0 = η with r0 = 0. We want to eη ϕ ∈ l
construct a (level preserving) valuated homomorphism g : Tη −→ Tη0 . Hence Tη , Tη0 are not rigid, and this  would contradict the implication of Theorem 14.4.1. We will construct g = k<ω gk as the union of an ascending chain of valuated homomorphisms gk : Tη ∩ k≥ λ −→ Tη0 ∩ k≥ λ.

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Let g0 (⊥) = ⊥ and suppose that gk is defined subject to the following condition which we carry on by induction.   (14.4.3) If ν1 ∈ Tηk , then η 0 ∧ gk (ν1 )) ∈ ρν1 l ∧ σν1 l  : l < lν1 , then gk (ν1 ) ∈ Tη0 for η 0 = ρν1 l . Note that (14.4.3) is satisfied for k = 0 by the assumption on ϕ. Thus we can proceed. If now ν1 ∈ Tηk+1 and ν0 = ν1  k, then gk (ν0 ) ∈ Tη0 k is given and we want to determine gk+1 (ν1 ). By induction hypothesis we have some l∗ < lν0 with ρν0 l∗ = η 0 and gk (ν0 ) = σν0 l∗ ∈ Tη0 . We must find l < lν1 (see (14.4.2)) such that ρν1 l = η 0 and σν0 l∗ = σν1 l  k. The second condition ensures that g will be the union of an ascending chain of gk s and also level preserving. The first assertion is our induction–bag which we must carry along. It is also the link to the undesired map ϕ. By the displayed equation for length k, there is some i (perhaps more than one) such that swi = 0 and ρwi = η 0 and νwi  k = σν0 l∗ . Then the other displayed equation of length k + 1, by picking one of the preceding is, yields the desired l . We now have l < lν1 with ρν1 l = η 0 and σν1 l ∈ Tη0 of length k + 1 with σν1 l  k = σν0 l∗ . So we can map gk+1 (ν1 ) ∈ Tη0 . If vη (ν1 ) = k, then (using lg(η) = n) eη∧ ν1  ∈ Fnk and (by the distinguished submodules in 4.) also eη∧ ν1  ϕ ∈ Fnk . Hence vη0 (gk+1 (ν1 )) = k = vη (ν1 ) follows, and valuation is preserved. We argue like this for all ν1 ∈ Tη of length k + 1. This completes the definition of gk+1 . Thus a valuated homomorphism g : Tη −→ Tη0 exists, a contradiction. 2

Proof of Theorem 14.4.3 From Lemma 14.4.4 follows e⊥ ϕ = re⊥ , eη ϕ = rη eη for some r, rη ∈ R and all ⊥ = η ∈ S. Moreover, (e⊥ − eη ) ∈ F0 , and therefore (e⊥ − eη )ϕ ∈ F0 and (e⊥ − eη )ϕ = re⊥ − rη eη ∈ R(e⊥ − eη ) by support (in the direct sum). Hence re⊥ − rη eη = r (e⊥ − eη ) for some r ∈ R and r = r , rη = r implies rη = r for all η ∈ S. Thus ϕ = r ∈ R. 2

Extension to fully rigid systems We want to strengthen Theorem 14.4.3, showing the existence of fully rigid systems of Rω –modules on λ. This is a family FU (U ⊆ λ) of Rω –modules such that the following holds.  R, if U ⊆ V HomR (FU , FV ) = 0, if U ⊆ V. This result will be the starting point for realizing R–algebras A as endomorphism algebras EndR F = A which are also absolute (see Fuchs, Göbel [176]).

14.4 Absolutely indecomposable modules

571

We first extend the countable indexing set W for F by one more element and let W  = {2} ∪ W . Hence W  and W are both countable, but W  has virtually one more element 2 added for a new definition replacing F from Theorem 14.4.3 by FU . The slot 2 is reserved for the only new distinguished submodule. F2 := FU :=



eR for any U ⊆ S.

e∈U

Thus FU = (F, F2 = FU , Fi : i ∈ W ). From Theorem 14.4.3 follows HomR (FU , FV ) ⊆ R for any U, V ⊆ S. Clearly, on the one hand, HomR (FU , FV ) = R, if U ⊆ V . On the other hand, if u ∈ U \ V , then eu ϕ = reu by the displayed formula. But reu ∈ FV , only if r = 0. Hence HomR (FU , FV ) = 0 whenever U ⊆ V . Finally note that |S| = λ and let W = ω. We have established the existence of fully rigid systems. Theorem 14.4.5. If R is any commutative ring with 1 = 0 and λ, |R| < κ(ω), then there is a fully rigid system FU (U ⊆ λ) of free Rω –modules with the following properties. (a) F is free of rank λ and FU = (F, FU , Fi : i < ω), and only FU depends on U. (b) The family FU (U ⊆ λ) is absolute, i.e. if the given universe is replaced by a generic extension, then the family is still fully rigid. The last theorem and Theorem 14.4.2 immediately characterize the first ω– Erd˝os cardinal. For clarity we restrict ourselves to countable rings R. Corollary 14.4.6. Let R by any countable commutative ring. Then the following conditions for a cardinal λ are equivalent: (a) There is an absolute Rω –module X of size λ with EndR M = R. (b) There is a fully rigid family FU (U ⊆ λ) of free Rω –modules. (c) There is a family of Rω –modules of size λ with only the zero-homomorphism between two distinct members. (d) λ < κ(ω) with κ(ω) the first ω–Erd˝os cardinal.

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Passing to absolutely fully rigid systems of R5 –modules Consider the Rω –modules MU = (MU , MUi | i < ω) (with U ⊆ λ) given by Theorem 14.4.5, where R = 0 is any commutative ring. For each U ⊆ λ, we define an R5 –module NU in the following way. Let E, G be free R–modules as defined above, and let N U = E ⊗ G ⊗ MU with the following five distinguished submodules. Choose an infinite set B = {ri | i < ω} ⊂ Im ρ \ Im ν, and set NUj = U j ⊗ MU

(j = 1, 2, 3, 4)

and NU0 =



R(e1 ⊗ gi + e2 ⊗ gi+1 ) ⊗ MU ⊕

i<ω

i<ω

(R(e3 ⊗ gν(i) ) ⊗ MU ) ⊕

(R(e3 ⊗ gri ) ⊗ MUi ). i<ω

Thus we have the inclusion relations U 0 ⊗ MU ⊂ NU0 ⊂ U 0 ⊗ MU that guarantee that NU satisfies the hypotheses of Lemma 14.1.5. Therefore we can claim: Theorem 14.4.7. Assume R = 0 is a commutative ring and NU = (NU , NUj (j = 0, 1, 2, 3, 4)) (U ⊆ λ) is a set of R5 –modules as just defined. If X, Y are any faithful R–modules, then the following holds in any generic extension of the universe.  1U ⊗ Hom(X, Y ), if U ⊆ V HomR (NU ⊗ X, NV ⊗ Y ) = 0, if U ⊆ V.

Proof. An application of Lemma 14.1.5 shows that every R–homomorphism NU ⊗ X → NV ⊗ Y must carry Rgn ⊗ NU ⊗ X into Rgn ⊗ NV ⊗ Y , for every n < ω. Consequently, the submodule NU0 ⊗ X must be mapped into NV0 ⊗ Y . In view of the definition of the gri s, it follows that MUi ⊗ X is mapped into MVi ⊗ Y for each i < ω. This shows that φ is an Rω –homomorphism (MU ⊗ X, MUi ⊗ X | i < ω) → (MV ⊗ Y, MVi ⊗ Y | i < ω). The conclusion now follows at once from Theorem 14.4.5. In the important special case X = Y = R, we have

2

573

14.4 Absolutely indecomposable modules

Corollary 14.4.8. For every commutative ring R = 0 of cardinality < κ(ω) and for every cardinal λ < κ(ω), there exists an R5 –module M of rank λ such that End M = R in every generic extension of the universe.

Proof. This follows from the preceding theorem once we observe that End M does not change under generic extensions. 2 ∼ K) the preceding In particular, if R = K is a field, then (with the choice X = result confirms the existence of absolutely indecomposable K5 –vector spaces of dimension < κ(ω). It is likely that the same will hold for K4 –vector spaces, but it is certain that the result cannot be improved to K3 –vector spaces. Indeed, K3 –vector spaces are of finite representation type (see Simson [371]), and from a theorem by Ringel-Tachikawa [343] (see also Simson [373]) it follows that all infinite–dimensional K3 –vector spaces decompose; cf. also [56, p. 137, Theorem 4.1]. Our next purpose is to extend the results from R–modules to A–modules, where A is an R–algebra. We are going to adjoin an additional distinguished submodule to the countably many distinguished submodules considered so far in such a way that it will guarantee that the R–homomorphisms will automatically become A–homomorphisms. First we settle the countably generated case. Let A be a countably generated faithful R–algebra, say, generated by the set {ai | i < ω}. We impose an additional condition on the choice of the subset B ⊂ Im ρ \ Im ν (see B at the beginning of this section) by assuming that its complement is infinite as well. So we can select infinite sets B1 = {si | i < ω} and B2 = {ti | i < ω} such that B, B1 , B2 are disjoint and B ∪ B1 ∪ B2 ⊆ Im ρ \ Im ν. For a right A–module M , put Mi = {e3 ⊗ gsi ⊗ x + e3 ⊗ gti ⊗ xai | x ∈ M }

(i < ω).

Each Mi is an A–submodule of E ⊗ G ⊗ M , R–isomorphic to M . The modules Mi (i < ω) are evidently independent in E ⊗ G ⊗ M , so we can form their direct sum

% & M∗ = Mi ∼ M . = 

i<ω

ω

Let us point out that i<ω R(e3 ⊗ gsi ) ⊗ M is obviously contained in a complement of M ∗ in E ⊗ G ⊗ M . Noticing that U 0 ⊗ M ⊂ M ∗ ⊂ U 0 ⊗ M, Lemma 14.1.5 applies. We can thus use it to verify:

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Modules with distinguished submodules

Lemma 14.4.9. With the notation and hypotheses as above, let M and N be right A–modules and φ ∈ HomR (E ⊗ G ⊗ CM , E ⊗ G ⊗ C N ). Then φ ∈ 1E ⊗ 1G ⊗ HomA (M, N ) ⇐⇒ M ∗ φ ⊆ N ∗ . The lemma and its proof differ only in notations from Lemma 14.1.4. For convenience we include the short argument in the new setting.

Proof. If ψ ∈ HomA (M, N ), then it is immediate that M ∗ (1E ⊗1G ⊗ψ) ⊆ N ∗ . Therefore assume that M ∗ φ ⊆ N ∗ . Owing to Lemma 14.1.5, we have φ = 1E ⊗ 1G ⊗ ψ for some ψ ∈ HomR (M, N ). Thus for x ∈ M e3 ⊗ gsi ⊗ xψ + e3 ⊗ gti ⊗ (xai )ψ = (e3 ⊗ gsi ⊗ x + e3 ⊗ gti ⊗ xai )φ ∈ N ∗ (here si ∈ B1 , ti ∈ B2 ) implies xψ ∈ N and (xai )ψ = (xψ)ai . This holds for every i < ω, and since the ai generate A, we can confirm that ψ is an A– isomorphism. 2 The following result is now a consequence of Theorem 14.4.5 and Lemma 14.4.9. Observe that we consider R–homomorphisms between A–modules. Theorem 14.4.10. Let λ < κ(ω) be any infinite cardinal and A any countably generated faithful algebra over the commutative ring R = 0. Then there exists a family of A5 –modules MU = (MU , MU0 , MU1 , MU2 , MU3 , MU4 )

(U ⊆ λ),

where M, MUj , M/MUj are free A–modules of rank λ for all j < 5 such that the following holds in any generic extension of the universe:  HomR (MU , MV ) =

A, if 0, if

U ⊆V U ⊆ V.

The preceding result can be extended to uncountable R–algebras A which are generated by κ (≤ λ) elements {ai | i < κ}, if we allow a sixth distinguished submodule. Indeed, if we choose two disjoint sets K, L of ordinals < λ, both indexed by κ, then our sixth distinguished submodule will be chosen as M ∗ , where M ∗ = ⊕α<κ Mα

  with Mα = e3 ⊗gsα ⊗x+e3 ⊗gtα ⊗xaα | x ∈ M

where sα ∈ K and tα ∈ L. Consequently, we get

(α < κ),

575

14.5 Passing to R–modules

Theorem 14.4.11. Let λ < κ(ω) be any infinite cardinal and A any faithful algebra over the commutative ring R = 0 such that A has fewer than λ generators. Then there exists a family of free right A6 –modules MU = (MU , MU0 , MU1 , MU2 , MU3 , MU4 , MU5 )

(U ⊆ λ),

where M, MUj , M/MUj are free A–modules of rank λ for all j < 6 such that  A, if U ⊆ V HomR (MU , MV ) = 0, if U ⊆ V holds in any generic extension of the universe.

14.5 Passing to R–modules Now we are ready to apply our results on the existence of rigid families of R5 – modules (or R4 –modules) which are arbitrarily large or absolute, but bounded by the first Erd˝os cardinal κ(ω) simultaneously, in order to find the corresponding families of (ordinary) R–modules (see Theorem 14.1.6, Theorem 14.2.12 and Theorem 14.4.11). We will follow the arguments (on absolute modules) in Fuchs, Göbel [176]. First we assume that there are enough primes (or other objects needed) in the ring, and finally we will discuss other cases, including the case of local rings. In each case, we have to specify how the R–algebra A and the A–modules in Definition 14.5.1 have to be chosen. We start with the case where the R–algebra A admits particular modules. Definition 14.5.1. Given an R–algebra A and a natural number n, we say that the family (W, W j , W | j < n) is almost fully rigid (for A), if the following properties hold. (i) W ⊂ W are faithful A–modules. (ii) W ⊂ W j ⊂ W for j < n.  (iii) W = j
576

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Modules with distinguished submodules

For suitable numbers n and algebras A these modules are used to create distinguished submodules by making them fully invariant. Fully rigid systems are almost fully rigid, but notice that Definition 14.5.1 is weaker than a fully rigid system. Recall that a family of A–modules MU (U ⊆ λ) of cardinality λ is A–rigid for some R–algebra A, if  A, if U ⊆ V HomR (MU , MV ) = 0, if U ⊆ V . We say that it is absolutely A–rigid, if it is A–rigid in any generic extension of the universe. Evidently, the modules in an absolutely A–rigid system are absolutely indecomposable, whenever A has no idempotents = 0, 1. Several applications are based on the following lemma. We use n = 5 in case A is countably generated over R and n = 6 otherwise. By the (stronger) R4 –module Theorem 14.2.12, we could use n − 1 to construct R–modules that are A–rigid but (not absolutely A–rigid). For clarity we will restrict ourselves to n = 6. Lemma 14.5.2. Suppose R is a commutative ring = 0 admitting an R–algebra A satisfying Definition 14.5.1. (a) Let λ < κ(ω) be any infinite cardinal. If A is ≤ λ–generated, there exists an absolutely A–rigid family of A–modules MU (U ⊆ λ) of cardinality λ. (b) For every infinite cardinal λ for which A is ≤ λ–generated, there exists an A–rigid family of A–modules MU (U ⊆ λ) of cardinality λ.

Proof. (a) We use the free right A–modules NU for U ⊆ λ as stated in Theorem 14.4.10. Define MUj = U j ⊗ NU ⊗ W j (j = 1, 2, 3, 4),

MU5 = NU∗ ⊗ W 5

(NU∗ ⊆ E ⊗ G ⊗ N denotes the module constructed from NU corresponding to M ∗ as described in the preceding section) and

%

MU0 = R(e1 ⊗ gi + e2 ⊗ gi+1 ) ⊗ MU ⊕ R(e3 ⊗ gν(i) ) ⊗ MU ⊕ i<ω



R(e3 ⊗ gri ) ⊗

MUi

&

i<ω

⊗W . 0

i<ω

Finally set MU =

j<6

MUj ⊆ E ⊗ G ⊗ NU ⊗ W .

14.5 Passing to R–modules

577

We consider the case that U ⊆ V ⊆ λ and let ξ be a homomorphism mapping (MU , MUj | j < 6) into (MV , MVj | j < 6), we conclude that every homomorphism ξ : MU → MV has to map each of the submodules MUj into MVj for j < 6. Since the modules MUj are direct sums of the same copies of W j by Theorem 14.4.7 and Definition 14.5.1(iv) the map ξ acts as scalar multiplication by an element from A. If U ⊆ V , then by a similar argument we have ξ = 0. Thus (MU , MUj | j < n) (U ⊆ λ) is an absolutely A–rigid family. (b) The parallel proof is left as an exercise; use Main Theorem 14.1.6 (with n = 6 in Definition 14.5.1) or Theorem 14.2.12 (with n = 5 in Definition 14.5.1). 2 Next we apply this lemma to special cases. In each case we have to specify how the algebra A and the A–modules in Definition 14.5.1 have to be chosen.

Case A. Let R be a domain with at least 4 prime elements p1 , p2 , p3 , p4 that are pairwise comaximal, i.e. Rpj + Rpk = R if j = k. Consider the following R–submodules of the quotient field Q of R: −∞ W 0 = p−∞ 1 p2 R,

−∞ W 1 = p−∞ 1 p3 R,

−∞ W 2 = p−∞ 1 p4 R,

−∞ W 3 = p−∞ 2 p3 R,

−∞ W 4 = p−∞ 2 p4 R

 (where the symbol p−∞ is an abbreviation for k<ω p−k ). It is straightforward to see that conditions in Definition 14.5.1 are satisfied with A = R = W , so we can apply the preceding Lemma 14.5.2 and claim: Corollary 14.5.3. Let R be a domain with at least 4 pairwise comaximal prime elements. (a) For every infinite cardinal λ < κ(ω), there exists an absolutely R–rigid family of torsion–free R–modules MU (U ⊆ λ) of cardinality λ. (b) For every infinite cardinal λ there exists an R–rigid family of torsion–free R–modules MU (U ⊆ λ) of cardinality λ.

Case B. Let R = Z and W = Z, W = H, where H ⊂ Qa0 ⊕ · · · ⊕ Qa5 is a rank 6 indecomposable torsion–free group (of Pontryagin type), constructed e.g. as in Fuchs [173, Vol. 2, p.125, Example 5], by using algebraically independent p–adic units πj (j < 5). We set W j = H ∩ (Qa0 ⊕ Qaj ), and observe that all these groups are indecomposable and homogeneous of type Z; furthermore, their endomorphism rings are Z. It is easily seen that Definition 14.5.1 will be satisfied, so we can conclude:

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Corollary 14.5.4. (a) For every infinite cardinal λ < κ(ω), there exists an absolutely Z–rigid family of cardinality λ of homogeneous torsion–free abelian groups MU (U ⊆ λ) of type Z. (b) For every infinite cardinal λ, there exists a Z–rigid family of homogeneous torsion–free abelian groups MU (U ⊆ λ) of type Z and of cardinality λ. Let us point out that the preceding two examples can easily be extended to the case where the endomorphism rings of the modules are A for algebras A considered. Case C. Let R be a domain with quotient field Q such that Q/R has a summand of the form W /R = W 0 /R ⊕ W 1 /R ⊕ W 2 /R ⊕ W 3 /R ⊕ W 4 /R with non–zero components. From Fuchs-Salce [181, p. 504] it follows that Definition 14.5.1 is satisfied with A = R and W = R. Hence we can state: Corollary 14.5.5. Let R be an infinite domain such that Q/R has a summand as stated. (a) For every infinite cardinal λ < κ(ω), there exists an absolutely R–rigid family of torsion–free R–modules MU (U ⊆ λ) of cardinality λ. (b) For every infinite cardinal λ, there exists an R–rigid family of torsion–free R–modules MU (U ⊆ λ) of cardinality λ. The hypotheses of the preceding corollary are satisfied, for instance, by an h– local domain with at least 5 maximal ideals. (Actually, only 4 of them are needed if we modify the construction following the pattern of Case A.)

Case D. Let R = Z and assume W is a torsion–free abelian group with endomorphism ring A. We may view W as a right A–module. Assume there are four primes pj (j < 4) for which W is reduced, i.e. k<ω pkj W = 0 for each j < 4. Define −∞ W k, = p−∞ W k p as a subgroup in the divisible hull Q ⊗ W of W , where the pair (k, ) ranges over the 2-element subsets of {0, 1, 2, 3}. Evidently, the conditions in Definition 14.5.1 is satisfied, so we can claim: Corollary 14.5.6. Let W be a torsion–free abelian group with endomorphism ring A such that W is reduced for at least 4 primes.

14.5 Passing to R–modules

579

(a) For every infinite cardinal λ < κ(ω), there exists an absolutely A–rigid family of torsion–free groups MU (U ⊆ λ) of cardinality λ. (b) For every infinite cardinal λ, there exists an A–rigid family of torsion–free groups MU (U ⊆ λ) of cardinality λ.

Case E. Let A be a countable, torsion–free and reduced algebra over a countable principal ideal domain R, and let Q denote the field of quotients of R. In Section 12.2 we proved the existence of a (rigid) family of ℵ1 –free A–modules Bγ (γ < 2ℵ0 ) of cardinality ℵ1 such that HomA (Bβ , Bγ ) = δβγ A, where δ stands for the Kronecker delta. Pick 5 modules B 0 , B 1 , B 2 , B 3 , B 4 in this set. Manifestly, they can be identified as submodules in an ℵ1 –dimensional Q–vector space, say with basis {aα |jα < j j j ω1 } ⊂ B for each j < 5. We set W = B (j < 5) and W = j<5 B . A straightforward  application of the well–known Pontryagin criterion convinces us that W = j<5 B j is again ℵ1 –free. From Lemma 14.5.2, we conclude: Corollary 14.5.7. Let A be a countable algebra over a countable principal ideal domain R and B j (j < 5) a rigid family of ℵ1 –free A–modules of cardinality ℵ1 . (a) Then for every infinite cardinal λ < κ(ω), there exists an absolutely A–rigid family of ℵ1 –free as A–modules (and also ℵ1 –free as R–module, if A is free as R–module) MU (U ⊆ λ) of cardinality λ. (b) For every infinite cardinal λ, there exists an A–rigid family of ℵ1 –free as A–modules (and also ℵ1 –free as R–module, if A is free as R–module) MU (U ⊆ λ) of cardinality λ.

Proof. As the modules are contained in a direct sum of copies of W , and since direct products and submodules of ℵ1 –free modules are again ℵ1 –free over a principal ideal domain, the claim is obvious. 2 Let R be a commutative ring furnished with a linear topology in which R is Hausdorff with a countable basis of open neighbourhoods of 0. Assume that  such that there exist units A is an R–algebra between R and its completion R  (γ ∈ Γ) with |Γ| ≥ 6 that are algebraically (or at least quadratically) πγ ∈ R independent over A.

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Modules with distinguished submodules

Theorem 14.4.11 implies the existence of an R6 –module M = (M, M 0 , M 1 , M 2 , M 3 , M 4 , M 5 ) of rank λ < κ(ω) with End M = A. Choose six algebraically (or quadratically) independent elements πj (j < 6) and form the R–module  := ⊕  H = M, πj M j | j < 6 ⊆ M λ A. Lemma 14.5.8. With the indicated notations, EndR H = A absolutely.

Proof. Let φ ∈ EndR H. By continuity, it admits a unique extension to an endo which is an R–homomorphism.  morphism of M No ambiguity arises, if φ denotes this extension as well. We focus our attention on the submodule π1 M 1 and consider h  ∈ (π1 M 1 )φ = 1  π1 (M φ) ⊆ H ∩ π1 H. Write h = x + j πj xj = π1 (x + j πj xj ) with x, xj , x , xj ∈ M , whence quadratic independence guarantees that x = x and xj = 0 = xj for j < 5. Hence φ maps π1 M 1 into itself and the same holds for each πj M j . Therefore, φ ∈ A, completing the proof of the lemma. 2 We now illustrate cases where the hypotheses of the preceding lemma are satisfied.

Case F. Let R = Z, furnished with the p–adic topology for a prime number p. Choose a pure subring A ⊇ Zp in the ring of the p–adic integers Jp such that there are at least 6 algebraically independent elements (in Jp ) over A. This condition is certainly satisfied whenever |A| < 2ℵ0 (see Theorem 1.1.20). In view of the preceding lemma we can state: Corollary 14.5.9. Let A be a pure subring of the p–adic integers as stated. (a) For every infinite cardinal λ < κ(ω), there exists an absolutely A–rigid family of A–modules MU (U ⊆ λ) of cardinality λ. (b) For every infinite cardinal λ, there exists an A–rigid family of A–modules MU (U ⊆ λ) of cardinality λ.

Case G. Let R be a commutative ring of cardinality < 2ℵ0 and S a countable multiplicatively closed subset of regular elements of R. Assume A is an R–algebra with |A| < 2ℵ0 that is S–reduced and S–torsion–free such that A is an S–pure  of R. From Theorem 1.1.20 it submodule between R and the S–completion R  of R contains elements πγ (γ < 2ℵ0 ) that are follows that the S–completion R algebraically independent over A. Consequently, we have:

14.6 A topological realization from Theorem 14.2.12

581

Corollary 14.5.10. Let R, A be as stated. (a) Then for every infinite cardinal λ < κ(ω), there exists an absolutely A–rigid family of A–modules MU (U ⊆ λ) of cardinality λ. (b) For every infinite cardinal λ, there exists an A–rigid family of A–modules MU (U ⊆ λ) of cardinality λ. Case H. Suppose R is a local ring with maximal ideal P such that n<ω P n =  of R in the P –adic topology contains at least 5 algebraically 0. If the completion R independent units, then Lemma 14.5.8 applies, and we are led to the following conclusion: Corollary 14.5.11. Let R be a local ring as stated. (a) Then for every infinite cardinal λ < κ(ω), there exists an absolutely R–rigid family of R–modules MU (U ⊆ λ) of cardinality λ. (b) For every infinite cardinal λ < κ(ω), there exists an R–rigid family of R– modules MU (U ⊆ λ) of cardinality λ.

14.6 A topological realization from Theorem 14.2.12 The following result (from Göbel, Ziegler [251]) is a prerequisite for Theorem 15.3.7. It complements (and overlaps) the realization theorems derived with the Black Box in Chapter 12, where the size of the constructed modules is never cofinal to ω (because cf(λℵ0 ) > ω). Here we apply Theorem 14.2.12 to derive the following theorem. Theorem 14.6.1. Let R be a domain and Pj (1 ≤ j ≤ 5) be five multiplicatively closed subsets of R \ {0}. Moreover, let A be a complete Hausdorff R–algebra whose topology has a basis Ni (i ∈ I) of right ideals such that (a) A/Ni is torsion–free and Pj –reduced for all 1 ≤ j ≤ 5 and P1 , . . . , P5 are pairwise A/Ni –coprime, (b) |I| ≤ λ and |A/Ni | ≤ λ for some infinite cardinal λ. Then A is topologically (and algebraically) isomorphic to the endomorphism algebra (EndR M, fin) with the finite topology fin of an R–module M of cardinality |M | = λ.

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See Proposition 9.3.2 for the finite topology. Recall that an R–module M is P –reduced (for the multiplicatively closed subset P of R \ {0}), iff pM = 0 p∈P

and P, P  are M –coprime (or M –comaximal), if pM ∩ qM = pqM for all p ∈ P , q ∈ P  or, equivalently P −1 M ∩ P −1 M = M . It is very likely that the number 5 in Theorem 14.6.1 (and in Theorem 14.6.3) can be replaced by 4. From the discussions at the beginning of this section it is clear that 4 then will be minimal. First we prove an extension of the Main Theorem 14.1.6. This comes from [96, p. 158, Theorem 2.3] and the missing cardinals were added in [197]. The answer to the question in [96], whether all closed subalgebras of endomorphism algebras come from subalgebras fixing distinguished submodules (as in the next theorem), was given in [197]. For the extension we will not use the Main Theorem 14.1.6, but apply the stronger Theorem 14.2.12. We will also adopt a simplified version of the topological rank from [96]. Definition 14.6.2. Let M be an R–module over a commutative ring R and consider bases of neighbourhoods of 0 ∈ EndR M (generating fin) which are families of ideals AnnEndR M X (with X ∈ B) for collections B of finite subsets of M . The topological rank of M is the least cardinal rkt M = |B|. Clearly rkt M ≤ |M | (and often rkt M ≤ rk M ). Then Theorem 14.6.1 will follow (see below) by the obvious arguments used in Section 14.5, Proposition 14.6.6 and the next characterization of closed algebras. Theorem 14.6.3. For an R–algebra A over a commutative ring R are equivalent: (a) There is an R–module M such that A is (algebraically and topologically) isomorphic to a closed R–subalgebra of EndR M in the finite topology. (b) If F is a free R–module of infinite rank λ ≥ rkt M , then there are 5 distinguished summands Ui of M ⊗ F such that the R5 –module M ⊗ F := (M ⊗ F, U1 , . . . , U5 ) satisfies EndR (M ⊗ F) = A. (c) If F is a free R–module of infinite rank λ ≥ rkt M , then there is an A–rigid λ–family Mα (α ⊆ λ) of R5 –modules. Remark 14.6.4. Recall the definition of an A–rigid λ–family of R5 –modules from page 550. Thus (c) provides 2λ such R5 -modules Mα constructed on M ⊗ F with endomorphism algebra Hom(Mα , Mβ ) = A, if α ⊆ β ⊆ λ, and Hom(Mα , Mβ ) = 0 otherwise.

14.6 A topological realization from Theorem 14.2.12

583

We will begin with a reduction lemma which establishes the link to Theorem 14.2.12. The following lemma (from Corner [96, p. 161, Theorem 4.2]) is similar to Lemma 14.1.4 and Lemma 14.2.10. Lemma 14.6.5. Let A be a closed subalgebra of (EndR M, fin) (equipped with the finite topology) and F a free R–module of rank rk F ≥ rkt M . Then there is (an R–summand) M ∗ ⊆ M ⊗ F such that A = EndR (M, M ∗ )F , where EndR (M, M ∗ )F := {σ ∈ EndR M | M ∗ (σ ⊗ 1F ) ⊆ M ∗ }.

Proof. We may assume that rk F = rkt M (delete some of the summands of F if necessary) and can enumerate a basis of F by {uX,x | x ∈ X ∈ B} for |B| = rkt M and a basis of neighbourhoods N (X) = AnnEndR M X (X ∈ B) of 0 ∈ EndR M . Thus 



M ⊗F = F ⊗ RuX,x . X∈B

Take

M∗ =



X∈B

x∈X

 (x ⊗ uX,x )(A ⊗ 1F ) .

x∈X

End(F, M ∗ )F ,

it remains to show that ϕ ∈ End(F, M ∗ )F is of the Since A ⊆ form a ⊗ 1F for some a ∈ A. From the last two displayed equations we have M∗ ∩



   F ⊗ RuX,x = x ⊗ uX,x (A ⊗ 1F ) (X ∈ B).

x∈X

x∈X

Consequently, for each X ∈ B we have     (xϕ) ⊗ uX,x = x ⊗ uX,x (ϕ ⊗ 1F ) ∈ x ⊗ uX,x (A ⊗ 1F ); x∈X

x∈X

x∈X

that is, there exist endomorphisms aX ∈ A such that x∈X

(xϕ) ⊗ uX,x =



(xaX ) ⊗ uX,x

(X ∈ B).

x∈X

Equating coefficients, we find that xϕ = xaX for (x ∈ X, X ∈ B) and ϕ is represented by the Cauchy sequence (aX )X∈B in A. Hence ϕ = a ⊗ 1F because 2 A is closed in (EndR M, fin).

Proof of Theorem 14.6.3. (c) ⇒ (b) follows trivially by the definitions (see the remark after Theorem 14.6.3).

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(b) ⇒ (a) follows more generally for Rk –families for any cardinal k. However we will restrict ourselves to k = 5 as stated in (b) and use the notations from (b). If ϕ is an endomorphism in the closure of A, then we want to show that ϕ ∈ A. Let U be one of the distinguished submodules Ui and u ∈ U . First we want to see that uϕ ∈ U .  We may write u = ni=1 mi ⊗ fi for suitable elements mi ∈ M and fi ∈ F . By definition of the finite topology, there exists an endomorphism a ∈ A (scalar multiplication ar by a) such that mi ϕ = mi a for (1 ≤ i ≤ n), and we have n n (mi ϕ) ⊗ fi = (mi a) ⊗ fi = u(a ⊗ 1F ) ∈ U. u(ϕ ⊗ 1F ) = i=1

i=1

Thus ϕ ⊗ 1F ∈ EndR (M ⊗ F) = A ⊗ 1F (=: A) by (b); whence ϕ ∈ A, as required. It remains to show that (a) ⇒ (c) and this is where we need the preceding work. Let A be the closed R–subalgebra of EndR M and F be a free R–module of rank rk F = λ ≥ rkt M . Then MA is an A–module and we may apply Theorem 14.2.12. There is an A–rigid λ–family of R4 –modules: We find R4 –modules Mα = M ⊗ Fα for α ⊆ λ with free R–summands Fα ⊆ F and direct summands Mαk of Mα for (1 ≤ k ≤ 4). Then Mα = (Mα , Mα1 , . . . , Mα4 ) satisfies HomR (Mα , Mβ ) ⊆ EndR M ⊗ 1αβ for α ⊆ β and is 0 otherwise. Here 1αβ is the canonical embedding of Fα into Fβ (see page 550). Finally, by Lemma 14.6.5 (in a trivially modified form) there is an R–summand ∗ Mα ⊆ M ⊗ Fα with the property that every homomorphism of the form ϕ ⊗ 1αβ with ϕ ∈ EndR M and Mα∗ (ϕ ⊗ 1αβ ) ⊆ Mβ∗ (α ⊆ β) is scalar multiplication by an element of A. We now replace the R4 –modules Mα = (Mα , Mα1 , . . . , Mα4 ) by the R5 –module Mα = (Mα , Mα1 , . . . , Mα5 ) where Mα5 = Mα∗ . Thus HomR (Mα , Mβ ) = A ⊗ 1αβ , if α ⊆ β and HomR (Mα , Mβ ) = 0 if α ⊆ β from above. Hence (c) is shown.

2

Proposition 14.6.6. Let R be a domain and let A be a complete Hausdorff R– algebra with a basis Ni (i ∈ I) of right ideal. Then there is an R–algebra B and a left B–module M0 such that A ∼ = EndB M0 as topological R–algebras. Under the hypothesis of Theorem 14.6.1 we find M0 torsion–free and Pj –reduced (1 ≤ j ≤ 5) and B of at most the cardinality λ. If A/Ni is free for all i ∈ I, then also M0 is free.

14.6 A topological realization from Theorem 14.2.12

585



A/Ni . M0 is a right A–module. The natural map A −→ Ni = 0. It is also easy to see that the topology EndR M0 is injective, since

Proof. Set M0 =

i∈I

i∈I

of A is induced by the finite topology of EndR M0 . Thus it remains to define a left B–module structure on M0 such that A = EndB M0 . We define B as an R–subalgebra of EndA M0 : let ε

π

i i → M0 −→ A/Ni → 0 0 → A/Ni −

be the natural maps. Let b ∈ A and i, j ∈ I such that bNi ⊆ Nj . Then left multiplication by b yields an A–homomorphism bij : A/Ni −→ A/Nj . We let B ⊆ EndA M0 be generated by all bij = πi bij εj . It remains to show that every ϕ ∈ EndB M0 is right multiplication by an element a ∈ A. Since ϕ commutes with the canonical projections 1ij , we know that (A/Ni )ϕ ⊆ A/Nj . Write (1 + Nj )ϕ = aj + Nj . Now suppose Ni ⊆ Nj , then ϕ must commute with 1ij , which gives ai − aj ∈ Nj . Whence by completeness we find a ∈ A such that a − aj ∈ Nj for all j ∈ I. We have now (1 + Ni )ϕ = a + Ni . We claim that ϕ = ar is multiplication by a on the right. To show this, we take an element b + Nj ∈ A/Nj and prove (b + Nj )ϕ = ba + Nj . Choose i ∈ I such that bNi ⊆ Nj . Then ϕ commutes with bij , whence (b + Nj )ϕ = (bij (1 + Ni ))ϕ = bij ((1 + Ni )ϕ) = bij (a + Ni ) = bij a + Ni . The algebra B has a cardinality ≤ λ because its generators bij and bij are the same, if and only if b − b ∈ Ni . 2

Proof of Theorem 14.6.1 We can reformulate the last Proposition 14.6.6 in terms of the finite topology on EndR M0 . The isomorphism from the complete Hausdorff R–algebra A can be seen as a topological embedding A −→ EndR M0 . Thus the image of A (which we now identify with A) is a closed R–subalgebra of End M0 . Thus A ⊆ End M0 satisfies the hypothesis (a) of Theorem 14.6.3, and Theorem 14.6.1 follows.

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Open problems 1. Try to replace 5 by 4 in Theorem 14.6.1. 2. Analyse other module categories of tame or wild representation type and prove the result corresponding to Theorem 14.3.1 and Theorem 14.2.12. 3. Replace ‘rigid’ (‘prescribed endomorphism algebra’) by other crucial algebraic properties (of the modules) like ‘being indecomposable’. Can the border line be strictly larger than κ(ω) (or perhaps one of the other large cardinals)? In Theorem 14.4.5 (and Corollary 14.4.6) we show that κ(ω), the first Erd˝os cardinal, is the precise border for the size of absolutely rigid systems of modules (or of modules with absolutely prescribed endomorphism algebras). Note that for slenderness the first measurable cardinal is another example of a large cardinal characterized by algebraic properties. This follows from Theorem 1.4.13 and the easy remark in Fuchs [173, Vol. 2, p. 161, Remark].

Chapter 15

Some useful classes of algebras

15.1 Leavitt type rings: the discrete case In this chapter we will provide constructions of Leavitt type algebras which – when realized – yield pathological decompositions of modules. We would like to emphasize that this elegant way of obtaining modules from properties reflected in the endomorphism ring has its origin in the work of Corner’s. This fruitful method has attracted many algebraists (including the authors of this monograph) and stimulated further research. Several of these examples come from Corner, some of them were also modified to find decompositions of E(R)–algebras (see e.g. Fuchs, Göbel [174]). We also present a recent result (without proof) by Braun [63] which can be applied to give further unexpected decomposition properties of modules. Constructions of some algebras We begin with an easy construction of an algebra A which can be used to produce super–decomposable modules. A module M = 0 is super–decomposable, if M has no indecomposable summands = 0. Example 15.1.1. Let R be a domain and G = {q | q ∈ Q, q ≥ 0} be the semigroup with multiplication q r = max(q, r) for all q, r ∈ G. The semigroup algebra A = RG has free R–module structure and no regular idempotents.

Proof. A is a commutative R–algebra, and as an R–module it is freely generated by the elements of G. First we show that idempotents 0 = ε ∈ A are very special elements. Write ε = rj qj for 0 = rj ∈ R and q1 < · · · < qk . We have ε2 = ε j≤k

and thus ε2 =

j≤k

rj (2r1 + · · · + 2rj−1 + rj ) qj =

j≤k

rj q j .

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Some useful classes of algebras

And equating coefficients we get 2r1 + · · · + 2rj−1 + rj = 1 for all j ≤ k. Hence r1 = 1 and recursively we get rj = (−1)j−1 for all j ≤ k and therefore ε = q1 − q2 + · · · (−1)k−1 qk is a typical idempotent satisfying again ε2 = ε. We must show that such an idempotent 0 = ε ∈ A decomposes into a proper orthogonal sum of idempotents. If k > 1, we choose a rational number t such that q1 < t < q2 and if k = 1, then let q1 < t. In the first case (k > 1) the idempotent ε = (q1 − t) ⊕ (t − q2 + − · · · (−1)k−1 qk ) is an orthogonal sum of two non–trivial idempotents and if k = 1, then (q1 − t) ⊕ t is a proper orthogonal sum of idempotents, hence ε is not regular. 2 The following theorem (the Leavitt Theorem) has several known proofs, due to Leavitt [307], Cohn [85] and Corner [94, 97] (see also Fuchs [173, pp. 146, 147]). We present a proof for pedestrians (which still needs some work) of this theorem on the existence of a suitable algebra given by A. L. S. Corner in a seminar at Essen University in 1983. One of the difficulties in constructing suitable R–algebras A (with particular idempotents) is to ensure that A is freely generated as an algebra modulo an ideal of well–defined relations and that at the same time the R–module R A is free, thus A can be used for realization theorems from Chapter 12 (see the next Theorem 15.1.2 (a), (b)). Beginning with Newman’s Diamond Lemma Bergman [50] developed a general method showing that for free rings modulo an ideal of special relations a reduction argument can be given which provides normal forms of elements and thus a basis exists. This is also reflected in the proof of the next theorem in showing that Σ is a basis, and the reader is invited to compare the arguments below with Bergman’s reductions. Theorem 15.1.2. Let q be a positive integer and R be a commutative ring such that R/qR = 0. Then there is an R–algebra A such that the following holds. (a)

RA

is a free R–module of countable rank.

(b) The algebra A is freelygenerated by σi , σ i (i ≤ q) subject to the only relations σ i σj = δij , σi σ i = 1. 0≤i≤q

(c) There is a ‘trace’–homomorphism T : A −→ R/qR such that the following holds for any ϕ, σ ∈ A.

589

15.1 Leavitt type rings: the discrete case

(i) (ϕ + σ)T = ϕT + σT, (ii) (ϕσ)T = (σϕ)T , (iii) 1T = 1 + qR.

Proof. The key to finding A is a suitable definition of a free R–module F such that A can be expressed as a subalgebra of EndR F . Thus let B = {b = (bi )i<ω | bi ∈ Z, 0 ≤ bi ≤ q, bi = 0 for almost all i < ω}  Rb be a free R–module with basis B. As usual, define the support and F = b∈B

for b = (bi )i<ω ∈ B as [b] = {i < ω : bi = 0} ⊂ ω  rb b ∈ F , where which extends naturally to v = b∈B

  [b] : b ∈ B, rb = 0 . [v] = Then we let  v  = sup [v] be the norm of v.  Bn is a partition of B. If Bn = {b ∈ B :  b  = n}, then B = n<ω

Let I be the set of all finite (possibly empty) sequences i = (i0 , . . . , im−1 ) = i0 ∧ . . . ∧ im−1 with 0 ≤ ij ≤ q and 0 ≤ j < m < ω. Then  i  = m is the norm of i and  ∅  = 0 as above. For 0 ≤ i ≤ q define the shift homomorphism σi : F −→ F as  0 if b0 = i (b0 , b1 , b2 , . . . )σi = (b1 , b2 , . . . ) if b0 = i, and the substitution homomorphism σ i : F −→ F as (b0 , b1 , . . . )σ i = (i, b0 , b1 , . . . ) and σ ∅ = 1 = idF by the action on the basis elements b = (b0 , b1 , . . . ) ∈ B. Let A be the subalgebra of EndR F generated by σi , σ i (i ≤ q). (15.1.1) It is easy to check that σ i σj = δij and

0≤i≤q

σi σ i = 1.

(15.1.2)

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Some useful classes of algebras

More generally, if i = (i0 , . . . , im−1 ), j = (j0 , . . . , jn−1 ) ∈ I,

(15.1.3)

then let σij ∈ EndR F be defined by (b0 , b1 , . . . )σij =



0 (j0 , . . . , jn−1 , bm , bm+1 , . . . )

if i = (b0 , . . . , bm−1 ) if i = (b0 , . . . , bm−1 ).

Let σi = σi∅ , σ j = σ∅j ; the shift homomorphism σi and the substitution homomorphism σ j are special cases. We call i the covariant and j the contravariant component of σij . Next we want to discuss the arithmetic of the elements σij of the R–algebra A. With i and j, as above the following formulas are obvious: σi = σi0 · · · σim−1 and σij = σi σ j .

σ j = σ jn−1 · · · σ j0 ,

(15.1.4)

From (15.1.4) follows immediately that the R–module R A is generated by the σij s, thus A = Rσij | i, j ∈ I. Obviously

A ⊆ HomR (F, F ) ⊆ HomR (F, Rω ) ∼ = Rω

and if R is noetherian, then Rω is ℵ1 –free by Theorem 1.3.8, hence R A is a submodule of a free R–module of countable rank. However, we can say more, which is needed to prove that A is freely generated by the σ i , σi (i ≤ q) with the only relations in (b) of the theorem: We use (15.1.2) and (15.1.4) to calculate   σij = σi σ j = σi σk σ k σ j = (σi σk )(σ k σ j ) =



0≤k≤q

σ(i∧ k) σ

(j∧ k)

=

0≤k≤q

Thus

0≤k≤q



∧

σij∧ kk .

0≤k≤q

∧

σij∧ 00 = σij −



∧

σij∧ kk ,

1≤k≤q

and so the σij s are not linearly independent. We have seen that the σij s generate ∧ the R–module R A and if we discard all homomorphisms of the form σij∧ 00 from

15.1 Leavitt type rings: the discrete case

591

the σij s and let Σ be this reduced subset of EndR F , then clearly still A = R Σ. Moreover, Σ freely generates R A.

(15.1.5)

To see this, we order Σ by the norms: first we note that ω × ω is well–ordered  by the lexicographical ordering induced from the ordinal ω. If σij , σij ∈ Σ, then let 

σij < σij ⇐⇒ ( i ,  j ) < ( i ,  j ) in ω × ω.  Now consider any sum σ = rji σij with σij ∈ Σ and coefficients rji ∈ R almost all 0. Let k ∈ ω be a fixed integer which is larger than any of these norms contributing to σ. For σ = 0 we must show that all rji are 0, and work from the bottom to the top. 

Let rji be a first non–trivial coefficient of σij with respect to the ordering defined by the norms, thus 

i = (i0 , . . . , im −1 ), j = (j0 , . . . , jn  −1 ) and (m , n ) is minimal among the (m, n) with i and j as in (15.1.3) and rji = 0 in σ. We consider the element i = (i , 0, . . . , 0, 1, 0, . . . ) ∈ B with an extra coordinate 1 and k zeros between i and 1; similarly let j = (j , 0, . . . , 0, 1, 0, . . . ) ∈ B also with an extra 1 and k zeros between j and 1. It follows immediately that  i σij = j . Conversely we claim i σij = j =⇒ i = i and j = j .

(15.1.6)

By minimality we have m ≤ m and from i σij = 0 follows i = (i , 0, . . . , 0) with m − m zeros after i . The hypothesis of (15.1.6) now reduces to i σij = (i , 0, . . . , 0, 1, 0, . . . )σij = (0, . . . , 0, 1, 0, . . . )σ j = (j, 0, . . . , 0, 1, 0, . . . ) = j . The last bracket has k − (m − m ) zeros between j and 1, while j has k zeros between j and 1. It follows that (j, 0, . . . , 0, 1) = (j , 0, . . . , 0, 1), hence j = (j , 0, . . . , 0) with m − m zeros at the end of the last bracket. If m < m, then jn−1 = 0 from the last expression and in−1 = 0 from the parallel result on i above. However, elements σij with such end terms are excluded from Σ. Therefore m = m and both implications in (15.1.6) follow. Now it is clear that the only contribution of i σ to the basis element j comes    from the summand rji σij of σ. Thus rji = 0 from σ = 0, and the set Σ freely

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Some useful classes of algebras

generates the R–module A. Hence (15.1.5) is shown and (a) follows for arbitrary (not only for noetherian) commutative rings R. The assertion (b) of Theorem 15.1.2 follows immediately from the proof of (a). By the relations mentioned in (b) any algebra relation σ = 0 in A reduces to an q  i j  ∧ ∧ R–linear equation σ = rj σi with i, j ∈ I. Using σij∧ 00 = σij − σij∧ kk above k=1

(which follows from the relation mentioned in (b)), we may assume that σij ∈ Σ, and if σ = 0, then all coefficients rji must be 0. (c) Next we construct the ‘trace’–homomorphism T : A −→ R/qR satisfying (i), (ii) and (iii) of Theorem 15.1.2 (c). The elements σ ∈ A are ω × ω– matrices with respect to B; thus we will extract the trace mod qR from σ. We must do this with care, because σ represents an infinite matrix. If b ∈ B, then let   πb : F −→ R/qR j → jπb = δbj (1 + qR) be the natural projection onto Rb mod qR, where δbj denotes the Kronecker symbol. If σ ∈ A and s ∈ ω, we define kσπk , (15.1.7) σTs = k∈Bs

which counts the multiplicity of the eigenvalue 1 of σ mod qR in Bs . The right– hand side is a finite sum because Bs is a finite set, thus Ts : A −→ R/qR is a well–defined homomorphism, which we want to extend to   (15.1.8) T : A −→ R/qR σ → σTs . s<ω

Hence we must show that σTs = 0 for almost all s ∈ ω. It is enough to consider the action of Ts on the generators σji , that is σji Ts =



(kσji πk ).

k∈Bs

Let i, j be as in (15.1.3) and note that kσji πk takes only the values qR or 1+qR. Moreover, bσji πk = 1 + qR, if and only if i  m = b  m, j  n = k  n and bm+t = kn+t for all t ∈ ω. We now assume that s > min(m, n) and consider only summands kσji πk . Thus s =  k  > min(m, n), and the last equations force m = n. Hence i = j and k  n = i, and now we count the non–trivial

15.1 Leavitt type rings: the discrete case

593

summands kσii πk for k ∈ Bs . The entries k  n = i are fixed and ki (n ≤ i < s − 1) can take all values 0 ≤ ki ≤ q, while 0 < ks−1 ≤ q because  k  = s and therefore ks−1 = 0. Hence the number of non–trivial summands from Bs is q(q + 1)s−m summand contributes 1 + qR, and the corresponding sum is  . Each i i (kσj πk ) = q(q + 1)s−m (1 + qR) = qR as well. Hence σji Ts = 0, σj Ts = k∈Bs

for all s < ω with s > min(m, n). Thus (15.1.8) is well–defined and T is a homomorphism, hence (i) holds. It remains to show (ii) and (iii) of the theorem. (iii): if m = n = 0, then σji = σ∅∅ = id and σji Ts = 0 for all s > 0 from the last few lines of the proof. Moreover,  k  = 0 ⇐⇒ k = (0, 0, . . . ), hence B0 = {(0, 0, . . . )} and id T = σ∅∅ T = σ∅∅ Ts = k id πk = (0, 0, . . . , )π(0,0,... ) = 1 + qR k∈B0

s<ω

and (iii) holds.  (ii): if σ, ϕ ∈ A and k ∈ B, then k = sc c is a finite sum. Multiplication by c∈B

πb shows that sb = kσπb . It follows (by matrix multiplication) that kσϕπk = sb (bϕπk ) = (kσπb )(bϕπk ) b∈B

=

b∈B



s<ω

 (kσπb )(bϕπk ) .

b∈Bs

In the last equation we rearranged the basis B using the finite subsets Bs . By definition of T it follows (kσπb )(bϕπk ) (1σϕ)T = t<ω k∈Bt s<ω b∈Bs

=

 t<ω s<ω

 (kσπb )(bϕπk ) .

k∈Bt b∈Bs

If ast denotes the sum in the large bracket, then we want to show that almost all ast are 0. If this is the case, we can rewrite the sum as   (σϕ)T = (kσπb )(bϕπk ) =

t<ω s<ω

k∈Bt b∈Bs

s<ω t<ω

k∈Bs b∈Bt



 (bϕπk )(kσπb ) = (ϕσ)T

594

15

Some useful classes of algebras

and (ii) is shown.  We may assume that σ = σij and ϕ = σij are generators of A and let s > max{ i ,  j } or t > max{ i ,  j } and claim that ast = 0. We only consider non–trivial summands of ast . Therefore we must have 

kσij = b and bσij = k. Hence k = i∧ c for a suitable c ∈ B and b = (i )∧ d for a suitable d ∈ B. Now it follows (as in the proof that Ts is 0 for almost all s) that the number of contributing summands of ast is a multiple of q and each value is 1 + qR, thus the total sum must be 0. 2

Corollary 15.1.3. Let A = σ i , σi : i ≤ q be the R–algebra from Theorem 15.1.2 and R/qR ∼ = Z/qZ. If X is an r × s–matrix and Y an s × r–matrix over A with XY ≡ Ir , Y X ≡ Is mod qA, where Ir , Is are the unit matrices of rank r and s respectively, then r ≡ s mod qZ.

Proof. Let X, Y be matrices as above with entries xij and yij from A, respectively. Then we apply the ordinary trace–homomorphism tr to the products XY =



 xij yjk

j<s

ik

and Y X =

 i
 yji xik

jk

  respectively, thus (XY )tr = ij xij yji and (Y X)tr = ij yji xji . The right– hand side is an element of A and we can apply the new trace–homomorphism T from Theorem 15.1.2 to get (XY )trT =

i,j

(xij yji )T =

(yji xij )T = (Y X)trT. i,j

The values of trT are the same on XY and Y X, respectively. Using the hypothesis of Corollary 15.1.3 and Theorem 15.1.2 (c) it follows that r = Ir tr ≡ (XY )tr T = (Y X)tr T ≡ Is tr = s mod qR, hence r ≡ s mod qZ.

2

15.2 Automorphism groups of torsion–free abelian groups

595

15.2 Automorphism groups of torsion–free abelian groups Another application of results from Chapter 12 immediately leads to automorphism groups of torsion–free abelian groups G or equivalently the groups of units of endomorphism rings EndZ G. The following rings are basic for a characterization of finite groups which appear as automorphism groups of torsion–free abelian groups. Example 15.2.1. The following rings are the so–called primordial rings: Z, Z[x]/(x2 + 1), Z[x]/(x2 + x + 1) and Rεδ = Z[x, y]/(x2 + εx + 1), (y 2 + δy + 1), ((xy)2 + 1) for (ε, δ) ∈ {(0, 0), (1, 0), (1, 1)}. These classical rings have individual names: • Z[x]/(x2 + 1) = Gaussian integers, • Z[x]/(x2 + x + 1) = Eisensteinian integers, • R00 = quaternion integers and • R11 = tetrahedral integers, and the corresponding groups are the so–called primordial groups Z2 , Z4 and Gεδ = a, b | a2+ε = b2+δ = (ab)2 , where • Q = G00 is the quaternion group, • D = G01 is the dicyclic group and • G11 is the tetrahedral group. It is not hard to see that these groups are groups of units of primordial rings: • U (Z) = Z2 , • U (Z[x]/(x2 + 1)) = Z4 and • U (Rεδ ) = Gεδ for (ε, δ) ∈ {(0, 0), (1, 0), (1, 1)}. These primordial rings are the building blocks for the following

596

15

Some useful classes of algebras

Theorem 15.2.2. [Corner] A finite group G is the automorphism group of some torsion–free abelian group, if and only if the following holds: (a) G is the subdirect product of finitely many primordial groups. (b) If G has no direct factor of order 2, then there is an element in G whose centralizer is a 2–group, and g ∈ G | g 3 = 1 is a direct product of groups whose 3–Sylow–subgroups are either 1 or Z3 . One direction of the characterization is closely related to our results in Chapter 12 and to the above primordial rings. Proposition 15.2.3. (a) The automorphism group Aut G of some torsion–free abelian group G is (isomorphic to) the group U := U (R) of units of a subring R of some Q– algebra A. If Aut G is finite, then R is an order of A. (b) Conversely, if U is a finite group of units of a subring R of a Q–algebra, then U is the automorphism group of some torsion–free group G of finite rank. Moreover, U is the automorphism group of some (proper) class of torsion–free abelian groups.

Proof. (a) If G is a torsion–free abelian group, then Aut G ∼ = U := U (End G) where U is the group of units of End G. Moreover, R := End G ⊆ Q ⊗ End G and Aut G is the group of units of the subring End G of the Q–algebra Q⊗End G. (b) If U ∼ = U (R) for a subring R of a Q–algebra, then let R = U (R) be the subring of R generated by the units. Thus also U (R) = U (R ), and R is free of finite rank (because U is finite). By Zassenhaus’ Theorem 12.1.6 there is a group G of the same rank, by Corner’s Theorem 12.1.1 there is a group of twice this rank, and by Theorem 12.3.27 there is a proper class of ℵ1 –free groups G with End G = R . Hence Aut G ∼ = U (R ) ∼ = U.  If U is finite, then R is an order of the Q–algebra Q ⊗ R . Thus the proposition is shown. 2 By Proposition 15.2.3 the problem of characterizing all finite groups that are automorphism groups of torsion–free abelian groups is reduced to the problem of characterizing certain orders. Some crucial steps are due to Hirsch, Zassenhaus [271]. However, the main part of the proof of Theorem 15.2.2 (now a theorem on orders) is in Corner’s Padova Lecture Notes [100] (that also overcome flaws in [261, 262, 263]).

15.3 Algebras with a Hausdorff topology

597

15.3 Algebras with a Hausdorff topology The following two constructions of rings are due to Corner [99]. Let R be an S– cotorsion–free S–ring (Definition 1.1.1) and κ be a cardinal such that |R| ≤ κ. If Λ = Pℵ0 (κ × κ) and P = Pℵ0 (κ) denote the sets of all finite subsets of κ × κ and κ, respectively, then |Λ| = |P | = κ. Take a free R–module A with basis fα (α ∈ Λ) and define an R–bilinear multiplication on A by setting fα fβ = fα∪β . This makes A a commutative R–algebra with f∅ = 1, and all the fα s are idempotents. We observe that the algebra A of Example 15.1.1 and A above are very similar. Both are commutative semigroup algebras. In the first case the multiplication of the semigroup is defined as the supremum of its elements with respect to the ordering on Q, and in this case multiplication αβ is also the supremum α ∪ β in the poset Λ. For each ρ ∈ P we have a split extension of the R–module A = Aρ ⊕ Nρ , where Aρ = fα | α ⊆ ρ × κ and Nρ = fα | α ⊆ ρ × κ, respectively. Clearly Aρ is an R–subalgebra and Nρ an ideal of A; moreover, 

ρ ⊆ σ ⇐⇒ Nρ ⊇ Nσ , and

Nρ = 0.

ρ∈P

Given a discrete R–algebra D of characteristic 0 (which will not be commutative), take the tensor product DA = D ⊗R A and identify D and A in the natural way as subalgebras of DA. Then the last split extension becomes DA = DAρ ⊕ DNρ (ρ ∈ P ). Again ρ ⊆ σ ⇐⇒ DNρ ⊇ DNσ , and



DNρ = 0.

ρ∈P

Thus the ideals DNρ generate a Hausdorff topology on the algebra DA, and DA becomes a dense subalgebra of the corresponding completion CD = DA. This R–algebra CD is a complete Hausdorff R–algebra which has the closed ideals

598

15

Some useful classes of algebras

ρ (the closure of the DNρ s) as a basis of neighbourhoods of 0. We Cρ = DN obtain a split extension CD = DAρ ⊕ Cρ

(ρ ∈ P ),

and the subalgebra DAρ is therefore of course discrete. Note that A and its com are central subrings of CD (remember that D may not be commutative), pletion A  are thus central in CD . all idempotents in A Now consider any non–zero x ∈ CD . Choose σ0 ∈ P such that x ∈ Cσ0 , and put σ(i) = σ0 ∪ {i} (i ∈ κ \ σ0 ). For each i ∈ κ \ σ0 there is a unique non–zero xi ∈ Cσ(i) such that x ≡ xi mod Cσ(i) , and since xi is a finite linear combination over D of the fα (α ⊆ σ(i) × κ), there is a finite subset τ (i) of κ such that

0 = xi ∈ Rfα . α⊆σ(i)×τ (i)

Now choose an injection p : κ \ σ0 −→ κ such that p(i) ∈ τ (i) (i ∈ κ \ σ0 ): note that |α| < κ for all α ∈ κ, then by an obvious induction such a p exists, for at each stage fewer than κ values in κ must be avoided. Define  gi = f{i,p(i)} , ei = gi (1 − gj ) (i, j ∈ κ \ σ0 ). j =i

Evidently

ρ ⇐⇒ i ∈ ρ; gi ∈ N

it is clear that the products converge in CD to a summable family of idempotents; and these idempotents are manifestly orthogonal. Finally, suppose for contradiction that for some i one has xei = 0. Then on projecting modulo Cσ(i) into DAσ(i) one finds from the last three displayed equations that xi f{i,p(i)} = 0. But xi is a non–zero linear combination over R of the fα with α ⊆ σ(i) × τ (i), and since p(i) ∈ τ (i), the map (α → α ∪ {i, p(i))} maps these α injectively. Multiplication by f{i,p(i)} therefore cannot swallow xi . The contradiction proves that the xei (i ∈ κ \ σ0 ) are all non–zero. Using D = R and C = CR we get our first Example 15.3.1. Let R be an S–ring as above. If κ is a cardinal, then there exists a commutative, topological complete R–algebra C of cardinality |R|κ with a Hausdorff topology generated by the ideals Cρ (ρ ∈ Pℵ0 (κ)) constituting a system of neighbourhoods of 0 such that C = Aρ ⊕ Cρ is a ring split extension

15.3 Algebras with a Hausdorff topology

599

with R Aρ a free R–module of cardinality κ. Moreover, assuming that R has only the idempotents 0 and 1, then any non–zero idempotent of C is a (topological) direct sum of κ non–trivial idempotents. The second example serves for constructing strange direct sums of indecomposable modules. In this case we use an R–algebra D over a principal ideal domain R which is freely generated by two elements a and b with relations a2 = a, b2 = b. Thus DR is an R–module freely generated by all reduced words in a and b having no factors a2 , b2 , respectively. It follows that aDa is a ring with identity element a in which every element is a unique linear combination over D of powers of aba. Thus aDa is isomorphic to the polynomial ring R[t] and, since the only non– zero idempotent in R[t] is 1, it is immediate that the only non–zero idempotent in aDa is a. But from its definition D admits automorphisms carrying a to each of 1 − a, b, 1 − b, so we have proved Lemma 15.3.2. The above R–module R D is free of countable rank and, for c ∈ {a, 1 − a, b, 1 − b}, the only non–zero idempotent in cDc is c. This lemma and the definition of CD are used to show the next Lemma 15.3.3. For c ∈ {a, 1 − a, b, 1 − b}, every idempotent in the subalgebra  cCD c of CD lies in cA.

Proof. To see this, note that it follows from the definition  of CD that every element x ∈ CD is expressible as a convergent sum x = xα fα , where the coefα

ficients xα (α ∈ Λ) lie in D, and for each ρ ∈ Λ almost all xα with α ⊆ ρ × κ vanish. It is easy to see that for each μ ∈ Λ the map Φμ : CD −→ D (x → xα ) α⊆μ

is a continuous algebra homomorphism. Given that x is an idempotent in cCD c, it follows that xΦμ is an idempotent in cDc and as such is equal to 0 or to c for each μ ∈ Λ; hence the xα s all lie in cRc = c2 R = cR, for otherwise we should get a contradiction by considering xΦμ , where μ is a minimal element of Λ for which   as asserted. xμ ∈ / cR. Therefore x = xα fα ∈ cA, 2 α∈Λ

 be the Now we choose a partition κ = I ∪ I  with |I| = |I  | = κ, and let G closure in CD of the R–subalgebra G = a, 1 − a, bfα , (1 − b)fβ | α ⊆ I × κ, β ⊆ I  × κ ⊆ CD .

600

15

Some useful classes of algebras

Lemma 15.3.4. The only non–zero idempotents in the subalgebras  aGa,

 − a) (1 − a)G(1

are a and (1 − a), respectively.  By Lemma 15.3.3, Proof. For this we consider a non–zero idempotent x ∈ aGa.  x ∈ aA. Now the R–algebra D admits endomorphisms Θ and Θ such that aΘ = aΘ = a and bΘ = 0, bΘ = 1. And Θ, Θ extend to endomorphisms of CD , where (xα Θfα ) and xα fα → (xα Θ fα ), respectively. xα fα →  and map G  into DA I  , reI , DA These endomorphisms fix every element of aA spectively, where AI and AI  are the subalgebras of A generated by the fα with α ⊆ I × κ and with α ⊆ I  × κ, respectively. Therefore  ∩ DA I ∩ DA I  = Ra x ∈ aA and, as x is a non–zero idempotent we have x = a, as required. The other case is similar. 2 Also note the obvious  (1 − b)G(1  − b) lie in the subalgebras Lemma 15.3.5. The idempotents in bGb, I  , respectively, which are topologically isomorphic to A.  I , (1 − b)A bA  is complete and Hausdorff and admits the intersections Uρ = The R–algebra G   ρ are G ∩ DNρ (ρ ∈ P ) as a basis of neighbourhoods of 0. The quotients G/U isomorphic to subalgebras of DAρ , hence free R–modules of rank at most κ. We can summarize our results as Example 15.3.6. Let R be an S–ring which is also a principal ideal domain and let D be as above. If κ is an infinite cardinal, then there exists a commutative,  of CD of cardinality |R|κ with a Hausdorff topological, closed R–subalgebra G topology generated by the ideals Uρ (ρ ∈ Pℵ0 (κ)) constituting a system of neighbourhoods of 0 such that  = DAρ ⊕ Uρ is a ring split extension G with (DAρ )R a free R–module of cardinality κ. Moreover, DA has two regular orthogonal idempotents with e1 ⊕ e2 = 1 and two further orthogonal idempotents f1 ⊕ f2 = 1 such that any non–trivial orthogonal idempotent summand of fi is a (topological) direct sum of κ non–trivial idempotents.

15.3 Algebras with a Hausdorff topology

601

The following complete ring comes from Göbel, Ziegler [251]. We will consider its realization as an endomorphism ring of abelian groups in the next section. Theorem 15.3.7. For every infinite cardinal λ there is a topological ring A with the following properties. (a) A is complete and Hausdorff. The topology of A has a basis consisting of λ two-sided ideals N for which the abelian group A/N is free of countable rank. (b) For each κ < λ there is a summable system aκα (α < κ) of orthogonal non–zero idempotents. (c) There is no orthogonal system of λ idempotents.

Proof. Let A be a ring freely generated by A := {aκα | α < κ < λ}, where the generators are subject to the relations a2κα = aκα and aκα aκβ = 0 (α = β). The additive group of A is free with the set A∗ of (possibly empty) non–zero products of elements of A as a basis. For any finite subset s of A let As denote the subring generated by s and Ns the two–sided ideal generated by A \ s. Then As and Ns as abelian groups are freely generated by s∗ (which is defined similarly) and A∗ \ s∗ , respectively. Thus we have A = As ⊕ Ns . The Ns s form a basis of a Hausdorff topology on A. For A we take the completion of A. The closures N s are a basis of the topology of A. We have A = As ⊕N s . This proves (a). (b) is true, since even A is summable: for every s almost all elements of A lie in N s. Our main task is to show the validity of (c). First we note that A = Z ⊕ N ∅ implies that A \ N ∅ cannot contain two orthogonal elements. Thus we can restrict ourselves to the set of idempotents e from N ∅ , in the sequel denoted by M . It is easy to see that every element e ∈ M can uniquely be written as an infinite sum zw w, e= 1 =w∈A∗

where for all s ∈ A almost all integers zw (w ∈ s∗ ) are zero.

602

15

Some useful classes of algebras

Claim 1: If e is a non–zero idempotent, there is an a = a(e) ∈ A such that za = 0. Define the length lg(w) of w ∈ A∗ as the length of the shortest product representation of w. The claim follows easily from the inequality lg(w1 w2 ) + 1 ≥ lg(w1 ) + lg(w2 ), which holds for all wi ∈ A∗ such that w1 w2 = 0. Aκ in the following claim means {ακα | α < κ}. Claim 2: Let e and f be two orthogonal idempotents in M \ {0}. Then a(e) ∈ Aκ implies a(f ) ∈ Aκ . / Aκ . Then e and f are projected onto Assume a = a(e) ∈ Aκ and b = a(f ) ∈ two non–zero orthogonal idempotents in A{a,b} . Since A{a,b} is freely generated by the two idempotents a and b, the next claim gives the desired contradiction. Claim 3: Let B be the ring freely generated by the idempotents a and b. Then   N := zw | z1 = 0 w∈{a,b}∗

does not contain two non–zero orthogonal idempotents. Every e ∈ N can uniquely be written as p(ab) + bq(ab) + r(ab)a + bs(ab)a with polynomials p, q, r, s ∈ Z[x], where p(0) = 0. The equation e0 = e1 e2 transforms to p0 = p1 (p2 + q2 ) + r1 (p2 + xq2 ), q0 = q1 (p2 + q2 ) + s1 (p2 + xq2 ), r0 = r1 (r2 + xs2 ) + p1 (r2 + s2 ), s0 = s1 (r2 + xs2 ) + q1 (r2 + s2 ). Whence e2 = e means p(1 − p − q) = r(p + xq), q(1 − p − q) = s(p + xq), r(1 − r − xs) = p(r + s), s(1 − r − xs) = q(r + s).

15.4

Realizing particular algebras as endomorphism algebras

603

Since 1 − p − q and p + xq cannot both be zero (set x = 1), the first of the two equations in the second system yields ps = qr. Hence the second system of equations transforms to p = pt, q = qt, r = rt, s = st where t = p + q + r + xs. Thus e ∈ N \ {0} is idempotent, if and only if ps = qr and p + q + r + xs = 1 hold. Let eσ now be the image of e under the homomorphism σ : B −→ Z defined by aσ = bσ = 1. If e is an idempotent from N \ {0}, then p + q + r + xs = 1 implies eσ = p(1) + q(1) + r(1) + s(1) = 1. This proves Claim 3. Property (c) is now immediate by the three claims, and Theorem 15.3.7 is shown. 2

15.4 Realizing particular algebras as endomorphism algebras The discrete case In 1954 Kaplansky [289] posed the following ‘test problems’ for abelian groups: (a) Is G ∼ = H if G ⊕ G ∼ = H ⊕ H? (b) Is G ∼ = H if each is isomorphic to a direct summand of the other? If G is a countable p–group (or more generally a totally projective p–group), then Ulm’s theorem provides an affirmative solution to both problems. If G and H are torsion–free, negative answers were given by Jónsson [287] and Sa¸siada [354]; in fact, these counterexamples can be chosen to be of rank 4. Corner [92] substantially improved this result by showing that for any integer q > 0, there is a countable, torsion–free group M with the property that for all r, s ∈ N the following holds: Mr ∼ = M s ⇐⇒ r ≡ s mod q. Using ideas from Corner [92, 94, 97], we will derive this result as the main part of the following general theorem. We have constructed suitable R–algebras (Theorem 15.1.2) and will realize them as endomorphism algebras of R–modules (with extra properties) from Chapter 12. Using the same algebras and realization theorems for abelian p–groups (Corollary 12.3.47) or for separable abelian groups (Corollary 12.3.40) we get similar counterexamples for these classes of groups (for example, in contrast to well–behaving countable, abelian p–groups). A first counterexample for p–groups was given by Crawley [106] and extended (similar to the torsion–free case) by Corner [95].

604

15

Some useful classes of algebras

Theorem 15.4.1. Let R be a principal ideal domain which is also an S–cotorsion– free S–ring. Moreover, let q be a positive integer such that R/qR ∼ = Z/qZ. Then there is an S–torsion–free R–module M such that the direct sum of r copies of M is isomorphic to the direct sum of s copies of M , if and only if r ≡ s mod q. The size λ = |M | of M (depending on R) can be prescribed as well, subject to one of the following possibilities: (a) If R is as in the theorem, then λ = λℵ0 ≥ |R|. (b) If, in addition, R has at least four distinct primes, then λ ≥ |R|. (c) If, in addition, |R| < 2ℵ0 , then |R| ≤ λ ≤ 2ℵ0 . We remark that in particular M ∼ = M s ⇐⇒ s ≡ 1 mod q, and the minimal non–trivial direct sum isomorphic to M is M q+1 . This already provides counterexamples to Kaplansky’s test problems for various classes of modules, in particular for torsion–free abelian groups. The module M can have additional algebraic properties (depending on the choice of the realization theorem we apply). This can be seen from the references in the proof; we only mention that we can choose absolute modules M .

Proof. Let A be the R–algebra given by Theorem 15.1.2. Depending on the cardinality λ and the type of the R–module M we want, we can choose Theorem 9.1.19, Corollary 9.1.23, Theorem 12.3.4, Theorem 12.1.1, Theorem 12.2.1, Theorem 12.3.27, Corollary 14.2.1 or Theorem 14.4.10 (in conjunction with Corollary 14.5.7) to obtain an S–cotorsion–free R–module M with EndR M = A, |M | = λ and various additional properties. Theorem 15.1.2 (b) provides a family εi = σi σ i (0 ≤ i ≤ q) of pairwise  orthogonal idempotents of A with εi = 1, thus M = M εi , but σ i σi = 0≤i≤q

0≤i≤q

σi, A

1. Put ψ = σi , ϕ = = M εi , C = M and apply Lemma 1.3.7 to get M εi ∼ = M . Hence M ∼ = M q+1 . Conversely it remains to show that M r ∼ = Ms r s implies r ≡ s mod q: let α : M → isomorphism and β :  M be the given  s r r s M ei and M = M fi , then we M → M its inverse. If M = 0≤i
0≤i<s

consider a composition of α with the canonical embedding and projection: xij : M = M ei → M r → M s → M fj = M, with xij ∈ End M = A; similarly, yij : M = M ei → M s → M r → M fj = M

15.4

Realizing particular algebras as endomorphism algebras

605

with yij ∈ A. Thus X = (xij ) represents an r × s–matrix over A, and Y = (yij ) is an s × r–matrix over A such that XY = Ir and Y X = Is are the r × r respectively s × s unit matrices, because α and β are inverse isomorphisms. Now the decompositions are written in terms of (matrices over) the endomorphism algebra A, and Corollary 15.1.3 applies. We get the final result r ≡ s mod q. 2 If we want to find separable R–modules which are counterexamples to Kaplansky’s test problems, then we can argue similarly but need more care due to the unavoidable endomorphisms of finite rank; see Corollary 12.3.40. Nevertheless we will establish the following result (as shown in [124]). Theorem 15.4.2. Let R be a principal ideal domain which is also an S–ring. Let q be any positive integer such that R/qR ∼ = Z/qZ. For any cardinal λ = λω ≥ |R| there is an S–separable R–module M of size λ such that the following holds for all r, s ∈ N: Mr ∼ = M s ⇐⇒ r ≡ s mod q.

Proof. We apply again Theorem 15.1.2 and get an R–algebra A with free R– module R A of countable rank. Thus Proposition 12.4.1 and Corollary 12.4.2 hold and we can use Corollary 12.3.41. There is an S–separable R–module M with EndR M ∼ = A ⊕ Fin M.  σi σ i = 1. If εi = σi σ i , then we obtain By Theorem 15.1.2 (b) we have 0≤i≤q

M∼ = M εi from Lemma 1.3.7. Hence M r ∼ = M s , if r ≡ s mod q. r s ∼ where r ≡ s mod q, and let 0 < r, s < q w.o.l.g. Now assume  M = M , This implies i≤s M εi ∼ εi . Using Lemma 1.3.7 = i≤r M  once more, we find εi and ψϕ = εi . If 1q = 1 + qZ, ϕ, ψ ∈ EndR M such that ϕψ = 0≤i≤r

0≤i≤s

then we obtain from Theorem 15.1.2 (c) that (εi T ) = (ϕψ + Fin M )T = (ψϕ + Fin M )T = (εi T ) = s1q , r1q = i≤r

i≤s

and s ≡ r mod q as required.

2

Finally, with the help of Corollary 12.3.47, we apply Theorem 15.1.2 to abelian p–groups. Using the same arguments as in the proof of Theorem 15.4.2 we get the following Theorem 15.4.3. Let p be a prime, λ = λℵ0 be a cardinal and q be a positive integer. Then there is a separable, abelian p–group M of size λ such that the following holds for all r, s ∈ N. Mr ∼ = M s ⇐⇒ r ≡ s mod q.

606

15

Some useful classes of algebras

Remark 15.4.4. Recall that M is a separable, abelian p–group, if its Ulm invari ants are finite, i.e. n∈ω pn M = 0. Moreover, note that the abelian p–group M in the theorem has a size ≥ 2ℵ0 , so we are not in conflict with Ulm’s theorem. Braun [63] considers multiplicatively closed equivalences ≡ on the positive integers (i.e. n ≡ m ⇐⇒ nk ≡ mk for all positive integers n, m, k). It is easy to see that for each such equivalence there is a unique subgroup H of (Q>0 )× (the n ∈H multiplicative group of positive rational numbers) such that n ≡ m ⇐⇒ m ℵ (see [63, Theorem 1.2]). So (using types) it is also easy to construct 2 0 examples for multiplicatively closed equivalences. This is used to derive the main result in [63, Theorem 1.1 and Corollary 1.3]. There is a suitable (ultramatricial) ring A (with free additive group) which can be realized as endomorphism ring of an abelian group G (as above). The following result then is an immediate Corollary 15.4.5. Let ≡ be a multiplicatively closed equivalence on the positive integers. Then there is a torsion–free abelian group G such that the endomorphism rings of Gn and Gm are isomorphic, if and only if m ≡ n. Moreover, Gn and Gm are isomorphic, if and only if m = n. Next we consider super–decomposable modules. Recall that a module is super– decomposable, if any non–trivial summand decomposes into a proper direct sum. Thus we also use the algebra A over a domain established in Example 15.1.1. Its idempotents = 0 decompose into non–trivial orthogonal sums of idempotents and R A is free. If A is the endomorphism algebra of an R–module, then this idempotent property translates into a non–trivial decomposition of summands = 0. Moreover, R A is free of countable rank, hence A can be realized as A = EndR M for various modules M with the help of Theorem 9.1.19, Corollary 9.1.23, Theorem 12.3.4, Theorem 12.1.1, Theorem 12.2.1 or Theorem 12.3.27. We choose one of these examples. Theorem 15.4.6. Let λ be an infinite cardinal, R a principal ideal domain which is an S–ring, and let A be the countably generated R–algebra from Example 15.1.1. Then there exists an R-module M of rank λ with EndR M = A. Thus M is super–decomposable. Note that we can also require that the super–decomposable module M in Theorem 15.4.6 is in addition ℵ1 –free. Super–decomposable abelian groups of cardinality 2ℵ0 were also constructed (without considering endomorphism rings) in Birtz [54] and the existence of non–free but ℵn –free groups of cardinality ℵn (n ∈ ω) follows from Griffith [259].

15.4

Realizing particular algebras as endomorphism algebras

607

The topological case Next we want to find similar examples as in the last ‘discrete’ subsection, but whenever possible we want to replace finite direct sums of summands by infinite direct sums (of a given cardinality of summands). This will strengthen the expectation that decompositions do not behave well. In order to control these infinite decompositions in the endomorphism algebra of the module we need its finite topology. Thus we are forced to consider topological and algebraic realization theorems from Chapter 12 and will apply topological algebras as obtained in Section 15.3. Since the ground is already prepared, we will immediately get our main results. In particular, we will derive the existence of very super–decomposable modules and say that a module M is κ–super–decomposable, if for any cardinal ρ < κ any non–trivial summand of M is a direct sum of ρ non–trivial summands. Moreover, M is very super–decomposable, if, in addition, κ = |M |. Theorem 15.4.7. Let κ be an (infinite) cardinal. (a) Then there exists an abelian group G = 0 such that every non–zero direct summand of G is a direct sum of κ non–zero summands, hence G is κ+ – super–decomposable. (b) There exists an abelian group G with direct decompositions G = G1 ⊕ G2 = H1 ⊕ H2 , where H1 and H2 are indecomposable, but G1 and G2 are κ+ –super– decomposable.

Proof. Let A be either as in Example 15.3.1 (for case (a)) or as in Example 15.3.6 (for case (b)). By the topological realization theorems (Theorem 12.1.1, Theorem 12.3.12 and the following remarks) we find an R–module M with A ∼ = EndR M as topological R–modules. The correspondence between idempotents of the algebra A and (infinite) decompositions of the module establishes the results in (a) and (b), respectively. 2 Finally, we apply the existence of another complete ring constructed in Theorem 15.3.7 to answer a related question on infinite decompositions of abelian groups. We leave it for the reader to replace the basic ring Z by a suitable, more general S–ring. We say that an abelian group G is λ–decomposable, if we can find κ–decompositions G = Gi (with Gi = 0) for each cardinal κ < λ. In particular we i<κ

608

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Some useful classes of algebras

will say that G is almost decomposable, if G is |G|–decomposable, and G will be called decomposable, if it is |G|+ –decomposable. Clearly every decomposable group is almost decomposable and the converse almost decomposable ⇒ decomposable

(15.4.1)

is occasionally true, which relates to classical results on countable groups like Pontryagin’s theorem or similar results for abelian p–groups (see Fuchs [173]). It is more likely that (15.4.1) holds, if cf(|G|) = ω. We illustrate this by a result from Soifer and Shelah (see [368, 376]) and need the following Definition 15.4.8. A group G is DSP (for decompositions into smallerpieces), if Gi ⊕ H and only if for any cardinal κ < |G| there is a decomposition G = i<κ ! ! ! ! with Gi = 0 for all i < κ and !! Gi !! < |G|. i<κ

Theorem 15.4.9. DSP–groups of cardinality with cofinality ω are decomposable, i.e. (15.4.1) holds.

Proof. (From [251]) We begin with a trivial Claim: If G = F ⊕ G is DSP and |F | < |G| where |G| is a limit cardinal, then G is DSP.  Gi ⊕ H verifying We may assume |F | < κ < |G| = λ, and choose G = i<κ

DSP. After adding at most |F | terms Gi to H, we may assume F ⊆ H. Then

(Gi + F )/F ⊕ H/F = G/F ∼ = G i<κ

gives the desired decomposition

 i<κ

Gi ⊕ H  of G .

Remark: If B ⊆ G is any subset of cardinality < λ, then we may assume B ⊆ H and obtain a decomposition of G with B ⊆ H  . Let G be a DSP–group of cardinality λ with cf(λ) = ω. Moreover, choose  Ai ⊆ G (i < ω) such that G = Ai , and |Ai | = κi < λ. We inductively i<ω

construct a sequence of decompositions H0 = G = H1 ⊕ F1 , Hi = Hi−1 ⊕ Fi−1 (i ∈ ω) such that Fi is decomposable into κi non–trivial summands, with the aim that 

G= Fi ⊕ H where H = Hi . (15.4.2) i<ω

i<ω

15.4

Realizing particular algebras as endomorphism algebras

609

The last equation is obviously a λ+ –decomposition of G which implies Theorem 15.4.9. To this end we simultaneously construct an increasing sequence of subsets Bi ⊆ Hi and assume |Bi | < λ and |Fi | < λ. We start with G = H0 and B0 = ∅. If Bi−1 ⊆ Hi−1 are constructed, then the claim and the remark after the claim give a decomposition Hi−1 = Hi ⊕ Fi such that Bi−1 ⊆ Hi . Clearly (15.4.2) and |Fi | < λ are satisfied.If π denotes the projection of G onto Hi with respect to the decomposition G = Fj ⊕ Hi , then j≤i

set Bi = Bi−1 ∪ Ai−1 π. Clearly |Bi | < λ is true, Bi is a subset of Hi , and the sum in (15.4.2) is direct. It remains to show equality. If g  ∈ G, then g ∈ Ai−1 for Fj . Our construction some i and we can write g = f + h with h ∈ Hi and f ∈ j≤i  implies h ∈ Bi ⊆ Hi = H, hence g ∈ Fi ⊕ H. 2 i<ω

i<ω

Observe that any countable, decomposable abelian group G must have an endomorphism ring of cardinality 2ℵ0 . Hence any countable, almost decomposable abelian group G with countable End G provides a counterexample for (15.4.1), hence this implication fails for the countable groups in Theorem 15.4.6. The ring A constructed in Theorem 15.3.7 is designed to give further examples in order to answer the question, if almost decomposable groups G with |G| = λ and cf(λ) = ω are also decomposable. This also requires a topological realization theorem for such singular cardinals λ. Note that the groups obtained from a realization theorem based on the Black Box all have size λκ with κ ≥ ℵ0 . Thus their cofinality is > ω. However a suitable realization theorem (e.g. Theorem 14.6.1 based on the Shelah Elevator) is prepared for cardinals λ with cf(λ) = ω. Corollary 15.4.10. For every infinite cardinal λ there exists an almost decomposable abelian group G of cardinality λ which is not decomposable. The corollary will follow from a more general result. Theorem 15.4.11. Let R be a domain with five comaximal multiplicatively closed subsets Pj (1 ≤ j ≤ 5) and λ ≥ |R| be any cardinal. Then there is an almost decomposable R–module G of cardinality λ which is not decomposable.

Proof. The algebra A from Theorem 15.3.7 built on the cardinal λ is complete and Hausdorff with the properties (a), (b) and (c) mentioned there. By Proposition 14.6.6 it can be seen as a closed subalgebra of EndR M with a free R–module M of rank λ. Moreover, by Theorem 14.6.3 there is an R5 –module M ⊗ F (in fact a fully rigid family) built (with summands Ui ) on M ⊗ F of rank λ with EndR (M ⊗F) = A. Now we restrict ourselves to domains R with five comaximal

610

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Some useful classes of algebras

multiplicatively closed sets  Pi of non–trivial elements from R. Hence the R– module G := F ⊗ M + i≤5 Pi−1 Ui ⊆ Q ⊗ F ⊗ M is as required. From the topological isomorphism EndR G ∼ = A and the idempotent properties (b) and (c) in Theorem 15.3.7 the required decompositions follow. 2

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Index

A absolutely A–rigid module, 576, 577, 579– 581 indecomposable module, 564, 576 rigid module, 571 rigid valuated tree, 565, 566 admissible of type τ , 328 pair, 334 sequence, 327 of homomorphisms, 318 algebra E(R)–algebra, 324, 462, 464, 466, 468, 471, 472, 474, 483, 487, 491, 493 generalized, 462, 467, 468, 488, 493, 498, 500–503, 506, 507, 517, 521, 526, 530 with cotorsion, 496, 503 with torsion, 487, 492, 493 approximation left, 96 right, 96 B Bass invariant, 141 dual, 142, 286 bimodule cotilting, 278 Morita, 278 tilting, 189, 278 Black Box General Black Box, 9, 10, 13, 333, 340, 345, 349, 351 Strong Black Box, 10, 11, 316, 324, 326, 330, 333

version one, 348 version three, 340 version two, 349 body, 516, 517, 518–522, 524, 530, 531 of skeleton, 520, 522 branch, 341, 567 –like element, 303, 304 branching point, 412, 420 constant, 443 element, 303, 304 divisibility sequence, 438 family Br(TX ), 341 finite, 341 infinite, 341 length, 341 node, 341 stretched, 342, 342, 346 C canonical homomorphism, 316 homomorphism of type, 317 pair, 332 pair of type, 334 summand, 315, 332 cardinal ω–Erd˝os, 565, 571 measurable, 48 regular, 294, 301 category equivalence Matlis, 88, 89 Morita, 188 tilting, 189, 225 Cauchy equivalence, 3 net, 2

635

Index sequence, 74 cellular covering, 467 chain continuous, 63, 112 decreasing of cosets, 36 descending condition, 39 divisibility, 303 class contravariantly finite, 96, 262, 268, 272 coresolving, 104 cosyzygy closed, 104 cotilting, 275, 292 covariantly finite, 96, 179, 268 covering, 95, 111 definable, 116, 179, 202, 213, 222 enveloping, 95, 223 of cofinite type, 279, 280 of countable type, 201, 207 of finite type, 201, 214 precovering, 95 special, 98 preenveloping, 95 special, 98 pretorsion, 226 pretorsion–free, 281 resolving, 104 syzygy closed, 104 tilting, 189, 214, 223 coding map, 344 conjecture finitistic dimension first, 257, 260, 272, 273 second, 257, 259, 260, 268, 273 flat cover, 134 coresolution C–, 194 minimal, 141 cosyzygy, 104 cotorsion pair, 99 Cf , 257 closed, 106, 110, 123, 178, 181 closure, 178, 181

cogenerated, 99, 123, 124, 363, 370 complete, 102, 106, 110, 117, 123, 136, 137, 140, 141, 157, 164, 180, 363, 373 cotilting, 275, 276–278, 286 deconstruction, 134, 207, 235 Enochs, 101, 135 generated, 99, 117, 361 hereditary, 105, 135, 137, 140, 141, 157, 163, 363 kernel, 99, 168, 194, 276 lattice, 99, 101 Matlis, 164 perfect, 106, 110, 111, 123, 124, 135, 140, 141, 163, 183, 277 rank, 366 tilting, 189, 194, 199, 214, 221, 222, 227, 235 trivial, 100 Warfield, 101, 135 Whitehead, 102, 371, 373 cotorsion theory, 99 cotorsion–free, 305, 426, 500, 501, 503, 505 P –cotorsion–free, 500 S–cotorsion–free, 331, 481 p–cotorsion–free, 302, 307, 310 ultra–cotorsion–free, 331 cover, 95, 282 flat, 135, 172 projective, 96, 137 strongly flat, 173 torsion–free, 135, 167 cub ω–closed, 343, 346 on a cardinal, 343 D dimension C–coresolution, 194, 268 C–resolution, 193 finitistic big, 256 little, 256 global, 256

636 shifting, 104 direct limit of a direct system of homomorphisms, 22 of modules, 21 direct system continuous of exact sequences, 113 of modules, 112 of homomorphisms, 22 of modules, 21 of short exact sequences, 22 distinguished submodules, 353 HomR (M, M ), 353, 535 M = (M, M α : α < κ), 353, 535 divisible module S–divisible, 351, 470, 487, 488 domain almost perfect, 171, 173, 183 Dedekind, 231, 242, 284 IC, 365, 370 Matlis, 85, 170, 171, 253, 365 Prüfer, 231, 236, 240 valuation strongly discrete, 41, 289 duality standard, 125, 228 E element S–pure, 5 transcendental, 14 elevator, 353, 354, 535, 536 embedding S–pure, 4 pure, 23, 29, 32, 33, 136, 158 pure–essential, 39 envelope, 94 S–divisible, 250 cotorsion Enochs, 135 Matlis, 164, 165 Warfield, 135 divisible, 253 fp–injective, 253

Index epimorphism pure, 23, 25, 26, 32, 33, 180 equivalent cotilting modules, 275, 280 tilting modules, 189, 280 Ext–Tor relations, 28 F factorization property, B–, 208 family G(ℵ0 )–, 248, 250 filter Φκλ , 45 Fλ , 293 Fκλ , 45 κ–complete, 45 countably–complete, 45 Gabriel, 232 normal, 45 on a cardinal, 42 pure, 42 filtration κ–, 150, 154 of a set, 295 C–, 112 five–submodule theorem, 535 four–submodule theorem, 541 function, continuous, 150 functor, p–, 54 G generalized monomial, 509, 516–518, 521, 523 polynomial, 507, 516–518, 521, 523, 526, 528–530 H homomorphism left minimal, 94, 252 right minimal, 95 valuated, 566 hull cotorsion, 135 injective, 7, 95 pure–injective, 39, 95

Index hyper–cotorsion, 505 I intersection Δ–intersection, 45 inverse system continuous of exact sequences, 126 of modules, 126 J Jensen–functions, 295 L ladder on an ordinal α, 321 system, 321, 365 localization, 7, 218, 462 ϕ–localization, 463 Matlis, 85, 86, 242 M matrix pointed, 34 matrix subgroup, 34 Mittag–Leffler condition, 209, 211 module E(R)–module, 462, 492, 493 I–divisible, 232, 236, 239, 242 I–torsion–free, 284, 286 R–complete, 2, 164, 165 R4 –module, 542, 546, 547 R5 –module, 538, 541, 572, 573 Rκ –module, 535, 542 Rω –module, 536, 538, 568 S–bounded, 2, 3, 6 S–complete, 3, 4 S–completion, 3, 4 S–divisible, 4, 6, 190, 223, 241, 250 S–locally free, 8 S–reduced, 2, 3, 4, 6, 73 S–torsion–free, 2, 3, 4, 6, 18, 74 Σ–pure–injective, 23, 39, 40, 221 Σ–pure–split, 221, 261 ℵ1 –free, 8

637 S–separable, 43 κ–super–decomposable, 599, 607 λ–decomposable, 607 λ–free, 44, 46–48, 307 C–cofiltered, 126 C–coresolved, 194 C–filtered, 112, 119, 144, 148, 155, 207, 214, 235 C–resolved, 193 P ∞ -torsion, 193 h–divisible, 88 h–reduced, 88 absolutely pure, 136 almost decomposable, 608 Baer, 151, 193 base, 5 bounded, 6, 165 coherent, 232 completely decomposable, 43 completely reducible, 75 cotilting, 274, 283 n–, 275 Bass, 276, 292 cotorsion Enochs, 18, 101, 135, 167, 172, 373 Matlis, 18, 88, 89, 164, 167, 170, 172 Warfield, 18, 135, 167 cotorsion–free S–cotorsion–free, 19 S–precotorsion–free, 19 countably presented, 22, 25, 212 cp–filtered, 136 cyclically presented, 101 density character, 351 divisible, 6, 191 Lukas, 192, 231 Ringel, 192 dual, 29, 275 endofinite, 40 finendo, 219, 220, 230 finitely C–filtered, 112 finitely presented, 22, 25 finitely splitting, 50

638

Index flat, 7, 135, 167, 168, 172 fp–filtered, 136 fp–injective, 136 free, 7 Gorenstein flat, 141 injective, 141, 261 projective, 141 injective, 7 locally free, 8 locally projective, 8 of cofinite type, 279, 283, 291 of countable type, 201 of finite type, 201 of flat dimension ≤ n, 7, 135 of injective dimension ≤ n, 7, 137, 140 of projective dimension ≤ n, 7, 137, 138, 140 product–complete, 40, 221, 222, 261 projective, 7, 149, 188 projectively–separable, 43 pure–injective, 23, 34, 38–40, 43, 97, 101, 116, 121, 123, 124, 127, 128, 159, 166, 168, 169, 213, 274, 277, 285, 288, 289 reduced, 6, 164 S–reduced, 19 slender, 6, 43, 46, 46, 48, 57, 59–61, 63, 70, 71, 73–76, 78– 80, 83, 90, 91 stout, 71 strongly λ–free, 307 strongly flat, 164, 167, 170–173, 177 super–decomposable, 587, 606 tilting, 189 Tf , 259, 260, 271, 273 n–, 189, 275 Bass, 191, 242, 254, 276 classical, 189 Fuchs, 191, 223, 241, 366 Lukas, 193, 231 partial, 200

Ringel, 192, 231 Salce, 238, 240, 287 torsion–free, 18, 101, 135, 164, 167, 169 torsion–less, 43 tower, 208 transitive, 473 uniquely transitive, 473, 474, 475 W2 , 102, 366, 371 Whitehead, 102, 288, 362, 365, 371 morphism diagonal, 208 N norm, 325, 332, 342, 345 function, 351 of a branch, 342 of a module, 345 of a set, 10 of an ordinal, 10 on a tree, 10, 342 P precover, 95 special, 98 strongly flat, 164 weak, 130 preenvelope, 94, 219 special, 98 divisible, 136 fp–injective, 136 Principle Diamond ♦λ E, 294 Generalized Weak, 363 Generalized Weak Ψλ E, 301, 361 Minus ♦− λ E, 296 Weak Φλ E, 301 Shelah’s Uniformization, 365, 370 problem Whitehead, 365, 371 shifted, 366, 372 product F–, 42

Index F–reduced product, 45 progenerator, 188 pure S–pure, 4 S–pure, 4, 43, 312, 313, 332, 333, 335, 340, 470, 471 p–pure, 468 ideal pure, 58 pure–invertible algebra, 474, 475 R radical group radical, 69 rank, 317, 351, 462, 470–472, 493, 498 topological, 582 reduced module S–reduced, 305, 470, 471 reduced term, 512, 520 reduction of terms, 511, 513 refinement κ–, 121, 361 resolution C–, 193 minimal, 141 restriction, 244 rigid λ–family A–rigid λ–family, 403, 550, 551, 554, 576 of R4 –modules, 547 of R5 –modules, 572 of Rκ –modules, 542, 543, 544, 546, 548, 549, 551 of Rω –modules, 548, 570, 571 ring S–ring, 2 S–ring, 2, 5, 13, 14, 16, 19, 20, 43, 61, 85, 86, 89, 90, 302, 313, 316, 324, 326, 331, 333, 343, 349, 352, 470, 472, 481, 487, 506, 507, 516 κ–noetherian, 138 h–local, 79 coherent, 43, 231 strongly, 54 Eisensteinian integers, 595

639 Gaussian integers, 595 hereditary, 231 Iwanaga–Gorenstein, 140, 261 0–, 256 1–, 191, 254, 276, 283, 292 n–, 140, 193, 260 Morita equivalent, 188 of finite character, 79 of fractions, 236 primordial, 595 quasi–Frobenius, 256 quaternion integers, 595 semihereditary, 231 semiperfect, 97 tetrahedral integers, 595 tilted, 189 von Neumann regular, 41, 231 S sequence pure–exact, 23 set closed in a filtration, 142 coding, 347 non–reflecting, 294 sparse, 294 stationary, 294, 343 upper directed, 21 Signac–pair, 332 skeleton, 509, 511, 514, 515 socle, 75 Solovay partition theorem, 294 spectrum, 40, 241 maximal, 40, 242 Ziegler, 290 splitter, 99, 226, 227, 282, 289, 365 for C, 365 strong, 365 local, 366 Step Lemma, 302, 481, 526–529, 531 subcategory resolving, 216, 217, 280 subfiltration, 150 submodule S–pure, 4, 6 –stable, 75

640 pure, 6, 23, 26, 29, 31–33, 116, 121, 124, 157, 158, 160, 162, 163, 212 RD–, 6 tight, 232 torsion–free essential, 135 subset algebraically independent, 13 multiplicative, 1, 7, 85, 190, 218, 223, 240, 243, 250, 261 Σ, 243 saturated, 240 saturation of, 243 purification of, 33 summable family, 62 support of a module, 246 of an element, 9 Svenonius sentence, 318 system almost rigid, 289 localizing finitely generated, 236, 236, 240–242, 287 rigid, 289, 290, 292 syzygy, 103 T topology R–complete, 2 R–topology, 2 S–complete, 3 S–topology, 2 S–topology, 313, 524, 527 p–adic, 310 complete in the F–topology, 73 finite, 354, 354 Hausdorff, 73 metrizable, 72 Tor–pair, 100, 286 lattice, 100 torsion class tilting, 226 classical, 228 torsion pair, 180, 229 hereditary, 285

Index tilting, 226 torsion–free class cotilting, 281, 286 hereditary, 285 transcendence degree, 470, 475 S–transcendence, 14 trap, 343, 344, 347 admissible, 344 family of traps, 348 genuine, 347 partial, 343 tree for General Black Box, 340 nursery, 341 on X, 341 Signac, 333 subtree, 341 valuated, 566 with norm, 342, 342 V vocabulary for bodies, 516 for skeletons, 509, 516