Abelian Groups and Representations of Finite Partially Ordered Sets

Corollary 2.4.2 II A is a self-small abelian group , then each group in P(A) is isomorphic to a direct sum 01 copies 01 A if and only if each projective right End A -module is free . PROOF. A consequence of Theorem 2.4.1.

For an abelian group A, Pf(A) is the category with summands of finite direct sums of copies of A as objects and group homomorphisms as morphism sets. For a ring R, Pf(R) denotes the category of finitely generated projective right R-modules with right R-module homomorphisms as morphism sets. Modules in Pf(R) are precisely summands of finitely generated free R-modules. If R = End A / NEnd A with A a torsion-free abelian group of finite rank, then each M in Pf(R) is an R-lattice as defined in Section 2.2. This follows from the observation that the additive group of R is torsion-free with finite rank, since NEnd A = End A n J QEnd A is a pure subgroup of the finite rank torsion-free group End A. Consequently, a summand ofa finite direct sum ofcopies of R is a finitely generated R-module with finite rank torsion-free additive group . The next proposition demonstrates the intimate connection between torsionfree abelian groups of finite rank and lattices over subrings of finite-dimensional semisimple Q-algebras.

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71

Theorem 2.4.3 Assume that A is a torsion-free abelian groupoffinite rankand NEnd A = End A n lQEnd A. There is a functor H : Pf(A) ~ Pf(End AI' NEnd A) given by H(G) = HA(G)I HA(G)NEnd A such that ifG and K are in peA), then G and K are: (a) quasi-isomorphicifand only ifQH(G) andQH(K) are isomorphicas right

QEnd AI1QEnd A-modules; (b) isomorphic if and only if H(G) is isomorphic to H(K); (c) isomorphicat a prime p ifand only if H(G)p is isomorphicto H(K)p as a right (End AI NEnd A)p-module; (d) nearly isomorphic if and only if the End AI NEnd A-lattices H(G) and

H(K) are in the same genus class. PROOF. Notice that H is a functor, being the composite of the functor H A of Theorem 2.4.2 and the functor P(End A) ~ P(End AI NEnd A) given by M t-+ MIMNEndA. In particular, if f E Hom(G , K), then HA(f) : HA(G) ~ HA(K) induces H(f) : H(G) = HA(G)IHA(G)NEnd A ~ H(K) = HA(K)I HA(K) NEnd A. Assume that G, K E peA), f E Hom(G, K), g E Hom(K, G), and m is a nonzero integer with fg = mIx and gf = mIG· Then H(f)H(g) = mIH(x) and H(g)H(f) = mIH(G), since H is a functor . Conversely, assume that a': H(G) ~ H(K) and (3': H(K) ~ H(G) are End AINEnd A-homorphisms and m is a nonzero integer with a'B ' = mt ni«, and (3' a' = m IH(G). Since HA(G)andHA(K) are projective End A-modules, there exist a : HA(G) ~ HA(K) and (3 : HA(K) ~ HA(G) with a{3(HA(K» + HA(K)NEnd A = mHA(K) + HA(K)NEnd A. Now, NEnd A s;:; lEnd A and HA(G) is finitely generated as an End A-module, so, by Nakayama's lemma, a{3(HA(K» = mHA(K). Therefore, y = m- 1a{3 : HA(K) ~ HA(K) must be an automorphism. By Theorem 2.4.2, there are f : G ~ K and g : K ~ G with H A (f) = a and HA(g) = (3y-l . Hence, HA(gf) = HA(g)HA(f) = ({3y -l)a = (3{3-1a- 1 ma = m = HA(mIx) , and HA(fg) = HA(f)HA(g)=a({3y-l) = a{3{3-1a- 1 m = HA(mI G ) . But HA is a faithful functor, whence fg = mIx and gf = mIG. The proof is completed by choosing various values of m. For (a), any nonzero m will do, noticing that Q(End AI NEnd A) = QEnd AI lQEnd A. For (b), let m = 1, and for (c), let m be an integer prime to p . Finally, (d) is a consequence of (c) and the definitions of near-isomorphism and genus class of R-Iattices given in Section 2.2. Let R be an integral domain and M a torsion-free R-module, that is, if rm = 0 with r E Rand m E M , then either r = 0 or m = O. If M is a torsion-free Rmodule and N is a submodule of M, then N is a pure submodule if r M n N = r N for each r E R; equivalently, MIN is a torsion-free R -module, If N is a submodule of M, then {m EM: rm E N for some 0 =F r E R} is the pure submodule of M generated by N. The rank of M is the cardinality of a maximal R-independent subset of M .

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Theorem 2.4.4 Suppose R is a principal ideal domain and M is a countably generated torsion-free R-module. Then M is afree R-module if and only if each pure finite rank submodule of Mis afinitely generated free module. PROOF. (=}) Assume that M is a free R-module and N is a finite rank pure Rsubmodule of M. Then N is contained in, hence a pure submodule of, a finitely generated free summand of M . Thus, it is sufficient to assume that M is a finitely generated free R-module. In this case, M / N is a finitely generated torsion-free R-module, hence a free R-module (Exercise 2.4 .2). Because free modules are projective, N is a summand of M. Since M is a finitely gener ated free module and R is a principal ideal domain, N is a finitely generated free R-module. ( ~) Let {Xl , • •. , X n , .. . } be a countable set of generators for M . Define M; to be the pure submodule of M generated by {XI , . . . , x n }. Then MI ~ . . . ~ M; ~ . .. ~ M is an ascending chain of pure finite rank submodules of M with M = U{M n : n ~ I}. By assumption, each M; is a free R-module. For each n, Mn+iI M; is a finitely generated torsion-free R-module, hence free by Exercise 2.4.2. Since free modules are projective, Mn+l = Nn+l EB M; for some free R-module N; and n ~ 1. Define Ni = M, and observe that for each n , M; = N, EB . . . EB N n - l • Then M = EB{Nn : n ~ I} is a free R-module, since M = U{M n: n ~ I} and each N; is a free R-module.

Corollary 2.4.5 [Pontryagin 34] Assume that G is a countable torsion-free abelian group offinite rank. Then G is afree group if and only if each purefinite rank subgroup is afree group. PROOF. Apply Theorem 2.4.4 with R = Z.

Theorem 2.4.6 [Comer 63] Suppose R is a ring with identity and that the additive group ofR is countable, reduced, and torsion-free. There is a countable torsion-free abelian group G with End G isomorphic to R. Moreover. if the additive group of R has rank n, then G may be chosen with rank G = 2n. PROOF. Only an outline of this proof is given. See, for example, [Fuchs 73] for more detailed arguments. Suppose that the additive group of R is torsion-free with rank n. The Z-adic topology is the topology on R with open sets r + nR such that r E Rand 0 #- n E Z. Since R is reduced, R is Hausdorff in this topology. Let R* be the completion of R in the Z-adic topology . Then R* is a commutative Z*-algebra with additive group uncountable and torsion-free and R ~ R*. In fact, Z* = DpZ; with Z; the p-adie integers, an uncountable integral domain, and the transcendence degree of the quotient field of Z; over the algebraic closure of Q is uncountable. First, assume that the additive group of R has finite rank and let {I = rl , ... , r n } be a maximal Z-linearly independent subset of R. Since QR is countable, there is a

2.4 Self-Small Groups andEndomorphism Rings

73

subset {ai , . . . , an} of Z* that is algebraicallyindependent over QR , i.e., for each nonzero polynomial h(XI, . . . ,xn) with coefficients in QR , h(aj , . .. ,an) 1= 0. Define G to be the pure subgroup of R* generated by the group R + RfJ, where fJ = rial + ... + rnan E R*. Then rank G = 2n, and R is isomorphic to a subring of End G, via r t-+ left multiplication by r . On the other hand, let f E End G. Then f extends uniquely to a Z*-endomorphism g of R*. For some nonzero integer m, mf(fJ) = a + bfJ E R + RfJ and mf(ri) = ai + b,fJ E R + RfJ for each i. Hence, mf(fJ) = mg(fJ) = mg(rl)al

+ ...+ mg(rn)an = mf(rl)aj + ... + mf(rn)an = a + bfJ.

Substitution gives

Since the a/s are algebraically independent over QR, each b, equals 0, a = 0, and br, = ai for each i. Hence, mf(ri) = br, for each i, In fact, mf(l) = b, since rl = 1. This shows that mf(x) = bx for each x E R + RfJ with b E R. Since R is pure in R* and mf(l) = b E R, it follows that f is left multiplication by f(l) E R . Consequently, R is isomorphic to End G. For the general case, assume that the additive group of R has countably infinite rank. Define G to be the pure subgroup of R* generated by {R, RfJr : r E R}, where fJr = Pr lR + a.r and {Pr, ar : r E R} is a countable subset of Z* that is algebraically independent over QR. Then G is a countable group, and calculations like those above show that End G is isomorphic to R.

Theorem 2.4.7 [Beaumont,Pierce 61A] If R is a subring ofafinite-dimensional Q-algebra, then the additive group of R is quasi-isomorphic to the group NR EI7 R/NR. PROOF. See, for example, [Arnold 82].

EXERCISES

1. Prove thata functor F : X ~ X' is a category equivalence of additive categories if and only ifthere isa functor G : X' ~ Xwith F G naturally equivalent totheidentity functor lx' and GF naturally equivalent to theidentity functor Ix-

2. Show that if R is a principal ideal domain and M is a finitely generated torsion-free R-module, then M is a free R-module. 3. Assume that A is a torsion-free abelian group of finite rank and End A is a principal ideal domain. (a) Prove thatif G is a subgroup of$1 A with Hom(A , G)A = G, then G is isomorphic to $ J A for some subset J of I . (b) Prove that if G is a pure subgroup of $1 A with I finite, then G is a summand of $1 A. Conclude thatG is a finite direct sumof copies of A.

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2. Torsion-Free Abelian Groups

NOTES ON CHAPTER

2

Chapter 2 is a brief introduction to some fundamental techniques for countable torsion-free abelian groups. Further results and historical background, especially for groups of infinite rank, may be found in [Fuchs 73], a comprehensive exposition of the status of the subject up to 1973. Much of Chapter 2 is a revised and extended version of portions of [Amold 82], lecture notes focusing on torsion-free abelian groups of finite rank and their relationship to modules over subrings of finite-dimensional Ql-algebras. Included in [Arnold 82] are complete arguments for properties of orders and lattices over orders referenced in Chapter 2. The notion of quasi-isomorphism of torsion-free abelian groups of finite rank is due to [J6nsson 57, 59]. The categorical version here is as given in [Walker 64]. In [Jonsson 59], it is proved directly that frA
2.4 Self-Small Groupsand Endomorphism Rings

75

Theorem 2.4.3 is the crucial result for an interpretation of torsion-free abelian groups of finite rank as finitely generated projective modules over subrings of finite-dimensional Ql-algebras. Suchan interpretation provides justification for thecomplexity of thesegroups as well as for positive results for groups with restricted endomorphism rings. This point of view, as well as some far-reaching generalizations of Theorem 2.4.1, is exploited in [Albrecht 83, 89, 91], [Albrecht, Goeters 89, 94], and references therein. Comer's theorem, Theorem 2.4.6, can be used to construct pathological direct sum decompositions [Comer69]. This theorem has also provided motivation for a largenumber of papers on realization of endomorphism rings; see, for example, [Brenner 67], [Dugas, Gobel 82,97], [Comer, Gobel85], [Gobel, Shelah 95], and references therein.

3 Butler Groups

3.1 Types and Completely Decomposable Groups Finite direct sums of torsion-free abelian groups of rank I, called completely decomposable groups, can be classified in terms of types . Included in this section is a compilation of some of the fundamental properties of types and fully invariant subgroups determined by types . The isomorphism class of a rank-I torsion-free abelian group A is denoted by type A. Types of rank-l groups are represented by isomorphism clas ses of subgroups of Q . If X and Y are subgroups of Q and f : X --+ Y is a nonzero homomorphism, then f extends to a unique Q-endomorphism g ofQ, with g(qx) = qf(x)foreach q E Q and x E X. Since Q-endomorphisms of Q'are multiplication by elements of Q, Hom(X, Y) is isomorphic to a subgroup of Q, each 0 =f=. f E Hom(X, Y) is a monomorphism, and End X is isomorphic to a subring of Q. Lemma 3.1.1 (a) If X and Yare torsion-free abelian groups of rank 1, then the following statements are equivalent: (i) Hom(X , Y) and Hom(Y, X) are both nonzero; (ii) X and Yare quasi-isomorphic; (iii) X and Yare isomorphic. (b) The set oftypes form a distributive lattice, where for subgroups X and Yof Q, (i) type X s type Y ifand only ifHom(X, Y) =f=. 0; (ii) the meet, type X n type Y, of type X and type Y is type X n Y; and (iii) the join, type X U type Y, of type X and type Y is type X + Y.

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77

PROOF. Exercise 3.1.1. A consequence of Lemma 3.1.1(a) is that if X and Yare nonzero subgroups of = type Y if and only if (X + Y)/X and (X + Y)/ Yare finite. Let G be a torsion -free abelian group. Recall from Section 2.1 that a subgroup K of G is a pure subgroup of G if K n n G = n K for each nonzero integer n, equivalently, G/ K is torsion-free. It is routine to confirm that if K is a pure subgroup of Hand H is a pure subgroup of G, then K is a pure subgroup of G. Let H be a subset of G and define (H) to be the subgroup of G generated by H and define (H)* = {x E G : nx E (H)} for some 0 =1= n E Z}, the pure subgroup of G generated by H. In particular, (H)* is a pure subgroup of G . If x is a nonzero element of a torsion-free abelian group G and X is the pure rank-1 subgroup of G generated by x , then type x is the isomorphism class of X. Define typeset G = {type x : 0 =1= x E G}. If K is a pure subgroup of G and x is a nonzero element of K, then x has the same type in K as it does in G . Pure rank-I subgroups of a torsion-free group G are like simple modules in the sense that if X and Yare pure rank-I subgroups of G, then either X = Y or X n Y = O. Types can be viewed as equivalence classes of height sequences h = (hp)n, indexed by the set of primes IT, where h p is a nonnegative integer or 00 for each prime p . The terminology arises from the notion of p-heights. Let G be a torsionfree abelian group and x a nonzero element of G . Define h p(x), the p-height ofx in G, to be 00 if x E pWG and n if x E pnG\pn+JG . Then hex) = (hp(x»n is a height sequence. Let X be the pure subgroup of G generated by x . Since X is a pure subgroup of G, the p-height of x in G is exactly the same as the p-height of x in X for each prime. Hence, rank-l groups give rise to height sequence s. There is an equivalence relation on height sequences that determ ines precisely when associated rank-I groups are isomorph ic. Two height sequences h and h' are equivalent if they differ in at most finitely many finite entries. For example , (1,1 , . . . ,1, .. .) is equivalent to (0,0,0,1,1, .. .) but not to (0,00,1,1 , . . .). A routine verification shows that this is an equivalence relation . Observe that if m and n are nonzero integers and 0 =1= x E G, then h(mx) is equivalent to h(nx). It follows that if G is a torsion-free group, =1= x E G, and X = (x)*, then hey) is equivalent to h(x) for each =1= y EX .

Q, then type X

°

°

Theorem 3.1.2 There is a one-to-one correspondencefrom the set of types to the set of equivalence classes ofheight sequences. PROOF. The correspondence is given by ¢(type X) = [hex)], where [hex)] is the equivalence class of the height sequence of 0 =1= x EX. To see that ¢ is welldefined, assume that X and Y are rank-l groups, 0 =1= x EX, 0 =1= y E Y , and f : X -+ Y is an isomorphism. Then hex) = h(f(x». Since Y = (f(x»)* = (y)*, h(f(x» is equivalent to hey), and so hex) is equivalent to hey). As for one-to-one, suppose that hex) and hey) are equivalent. Then there are nonzero integers m and n with h(mx) = h(ny) . Define f: Qmx -+ Qny by

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f(qmx) = qny for each q E Q. Then f: X ~ Y is a well-defined isomorphism, since h(mx) = h(ny), X = (mx)*, and Y = (nYk Finally, let h = (hp)ll be a height sequence and define X to be the subgroup of Q generated by {l / pi : i ::: h p , p E TI}. Then lQ E X, h(lQ) = h, and ¢(type X) = [h] . This completes the proof. The type of an element of a direct sum can be computed from the types of its components. Suppose G = A $ B is a direct sum of torsion-free abelian groups and x = (a, b) is a nonzero element of G. Then type x = (type a) n (type b), the meet of type a and type b. This is because hp(x) = min{hp(a) , hp(b)} for each prime p and the type of an element is determined by the equivalence class of its height sequence. More generally, if G = A + B, not necessarily a direct sum , and x = a + b with a in A and bin B, then type x ~ (type a) n (type b), since hp(x) ~ min{hp(a), hp(b)} for each prime p.

Lemma 3.1.3 If G is a torsion-free abelian group with finite typeset, then the typeset of G is closed under meets. PROOF. If rank G = l , then each nonzero element of G has the same type. Assume that rank G ~ 2 and let x and y be Z-independent elements of G. Then {x + my: m nonzero integer} is an infinite subset of G . Since typeset G is finite, there are nonzero integers m and m' with m i= m' and T = typex + my = typex + m'y E typeset G. Now, T ~ typex n type my = typex n typey . On the other my) - (x m'y), so that type y = type (m - m')y ~ hand, (m - m')y = (x (type x + my) n (type x + m' y) = T. Similarly, type x ::: T, since (m' - m)x = m'(x + my) - m(x + m'y). Hence, type x n type y ~ T, and so type x n type y = T E typeset G.

+

Let

T

+

be a type and G a torsion-free abelian group. Define G(T)

= {x: type x 2: T},

called the t-socle of G, a pure subgroup of G. Define G*(T)

= ({x E G : type x >

T})

to be the subgroup of G generated by {x E G : type x > T} and G#(T)

= (G*(T))* ,

the pure subgroup of G generated by G*(T). For example, if X is a rank-l group, then X (T) = X if type X ~ T, and X(T) = 0 otherwise; X*(T) = X if type X > T, and X*(T) = 0 otherwise; and X#(T) = X*(T) for each T.

Lemma 3.1.4 Suppose G and H are torsion-free abelian groups and are types.

(1

and T

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79

(a) If t ~ a, then G(a) £; G(r). (b) G(r)(a) = G(r U a) = G(r) n G(a). (c) If H is a pure subgroup of G, then H(r) = H n G(r). (d) If f E Hom(G , H) then f(G(r» £; H(r) , f(G*(r» £; H*(r), and f(G#(r» £; H#(r). (e) (G $ H)(r) = G(r) $ H(r) , (G $ H)*(r) = G*(r) $ H*(r), and (G $ H)#(r) = G#(r) $ H#(r). PROOF. Assertions (a), (b), and (c) are consequences of the definitions , while (d)

follows from the fact that if X is a pure rank-I subgroup of G with f(X) =f:. 0, then type X ~ type f(X) , since hp(x) ~ hpf(x) for each x E X and prime p . To confirm (e), apply (d) to injections and projections for the direct sum G $ H . Recall that a completely decomposable group is a finite direct sum of rank-I torsion-free abelian groups . The r-socle of a completely decomposable group C is completely determined by the rank-I summands of C. Example 3.1.5 Suppose C = X I $ ... $ X; is a completely decomposable group with rank Xi = 1 for each i.

lf:

is a type, then CCr) = ${X i : type Xi ::: r}, C*(r) = ${Xi: type Xi > r ] = C#(r), and C(r)/C#(r) is isomorphic to ${X i: type Xi = r}, (b) The typeset of C is the meet closure of {type X i: 1 ~ i ~ n}. In particular, typeset C is finite. (a)

If x = Xl $ . . . $ Xn is a nonzero element of C with Xi in X i, then type x = n{typexi : Xi =f:. O] and type Xi = typer, if Xi is nonzero. This observation can be used to confirm both (a) and (b).

PROOF.

The critical typeset of a torsion-free abelian group G is Tcr(G) = {r : G(r)/ G#(r) =f:. O} . In view of Example 3.1.5(a), the critical typeset of a completely decomposable group records the types of rank-I summands of G . Specifically, if C = X 1$ · .. $ X; with rank Xi = 1 for each i , then Ter(C) = {type (Xi): 1 ~ i ~ n} . A torsion-free abelian group G is homogeneous if there is a type r with type X = r for each nonzero element X of G. In this case, G is called t-homogeneous . A rank-I group X is homogeneous. It follows from Example 3.1.5 that if G is homogeneous completely decomposable, then G is isomorphic to x n for some rank-I group X with type X = r , Homogeneous completely decomposable groups have some special properties. Lemma 3.1.6 [Baer 37] Assume that C is a r -homogeneous completely decomposable group.

(a)

If C = G / K for some torsion-free abelian group then K is a summand of G.

G and G( r ) + K

= G,

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3. Butler Groups

(b) Each pure subgroup of C is a summand. (c) Each summand of C is r -homogeneous completelydecomposable. PROOF. (a) It is sufficient to assume that C has rank I and type r . For the general case project onto a rank-I summand of C and induct on the number of rank-I summands of C. Let JT : G -+ C = G I K denote the canonical epimorphism with ker zr = K. Define I = {JTh: h E Hom(C , G)}, a right ideal of End C. Then IC = C . To see this, let 0 :f:. x E e. Since G(r) + K = G and JT is onto with ker zr = K, there is y E G(r) with x = JT(y). Because type y ::: r = typex and x E C with rank C = 1, there is h E Hom(C, G) with y E h(C). Thus, x = JT(y) E JTh(C) 5:; l C. Since End C is a principal ideal domain, I = fEnd C for some f E End C. Then f C = I C = C, whence f E I is an automorphism of C, and so I = End C . But Ie E I, so that there is h E Hom(C , G) with JTh = Ie . This shows that K = ker JT is a summand of G. Both (b) and (c) are consequences of Exercise 2.4.3. Following is a more direct argument. (b) It is sufficient to assume that C = n for some subgroup X of Q of type r with Z 5:; X, since a homogeneous completely decomposable group C is a direct sum XI EB· . .EB K; with each Xi a rank-I group isomorphic to a common subgroup X of Q containing Z. Write C = X" = X (Z") with C IZn a torsion group. If H is a pure subgroup of C , then H nzn is a pure subgroup, hence a summand, of Z", This is because zn I H nzn is free, being finitely generated and torsion-free. Write Hn EB D . Then H = X(H nzn), since H I(H nzn) is torsion and X(H nzn) is a pure subgroup of C. Thus, C = x(zn) = X(HnZn)EBX D = H EBX D, and H is asummandofC. (c) Assume that H is a summand of C. As in (b), C = H EB H' , where zn = (H n zn)EBD with X(HnZn)=H and H'=XD . Then H is r-homogeneous completely decomposable, since H n is a finitely generated free group, being a summand of the free group zn, and X is a rank-I group of type r ,

x

zn =

zn

zn

Theorem 3.1.7 [Baer 37] (a) Summands of completelydecomposable groups are completelydecompos-

able. (b) Let C = XI EB .. . EB X m and D = YI EB ... EB Yn be two completely decomposable groupssuch that Xi and Yj are rank-l groups. Thefollowing statements are equivalent: (i) C and D are isomorphic; (ii) C and D are quasi-isomorphic; (iii) m = n and there is a permutation a of {l , 2, .. . , n} with type Xi = type Yu(i)/or each i; (iv) rank C(r)/C#(r) = rank D(r)1 D#(r)for each type r .

PROOF. (a) Let G be a summand of a completely decomposable group C, say C = G EB H. If r E Ter(C), then C(r) = C r EB C*(r) with C r a r-homogeneous

3.1 Types and Completely Decomposable Groups

81

completely decomposable group by Example 3.1.5(a) . Moreover, C = EB{ C r : r E Tcr(C)}. The proof is an induction on n, the cardinality of Tcr(C). If n = I, then C is homogeneous and G is completely decomposable by Lemma 3.1.6(c). Assume that n » I and that r is a maximal element of Tcr(C). Then C r = C(r) is homogeneous completely decomposable and a summand of C. But C(r) = G(r) EB H(r), by Lemma 3.1.4(e) . Thus, G(r) and H(r) are homogeneous completely decomposable summands of C by Lemma 3.1.6, hence of G and H, respectively. Write G = G(r) EB G' and H = H(r) EB H' for some groups G' and H'. Then EB{Cu:a E Tcr(C)} = C = G EB H = C r EB G' EB H', whence G' EB H ' is isomorphic to EB{Cu : a E Tcr(C), a =f:. r ]. By induction on n, G' is completely decomposable, as is G = G(r) EB G' . (b) (ii) ~ (iii) Assume that C and D are quasi-isomorphic and let r be the type of some C, or D]. By Example 3.1.5(a), C(r) = C, EBC*(r) and D(r) = Dr EB D*(r) with C r and Dr r-homogeneous completely decomposable groups . Then C(r) is quasi-isomorphic to D(r), C*(r) is quasi-isomorphic to D*( r ), and so C, is quasiisomorphic to Dr. In particular, rank C, = rank Dr for each r E critical typeset C. Thus , C, and Dr are isomorphic, both being isomorphic to X" for some rankI group X of type r. It follows that C = EB{C r : r E Tcr(C)} is isomorphic to D = EB{D r : r E Tcr(D)}. The implications (iii) =} (iv) =} (i) =} (ii) follow immediately from the definitions and Example 3.1.5(a) . Recall from Chapter 2.2 that if G is a torsion-free abelian group of finite rank, then NEnd G = (JiQEnd G) n End G is a nilpotent ideal containing each nilpotent ideal of End G.

Proposition 3.1.8 If C is a completely decomposable group, then End C/ NEnd C is a finite product of matrix rings over subrings ofiQ. Let C = X I EB . .. EB X n be a completely decomposable group with rank Xi = I for each i. By grouping together those Xi 'S that are isomorphic, write C = Al EB · ·· EB Am, where each Ai is ri-homogeneous completely decomposable and the ri's are distinct types . Observe that HomrA}, A j) =f:. 0 if and only ri :::: rj. If Horn/A, , A j) =f:. 0 and Hom(A l> A i) =f:. 0, then i = j, since the ri's are assumed to be distinct. Since End C = EB{Hom(A j , A j ) : 1 :::: i, j :::: m} , it is sufficient to prove that if I = EB{Hom(A i, A j ) : I :::: i =f:. j :::: m}, then I is a nilpotent ideal of End C. In this case, I ~ NEnd C and (End C)/ I = End Al x ... x End An is isomorphic to Matn(l)(End X t> x -.. x Matn(m)(End X m ) with each End Xi asubring ofiQ and rank Ai = nU) for each i, Consequently, N«End C)/ l) = 0 and NEnd C ~ I, since NEnd C is a nilpotent ideal of End C. Then NEnd C = I and End C/ NEnd C is a finite product of matrix rings over subrings of iQ. To see that I is an ideal, let 0 =f:. f E Horn/A}, A j ) and 0 =f:. g E Hom(A n As) ~ I with r =f:. s . Then t; < rs . If 0 =f:. fg E Homt.A,, A j ) , then t, < r s :::: ri :::: rj, whence r =f:. j . Thus, f gEl, and I is a left ideal of End C. A similar argument shows that I is also a right ideal of End C. PROOF.

82

3. Butler Groups

If n is a positive integer and x E In, then x is a finite sum of a composition of homomorphisms Iu E Hom(A i, A j) ~ I with i i= j. Since m, the number of Ai'S, is finite, there is a sufficiently large n such that any such composition has some subscript i repeated. Each such composition is 0; otherwise, there is some j i= i with both Hom(Ai, A j ) and Hom(A j , Ai) nonzero, a contradiction. Consequently, In = O.

The typeset of a torsion-free abelian group G records the isomorphismclasses of pure rank-I subgroups of G. There is also a set of types that records the isomorphism classes of rank-l torsion-free homomorphic images of G. Up to isomorphism, a rank-l torsion-freehomomorphic image of G can be written as f(G) for some 0 i= f : G ~ Q. For a finite rank torsion-free abelian group G, define cotypeset G to be {type f(G): 0 i= f: G ~ Q}. Equivalently, cotypeset G is the set of all types of rank-l homomorphic images of G. For a type r, define G[r] = n{kemelf: f

s r},

E

Hom(G, Q) , type f(G)

E

Hom(G, Q), type f(G) < r}.

called the r -radical of G and let G*[r]

= n{kemelf : f

° °

Notice that G*[r] = n{G[a] :a < r}, G[r] is a subgroup of G*[r], and if i= x E G, then there is a pure subgroup K of G with rank G/ K = 1 and x + K i= 0. Hence, G[r] = G if and only if a 1:. r for each a E cotypeset G, and G[r] == if and only if r ::: a for each a E cotypeset G. For example, if X is arank-l group and r is a type, then X[r] = X if type X 1:. r and if type X :::: r , Moreover, X*[r] = X if type X -{:. rand 0 if type X < r , In particular, if r = type X, then X[r] = 0 and X*[r] = X. The next proposition shows that properties of r-socles have analogues for rradicals.

°

Lemma 3.1.9 Suppose G and Hare torsion-free abelian groups offinite rank.

(b) (c)

(d) (e)

s

=

=

r, then G[r] ~ G[a]. Moreover, (G/ G[ rDl r] 0, and if H[ r] 0, then each nonzero element of H has type less than or equal to r . /f g E Hom(G, H) and Tis a type, then g(G[r]) ~ H[r], and g(G*[ rD ~ H*[r]. /f H is a pure subgroup ofG[r], then (G/ H)[r] G[r]/ H. For each type r, (G E9 H)[ r] G[r] E9 H[ r] , and (G E9 H)*[ r] G*[ r] E9 H*[r]. If cotypeset G is finite, then it is closed under joins.

(a) Ifa

=

=

=

(a) Assume that a :::: r. Then G[r] = n{kerh: h : G ~ Q, typeh(G) :::: G[a] = n{kerh : h: G ~ Q, typeh(G) :::: a}, since if h: G ~ Q with type h(G):::: a, then typeh(G):::: r .

PROOF.

r]

~

3.1 Types and Completely Decomposable Groups

83

Next, let g : G ~ G j G [T] be the canonical epimorphism defined by g(a) = a + G[ T] . If x + G[ T] E (G jG[T])[T] = n(ker h : h : G jG[T] ~ Q, type h(G) :::: T}, then x E n(kerhg : hg : G ~ Q, typeh(G):::: T}. On the other hand , if f: G ~ Q with type f(G) :::: T, then f(G[T]) = 0 and f = hg for some h: GjG[T] ~ Q with typeh(G):::: T. Thus ,x E G[T] = n{kerf : f :G ~ Q ,typef(G):::: T} and so (G j G[ T])[T] = O. Finally, let X be a pure rank-I subgroup of H . If type Xi T, then f (X) = 0 for each f E Hom(H, Q) with type f(H) :::: To Hence, X ~ H[T] = 0, a contradiction. (b) If h : H ~ Q with type h(H) :::: T, then hg: G ~ Q with type hg(G) :::: typeh(H) :::: To Let x E G[T] . Then hg(x) = 0 and g(x) E kerh for each h: H ~ Q with type h(H) :::: T, i.e., g( x) E H[T]. A similar argument, using thoseh:H ~ Q with type h(H) < T,showsthatg(G*[T]) ~ H*[T] . (c) Let K be the subgroupofG with K j H = (G j H)[T] .As a consequence of(b), G[T]jH is contained in (GjH)[T], and so G[T] is contained in K . Conversely, if h: G ~ Q with typeh(G) :::: T, then h(H) = 0, since H ~ G[T]. Thus, h : GjH ~ Q with type h(GjH):::: T, and h(K) = h(KjH) = h«GjH)[T]) = O. Therefore, K ~ n{kerh : h: G ~ Q , typeh(G) :::: T} = G[T]. (d) This follows from (b) using projections and injections of G E& H . (e) Let 0 i= f , g E Hom(G , Q) with type f(G) i= typeg(G). For each integer n , n; = f+ng: G ~ Qisanonzerohomomorphismwithhn(G) ~ f(G)+g(G) . In particular, type hn(G) :::: type f(G) U type g(G) for each n. Since cotypeset G is finite, there are integers n i= m with type hn(G) = typehm(G). Now, b; h m = (n - m)g, and -mh n + nh m = (n - m)f with n - m i= O. Hence , (n - m)(f(G) + g(G)) ~ hn(G) + hm(G), and so type f(G)Utypeg(G) :::: typehn(G) U type hm(G) = typehn(G). Consequently, type f(G) U typeg(G) = typehn(G) E cotypeset G, as desired . The critical cotypesetofa torsion-free abelian group G is CTcr(G) = {T : G*[T]j G[ T] i= OJ. The critical cotypeset is, in fact, a subset of the cotypeset of G, since if G*[T]jG[T] i= 0, there must be h : G ~ Q with type h(G) = T. Just as for typesets, the critical cotypeset records the types of rank-I summands of a completely decomposable group . Example 3.1.10 (a) IfC = XIE&· "E&X n isa completely decomposable group with rank Xi for each i and T is a type, then C[ T]

= E&{X i : type X;f. T},

C*[T]

=1

= E&{Xi : type XiI T},

C j C[ T] is isomorphic to E&{X i : type Xi :::: T}, and C*[T]j C[T] is isomorph ic to E&{X i : type Xi T}. Moreover, cotypeset C is the join closure of {type Xi : 1 :::: i :::: n}, ajinite set. (b) If C and D are completely decomposable groups, then C and D are isomorphic if and only ifrank C*[T]jC[T] = rank D*[T]j D[T]for each T.

=

84

3. Butler Groups

PROOF. (a) The first statement follows from Lemma 3.1.9(d) and the computations of X[ r] and X*[r] for a rank-l group X. As for the latter statement, let r E cotypeset C and 0 =f- f : C ~ iQ with type f(C) = r. Then f(C) = :E{f(X;) : f(X;) =f- 0, type f(X;) ::: r}, whence r is the join of {type X; : f(X;) =fOJ. Conversely, cotypeset C is closed under finite joins by Lemma 3.1.9(e) . Assertion (b) is a consequence of (a) and Lemma 3.1.9(b).

Afinite Boolean algebra T is a finite complemented distributive lattice (T, :::) with a least element 0 and a greatest element 1. An atom of T is an element t of T with 0 < t such that if sET and 0 < s ::: t, then s = t. A finite Boolean algebra T is lattice isomorphic to the Boolean algebra peS) of all subsets of a set S with n elements. In Pt S], 0 is the empty set, 1 is the set S, and the atoms are precisely the l-element subsets of S. For an example of a Boolean algebra of types, let S3 = {type Z[l/ p], typeZ[I/ q], type Z[l/r]}, where p, q , and r are distinct primes and Z[1/p] = {m/ pj : m E Z, j ::: OJ, the subring of Q generated by lip. The sublattice T3 of the lattice of types generated by S3 is a Boolean algebra with S3 as the set of atoms. Example 3.1.11 Let T3 be afinite Boolean algebra oftypes with exactly 3 atoms rl, r2, r3, and X; a subgroup ofiQ with type X; = r; for i = 1,2,3. Define G = XI E17X2+X3(l+ I) f iQEI7iQ, a torsion-free abelian group ofrank 2. Then: (i) (ii) (iii) (iv)

typeset G = {ro, 1'1, r2, 1'3}, where 1'0 is the least element of Ti. cotypeset G = {rl U r2, r2 U r3, 1'1 U r3, rl U r2 U r3}' G(1'o) = G, G(1';) = X; for i = 1,2, and G(1'3) = X3(l + 1). G*[1'1 U r2 U r3] = G[rl U r2 U 1'3] = 0, G[r; U rj] = x, with ii, j, k} = {I, 2, 3}, and G*[1'; U 1'j] = G whenever i =1= j.

PROOF. Exercise 3.1.2.

As noted in Exercise 3.1.3, the lattice of types is not a complete lattice. Specifically, infinite meets and infinite joins need not exist. Nevertheless, there are types that bound all of the elements of the typeset and, respectively, the cotypeset of a finite rank torsion-free abelian group G even if the typeset of G, respectively the cotypeset of G, is infinite. Let S = {XI ," " x n } be a maximal Z-independent subset of G. Define the inner type of G to be IT(G) = n{typex; : 1 ::: i ::: n}. The outer type of G is OT(G) = U{typeG/Y;: 1 ::: i ::: n}, where Y; is the pure subgroup of G generated by {x j : 1 ::: i =f- j ::: n}, noticing that rank Y; = n - 1 and rank G/ Y; = 1 for each i . Lemma 3.1.12 [Warfield 68] Let G be a torsion -free abelian group offinite rank. Then (a) IT(G) and OT(G) do not depend on the maximal Z-independent set S;

3.1 Types and Completely Decomposable Groups

85

(b) IT(G ) ::: r for each r E typeset G and OT(G) ::: a for each a E cotypeset

G; (c) iftypeset G isfinite, then !T(G ) E typeset G, and ifcotypeset G isfinite, then OnG ) E cotypeset G; and (d) IT(G ) ::: OT(G ). x n} and 5' = {x ; , . . . , x~ } be two maximal Zindependent subsets of G . For each i , there is a nonzero integer m with mx; = k\ x\ + ... + knx n for some k, E Z. Then type x; = type m x ; ::: n{type k.x , : k, =I O} ::: n {typexi ; 1 ::: i ::: n} . Hence, n{type x;; I ::: i n} ::: n{typexi: 1 i n}. The reverse inclusion is proved by a similar argument. The fact that OnG) does not depend on 5 follows from the proof of (b). (b) If r E typeset G, then r = typex for some 0 =I x E G. Extend x to a maximal independent subset 5, so that type x ::: n{type s: s E 5} = !T(G). Next, let X be a rank-I group with OT(G) = U{type G I Y i : 1 ::: i ::: n} = type X. There is a monomorphism h : G ~ G I Y\ EEl • • . EEl G I Yn defined by h(x) = (x + Y I , . •• , x + Yn). Since type G I Yi ::: type X, there is a monomorphism g. : GIYi ~ X for each i . The composite g = (g l, . .. , gn)h : G ~ K" is a monomorphism with rank G = n = rank X", Let a E cotypeset G and f: G ~ Q with type f (G ) = a . Define C to be the pure subgroup of x n generated by g( ker f) . Then rank C = n - 1, and so X n I C is a rank-1 group isomorphic to X by Lemm a 3. 1.6(b) and (c). Thus, g induces anonzerohomomorphismG/ker f ~ Xnl C ~ X. By Lemma 3.1.1(b), a = type f( G) ::: type X = OT(G ), as desired. Assertion (c) follows from (b) and Lemmas 3.1.3 and 3.1.9(e). As for (d), let o =I x\ E G, 5 = {XI , X2, . • • , x n } a maximal Z-independent subset of G, and Yi the pure subgroup of G generated by { Xj : I ::: i =I j ::: n}. Then type Xi ::: type G I Yi ::: OT(G ), since 0 =I X i + Yi E G I Yi . It follows that IT(G ) ::: OT(G ). PROOF. (a) Let

5 =

{XI , . .. ,

s

s s

There are relation ships between inner types and r -socles and between outer types and r -radicals, Proposition 3.1.13 [Lady 79] IfG is a torsion-free abelian group of fi nite rank, then: (a) G(r) = G if and only if Yi'(G'y > t ; (b) G[r] 0 ifand only ifOT(G)::: t ; and (c) the followin g statements are equivalent: (i) r = !T(G ) = OT(G ); (ii) Gis t-homogeneous completely decomposable; (iii) G(r ) = G and G[r] = O.

=

PROOF. Statements (a) and (b) are consequences of the definitions.

(c) (i) ~ (ii) Assume that IT(G ) = OT(G ) = r. Let 0 =I f E Hom(G , Q). Then G(r ) = G , since InG ) = r , and type f (G ) = r , since OT(G ) = r = !T(G ). By Lemma 3. 1.6(a), G = K EElY with K = ker f and Y isomorphic to

86

3. Butler Groups

f(G) . Then IT(K) = r = OT(K) so, by induction on the rank of G, K is rhomogeneous completely decomposable. Since Y is a rank-l group with type r , G is r-homogeneous completely decomposable. (ii) => (iii) If G is r-homogeneous completely decomposable, then G(r) = G by Example 3.1.5(a) and G[r] = 0 by Example 3.1.10(a). (iii) => (i) By (a) and (b), IT(G) ::: OT(G). Since IT(G) ~ OT(G) by Lemma 3.1.12(d), the conclusion is that IT(G) = OT(G).

Proposition 3.1.14 [Warfield 68] Suppose G and H are torsion-free abelian groups offinite rank. Then rank Hom( G , H) = (rank G)(rank H) if and only if IT(H) ::: OT(G). PROOF. Observe that Hom( G, H) is contained in Hom(iQG, iQH) , a iQ-vector space

with dimension equal to the product of the dimension of iQG and the dimension of iQH. Then Hom(G, H) is a torsion-free abelian group with rank ~ (rank G) (rank H) . Assume IT(H) ::: OT(G). Let S (x), . . . , x n } be a maximal Z-independent subset of H with IT(H) = n{type Xi r l ~ i ~ n}, Xi the pure subgroup of H generated by Xi, and T = {Yl, " " Ym} a maximal Z-independent subset of G with OT(G) = U{type G j Yj : } ~ j ~ n}, where Yj is the pure subgroup of G generated by {Yi :} ~ i #- j ~ n} . Then type Xi ::: IT(H) ::: OT(G) ::: type G[Y], Hence, there are nonzero homomorphisms Ii i : G ~ Gf Y, ~ Xi. Now, Xl $ . .. $ K; S; H, so that {fji:} ~ j ~ m, } ~ i ~ n} is a Z-linearly independent subset ofHom(G, H). It now follows that rank Hom(G , H) mn (rank G)(rank H). Conversely, assume rank Hom(G, H) = (rank G)(rank H). Then iQHom (G , H) = Hom(iQG, iQH), since iQHom(G , H) is a subspace of Hom(iQG, iQH) with the same dimension. It suffices to prove that if G j K is a rank-I quotient of G and C is a pure rank-I subgroup of H , then Hom( G j K, C) #- O. The composite iQG ~ iQ(Gj K) ~ iQC S; iQH is an element f E Hom(iQG, iQH) = iQHom(G, H) with iQK = ker f . Choose 0 #- k E Z with kf E Hom(G, H) . Then kf(G) S; iQC n H = C, since C is a pure subgroup of H, and so kf induces a nonzero homomorphism G j K ~ C , as desired.

=

=

=

Conditions on the r -radical and r -socle influence the structure of subgroups and torsion-free quotients of a finite rank torsion -free abelian group G. Notice that (c) of the next proposition is a partial "dual" to Lemma 3.}.6(a). Proposition 3.1.15 [Lady 79] Let G be a torsion-free abelian group offinite rank, r a type , and K a pure subgroup of G. (a) If(GjK)(r) = GjK andG[r] = 0, then GjK isa x-homogeneous completely decomposable group. (b) If K(r) K and G[r] 0, then K is a x-homogeneous completely decomposable summand of G.

=

=

3.1 Typesand CompletelyDecomposable Groups

87

(c) If H is a t-homogeneous completely decomposable subgroup of G with H ED G[r] a pure subgroup of G, then G = H ED Lfor some subgroup L of G with G[r] ~ L. PROOF. (a) By Proposition 3.1.13(a) and (b), IT(G/K) 2: r 2: OT(G). Now, OT(G) 2: OT(G / K) 2: IT(G/ K) , observing that a rank-l quotient of G / K is also a rank-l quotient of G . Then OT(G/K) = r = IT(G/K). By Proposition 3.1.13(c), G / K is r -homogeneous completely decomposable. (b) Let p be a prime. By Proposition 3.1.13(a) and (b), IT(K) 2: r 2: OT(G). Either K p is divisible, or else G p is a free Zp-module. To see this, let G / K be a torsion-free rank -l quotient of G and X a pure rank-l subgroup of K. Since type G/K ~ OT(G) ~ IT(K) ~ type X, there is a nonzero f :G/K -+ X , hence a nonzero homomorphism fp : (G / K)p -+ X P ' If (G / K)p is not a free Zp -module, then (G/K)p is isomorphic to Q. This is because a rank-I p-Iocal group is isomorphic to either Z p or Q by Exercise 2.1.3. Hence, X p is isomorphic to Q, and so either G p is a free Zp-module or K p is divisible. Now, Hom(G , K)p is a p-pure subgroup of Hom(G p, K p), and as a consequence of Proposition 3.1.14 and the assumption that OT(G) ~ IT(K) , rank Hom(G , K) = rank Hom(G , K)p = (rank G)(rank K) = (rank Gp)(rank K p) . Thus , Hom(G, K)p = Hom(G p, K p). Since K p is divisible or G p is a free Zpmodule, the restriction homomorphism ¢p : Hom(G p, K p) -+ Hom(K p, K p) is onto . But Hom(G, K) = n{Hom(G , K)p : p E TI} and Hom(K, K) = n{Hom(K, K)p : p ETI} , by Exercise 3.1 .9, so that the restriction homomorphism ¢ : Hom(G , K)-+Hom(K, K), is onto . Choosing g E Hom(G , K) with ¢(g) = I K shows that K is a summand of G . Finally, write G = K ED H . Then 0 = G[r] = K[r] ED H[r] by Lemma 3.1.9(d), whence K[r] = O. Since K(r) = K, K is r -homogeneous completely decomposable by Proposition 3.1.13(c). (c) In this case, K = (H ED G[r])/G[r] is a pure r-homogeneous completely decomposable subgroup of G/G[r] with K(r) = K and (G/G[r])[r] = O. By (b), K is a summand of G / G [r], say f : G/ G [r] -+ K is a projection. Define g = f3fex : G -+ H , where ex : G -+ G/G[r] is the quotient map and f3 : K -+ H is the natural isomorphism. Then G[r] is contained in ker g and g(x) = x for each x E H. Therefore, G = H ED ker g with G[r] ~ ker g , as desired.

EXERCISES 1. [Fuchs 73] or [Arnold82] Providea proof for Lemma 3.1.1. 2. Verify Example 3.1.11. 3. (a) Show that the lattice of types is not a completelattice. (b) Prove that there are infinitechains in the lattice of all types with no least element and no greatest element. (c) Prove that the lattice of types is uncountable.

88

3. Butler Groups

4. Let T be a finite Boolean algebra of types with n atoms . Prove that if {PI, . .. , PnI is a set of n distinct primes, then T is lattice isomorphic to the lattice of types generated by (type Z[I/ pd , . . . , typeZ[1/ Pnll. 5. Let G be a torsion-free abelian group of finite rank and r a type. Prove that if H is a pure subgroup of G with H(r) = Hand H EB G[r] is a pure subgroup of G, then H is a summand of G and G/ H is homogeneous completely decomposable. Give examples to show that the hypothesis that H EB G[ r] is a pure subgroup of G is necessary for this exercise and Proposition 3.1.15(c). 6. (a) Give an example of a rank-2 torsion-free abelian group G with infinite typeset. (b) Give an example of a rank-2 torsion-free abelian group G such that IT(G) fI. typeset G. (c) Give an example of a rank-2 torsion-free abelian group G such that OT(G) fI. cotypeset G. 7. Let X be a subgroup of lQ,0 =1= x EX, and h(x) the height sequence of x in X. (a) Give necessary and sufficient conditions on h(x) for X to be a subring of lQ. (b) Let Y be another subgroup oflQ and 0 =1= y E Y. Show that Hom(X , Y) is a torsionfree abelian group of rank I and compute a height sequence for Hom(X, Y) from h(x) and h(y). 8. Let G be a torsion-free abelian group of finite rank and F a free subgroup of G with rank F = rank G. Define the Richman type of G , RT(G) , to be the quasi-isomorphism class of the torsion group G/ F, where two torsion groups T and T' are quasi-isomorphic if there is a homomorphism f : T --+ T' with ker f and T' /image f finite groups . (a) Prove that RT(G) does not depend on the choice of F . (b) Compute the Richman type ofa rank-I group G in terms ofa height sequence h(x) for G. (c) Write G/ F = EB{Tp : P E ill, Tp = Z(p i(p.I»EB · . ' EB Z(pi(p,n» a direct sum of pgroups with 0::: i(p, I) ::: . . . ::: i(p, n) ::: 00 . Prove that IT(G) is the equivalence class of the height sequence (i(p, I» and OT(G) is the equivalence class of the height sequence (i(p,

n» .

9. Prove that if G is a torsion-free abelian group of finite rank and Il is the set of primes of Z, then G = n{G p : p E ill . Conclude that if H is another torsion-free abelian group of finite rank, then Hom(G, H) = n{Hom(G, H)p : p E ill.

3.2 Characterizations of Finite Rank Groups The class of finite rank Butler groups arises naturally . For instance, it is the smallest class of torsion -free abelian groups that contains all rank-I groups and is closed under isomorphism, finite direct sums, pure subgroups, and torsion-free homomorphic images. Furthermore, the class of Butler groups is precisely the class of finite rank torsion-free abelian groups that arise as representatives of Q-representations of finite posets, as demonstrated in Section 3.3. The first example illustrates that pure subgroups or homomorphic images of completely decomposable groups need not be completely decomposable.

3.2 Characterizations of FiniteRank Groups

89

Example 3.2.1 Let PI , P2, and P3 be three distinct primes, Xi the localization of Z at Pi, and B the pure subgroup of the completely decomposable group Xl EI3 X2 EI3 X3 generated by (1, I , 0) and (0, 1, 1). Then B is a strongly indecompo sable group of rank 2 and a homomorphic image of the completely decomposable group (XI n X 2) EI3 (X 2 n X3) EI3 (XI n X 3). PROOF. Clearly, rank B = 2. Now, type (1, 1,0) = type XI n X2, type (0, 1, I) = type X2 n X 3 , and type (1,0, -I) = type (XI nX 3) . Since (1,0, -I) = (1,1,0)(0, I, I) E B , the typeset of B contains three pairwise incomparable elements. Consequently, B is strongly indecomposable; otherwise, B is quasi-isomorphic to a completely decomposable group of rank 2, which can have at most two pairwise incomparable types by Example 3.1.5. A routine calculation shows that B = «(1, 1,0))* + «0, I , 1))* + «(1 ,0, -1))*, whence B is a homomorphic image of (XI n X2) EI3 (X2 n X 3) EI3 (XI n X3)·

Theorem 3.2.2 [Butler 65] Suppose G is a torsion-free abelian group offinite rank. Then G is a pure subgroup of a completely decomposable group if and only ifG is a homomorphic image of a completely decomposable group. PROOF. A consequence of the more general Lemma 3.2.3. A group epimorphism f : G ---+ H of torsion-free abelian groups is balanced if

f (G (r)) = H (r ) for each type r. In this case, typeset H is contained in typeset G, and f induce s an onto homomorphism G(r)/G#(r) ---+ H(r)/ H#(r) for each

type r . As an example, the homomorphism in Example 3.2.1, (XI n X2) EI3 (X2 n X3) EI3 (XI n X 3) ---+ B, is balanced. A monomorphism f : G ---+ His cobalanced if f(G) is a pure subgroup of H and f induces a monomorphism fr : G/G[r] ---+ H/ H[r] with pure image for each type r , In this case, cotypeset G is contained in cotypeset H . Notice that if f : G ---+ H is a cobalanced monomorphism, then fr(G*[rJ/G[r]) is a pure subgroup of H* [r J/ H [r] for each r , The proof of Lemma 3.2.3 is constructive in the sense that if G is a pure subgroup of XI EI3 ... EI3 X; with rank Xi = I, then there are rank-I groups Yj , constructed from the X/s, such that G is a homomorphic image of YI EI3 .•. EI3 Ym .

Lemma 3.2.3 Let G be a torsion-free abelian group offinite rank. (a) IfG is a pure subgroup ofa completely decomposable group, then there is a completely decomposable group D and a balanced epimorphism f : D ---+ G.

(b) If G is the homomorphic image of a completely decomposable group, then there is a completely decomposable group D and a cobalanced monomorphism f : G ---+ D.

PROOF. (a) Assume that G is a pure subgroup of C = XI EI3 ..• EI3 X; with each Xi a rank-l group . Then typeset G is finite, as a consequence of Example 3.1.5(b),

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and closed under meets by Lemma 3.1.3. Given x = Xl E9 ... E9 X n in C, let support X = {i : Xi # O}, support G = {support x : 0 # X E G} ordered by inclusion , and S the set of minimal elements in support G . Then S is a finite set, and for each I in S, there is a nonzero X/ in G with support X/ = I. Let X/ be the pure rank-I subgroup of G generated by X/ and notice that support y = I for each nonzero y in XI. Define D = E9{X/ : I E S} and f : D ~ G by f(E9x/) = :Ex/ for X/ E XI' It is sufficient to show that f(D(r» = G(r) for each type r , in which case f(D) = f(D(r» = G(r) = G for r = IT(G), the inner type of G . If r is a type, then D(r) = E9(X/ : type X/ ?: r}, by Example 3.1.5(a) , and H = f(D(r» = :E{X/ : type X/ ?: r] ~ G(r). The proof is concluded by showing that G(r )p is contained in H p for each prime p. In this case, via Exercise 3.1.9, G(r) = n{G(r)p : p prime} ~ H = n{Hp : p prime}, and so H = G(r). Let X = ql X\ + .. . + qnxn be a nonzero element of G(r)p with qi in Zp and Xi in Xi and choose I in S with I ~ support x . Next choose a nonzero X/ = k\x\ + ... + knxn in X/ with each k, in Z, support X/ = I, and k j prime to p for some j in I, recalling that X/ is a pure subgroup of G. Then type xi = n{type Xi: i E I} ?: n{type Xi : qi # O} ?: r, since I ~ support x andx E G(r)p. This shows that X/ E G(r)p. Furthermore, y = X - (qj/kj)x/ E G(r)p with support y ~ support x\ {j } C support x . By induction on the cardinality of support x, Y E Hp, observing that if support X has one element, then X = xi E H . Therefore, X = (qj/ kj)x/ + y E Hp and G(r)p ~ H p as desired. (b) Assume that n : C ~ G is an epimorphism for C = X\ E9 . . . E9 Xn a completely decomposable group with rank Xi = I for each i, Given a nonzero f : G ~ Q, define cosupport f = Ii : frr(X i ) = OJ, cosupport G = {cosupport f : 0 # f E Hom(G, Q)} ordered by inclusion , and S the set of maximal elements of cosupport G. For each I E cosupport G, choose j) E Hom(G, Q) with cosupport !J = I. Define D = E9{!J(G) : I E S} and ¢ : G ~ D by ¢(x) = E9{!J(x): I E S}. If r is a type, then D/ D[r] = E9{f/(G): type !J(G) ::: r ] by Example 3.1.10(a). For each type r , ¢ induces a homomorphism ¢r : G/ G[ r] ~ D / D[ r] with ¢r(x + G[r]) = ¢(x) + D[r] . It suffices to prove that ¢r is a monomorphism with pure image, in which case ¢ is a cobalanced monomorphism. In particular, cotypeset G has a largest element r = OT(G) by Lemma 3.1.9(e), since cotypeset G ~ cotypeset C is finite by Example 3.1.10(a) . By Proposition 3.1.13(b) , G[r] = O. Hence, G/G[r] = G, and ¢ = ¢r is a monomorphism with pure image. To see that ¢r is a monomorphism, it is sufficient to prove that if 0 # X + G [r] E G/G[r], then ¢r(x) # O. Since x tf. G[r], there is some f: G ~ Q with type f(G) ::: rand f(x) # O. Choose I E S with cosupport f ~ I . Then type !J(G) = U{type Xi : i tf. I} ::: U{type Xi: i tf. cosupportf} = type f(G) ::: r. If !J(x) # 0, then ¢r(x+D[r]) # 0 by the definition of¢r, and the proofis complete. Now suppose that !J(x) = O. Since !J # 0, there is 1 ::: i ::: n with i tf. I ; hence i tf. cosupport [ . Choose 0 # q E Q such that 0 # h = q!J - f : G ~ Q with i E cosupport h, i.e., hrr(X i) = O. Then type h(G) ::: (type q!J(G»U(type f(G» ::: r with cosupport h C cosupport f, since i E cosupport h \ cosupport f. Furthermore,

3.2 Characterizations of Finite Rank Groups

91

=

h(x) # 0, since f(x) # 0 and h(x) O. By induction on the cardinality of cosupport f, there is some I' E S with fl'(x) # O. Hence, ¢r(x) # 0, as desired. It remains to prove that image ¢r is a pure subgroup of D / D[ r]. Replace G[r] with G/G[r] and apply Lemma 3.1.9(a) to safely assume that G[r] 0 and ¢ ¢r. It suffices to prove that if p is a prime and x E G\pG, then ¢(x) ¢: pD . There is 0 # f: G 4- iQ with f(x) ¢: pf(G) and f(G)p = 7l pf(x) = f(G)p. This holds because G p is a finite direct sum of copies of tl p and iQ, being an epimorphic image of a direct sum C p = (X\)p $ ... $ (Xn)p of rank-l p-Iocal groups, and if X = (x)*, then X p = ZpX is a pure submodule, hence a summand of, G p. As above, there is I E S with cosupport f S; I and type h(G) type f(G) r . If h(x) ¢: ph(G), then ¢(x) ¢: pD by the definition of ¢ . Now assume 7lpf(x), and that h(x) E ph(G). Observe that h(x) E ph(G)p S; f(G)p if i ¢: I, then f(Jr Xi)p # O. Thus, there is some i ¢: I and integers a and b such that 0 # h = ah - bf:G 4- Q,h(JrX i) = 0, the p -height of bf(x) is 0 in bf(G), and the p-height of ah(x) is not equal to 0 in ah(G). Hence, h(x) ¢: ph(G), h(G)p = 7l ph(x) , and cosupport h C cosupport f. By induction on the cardinality of the cosupport of f, h '(x) ¢: pfl'(G) for some I' E S. Thus, ¢(x) ¢: pD, as desired.

=

:s

=

:s

=

A finite rank Butler group is a pure subgroup of a completely decomposable group, equivalently, a torsion-free homomorphic image of a completely decomposable group. Corollary 3.2.4 [Butler 65] The class of finite rank Butler groups is closed underfinite direct sums, pure subgroups, and torsion-free homomorphic images. Moreover, if G is a torsion-free abelian group offinite rank and H is a finite rank Butler group quasi-isomorphic to G, then G is a Butler group. PROOF. Suppose G is quasi-isomorphic to H. Then there is a monomorphism

f: H 4- G with G/f(H) finite, say G = H + 7lXl + .. . + 7lx n. Since H is a Butler group, there is a completely decomposable group C and an epimorphism g : C 4- H. There is also an epimorphism h : tl n 4- 7lxI + ... + 7lx n defined by hia», ... , an) a\x\ + ...+ anx n. Then (fg, h) : C $ tl n 4- G is an epimorphism and C $ tl n is a completely decomposable group. The remainder of the corollary follows from Theorem 3.2.2.

=

The next theorem, a characterization of Butler groups among the class of torsionfree groups of finite rank, is one of the most useful tools in the subject. This is so because the finiteness conditions of this theorem are amenable to induction arguments. Theorem 3.2.5 Let G be a torsion-free abelian group offinite rank. The following statements are equivalent:

(a) The group G is a Butler group;

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(b) [Butler 65] (i) The typeset of G is finite and (ii) for each type r in typeset G, there is a decomposition G(r) = G, EB G#(r) with G, a t-homogeneous completely decomposable group and G#(r)jG*(r) bounded; (c) (i) ThecotypesetofG is finite and (ii) for each type r E cotypeset G, G*[r]jG[r] is a t-homogeneous completely decomposable summand ofGjG[r] and if H is the image of G jG*[r] in EB {G j G[a] : a < r ], then (H) *j H is bounded. PROOF. (a) :::} (b)

(i) If G is a pure subgroup of a completely decompo sable group, then typeset G is finite by Example 3.1.5(b) . (ii) By Lemma 3.2.3, there exist C = XI EB· ·· EB X n , with each Xi ofrank 1, and a balanced epimorphism f: C ---+ G . As a consequence of Example 3.1.5(a), C(r) = Cr EB C*(r) with Cr = EB{X i : type Xi = r ] a r-homogeneous completely decomposable group . Then Dr = f(C r ) is r-hornogeneous completely decomposable as a consequence of Lemma 3.1.6. Moreover, Dr n G#(r) is a pure subgroup, hence a summand of Dr. Write Dr = G« EB(D r nG#(r» with G r r-homogeneous completely decomposable . Then G(r) = f(C(r» = Dr + f(C*(r» = Dr + G*(r ) = Dr + G#(r) = G; EB G#(r). Furthermore, G#(r) is a Butler group , being a pure subgroup of the Butler group G. Hence, G#(r ) = Y1 + . . . + Ym +G* (r), with each Yi a pure rank-I subgroup of G of type r. Then G#(r)jG*(r) is torsion , since G#(r ) = (G*(r ») *. Now, Yi n G*(r ), being a pure subgroup of G*(r ), is a rank-l group of type r = type Yi . Consequently, each Yi] Yi n G*(r) is bounded , from which it follows that G#(r)jG*(r) is bounded. (b) :::} (a) The proof is an induction on the length of a chain from an element of typeset G to a maximal element of typeset G. By (i), such a chain must have finite length. Let r be a maximal type in typeset G. Then G(r) = G, is r-homogeneous completely decomposable, in particular a Butler group. Let r E typeset G and suppose, by way of induction, that G(a) is a Butler group for each a in typeset G with a > r . Then G*(r) is a Butler group, being a homomorphic image of the Butler group EB{G(a): a > r} . Moreover, G(r)j(G r EB G*(r» is bounded by (ii). As a consequence of Corollary 3.2.4, G( r ) is a Butler group for each r in typeset G. Let r = IT(G) E typeset G. Then G(r) = G is a Butler group. (c) :::} (a) Let r be a minimal element of cotypeset G . Then G*[r] = G, and by (ii), GjG[r] = G*[r]jG[r] is a r-homogeneous completely decomposable group , hence a Butler group. Now assume that r E cotypeset G and, by way of induction, that GjG[a] is a Butler group for each a E cotypeset G with a < r. By (ii), GjG*[r] is quasi-isomorphic to a pure subgroup of the Butler group EB{GjG[a]: r > a E cotypeset G}. Thus, GjG*[r] is a Butler group as a consequence of Corollary 3.2.4. Moreover, G*[r]jG[r] is r-hornogeneous completely decomposable and GjG[r] is isomorphic to G*[r]jG[r] EB GjG*[r] by (ii). This shows that GjG[r] is a Butler group for each r E cotypeset G. Let r = OT(G) E cotypeset G . Then G[r] = 0, and G = GjG[r] is a Butler group.

3.2 Characterizations of FiniteRankGroups

93

(a) => (c)

(i) Since G is a Butler group, there is a completely decomposable group C and an epimorphism f : C ~ G. Then cotypeset G S; cotypeset C, a finite set by Example 3.1.10(a). (ii) In view of Lemma 3.2.3, there is a completely decomposable group C such that G is a pure cobalanced subgroup of C. It follows that if T is a type, then G*[T]/G[T] is a pure subgroup of C*[T]/C[T]. But C*[T]/C[T] is a Thomogeneous completely decomposable group by Example 3.1.10(a) , so that G* [T]I G [T] is a r-homogeneous completely decomposable group by Lemma 3.l .6(b)and (c). Moreover, G*[T]/G[T] is a pure subgroup ofG I G[ T], since G*[T] is a pure subgroupofG . Now(G*[ T]I G[ T])(T) = G*[T]I G[ T] and (G I G[ T])[ T] = as a consequence of Lemma 3.1.9(a). Thus, G*[T]/G[T] is a r-homogeneous completely decomposable summand of GIG[T] by Proposition 3.1.15(b). As for the final statement, let H be the image of the monomorphism fa : G I G*[T] ~ EB{ G I G[a] : a < T} defined by fa(x+G*[ T]) = EB{x+G[a] : a < T}, recalling that G*[T] = n{G[a]:a < T}. Since G is a cobalanced subgroup of C, GIG[a] is a pure subgroup of CIC[a] for each type a . Hence, there is a monomorphismh : EB{GIG[a]:a < T} ~ EB{C/C[a] :a < T},inducedby hex + G[a]) = x + C[a], with pure image. Define g : GIG*[T] ~ CIC*[T] by g(x + G*[T]) = X + C*[T]. Then bfo = fcg , where Ic : CIC*[T] ~ EB{CIC[a] :a < T} is a monomorphism. By Example 3.1.10(a) , the image of fc is a pure subgroupofEB{CIC[a] :a < T}. It now suffices to show that if B is the image of g , then (B)*I B is bounded. In this case, since H = image fa , (H) *I H would also be bounded, as desired. There is a commutative diagram with exact rows and columns:

o

o o~

o~ o~

0 -l-

-l-

G*[T]/G[T]

~

-l-

C*[T]/C[T]

~

CIC[T]

-l-

g

-l-

-l-

o

~

-l-

GIG[T] GIG*[T]

0 -l-

~

CIC*[T]

-l0

A

~

0

~

0

~

0

-l-

~

K

-l-

~

L

-l0

with A a r-homogeneous completely decomposable group . Now, K[ T] = 0, since (C I C[ T])[T] = 0 by Lemma 3.1.9(a), and if A' is the pure subgroup of K generated by A, then A'(T) = A', since A(T) = A. By Proposition 3.1.15(b), A' is Thomogeneous completely decomposable. It follows that A'lA is bounded, whence the torsion subgroup of L is bounded, and since B = image g, (B)* I B must also be bounded, as desired. Corollary 3.2.6 Assume that G is a fin ite rank Butler group and T is the finite lattice of types generated by the typeset of G. (a) The typeset ofG is the meet closure ofthe critical typeset ofG.

94

3. ButlerGroups (b) The cotypeset of G is the join closure ofthe critical cotypeset of G. (c) The lattice T is the join closure ofthe typeset of G and the meet closure of the cotypeset of G.

PROOF. (a) The typeset of G is closed under finite meets by Lemma 3.1.3. Con-

versely, first assume that r is a maximal type in typeset G. Then r E critical typeset G . Next assume that r E typeset G, r ~ critical typeset G, and for each t < a E typeset G there is a subset Sa of the critical typeset of G with a = n{8 : 8 E Sa}. Since r ~ critical typeset G , G(r) = G#(r), whence G(r)jG*(r) is bounded. Consequently, there is x E G with type x = r and x = 1: {xu :a > r] for some Xu E G with Xu = 0 or type Xu = a. Then r = typex ~ n{typexu:a > r ] ~ n{a : a > r} ~ r , By induction on the length of a chain from r to a maximal type in typeset G, r = n{8 : 8 E S} for S = U{Su : a > r ], a subset of the critical typeset of G. Thus, typeset G is contained in the meet closure of the critical typeset ofG. (b) The proof is analogous to that of (a); just use induction on the length of a chain from an element of the cotypeset of G to a minimal element of the cotypeset ofG . (c) The join closure of the typeset of G and the meet closure of the cotypeset of G is contained in T . Conversely, the lattice of all types is a distributive lattice. It follows that elements of T are joins of meets of elements of the typeset of G . But the typeset of G is closed under meets by Lemma 3.1.3. Hence, T is the join closure of the typeset of G . Analogously, T is the meet closure of the cotypeset of G, since the cotypeset of G is closed under joins by Lemma 3.1.9(e). As a consequence of the following corollary, homogeneous Butler groups are completely decomposable. Corollary 3.2.7 Assume that Gis afinite rank Butler group.

= critical typeset G and G is completely decomposable. (b) If the critical cotypeset of G is linearly ordered, then cotypeset G = critical cotypeset G and G is completely decomposable. (c) Homogeneous Butler groups are completely decomposable. (d) If all rank-l torsion-free quotients of G are isomorphic. then G is a homogeneous completely decomposable group. (a)

If the critical typeset of G is linearly ordered, then typeset G

PROOF. (a) If the critical typeset is linearly ordered, then the critical typeset is equal

to the typeset by Corollary 3.2.6(a). Hence, the typeset of G is linearly ordered. Write typeset G = {ro < ... < rn}. Then G = G(ro) = Go EB G#(ro) with Go ro-homogeneous completely decomposable by Theorem 3.2.5(b) . Moreover, G#(ro) = G(rl), and typeset G(rl) = {rl < . . . < tn }. An induction on n shows that G(rl) is completely decomposable, noting that G(rn) is completely decomposable. Then G must be completely decomposable.

3.2 Characterizations of Finite Rank Groups

95

(b) The proof is analogous to that of (a) using Corollary 3.2.6(b) and Theorem 3.2.5(c). Assertions (c) and (d) are consequences of (a) and (b), respectively. Radicals and socles are intimately related for finite rank Butler groups . Proposition 3.2.8 [Lady 79] IfG is afinite rank Butler group and rand 0 are types, then (a) (b) (c) (d)

G[r] = (G(a) : a1:. r), andG*[r] = (G(a): aj:. r)* ; G(r) = n{G[a] : at r} and G#(r) = n{G[a]: air}; G#(r) £; G[r] and G(r) £; G*[r]; and G[r no] = (G[r] + G[a])* .

PROOF. (a) Define H = (G(a) :a1:.r)* . Then H £; G[r], since if f:G -+ Q with type f(G) :::: r , and a 1:. r , then f(G(a)) = 0; otherwise, a = typex :::: type f(x) :::: r , By Lemma 3.2.3(a), there is a completely decomposable group C and a balanced epimorphism g: C -+ G. Now, g(C/C[r]) £; G/G[r]' C[r] = ~{C(a) : af.:. r} by Examples 3.1.5 and 3.1.10, and g : C/C[r] -+ G/ H is onto, noting that g(C[ r]) £; H , since g(C(a)) £; G(a) for each a . Then OT(G / H) .s OT(C/C[r]) :::: r by Example 3.1.10(a). Hence, (G/H)[r] = 0 by Proposition 3.1.13(b). But G[ r]/ H £; (G / H)[r] , as a consequence of Lemma 3.1.9(b), and so H = G[r]. A similar argument shows that G*[r] = (G(a): a f:. r)*. (b) Observe that G(r) £; n{G[a] :a t r} , This holds because if x E G(r) and f : G -+ Q with type f(G) = at r, then f(x) = 0, otherwise, a = type f( x) ::: typex ::: r . Conversely, G is a pure cobalanced subgroup of a completely decomposable group C = XI E9 . . . E9 K; with rank Xi = 1 for each i by Lemma 3.2.3(b) . Thus, n{G[a] :a t r} £; n{C[a]:a t r} = C(r), as a consequence of Example 3.I.I0(a). Since n{G[a] : at r} ~ G n C(r) = G(r), the proof is complete. Analogously, G*(r) = n{G[a] : air}. Assertion (c) follows from (a) and (b), and (d) follows from (a) and Lemma 3.1.4(b). Corollary 3.2.9 Assume that Gis afinite rank Butler group. JfG has no rank-l summands, then: (a) the critical typeset of G has neither a least nor a greatest element; and (b) the critical cotypeset ofG has neither a least nor a greatest element.

PROOF. (a) Suppose r E critical typeset G with r :::: a for each a E critical typeset G . By Theorem 3.2.5(b), G = G(r) = G r E9 G#(r) with G; a nonzero r-homogeneous completely decomposable group . This is a contradiction to the assumption that G has no rank-I summands. Next suppose r E critical typeset G with r ::: a for each a in critical typeset G. In view of Corollary 3.2.6(a), Proposition 3.1.13(b), and Proposition 3.2.8(a), G[ r] = o and G*[r] =1= O. Then G*[r]/G[r] = G*[r] is a nonzero r-homogeneous

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completely decomposable summand of GIG[r] a contradiction. (b) The proof is analogous to that of (a).

= G, by Theorem 3.2.5(c), again

Socles and radicals can be used to identify completely decomposable groups among finite rank Butler groups. If G is a finite rank Butler group and r is a type , then G(r)/G#(r) and G#[r]IG[r] are both r-homogeneous completely decomposable. By Proposition 3.2.8(c), inclusion induces a homomorphism
(a) G isacompletelydecomposablegroupifandonlyifG*[r] = G(r)+G[r] for each type t ; and (b) G is quasi-isomorphic to a completely decomposable group if and only if G*[r]/(G(r) + G[r]) is boundedfor each type r . PROOF. If C is completely decomposable, then C* [r 1 = C( r) + C[ r] for each type r by Examples 3.1.5 and 3.1.10. Moreover, if C is a subgroup of G with GIC bounded, then G(r)/C(r), G*(r)/C*(r), G[r]/C[r], and G*[r]/C*[r] are all bounded. It now follows that G*[r]/(G(r) + G[r]) is bounded. Conversely, assume that G*[r]/(G(r) + G[r]) is bounded for each type r . In view of the hypotheses and Theorem 3.2.5, if r is a minimal element of critical cotypeset G, then G*[r] = G and G/(G(r) + G[rJ) = G/(G r $ G#(r) + G[r)) = G/(G r $ G[r]) is bounded for some nonzero r-homogeneous completely decomposable group Gr . If H = G[r], then rank H < rank G and H*[a]/(H(a)+ H[a]) is also bounded for each type a as a consequence of Proposition 3.2.8. By induction on the rank of G , G[ r] is quasi-i somorphic to a completely decomposable group . Hence, G is also quasi-isomorphic to a completely decomposable group. This proves (b). As for (a), the same argument shows that if, in addition, G*[r]/(G(r)+ G[ r]) = 0, then G is completely decomposable. Corollary 3.2.11 (a) Given a type r, there is afunctor F; from the category offinite rank Butler groupsto the categoryofabeliangroupsdefinedby Fr( G) = G*[r ]/( G( r)+ G[r]). (b) A finite rank Butler group G is completely decomposable if and only if Fr(G) = for each type r . Moreover, G is quasi-isomorphicto a completely decomposable group ifand only if Fr(G) is a bounded groupfor each r. (c) For each type r and finite rank Butler group G, image
°

maximal rank.

3.2 Characterizations of Finite Rank Groups

97

(d) A finite rank Butler group G has no rank-l quasi-summands if and only

if

G(r) is contained in G[r]for each type T. PROOF.

Exercise 3.2.4.

Let G be a finite rank Butler group. By Theorem 3.2.5(b), G(r) = G, ED G#(T) with Gr a homogeneous completely decomposable group of type T for each r in the critical typeset of G. Define a regulating subgroup of G to be B = 1:{Gr : r E critical typeset G}. Notice that B need not be unique, since the Gr 's are unique only up to isomorphism. The followingpropositionisdue to A. Maderand KJ. Krapf;see [Mutzbauer93]. Proposition 3.2.12 Let G be afinite rank Butler group.

(a) If B is a regulating subgroup ofG, then G/ B is bounded. (b) There are only finitely many regulatingsubgroups of G.

(a) Write B = 1:{Gr : T E critical typeset G}. The proof is by induction on the length of a chain from a type to a maximal type in the typeset of G. If r is maximal in typeset G, then G(r) = Gr. So assume that r is not maximal and G(a) / 1:{G8 : 8 ~ a} is bounded for each T < a E typeset G. Then G(r) = G; ED G#(T), G#(r)/G*(r) is bounded, and G*(T)/1:{G 8 : 8 > T} is bounded by induction. Hence, G(T)/1:{G u:a ~ T} is finite for each r E typeset G. Let T = IT(G). Then G = G(T) and G/ B is finite. (b) For each Tin typeset G, choose n(r) with n(T)(G#(T)/ G*(T» = 0. It suffices to show that if e = lcmlnt r): r E typeset G} , d is the length of a longest chain in typeset G, and B = 1:{Gr : r E critical typeset G} is a regulating subgroup of G, then ed G £ B £ G. In this case, there are only finitely many regulating subgroups, since G/ ed G is bounded, hence finite by Exercise 2.1.1. For =f:. x in G, let d(x) denote the length of a longest chain in typeset G from type x to a maximal type. It suffices to prove that ed(x)x E B. If d(x) = 0, then type x is maximal, G#(typex) = 0, and x E G, 5; B for T = typex. Now assume that d(x) > and assume, by way of induction, that ed(Y)y E B for each y in G with type y > typex = T. Then G(T) = G, ED G#(T) and x = c + y for some c in G r and y E G#(T). If type y > typex , then d(y) < d(x) and ed(x)x = ed(x)c + ed(x)y E B. Finally,assume T = typex = typeyandd(y) = d(x).Inviewofthechoiceofe , ey = 1:{yu . type y, = a> T} E G*(r).Byinduction,ed(y)y = :E{ed(y)-IYu:a > T} E B, since d(yu) < d(y) for each a. But d(y) = d(x) , so that ed(x)x = ed(x)c + ed(Y)y E B, as desired.

PROOF.

°

°

Givena Butler group G,definethe regulatorsubgroupR(G) to be theintersection of all regulatingsubgroupsof G. A finitepartially orderedset S is an invertedforest if for each s in S, {T E S: T ~ s in S} is linearly ordered. Part (c) of the next

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3. Butler Groups

corollary is an unpublished result of O. Mutzbauer, wherein an inverted forest is called a V-tree.

Corollary 3.2.13 Let G be afinite rank Butler group. (a) The group G/R(G) isfinite. (b) If f : G ~ H is an isomorph ism, then f(R(G)) = R(H) and G / R(G) is isomorphic to H / R(H). (c) If the critical typeset ofG is an invertedforest, then R(G) ~{G(r) : r E crit ical typeset G} is a unique regulating subgroup of G, and G(r) is completely decomposable with linearly ordered typesetfor each r in the critical typeset of G.

=

PROOF. Assertion (a) follows from Proposition 3.2.12, since G has only finite many regulating subgroups B and each G / B is bounded, hence finite . (b) Notice that f(B) is a regulating subgroup of H ifand only if B is a regulating subgroup of G. This follows from the fact that if f is an isomorphism, then f(G(r)) f(G r) EB f(G#(r)) f(G r) EB H#(r). Hence, f(R(G)) R(H), since the regulator subgroup is the intersection of all the regulating subgroups. (c) Let B = ~ { G r : r E critical typeset G} be a regulating subgroup of G . Then the typeset of G (r) is a chain for each r in the critical typeset by hypothesis. Hence, G(r) is completely decomposable by Corollary 3.2.7(a), and G(r) = EB{G u : a ~ r ] by Example 3.1.5(a). Thus, G(r) S; B for each r in the critical typeset of G. Since B S; B' = ~{G(r) : r E critical typeset G}, B = B' must be a unique regulating subgroup of G.

=

=

=

The following proposition is a "dual" of the definition of a regulating subgroup and Proposition 3.2 .12(a).

Proposition 3.2.14 Let G be a finite rank Butler group. For each type r in the critical cotypeset of G, choose an H with G/G[r] = (G*[r]/G[r]) EB H and let fr : G/G[r] ~ G*[r]/G[r] be a projection with ker fr = H . Then f = EB fr : G ~ EB{ G*[ r]/ G[ r] : r E critical cotypeset G} is a monomorphism with (f(G)}*/f(G) bounded, hence finite . PROOF. Exercise 3.2.3. This section concludes with another characterization of finite rank Butler groups that will be used in Theorem 3.4 .6. Let G be a torsion-free abelian group of finite rank. Given a nonempty subset S of Il , the set of primes of Z, define G s n{G p : pES} . Observe that G n{G p ; pEn} is a subgroup of Gs .

=

=

Proposition 3.2.15 [Bican 70] Let G be a torsion-free abelian group offinite rank. Then G is a finite rank Butler group if and only if there is a partition Il = U{S(i) : 1 ::: i ::: n} such that each G S(i ) is a completely decomposable group with linearly ordered typeset.

3.2 Characterizations of Finite Rank Groups

99

PROOF. Assume that G is a finite rank Butler group and write G = XI +...+ X; for pure rank-I subgroups Xi of G . Choose 0 #- Xi E X i ' Given a permutation a of n+ = {I, 2, . .. , n}, define 5(0') to be the set of primes p with p-height xO'(I) ::: . . . ::: p-height xO'(n). Then Il = U{5(O'): a permutation of n+}, since for each prime p there is a permutation a of n+ with p-height xO'(I) ::: .. . ::: p-height xO'(n). Now, GS(O') = (X ])S(O')+...+ (Xn)s(O') has a linearly ordered typeset, since type XO'(I) ::: . . . ::: typexO'(n). But G S(O') is a finite rank Butler group , so that G S(O') is completely decomposable by Corollary 3.2.7(a). To complete the proof, choose subsets 5'(0') of 5(0') such thatthe 5'(O')'s partition Il, and observe that each Gs'(O') is completely decomposable with linearly ordered typeset for the same reason that GS(O') is. Conversely, assume that Il = U{5(i) : I ~ i ~ n} is a partition such that each G S(i) is a completely decomposable group with linearly ordered typeset. Define a homomorphism f: G ~ EB{G S(i) : I ~ i ~ n} by f(x) = (x , .. . , x). Since Il = U{5(i) : I ~ i ~ n}, G = n{GS(i) : I ~ i ~ n} is a pure subgroup of the completely decomposable group EB{ G S(i) : I ~ i ~ n} . EXERCISES

1. Suppose G is a finite rank Butler group, C = A I ED . . . ED An with rank Ai = 1 for each i , and f : C --+ G is an epimorphi sm with f(A i) i 0 for each i, For each nonempty subset S of (I , 2, . . . , n), define is = n{type Ai : i E S}. Prove that if X is a pure rank-l subgroup of G, then type X = U{rs : X n (~(f(Ai) : i E S}) i O} . 2. Prove that if G is a finite rank Butler group , then G is completely decomposable if and only if G[r] = ~{G (a) : aj r} for each r E cotypeset G. 3. Give a proof for Proposition 3.2.14. 4. Give a proof for Corollary 3.2.11. 5. Prove that if G is a r-homogeneous completely decomposable group and H is a ahomogeneous completely decomposable group, then G ~z H is a 8-homogeneous completely decomposable group , where if r has height sequence (m p) and a has height sequence (n p) , then 8 has height sequence (m p + n p) , agreeing that n + 00 = 00 for each 0 ~ n ~ 00. 6. A torsion-free abelian group G is locally completely decomposable if G p is completely decomposable for each prime p. Prove that a finite rank Butler group is locally completely decomposable. Give an example of a locally completely decomposable group of finite rank that is not a Butler group . 7. [Koehler 65] Let G be a torsion-free abelian group of finite rank with finite typeset. (a) Prove that if T is a nonempty subset of typeset G and a = n{r : rET), then there is a maximal independent subset B of G(a) such that each b E B has type a in G. (b) Show that there is a finite rank Butler group H such that H is a subgroup of G, G(r)j H(r) is torsion for each type r , and typeset H = typeset G. Conclude that H is unique up to quasi-equality. 8. Compute the cotypeset of the group B given in Example 3.2.1. Is B --+ X I ED X2 ED X3 a cobalanced monomorphism?

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3.3 Quasi-isomorphism and Q-Representations of Posets There are equivalences between categories of finite rank Butler groups and repre sentations of finite posets over the field Q of rational numbers. These equivalences can be used to transfer properties of representations developed in Chapter I to the quasi-isomorphism category of finite rank Butler groups. If G is a finite rank Butler group, then typeset G is finite. Hence, the finite sublattice of the lattice of all types generated by typeset G is finite. Given a finite lattice T of types, define B(T) to be the category of finite rank Butler groups G with typeset G contained in T. Clearly, each finite rank Butler group is in B(T) for some T. As a consequence of the next lemma, B(T) is the smallest category containing all rank-l groups with types in T that is closed under isomorphism, finite direct sums, pure subgroups, and torsion-free homomorphic images.

Lemma 3.3.1 [Butler 65] Let T be a finite lattice of all types and G a finite rank Butler group. The following statements are equivalent: (a) G is in B(T); (b) G is a pure (cobalanced) subgroup ofa completely decomposable group in B(T); (c) G is a (balanced) homomorphic image ofa completely decomposable group in B(T).

(a) =? (c) By Lemma 3.2.3(a) , there is a balanced epimorphism C -+ G, where C is a direct sum of rank-I groups XI with type XI E typeset G £ T . Then typeset C £ T by Example 3.1.5(b) and the fact that T is closed under finite meets. This shows that C E B(T). (c) =? (b) In this case, G is a pure cobalanced subgroup of a finite direct sum C of rank-I quotients YI of G by Lemma 3.2.3(b). Then type YI E T, since T is closed under finite joins. Hence, typeset C £ T and C E B(T). (b) =? (a) follows from the fact that if G is a pure subgroup of C E B(T), then typeset G S; typeset C S; T .

PROOF.

Let T be a finite lattice of types with least element .0 and B(T)Q the quasiisomorphism category of B(T). Define JI(T) to be the poset of join irreducible elements of T , i.e., those. > .0 in T such that if . = a U 8, then either a = r or 8 = r. Notice that if.o < r E T, then a :s r for some a E lI(T), and if r E T\JI(T), then. = Uta E JI(T) :a < r} , Define ST to be JI(T)0P, the poset with elements those of lI(T) but with reverse ordering. In particular, if a, r EST, then a :s r in ST if and only if a ::: r as types .

Theorem 3.3.2 [Butler 68] Let T be afinite lattice oftypes. There is a category equivalence H: B(T)Q -+ rep(ST, Q) given by H(G) = (QG, QG(r) : r E ST)'

3.3 Quasi-isomorphism and Q-Representations of Posets

101

PROOF. The strategy is to define a fully faithful functor H : B(T)Q -+ rep(YOP, Q) by H(G) (QG, QG(r) : rET) and then prove that the image of H coincides with the image of the functor F-: rep(Sr , Q) -+ rep(YOP , Q) given in Proposition 1.3.I(b). Observe that H (G) E rep(YOP, Q), since if a and r are elements of T with a :::: r as types, then r :::: a as elements of TOP and G(r) S; G(a) by Lemma 3.1.4(a). For j E Hom(G, K) define H(f) : QG -+ QK to be the unique extension of j. Then H(f)(QG(r)) S; QH(r) for each r E Sr, since j(G(r» S; H(r) by Lemma 3.1.4(d). Hence, H(f) is a representation morphism. It follows that H is a faithful functor. To see that H is a full functor, let g : QG -+ QK be a representation morphism from H(G) to H(K). Since G is a finite rank Butler group , there are finitely many pure rank-l subgroups XI, . . . , X; of G with G = Xl + ...+ X n • Since g : QG(r) -+ QK(r)foreach type r , there is a nonzero integer m, with mig : Xi -+ K for each I :::: i :::: n . Let m mi '" m.: Then mg E Hom(G , K) with H(m!) = g , as desired. In order to confirm that image H is contained in the image of F- , let ro ::j:. rET . Then r = U{a E JI(T) : a :::: r in T}. As a consequence of Lemma 3.1.4(b), QG(r) = n{QG(a):a E JI(T),a :::: r in T}. Hence, (QG ,QG(a) :a E Sr) E rep(Sr, Q) with H(G) F-(QG, QG(a): a E Sr). It remains to prove that if U (Uo , U, : r E TOP) E rep(YOP , Q) is in the image of F-, then H(G) = U for some G E B(T). Observe that U is in the image of F- if and only if U r Uu = n for each a, rET . Choose a free subgroup Vo of Uo with QVo = Uo and let Vr = Vo n U; for Ur ' each rET . Each Vr is a pure subgroup, hence a summand of Yo, and Q Vr For each r E Sr , let X r be a subgroup of Q with type X r = r and let X0 be a subgroup of Q with type Xo roo Define

=

=

=

=

u, u,

=

=

G = Xo Vo

=

+ E{X r Vr : rEST},

=

a subgroup of Q Vo Uo with QG U«. There exist a completely decomposable group C , a finite direct sum of groups isomorphic to X o or X r for rEST S; T, and an epimorphism n : C -+ G as a consequence of the definition of G . By Lemma 3.3.1 , G E B(T). The proof is concluded by showing that QG(r) = U; for each rET. Now, tj, = QX r Vr S; QG(r) for each r E Sr, since type X r = r , Vr is a free group, and X r Vr is r-homogeneous completely decomposable. Because U is in the image of F- and the elements of Sr are the join irreducible elements of T, Uris contained in QG(r) for each rET. As for the reverse inclusion, let X be a pure rank-I subgroup of G with type X = rET. Then rr-I(X) is a Butler group, being a pure subgroup of C. Write rr-I(X) = YI +...+ Yn with each Yi a pure rank-I subgroup ofC and rr(Yi)::j:. O. Now, n restricts to an epimorphism YI EB · .. EB Yn -+ X and r = type X = a (l) U .. . U a(n), where aU) = type Yi E T . Then QX = Qrr(Yi) S; QG(ai) S; UU(i) for each i . Consequently, QX ~ U, = n{Uu(i) : I .::: i .::: n}, since U is in

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the image of F-. Therefore, QG(T) is contained in V T for each type T E T , as desired. Theorem 3.3.2 gives an explicit method of constructing groups in B(T) from representations in rep(ST , Q) as illustrated by the following example: Example 3.3.3 If T is a finite lattice, and ST = {type X I , type X 2, type X3} is an antichain with each Xi a subgroup ofQ and Xo = Xi n X j for each i =f. j, then a strongly indecomposable group in B(T) is quasi-isomorphic to one of the following rank-l groups with types in T, Xo, Xl , X 2 , X 3, XI

+ X 2 , X 2 + X 3, XI + X 3, XI + X 2 + X3 ,

or a rank-2 group

In each case, the endomorphism ring is isomorphic to a subring ofQ. PROOF. By Example 1.1.5, the indecomposable representations in rep(S3, Q) are (Q, 0, 0 , 0), (Q , Q, 0, 0), (Q, 0, Q, 0), (Q, 0, 0, Q), (Q, Q, Q , 0), (Q, 0 , Q , Q) , (Q , Q , 0, Q), (Q , Q, Q, Q), and (Q E9 Q, Q E9 0, E9 Q, (1 + l)Q). In each case, the representation endomorphism ring is isomorphic to Q. Now apply Theorem 3.3.2. For example, if V = (Vo, VI, V2 , V3) is an indecomposable representation with Q-dimension Vo = 1, then G = Xo + I:{Xi: Vi =f. O} is a rank-I group with type G = U{type Xi : Vi =f. OJ.

°

The category B(T)Q has finite, tame, or wild representation type as, respectively, rep(ST, Q) has finite, tame, or wild representation type . Since the functo r of Theorem 3.3.2 is a category equivalence, the representation type of B(T)Q is well-defined and is determined by the structure of ST.

Corollary 3.3.4 Suppose T is a finite lattice oftypes.

If B(T)Q has infinite representation type, then there are strongly indecomposable groups in B(T) ofarbitrarily large rank. (b) If B(T)Q has wild representation type, then for each finite-dimensional Qalgebra A, there is a G in B(T) with QEnd G isomorphic to A. (a)

PROOF. Assertion (a) is a consequence of Theorem 3.3.2, while (b) follows from Theorem 3.3.2, Proposition 1.4.5, and Corollary 1.4.3.

Corollary 3.3.5 Let T be afinite lattice oftypes. Then B(T)Q has: (a) finite representation type if and only if ST does not contain S4, (2, 2, 2), (1, 3, 3), (1, 2, 5), or (N, 4) as a subposet;

(b) wild representation type if and only if ST contains S5, (1, 1, 1,2), (2,2,3), (1,3,4), (N, 5), or (1, 2, 6) as a subposet;

3.3 Quasi-isomorphism and Q-Representations of Posets

103

if and only if ST contains S4, (2, 2, 2), (1, 3, 3), (1,2,5), or (N, 4) but does not contain S5, (1, 1, 1,2), (2,2,3), (1,3,4), (N, 5), or (1,2,6) as a subposet.

(c) tame representation type

PROOF. Apply Theorems 3.3.2, 1.3.6, 1.4.4, and 1.4.6. The tools of representation theory can be used to provide some striking results for finite rank Butler groups . Corollary 3.3.6 Suppose T is a finite lattice of types. Then B(T)Q has finite representation type if and only if End G is a subring of Q for each strongly indecomposable group G in B(T). In this case, each strongly indecomposable group G in B(T) has rank less than or equal to 6 and G is determined uniquely up to quasi-isomorphism by the vector (rank G, rank G(r) : r E JI(T)). PROOF. A consequence of Theorem 3.3.2, Corollary 1.3.9, and Theorem 6.2.7. The next corollary provides analogues of sincere representations and sincere posets. Corollary 3.3.7 Let T be afinite lattice oftypes. If B(T)Q hasfinite representation type, then there is a strongly indecomposable group G in B(T)Q with rank G(r)/G#(r) =I Ofor each rEST if and only if ST is one of

(1), (1,1), (1,1,1), (1,1 ,2), (1,2,2), (1,2,3), (1,2,4), (N,2), (N,3) , 4

3

6

8

3

~I I !X

5

4

7,

4

I

7

I

6,

6

8

~\I/I I

5

7

2

I 1

4

I

3

6

8

f/J~ 1

PROOF. Apply Theorems 3.3.2 and 1.3.7. Given a poset S with finite representation type, all ofthe indecomposable representations in rep(S, Q) can be found by listing the indecomposable representations for each sincere subposet of S. In view of Theorem 3.3.2, this same process carries over verbatim to B(T)Q.

Example 3.3.8 Let T be afinite lattice oftypes with ST = (l, 1,2) = {type XI, type X2, type X3 > type X 4 }, each Xi a subgroup of Q.

If G is a strongly

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3. Butler Groups

indecomposable group in B(T), then G is quasi-isomorphic to a rank-I group with type in T or one ofthe rank-2 groups

+ X4 ) EB X 3 ) + (I + I)X 2 , (XI EB X 2 ) + (I + I)X 3 , (XI EB X2) + (I + l)X 4 .

«XI

In each case, End G is a subring ofQ. PROOF. A consequence of Theorems 3.3 .2 and 1.3.7 and Examples 1.3.8 and U.5 . The rank-2 groups arise from a sincere 2-dimensional representation of (I , 1,2) and Example 3.3.3 , noticing that S3 can be embedded as a subposet of (I, I, 2) in two different ways.

Example 3.3.9 Let 5

j \ 3 4 j \ j

T=

I 2 \ j

o

be a lattice of types. Then Sr = {I , 4 < 2}. Moreover, the rank-I groups with type in T are the only strongly indecomposable groups in B(T). PROOF. Exercise 3.3.3.

Example 3.3.10 Assume that T is a finite lattice of types and Sr = S4 = {type X I, type X2, type X 3 , type X 4 }. There are strongly indecomposable groups ofarbitrarily large finite rank in B(T). ln fact.for each irrreducible polynomial g(x) E Q[x] and integer e :::: I, there is a strongly indecomposable group G in B(T) with QEnd G isomorphic to Q[x]j(g(x)").

I. 1.6. Specifically, given an n x n Z-matrix A that is indecomposable as a Q- matrix,

PROOF. Apply Theorem 3.3 .2 and Example

G = X~ EB X~

+ (I + I)X 3+ (I + A)X~

is a strongly indecomposable group in B(T) with QEnd G isomorphic to Q[x]j (g(x)"), where g(x)" is the minimal polynomial of A.

3.4 Countable Groups

105

EXERCISES

1. Let T be a finite lattice of types. Prove that W(ST) ::::: 2 if and only ifeach group in B(T ) is quasi-isomorphic to a completely decomposable group. 2. State and prove Theorem 3.3.2 for modules overan arbitrary principal ideal domain. 3. Confirm Examples 3.3.8 and3.3.9. 4. Prove that if S is a finite partially ordered set, then there is a finite lattice of types T with ST = S. Show that the elements of T may bechosento be isomorphismclasses of subringsof Q.

3.4 Countable Groups A countable torsion-free abelian group G is a Butler group if each pure finite rank subgroup of G is a Butler group. Such groups are also called finit ely Butl er groups [Fuchs, Metelli 92]. An equivalent characterization of countable Butler groups in terms of balanced exact sequences is given in Theorem 3.4.6. A countable torsion-free abelian group G is completely decomposable if G is a direct sum of rank-I groups. Proposition 3.4.1 Assum e that G is a countable torsion -free abelian group. (a) If G is a countable pu re subgroup of a completely decomp osable group, then G is a Butl er group. (b) The group G is a Butler group ifand only ifG is the union of an ascending chain 0 S; B, S; . . . S; Bn S; B n+l S; . . . ofpure subgroups such that each B; is a finite rank Butler group. The Bn's may be chosen with rank Bn+d B; = 1. (c) If G is a hom ogeneous Butler group, then G is completely decomposabl e.

PROOF. (a) Suppose G is a pure subgroup of a completely decomposable group C and H a pure finite rank subgroup of G. Since H has finite rank, H is a pure subgroup of a finite rank completely decomposable summand of C, hence a finite rank Butler group. (b) Assume that G is a Butler group and {Xl ,"" xn, ... j is a set of generators of G. Define B; to be the pure subgroup of G generated by {Xl , ... , Xn j. Then B, S; .. . S; B; S; . . . is an ascending chain of pure finite rank subgroups of G with G = U{Bn : n ::: I}. Since G is a Butler group, each Bn is a Butler group. Moreover, Bn+ll B; S; Q( Xn+ l + Bn ), so that rank Bn+ll B; = 1. Conversely, suppose B, S; ... S; Bn ~ • . . is an ascending chain of pure subgroups of G with G = U{ B; : n ::: l} and each B; a finite rank Butler group. If H is a pure finite rank subgroup of G, then H is a pure subgroup of some Bn • To see this, let {Xl , . . . , x m } be a maximal Z-independent subset of H . Choose n

106

3. ButlerGroups

sufficiently large with each X i E Bn . Then H = (Xl, ••• , x n )* is a pure subgroup of Bn , whence H is a Butler group by Corollary 3.2.4. (c) By (b), there is an ascending chain BI ~ ... ~ Bn ~ • . • of pure subgroups of G with G = U{Bn : n ::: I} and B; a finite rank Butler group. Since G is homogeneous, B; is homogeneous, hence completely decomposable by Corollary 3.2.7(a) . Each Bn is a summand of Bn+1 by Lemma 3.1.6(a), say Bn+1 = Bn EB Cn+ 1 for each n ::: I. Define C I = B I . For each i , C, is a finite rank homogeneous completely decomposable group, by Lemma 3.1.6(c), and so G = EB{Ci : i ::: I} is a homogeneous completely decomposable group. In contrast to finite rank groups, the converse of Proposition 3.4.1(a) does not hold for countable groups.

Example 3.4.2 There is a countable Butler group that is not isomorphic to a pure subgroup ofa completelydecomposable group. PROOF. Let S be the set of words on the alphabet {0,1}. The empty word is denoted by 0 . For each s E S, choose a proper subgroup X s of Q such that if t = s I and u = sO, then X s = XI n Xu. In other words, the Xs's for s E S form a tree with root X 0 . In addition, choose the Xs's such that if r is a type and for each n there is s E S with length s ::: nand r ::: type Xs , then r = type Q. To see that such a choice is possible, let Jro = {TI}, Jrl = {Po , PI}, Jrz = {Poo , POI, PIO, PII }, . . . , x; = {P s : s E S , length s = n}, . .. be successive partitions of the set TI of all primes such that for each s E S, P, is infinite, P, is the disjoint union of Pso and PsI. and if s(1), . . . , s(n), . . . is a collection of elements of S with length s(i) = i, then n{ps(i) : I ~ i} is empty . For s E S, let X s = n{Zp : pEPs}. These Xs's satisfy the above conditions. Given a nonnegative integer n, define G; = EB{X s : s E S, length s = n}. There is a homomorphism fn: G« ~ Gn+1 with pure image defined by f(x s ) = (x s , x s ) E Xso EB X s), since X, = X so n X sI' Define G to be the direct limit of {Gn, fn : n ::: I}. Then G is a countable Butler group by Proposition 3.4.1(b) and is reduced, since each X s is a proper subgroup of Q. Assume, by way of contradiction, that G is a pure subgroup of a completely decomposable group C = CI EB Cz EB .. . EB Cn EB . . . with rank Cn = 1 for each n. Since G is reduced, G is a pure subgroup of C/d(C), so that it is sufficient to assume that each C; is reduced. Then X0 is a pure subgroup of C l EB · . . EB Cm for some fixed m. Let a i= X E X 0 and n ::: I. Then X = EB{xs : length s = n} E G; with each x, i= a by the choice of the Xs's. Moreover, X s = c(1, s) EB . .. EB c(k, s) with c(i , s) E C, for some k ::: m. Now, type x, = n{typec(i,s): 1 ~ i

~

k}

~

n{typec(i ,s) : 1 ~ i

~

m}

3.4 CountableGroups

107

and x = EI7{xs : length s = n} = EI7{I:{c(i, s): length s = n}: 1

=:: i =:: k}

E C I EI7 .. . EI7

Cm'

Since 0 "# x, there is some I =:: j =:: m with c(j, s) "# O. In summary, for each n ::: 1 there exist s E S with length s = n and some I =:: j =:: m with type Xs =:: type C]. By the choice of the Xs's , type C, = type Q, a contradiction. The next theorem gives a method of constructing countable Butler groups of infinite rank with finite typesets. The proof is a modification of the original proof that a torsion-free homomorphic image of a finite rank completely decomposable group is a Butler group given in [Butler 65], wherein R = Z. Let R be a ring with free additive group such that QR is a division ring. Then Z ~ R and a torsion-free R-module is also torsion-free as an abelian group. For a torsion-free R-module M, R-rank(M) is the cardinality of a maximal Rindependent subset of M, equivalently the QR-dimension ofQM. Let R p = ZpR, the localization of R at a prime p of Z. If R/ pR has no zero divisors, then each element of R p is of the form p iu for u a unit of R p' To see this, let 0 "# x E R p and write x = pi r for some r E R p\pR p. This is possible, since the additive group of R is free, hence preduced. Since QR p = Q R is a division ring, there are s E R p and j ::: 0 with r(s/pj) = I = (s/p j)r . Then rs = p j = sr. But R/pR = Rp/pR p has no zero divisors, so that s = pj u for some u in R p. Hence, ru = 1 = ur, and r is a unit of R p' as desired.

Theorem 3.4.3 Suppose R is a countable ring with free additive group, QR is a division ring, and R/ pR has no zero divisors for each prime p ofZ. lfG is a torsion-free R-module with G = RYI + ... + RYnfor rank-I subgroups Yi of G. then G is a countable Butler group with finite typeset. PROOF. It is sufficient, via Proposition 3.4.l(a), to prove that G is a pure subgroup of a countable completely decomposable group with finite typeset. The proof is by induction on the R -rank of G . If R-rank G = I, then G = RYI is a countable direct sum of copies of YI , since R is a countable free abelian group. Hence, G is a countable homogeneous completely decomposable group with type G = type YI. Next assume that R-rank G > 1 and let N, = (Q RYi ) n G, a pure subgroup of G containing R Yi • Define an R -homomorphism f : G -+ H = G / N I EI7 .. . EI7 G / Nn by f(x) = (x + Nv , . .. , x + N n ) . Since N, is a pure subgroup of G, G/Ni is a torsion-free abelian group that is generated by the rank-I groups (Y j + Ni ) / N, for j "# i as an R-module. Moreover, ker f = N I n·· · n N; = 0, since R-rank G > 1, so that at least two of the R Yi's are R -independent.

108

3. Butler Groups

The next step in the proof is to prove, by contradiction, that the image of f is a pure subgroup of H. If not, there is a prime p, Y E G, and Yi E G such that P(YI + NI,"" Yn + Nn) = (y + NI,···, Y + Nn) and Y - PYi = Xi E Ni\pNi for each i. Now, (Ni)p = RpXi since Xi (j. p N] , Yi is a rank-1 group, and QR is a division algebra. Thus, G pi pG p = (Rpl pRp)(Y + pG p) is cyclic, since + (Nn)p = RpXl + ... + Rpxn, and Y - Xi E pG for each i. G p = (Nd p + Let 5 = {Xl, , x m } be a maximal Rp-independent subset of {XI, ... , x n} with m = Rp-rank G p = R-rank G > 1 and Y = XI. Given m + 1 .::: i .::: n, there is some r in Rp with r x, in K = RpXI ED·· · ED Rpxm , since {Xlt ... , x m } is a maximal Rp-independent subset of {Xl, .. . , Xn}, a set of Rp-generators for G r: Write r = pj u for some unit u of Rp. Then p! UX; E K ~ G p, whence p! Xi E K , since u is a unit. Consequently, there is some e with p' Xi E K for each i, and so peG p = pe(RpXI + ... + Rpxn) ~ K. Thus, K = G p, since GplpeGp is generated by Y + peG p and Y = Xl E K. In particular, Kl p K = GplpGp is cyclic, which contradicts K I pK = (Rpl pRp)xI ED·· 'ED(RplpRp)xmwithm > 1. By induction on the R -rank of G, each GIN; is a pure subgroup of a countable completely decomposable group with finite typeset. This shows that H, and consequently G, is a pure subgroup of a countable completely decomposable group with finite typeset. Let T be a finite lattice of types with 5 r = {Tl, T2} = 52. As a consequence of Theorem 3.3.2 and Example 1.1.3, each strongly indecomposable group in B(T) has rank 1. Hence, each group in B(T) is quasi-isomorphic to a completely decomposable group. The next example demonstrates that countable Butler groups with typeset contained in T are much more complicated. Example 3.4.4 Assume that T2 = {type XI n X 2, type XI , type X 2}, Xi is a subgroup of Q containing 1 with type X I and type X2 incomparable, p is a prime with lip (j. Xi , and R = End X, for i = 1,2. There is a strongly indecomposable Butler group G with countably infinite rank such that the typeset of G is contained in T2 and End G is isomorphic to R. PROOF. The first step is to construct V = (Vo, VI , V2) E Rep(52, R) with Vo a countable free R-module, each Vi a pure submodule of Vo, Vo/(VI ED V2) a p-group, and End V = (f: V o --+ U«: f(V i) ~ V;} isomorphic to R. A slight modification of the proof of Theorem 2.4.6 can be used to find a torsionfree R -module M of countably infinite rank containing a free R -submodule L with MIL a p-group, End M = R, and Hom(M, R) = O. Specifically, let {Jri : i 2: l} be a countably infinite subset of the p-adic integers R* that are algebraically independent over Q, L a free R-submodule of R* with basis {I, Jri: i 2: I}, and M the pure subgroup of R* generated by L. Then M is a torsion-free Rmodule with countably infinite rank, and MIL is a p-group. Since each f E End M extends uniquely to multiplication by an element of R* and {zr, : i 2: I} is an algebraically independent set over Q, it follows that End M = Rand Hom(M, R) = O.

°

3.4 Countable Groups

°

109

Let 4- VI 4- Vo 4- M 4- be a countable free R-resolution of M . Then Vo is a countably free R-module, and VI is a pure free R-submodule of Vo, since M is torsion-free, Vo is a free module, and R is a principal ideal domain . Let VI $ L' be the preimage of Lin Vowith L' isomorphic to the free R-module L , {e; : i ::: I} a basis of L', and ((n(i) , u(i» : i ::: I} an indexing of the countable set N x VI, N the set of natural numbers. Define Vz to be the pure R -submodule of Vo generated by {pn(i)e; +u(i): i ::: I}. Then VI nvz = 0, since L'nV 1 = 0, and Vo/(V I $Vz) is a p-group, since MIL is a p-group. To see that End V is isomorphic to R, let f E End V . There is r E R with g = f-r : U» 4- VI.Thisissobecausef : U» 4- V o, f (V I ) S; VI, M = VoIVJ, and End M = R. Assume, by way of contradiction, that g =f:. 0. Then g(Vd =f:. 0, since V, is a free R-module and Hom(M , R) = 0. But Vo is p -reduced so that =f:. g(pn(; )e; + u(i» E g(Vz) for some i. Hence, g(Vz) S; Uz n VI = 0, a contradiction. It follows that each endomorphism of V is multiplication by an element of R, whence End V is isomorphic to R. Define G = X I V , $ XzUz + U» . an R -submodule of QVo = QV I $ QVz. A routine computation, using the assumptions that V; is pure in Vo and lip fj. X;, shows that if r, = typeX}, then G(r;) = X iV;. Moreover, G is a Butler group, since if H is a pure finite rank subgroup of G, then H = X 1 VI $ X Z Vz + Vo, with each V; a finite rank free R-summand of V;. Hence, H is a Butler group by Corollary 3.2.4. Finally, since R = End X, = End x-, it follow s that R S; End G S; End V = R.

°

In the previous example, G I(G(rl) $ G(rz» is an unbounded group. However, if G is indecomposable with G/(G(r,) $ G(rz» bounded, then rank G ~ 2 as a consequence of the next theorem, the proof of which is not included. The finite rank case is proved in Corollary 5.1.12. Theorem 3.4.5 [Files , Gobel 98] If G is a countable Butler group with typeset G S; Tz = fro, rl , rz} and Gj(G(rd $ G(rz» is a bounded group , then G is a direct sum ofgroups of rank less than or equal to 2.

°

°

An exact sequence 4- K 4- H ~ G 4- of abelian groups with G torsionfree is balanced if rank-l torsion-free abelian groups are projective with respect to this sequence, i.e., if X is a rank-l torsion-free abelian group and g: X 4- G, there is h: X 4- G with fh = g. If H is also torsion-free, then this definition of balanced coincides with that of Section 3.2 for exact sequences of finite rank torsion-free abelian groups (Exercise 3.4.3).

°

Theorem 3.4.6 [Bican, Salce 83] IfG is a countable torsion-free abelian group, then G is a Butler group if and only if each balanced exact sequence 4- T 4H 4- G 4- of abelian groups with T a torsion group is split exact.

°

PROOF.

[Mines, Vinsonhaler 92] Assume that G is a Butler group and

04-T4-H~G4-0

110

3. Butler Groups

is a balanced exact sequence with T a torsion group. By Proposition 3.4.1(b), G is the union of an ascending chain 0 ~ G I ~ . . . ~ G n ~ G n+ I ~ . .. of pure subgroups such that each G, is a finite rank Butler group with rank(G i+l / G i) = 1. Assume, by way of induction, that f-I(G n) = T EB An. Let K = G n+l . By Proposition 3.2.15 , there is a partition IT = U{ S(i) : 1 :::: i :::: m} such that each KS(i) is a completely decomposable group with linearly ordered typeset and K is a pure subgroup of EB{Ks(i): 1 :::: i :::: mI. Since rank K/G n = 1, the partition may be refined to assume that KS(i) = (Gnh(i) EB DS(i) for some rank-I group D isomorphic to K/G n. The sequence 0 -+ TS(i) -+ U-I(K))s(i) -+ KS(i) = (Gn)s(i) EB DS(i) -+ 0 is balanced exact, hence split exact since KS(i) is completely decomposable (see Exercise 3.4.3) . Since T is a torsion group and the S(i)'s partition IT , T = EB{Ts(i) : I :::: i :::: m}. Use the composite of the homomorphisms (K) -+ EB{f-I(K)s(i) : 1 :::: i :::: m} -+ EB{TS(i): 1 :::: i :::: m} -+ T to see that f-I(K) = f-I(G n+d = TEBA n+1forsomeA n+1 2 An. By transfinite induction, H = TEBA for some A isomorphic to G. Conversely, assume that G is a countable torsion-free abelian group and each balanced exact sequence 0 -+ T -+ B -+ G -+ 0 with T a torsion group is split exact. Let {XI, . . . , Xi , .. .} be a maximal Z-independent subset of G . For each n, define F; = ZXI EB EB Zxn. Enumerate all pure rank-I subgroups X of G with X n Fn =f:. 0 by AnI , , Ani, ... by listing all linear combinations al Xl +- . -+anx n with ai E Z and gcd{al, . . . , an} = 1 and taking the pure subgroup of G generated by each such linear combination. Define F = E9{ZXi : I :::: i), M n •i = F; EB An.1 EB · .. EBAn.i and M = U{Mn.i : 1 :::: i, n}. Then M = FE9(E9{A n.i : 1 :::: i, n}). There is an exact sequence 0 -+ K -+ M -+ G -+ 0 with K = ker f and f : M -+ G induced by inclusion. Since each pure rank-I subgroup X of G with X n Fn =f:. 0 is equal to A n•i for some nand i, f is onto and, the sequence is balanced. For each nand i , there is an induced exact sequence 0-+ K n Mn.i -+ Mn.i -+ G n.i -+ O. Notice that each Gn.i is a finite rank Butler group, since Mn .i is a finite rank completely decomposable group . Furthermore, G I ~ ... ~ G; ~ .. . is an ascending chain of pure subgroups of G , where G n = U{ G n.i : i :::: 1), and Gn.i = Gn •i - l + A n.i = An.1 + ...+ An.i. The next step in the proof is to define a set of elements p(n, i) that are either 1 or a prime and aset of positive exponents e(n , i) . If Gn,;-I = G n.i , define pen, i) = 1 = e(n , i) . If Gn.i-I =f:. Gn,i, choose a prime p = pen, i) with (Gn,i- I)p =f:. (Gn.i)p and a least positive integer e = e(n, i) with pe(Gn,i_d ~ (Fn)p + d«G n.i-d p) ~ (Gn,i)p, where (Fn)p is a free Zp-moduleandd«Gn,i_l) p) is the divisible subgroup of (Gn.i-d p. Such a choice is possible because (An,j)p is isomorphic to either Zp or Q with (An.j)p n (Fn)p =f:. 0 for each j :::: i, and G n.i = An.1 + ...+ An,;. It is now sufficient to prove that given n, P; = {pen, i) : pen, ne(n.i) =f:. l} is finite . In this case, Gn.i-I =f:. G n.i for at most finitely many i, and so G; = Gn,i is a finite rank Butler group for some sufficiently large i. An application of Proposition 3.4.1(b) shows that G is a Butler group. For each i, n :::: 1, define mn,i = lcm{p(n , iie(n.j) : 1 :::: j :::: i} and L = ~{mn .i(K n Mn.i) : 1 s i, n}. Then K/ L is torsion, since K = U{K n Mn ,i : 1 :::: i, n} , and the induced sequence 0 -+ K / L -+ M/ L -+ G -+ 0 is a balanced exact

r:'

3.5

Quasi-Generic Group s

III

sequence. By hypothesis, this sequence is split exact, so choo se g : G --+ M / L with fg = IG· Because fg = IG, g(Xi) - (Xi + L ) E ker f = K/L for each i . Since KIL is torsion and {XI , " " x n } is a basis of Fn , there is a nonzero integer m with m (g(y ) - (y + L )) = 0 E K/L for each y E Fn. In this case, h c(my ) = h M/ dm y + L ), where hG(m y ) denote s the p-height of my in G . In view of the fact that L = b {mn,i(K n Mn,i ): I s i, n} and mn,i = lcm{p(n , i)ZE(n,j) : 1 ~ j s i }, it suffices to assume that all prime divisors of m are in Pn . It remai ns to show that each prime p E Pn divides m. In this case , P; must be finite, as desired . Let 1 =f. P E Pn. Then (Gn,i-d p =f. (Gn,i)p so there exist X E A n.i with px E G n.i - 1 and ph+l x = y + d E ( Fn)p + d «Gn ,i- d p) for some y E ( Fn)p and d E d «Gn,i-l) p) with h ~ e(n , i), h = hG(n ,i- l)(y ), and hG(n,i)(Xy ) ~ h + 1. Then hM/dm y + L ) = hG(n ,i)(my) ~ h'l (m ) + h + 1. Choo se a = al EB a2 E L with at E b{mn,j(K n Mn,j ) : j ~ i-I}, a2 E b{mn ,j(K n Mn,j) : j ~ i}, and hG(my + a) ~ h'l(m) + e(n, i) = h'l(m) + h + 1. By the definition of the mn,i's, h M(a2) ~ 2e(n, i ) > e(n, i) + h . It follows that peen , i ) divides m, as desired ,

EXERCISES

I. Let G and H be count able completely decompo sable groups. Prove that G and H are isomorphic if and only if rank G(r)/ G#(r) = rank H (r)/ H #(r ) for each type r . 2. Prove that if G is a countable completely decompo sable group and H is a summand of G , then H is a completely decomposable group. 3. Let 0 -l- K -l- H ~ G -l- 0 be an exact sequence of torsion-free abelian groups. Prove that f: H (r) -l- G (r) is onto for each type r if and only if for each rank-I torsion-free abelian group X and g : X -l- G, there is h : X -l- H with f h = g.

3.5 Quasi-Generic Groups There are countable Butler group s of infinite rank , called quasi-generic groups , that determine the representation type of quasi-homomorphism categories of finite rank Butler groups (Theorems 3.5.5 and 3.5.11). In fact, forthe tame representation type case, all strongly indecomposable finite rank Butler groups can be constructed from quasi-generic groups, Quasi-generic groups are related to generic representations (Section 1.5). However, the connection between countable Butler groups and count able Q- representations of finite posets is more complicated than the connec tion between finite rank Butler group s and finite-dimensional Q -representations given in Section 3.3. The first lemma con sists of examples of strongly indecomposable torsion-free abelian groups of countably infinite rank with finite typesets. For each critical poset ST and associated group G , in Lemma 3.5.1, (Q(x) G, Q(x) G( r): rEST ) is the repre sentation of ST given in Theorem 1.5.3 with k(x ) repla ced by Q [x ].

112

3. Butler Groups

Lemma 3.5.1 Let T be a finite lattice oftypes and ST the opposite ofthe poset ofjoin irreducible elements ofT. (a) For each of the following posets ST, there is a strongly indecomposable torsion-free abelian group G with countably infinite rank, typeset G ~ T,

and QEnd G = Q[x]. Write R for Z[x] and Ai for a subgroup of Q with 1 E Ai and type Ai = i. (i) ST = S4 G

= (AI R $ 0) + (0$ A2R) + «(1 + 1)A3R) + «I +x)A4R) ~ Q[xf. 246

I I I

(ii) ST = (2,2,2) = 1 3

5

+ «I + X)A2R $ A 2R) + «I + 1)A3R $ A4R $ 0) + (0 $ (1 + l)A sR) (1 + 1)A6R) ~ Q[x]3.

G = (0 $ 0 $ AIR)

+ (A 4R $ +(A6R $

0)

4 7

I I

3 6

I I

(iii) ST=(1,3,3)= G = «I

2

+ I)A I R $

+ (0 $

5

(1 + I)AI R) + (0 $ (l

+ I)A2R

$ 0)

A3R $ A 3R $ 0) + (0 $ A 4R $ A4R $ A4R)

+ (AsR E9 0 $ 0 $ 0) + (A6R $ 0 E9 0 $ A6R) + (A 7R E9 (l

+ x)A 7R $

A7R) ~ Q[X]4 .

8

I

7

1 3 (iv) ST G

~

NI

= (N, 4) = 2

4

5

= «(1 + 1)A2R $ (1 + l)A2R $O)+(AIR $ + (0 $ + (0 $

(l

+ I)A 4R $

0 $ 0)

+ (0 $

AIR $ AIR $ AIR $0)

A 3R $ A3R E9 (I

+ I)A3R)

0 $ 0 $ 0 $ AsR) + (A6R $ 0 $ 0 $ 0 $ A6R)

+ (A7R $ 0 $ 0 $ A7R $ A7R) + (AsR $ (I

+ x)AsR $

AsR $ AsR) ~ Q[x]s.

3.5 Quasi-Generic Groups

113

8

I

7

I

3 6

I I

2 5 (v) Sr=(l,2,5)=

I

4,

G = (0 + 1 + 1 + 0 + 0 + O)AI R + (0 + 0 + 1 + 0 + 0 + l)A,R

+ (0 + 0 + 0 + 1 + 1 + O)A, R + ((1 + l)A2R $ (1 + l)A2R $ 0 $ 0) + (A 3R $ A 3R $ A 3R $ A 3R $ 0 $ 0) + (0 $ 0 $ 0 $0 $ 0 $ A4R)+ (0 $ 0 $ 0$0$ AsR $ A sR)+(A6R $ 0 $ 0$ 0 $ A6R $ A6R) + (A 7R $0 $0$ A7R $ A7R $ A7R) + (AsR $ (1 (b)

+ x)AsR $

AsR $ AsR $ AsR) £; Q[xt

If n

:::: 2 is a positive integer, G is one of the groups constructed in (a) with typeset G £; T, and H = Gf Gx", then H is a strongly indecomposable group in B(T) with QEnd(H) = Q[x]j(x n }.

PROOF. (a) In each case, G is a torsion-free abelian group with countably infinite rank and Z[x] £; End G. It suffices to prove that End G is contained in Q[x], in which case G is strongly indecomposable, since QEnd G = Q[x] has no nontrivial idempotents. The proof is analogous to that of Theorem 1.5.3. As an illustration, let G be as defined in (i). A brief argument shows that if rj = type A j, then G(rd = A,Z[x] , G(r2) = A 2Z[x], G(r3) = A 3Z[x](1 , 1), and G(r4) = A4Z[x](1, x). Let f E End G. Since f : G(rj) -+ G(rj) for 1 s i s 3, f = (g, g) with g E End Q[x]. On the other hand, f : G(r4) -+ G(r4), so that g commutes with x . Thus, g E EndQ[xjQ[x] = Q[x] , as desired. (b) Since H is generated as a group by finitely many copies of the Aj's and type A j E Sr £; T , H E B(T) . The same argument as that for (a), shows that QEnd H = Q[x]j(x n } , and so H is strongly indecomposable. As an illustration, let Sr = S4 and R; = Z[x]j(x n }. In this case,

A torsion-free abelian group G is quasi -generic if G is strongly indecomposable with countably infinite rank and QG has finite length as a Q End G-module. This definition parallels that of a generic representation given in Section 1.5. The groups in Lemma 1 are not quasi-generic, since in each case, Q End G = Q[x] and QG has infinite length as a Q[x ]-module. This problem can be remedied

114

3. ButlerGroups

by replacing Z[x] with a principal ideal domain A such that QA = Q(x), a field. The ring A also satisfies the hypotheses of Theorem 3.4.3 and thereby establishes a method for constructing countable Butler groups with finite typesets that are quasi-generic. If f(x) E Z[x], the ring of polynomials with coefficients in Z , then the content of f, denoted by c(f), is the greatest common divisor of the coefficients of f . The polynomial f(x) is called a primitive polynomial if c(f) = 1. Since c(fg) = c(f)c(g), by Gauss's lemma [Hungerford, 74], the set S of primitive polynomials in Z[x] is a multiplicatively closed set. Define A = Z[x]s , the localization of Z[x] at S, a subring of the field of quotients Q(x) of Z[x] . Elements of A are of the form f(x)/g(x) with f(x), g(x) E Z[x] and g(x) a primitive polynomial. Lemma 3.5.2 The ring A has a countablefree additive group and is a principal ideal domain containing Z[x] withQA = Q(x) and 11./ pA afieldforeachprime pofZ. PROOF. (a) Observe that Z[x] is a subring of A and A is a subring of Q(x) with

QA contained in Q(x). On the other hand, if f(x)/g(x) E Q(x) with f(x), g(x) E Z[x], then g(x) = ah(x) with c(g) = a and hex) a primitive polynomial. Thus, f(x)/g(x) = (l/a)(f(x)/ hex»~ E QA. The additive group of A is countable and torsion-free. To prove that A is free as a group, it suffices, by Corollary 2.4.5, to show that each pure finite rank subgroup of A is free. Let H be a pure finite rank subgroup of A and suppose H is the pure subgroup of A generated by a finite number fl (x)/g, (x) , . .. , fn(x)/gn(x) of elements of A with each fi(X), gi(X) E Z[x] and gi(X) a primitive polynomial. Then g(x) = g, (x)· .. gn(x) is a primitive polynomial in Z[x] , since each gi(X) is a primitive polynomial, and c(g) = c(g,) . . . c(gn). Consequently, the subgroup of A generated by the fi(X)/gi(X)'S for 1 ::: i ::: n is a subgroup of B = (l/g(x»Z[x] . Observe that B is a free abelian group , being isomorphic to Z[x] . In fact, B is a pure subgroup of A. To see this, leta(x)/b(x) E A and n a nonzero integer with n(a(x)/b(x» = f(x)/g(x) E B. Then na(x)g(x) = b(x)f(x) E Z[x]. Taking the content of both sides yields nc(a) = c(f) E Z, recalling that c(g) = 1 = c(b), since g(x) and b(x) are assumed to be primitive polynomials. Thus, f(x) = np(x) for some p(x) E Z[x], and so a(x)/b(x) = p(x)/g(x) E B . Since B is a pure subgroup of A and each fi( X)/gi(X) E B, it follows that H is a subgroup of B. Because H is a subgroup of the free group B, H is also a free group, as desired. To see that A is a principal ideal domain , suppose that y = f(x)/g(x) is an element of A with f(x), g(x) E Z[x] and g(x) a primitive polynomial. Write f(x) = nh(x) for some primitive polynomial hex), where n = c(f). Then y = n(h(x)/g(x» with h(x)/g(x) a unit of A. It now follows that if I is a nonzero ideal of A, then I = nil., where n is the least positive integer such that nu E I for some unit u of A. Moreover, if p is a prime of Z , then 11./ p A is a field, since each yEA is an integer multiple of a unit of A.

3.5 Quasi-Generic Groups

115

Lemma 3.5.3 Let T be a finite lattice of types, G the group associated with ST as defined in Lemma 3.5.1, and A the ring given in Lemma 3.5.2. Then G' = GA is a quasi-generic group with End G' = A, QEnd G' = Q(x), and typeset G' 5; T. PROOF. In each case , G' is a countable Butler group with typeset contained in T, via Lemma 3.5.2 and Theorem 3.4.3. As in Lemma 3.5.1, A = End G' . Then QEnd G' = QA = Q(x) , and so G' is strongly indecomposable. Moreover, QG' is a finite-dimensional Q(x)-vector space, whence Q' has finite length as a QEnd G'-module and G' is a quasi-generic group. The quasi -generic groups G' given in Lemma 3.5.3 have some additional properties. Call a countable torsion-free abelian group G special if there are finitely many pure rank-l subgroups XI , " " X; of G such that Gf[(End G)X 1 + ... + (End G)X n ] is a bounded group. If G is a countable torsion-free abelian group with typeset G contained in a finite lattice T, then End UG = {f: QG ~ QG: f(QG("r» £ QG(r), rEST} is the representation endomorphism ring of UG = (QG, QG(r): rEST) E Rep(ST , Q) . Since each G(r) is fully invariant in G, QEnd G £ End UG • A special group G is central-special if there are finitely many pure rank-l subgroups Xi of G such that G/(CGX I + .. . + CGXn) is bounded as a group, where CG = (CEnd UG) n (End G) and CEnd UG is the center of End UG. Notice that CG is contained in CEnd (G), the center of End G. Each finite rank Butler group is central-special, being generated as a group by finitely many pure rank-l subgroups. For each countable Butler group G' in Lemma 3.5.3, End UG' = Q(x). Hence, G' is central-special, since CG' = Q(x) n End G' = End G' = A and G' is generated as a A-module by finitely many pure rank-l subgroups. Lemma 3.5.4 If G is a central-special group, then Q End G = End UG and CG = CEnd G. In particular; G is strongly indecomposable if and only if UG is indecomposable in Rep(ST, Q). PROOF. Since G is central-special, there are pure rank-l subgroups Xi of G such that G/(CGX I+ " , +CGXn)isboundedasagroup,saytG £ CGX I+" , +CGXn for some nonzero integer t. If f E End UG, then f(X i) 5; QG(ri), where type Xi = ri . Hence, there is a nonzero integer s with sf : Xi ~ G (ri) for each i. Now let x E G. Then tx = CIXI + . . . + Cn Xn for some c, in CG and Xi E Xi. Hence, tsf(x) = sf(tx) = sf(clx» + ... + sf(cnxn) = cl(sf(xl» + ... + cn(sf(xn E G, since sf E End UG and each c, E CG 5; CEnd UG. This shows that tsf E EndG, whence f E QEnd G and so End UG = QEnd G . Consequently, QEnd G has no nontrivial idempotents if and only if End UG does. In particular, G is a strongly indecomposable group if and only if UG is an indecomposable representation in Rep(ST , Q). Finally, CG = CEnd UG n End G= CQEnd G n End G = CEnd G .

»

116

3. ButlerGroups

For a finite lattice T of types, a torsion-free abelian group G is called a T-group if G is a countable Butler group with typeset G contained in T. The next theorem is the Butler group analogue of Theorem 1.5.3. Theorem 3.5.5 Let T be a finite lattice oftypes. The following statements are equivalent: (a) The category B(T)Q has finite representation type; (b) There are no central-special quasi-generic T -groups; (c) QEnd H= Qfor each strongly indecomposable H in B(T)Q. PROOF. (a) ~ (b) Suppose G is a central-special quasi-generic T-group. Then QEnd G = EndUe by Lemma 3.5.4. Now, UG = (QG,Q(r) :r E Sr) E Rep (Sr, Q) and rep(Sr, Q) has finite representation type in view of (a) and Theorem 3.3.2. By Theorem 1.5.3, UG is the direct sum of finite-dimensional Q-representations of Sr. This contradicts the fact that a quasi-generic group is strongly indecomposable with countably infinite rank. (b) ~ (a) Assume that B(T)Q has infinite representation type. By Corollary 3.3.5(a), ST contains one of the critical posets S4, (2, 2, 2), (I, 3, 3), (N, 4), or (1, 2, 5) as a subposet. For each of these critical posets Sr, there is a central-special quasi-generic T-group given by Lemma 3.5.3. This contradicts (b). (a) ~ (c) This follows from the category equivalence of B(T)Q with rep(Sr, Q) , Theorem 3.3.2, and the corresponding result for rep(Sr, Q), Corollary 1.3.9. (c) ~ (a) Suppose that B(T)Q has infinite representation type. By Corollary 3.3.5(a), it suffices to show that if 5r is one of the critical posets 54, (2, 2, 2), (1, 3, 3), (1, 2, 5), or (N, 4), then there is a strongly indecomposable H in B(T)Q with QEnd H not isomorph ic to Q. An application of Lemma 3.5.I(b) completes the proof.

The next two lemmas establish the framework for existence conditions for quasigeneric T -groups if B(T)Q has wild representation type. The groups listed in the next lemma are obtained from the list of Q-representations of the poset Sr given in Theorem 1.4.4. In each case, UG = (QG, QG(r) : E Sr) is the corresponding representation of ST . Lemma 3.5.6 Let T be a finite lattice oftypes. For each ofthe following posets Sr, there is a strongly indecomposable torsion-free T-group G with countably infinite rank and QEnd G = Q( x, y}. Let R = Z (x, y) and Ai a subgroup ofQ with I E Ai and type Ai = i.

5 (i) Sr

= S5 and(1,I,I, 1,2) = 1

2

3

I

4

3.5 Quasi-Generic Groups

117

G = (AIR EB AIR EBOEB 0) + (OEBOEB AzR EB AzR) +(1 +0+ 1 +0)A3REB(0+ 1 +0+ l)A3R) +(1 +O+x

+ I)A 4R

+«(1 + 0 + x + I)A sR EB (0 + 1 + Y + O)AsR). 5

I

246

I

(ii)

I

ST = (2,2, 3) = 1 3

G

I

7,

= (y+x +0+ 1 +0+ l)A IR EB(O+ 1 +

1 +0+ 1 +O)AIR

+ ((l + 0 + 1 + 0 + 0 + O)AzR EB (0 + 1 + 0 + 1 + 0 + O)AzR EB (y + x + 0 + 1 + 0 + l)AzR EB (0 + 1 + 1 + 0 + 1 + O)AzR) + (0 EB 0 EB 0 EB 0 EB A3R EB A 3R) + (0 EB 0 EB A4R EB A 4 R EB A4R EB A4R) + (A7R EBOEBOEBOEBOEBO) + (A6R EB A6R EB 0 EB 0 EB 0 EB 0) + (AsR EB AsR EB AsR EB AsR EB 0 EB 0). 8

I

4 7

I I

3 6 (iii)

ST=(l ,3 ,4)=1

I I

25,

G = «0 + 1 + 1 + 0 + 1 + 0 + 0 + O)AI R EB (y + x + 0 + 1 + 0 + 1 + 0 + O)AI R EB (1 + 0 + 1 + 0 + 0 + 0 + 1 + O)AI R EB(O+ 1+0+ 1+0+0+0+ l)A IR) + (0 EB 0 EB 0 EB 0 EB 0 EB 0 EB AzR EB AzR) + (0 EB 0 EB 0 EB 0 EB A 3R EB A3R EB A3R EB A3R) + (0 EB 0 EB A 4R EB A4R EB A4R EB A4R EB A4R EB A 4R)

+ (AsR EB 0 EB 0 EB 0 EB 0 EB 0 EB 0 EB 0) +

(A6R EB A6R EB A6R

EBO EB 0 EB 0 EB 0 EB 0) + (A 7R EB A7R EB A7R EB A7R EB A7R EB 0 EB 0 EB 0) + (AgR EB AgR EB AgR EB AgR EB AgR EB AgR EB AgR EB 0).

118

3. Butler Groups

9

I I 247 NI 1 3 6 I (iv) T = (N, 5) = 5 8

G

= (0$0$0$0$0$0$A1R$A1R$A1R$A1R) + «0 $ 0 $ A 2R $ A 2R $ 0 $ 0 $ A2R $ A2R $ A2R $ A2R) $ (0 + 0 + 0 + 0 + 1 + 0 + 0 + 0 + 1 + 0)A 2R $(0+0+0+0+0+ 1 +0+0+0+ l)A2R) +«0+0+0+0+ 1 +0+0+0+ 1 +0)A3R $(0+0+0+0+0+ 1 +0+0+0+ 1)A3R) + «0 + 1 + 1 + 0 + 1 + 0 + 0 + 0 + 0 + 0)A 4 R $(Y +x + 0+ 1 + 0 + 1 + 0+0+0+ 0)A4R

$(1 +0+ 1 +0+0+0+ 1 +0+0+0)A 4R $(0+ 1 +0+ 1 +0+0+0+ 1 +0+0)A4R $(0 + 0 + 0 + 0+ 1 + 0 + 0+0+ 1 + 0)A 4 R $(0+0+0+0+0+ 1 +0+0+0+ 1)A4R) +(A 5R $ 0 $ 0$0$ 0$ 0$0 $ 0 $ 0$ 0) + (A6R $ A6R $ A6R $ 0 $ 0 $ 0 $ 0 $ 0 $ 0 $ 0) +(A7R $ A7R $ A7R $ A7R $ A7R $0$0$0$0$0) +(AsR $ AsR $ AsR $ AsR $ AsR $ AsR $ AsR $ 0$0$ 0) +(A 9R $ A9R $ A9R $ A9R $ A 9R $ A 9R $ A9R $ A9R $0$0). 9

I

8

I

3 7

I I

126

I

5 (v) Sr

= (l, 2, 6) =

I

4

3.5 Quasi-Generic Groups G

119

= «0+0+0+ 1 +0+0+0+0+0+ 1 +0+ y)A)R $(0+0+ 1 +0+0+0+0+0+ I +0+ 1 +x)A)R $ (0 + 1 + 0 + 0 + 0 + 1 + 0 + 0 + 0 + I + 1 + O)A) R $(1 +0+0+0+ 1 +0+0+0+ I +0+0+ l)A)R $(0+0+0+0+0+0+0+ I +0+0+ 1 +O)A)R $ (0 + 0 + 0 + 0 + 0 + 0 + I + 0 + 0 + 0 + 0 + l)A) R) +(0 $ 0$ 0 $ 0$0 $0$ 0 $ 0 $ AzR $ AzR $ AzR $ AzR)

+(0$0$0$0$A3R $ A3R $ A3R $ A3R $ A3R $ A3R $A 3R $ A 3R)+(A 4R $ A4R $ 0$ 0$ 0 $ 0$ 0$ 0$ 0$ 0 $ 0 EB 0) + (AsR EB AsR $ AsR EB 0 $ 0 EB 0 $ 0 EB 0 EB 0 $ 0 $ 0 $ 0) + (A6R $ A6R $ A 6R $ A6R EB 0 $ 0 $ 0 $ 0 $ 0 EB 0 $ 0

EBO)+(A7R EB A7REB A7R EB A7R $ A7R $ A7R EB 0 $ 0 EB 0 EB 0 EB 0 $ 0) + (AgR EB AgR EB AgR EB AgR EB AgR $ AgR EB AgR $ AgR EB 0 $ 0 EB 0 $ 0) + (A 9R $ A9R EB A9R

$ A9R EB A9R EB A9R $ A9R $ A9R EB A9R EB A9R EB 0 EB 0). PROOF. In each case, G is a countable torsion -free group with typeset G S; T. It is sufficient to show that End G S; Q(x, y}, whence G is strongly indecomposable, since Q(x, y} = QEnd G has no nontrivial idempotents. The computations to show that End G S; Q(x, y} are as in Theorem 1.4.4. The idea is, using the invariance of the G(r;)'s and the fact that QlA;R = Ql(x, y) , to show that if f E EndG , then f E EndQQ(x, y} with fx = xf and yf = fy. In this case,

f

E

EndQ(x ,y)Q(x, y} = Q(x, y}.

A noncommutative version of Lemma 3.5.2 is needed to construct quasi-generic groups from the groups in Lemma 3.5.6 via Theorem 3.4.3.

Lemma 3.5.7 There is a ring l:1 with countablefree additive group containing Z( x , y} such that Ql:1 is a division algebra and l:1/ p S. has no zero divisorsfor each prime p ofZ. PROOF. Let R = Z[x, t, 2] be a skew polynomial ring over Z in two indeterminates x and t subject to the condition that tx = xt z . Induction arguments show that for positive integers m and n,

t nx=xt 2n , where rr(n)

= 2n - 1 +

tnxm=xmtUforu=2mn,

and

(xtt=x nt1f(nl,

2n - 2 + .. . + 2 2 + 2 + 1 if n ::: 2, rr(O) = 0, and rr(l)

= 1.

120

3. Butler Groups

It follows that a monomial xe(I)(xt)!(I) ... xe(k)(xt)!(k) is equal to Xe(I)+ !(I)+ ·+e(k)+!(klrlJ(f(l),e(2),...,e(k),j(k»,

where f3(f(1))

= JT(f(1)) = 2!(1)-1 + 2!(I)-2 + ... + 2 + 1

and

f3(f(1), e(2), . . . , e(k), f(k)) = 2!(k)+e(k)f3(f(l) , e(2) , . . . , e(k - I), f(k - 1)) + JT(f(k))

is defined recursively for k 2: 2. For example, with k = 2, f3(f(1), e(2), f(2)) = 2e(2)+!(2)JT(f(1)) + JT(f(2)).

Notice that f3(f(1), e(2) , . .. , e(k), f(k)) is a sum of powersof2. The next step is to show that Z(x , y) is isomorphic to the subring Z(x, xt) of R . To see this, view f3(f(1), e(2), ... , e(k), f(k)) as a 2-adic expansion of a natural number, i.e.,sumsofpowersof2 withcoefficients eitherO or I. Asa consequenceof the definition of f3(f(1), e(2), . . . , e(k), f(k)), this expansion has exactly k blocks of consecutive 1's as coefficients. Since 2-adic expansions of natural numbers are unique, k can be recapturedfrom f3(f(1) , e(2), ... , e(k) , f(k)). Now suppose that two monomialsxe(l)(xt)!(I) . . . xe(k)(xt)!(k) andx e'(I)(xt)f'(I) ... xe'(k')(xt)f'(k') are equal. From the precedingparagraph, f3(f(l) , e(2), ... , e(k), f(k)) = f3(f'(1), e'(2) , . . . , e'(k), f '(k))

and e(1)

+ f(1) + ...+ e(k) + f(k) = e'(1) + f'(1) + ...+ e'(k) + f '(k).

= k', e(i) = e'(i), and f(i) = f '(i) for each i, This shows that f : Z(x , y) -+ Z(x , xt) induced by f(x) = x and f(y) = xt is a ring iso-

Hence, k

morphism, as desired. Now, R is a right Ore domain, i.e., aR n bR #- (/) for each pair a, b of nonzero elements of R. This can be easily seen by observing that Z[t] is a right Ore domain and g : Z[t] -+ Z[t] induced by f(t) = t 2 is a one-to-one ring endomorphism. Hence, R has a right quotient ring D , a division algebra with each element of D of the form f(x , xt)g(x, xt)-I for f(x, xt), g(x , xt) E Rand g(x , xt) #- O. Note that !Q>(x, y) is isomorphicto !Q>(x, xt), a subringof D. Given g(x,xt) in R, define c(g(x,xt)) to be the gcd of the coefficients of g(x, xt). It is not difficult to confirmthat c(g(x, xt)f(x, xt)) = c(g(x, xt))c(f(x, xt)),just as in Gauss's theoremfor polynomials in onevariable. Let T = (g(x, xt) E R : c(g(x, xt)) = I}. Then T is a multiplicatively closedset. Hence, I:!,. = RT- 1 = (f(x, xt)g(x , xt)-I : f(x, xt) E R, g(x, xt) E T} is a subringof D . An argument

3.5 Quasi-Generic Groups

121

like that of Lemma 3.5.2 shows that QD. = D , the additive group of D., is a countablefree abeliangroup,and D. / PD. has no zerodivisorsforeachprime p of Z. The category B(T)Q is generically wild if there is a special quasi-generic Tgroup G such that QEnd G/ JQEnd G contains a copy of Q(x, y). Such a G cannot be central-special, since the center of Q(x , y) is Q and QG has infinite Q-dimension. The next theorem is a Butler group version of Theorem 1.5.4 for representations. Theorem 3.5.8 For the following statements, (a)

~

(b)

~

(c):

(a) B(T)Q has wild representation type ; (b) B(T)Q is generically wild; (c) B(T)Q is endowild.

If St n

~

= Sn, then

(a), (b), and (c) are all equivalent to the condition that

5.

(a) ~ (b) Assume that B(T)Q has wild representation type. By Corollary 3.3.5(b), s- contains one of the posets S5 , (1, I, 1,2), (2,2,3), (1, 3,4), (1,2,6), or (N , 5) as a subposet. By Lemma 3.5.7 there is a ring D. with countable free additive group such that QD. is a division algebra containing Q(x , y ) and D. / pI). has no zero divisors for each prime p of Z . Givena group G in Lemma3.5.6, define H = GD. . Then H is a countableButler groupof infiniterank withtypesetcontainedin T by Theorem3.4.3. Moreover, H is a special group from the definitionof G. Just as in Lemma 3.5.6,QEnd H = QD. is a division algebra containing Q(x, y). Thus, H is strongly indecomposable. Since H has finite dimension over QEnd H, H is a quasi-generic T-group. (b) ~ (c) Assume that B(T)Q is generically wild and let G be a special quasigeneric T -group withQEnd G/ J QEnd G a divisionQ-algebracontainingQ(x , y). Then Q(x, y) = QZ(x , y) is contained in QEnd G, so that Z(x, y) ~ End G. Since G is special, there are pure rank-l subgroups Xj of G with G/«End G)X) + . .. + (End G)X n ) bounded. Define H = Z(x, y)X) + ... + Z(x, y)X n • Then H is a countable torsion-free abelian group with typeset H contained in T and y ) ~ EndH. As a consequence of Example 1.1.7, for each finite-dimensional Q-algebra X, there are matrices A and B with X isomorphic to C(A , B). Suppose that A and B are two nonzero m x m Q-matrices with minimal polynomials f(xl and g(y)i, respectively, for irreducible polynomials f(x) E Q[x) and g(y) E Q[y). Let L = H/ K H, where K = (xy , f(xl, g(y) i) n Z(x , y) is an ideal of Z(x, y) . Then L E B(T)Q, since the Q-dimension ofQL is finite and H is generatedas a Z(x, y)-module by finitely manypure rank-I subgroupswithtypesin T. Moreover, C(A , B) is contained in QEn~(x . y)L. This is so because x acts on QH /QK H by A and y acts on QH /QK H by B . In particular, C(A, B) is a Q-subalgebra of QEnd L. It follows that B(T)Q is endowild.

PROOF.

z«.

122

3. Butler Groups

If ST = Sn and B(T)Q does not have wild representation type, then n ~ 4 by Corollary 3.3.5. Also , if G is a strongly indecomposable group in B(T)Q, then QlEnd G = End VG for VG an indecomposable in rep(Sn, Ql) by Theorem 3.3.2. However, each such End VG is a factor algebra of Ql[x] by Example 6.2.7. This contradicts (c). Hence , (c) => (a) if ST = S«. There is an interpretation of tame representation type of B(T)Q in terms of central-special quasi-generic T -groups . Suppose R is an integral domain with QlR = F, a field. A finite rank torsion-free R-module M is said to be afinite rank R-Butler module if there are finitely many pure rank-l subgroups Xi of M such that M/(RX I + ...+ RX n ) is bounded as a group. A finite rank R-Butler module M is strongly indecomposable if FEndRM / JFEndRM is a division F-algebra. These definitions directly generalize the definition of a finite rank Butler group. Quasi-generic Butler groups can be constructed from finite rank R-Butler modules via Theorem 3.4.3. An additional condition on the endomorphism ring yields a central-special quasi-generic group .

Lemma 3.5.9 Let R be an integral domain with countablyfree additive group of infinite rank such that p R is a prime ideal of R for each prime p of Z and QlR = F is a field. If G is a finite rank strongly indecomposable R-Butler module with R S; CG, then G is a central-special quasi-generic Butler group with End G = EndRG and QlEnd G/JQlEnd G a division F-algebra offinite F-dimension. By Theorem 3.4.3, G is a Butler group with countably infinite rank . Then G is a central-special group, since G is an R-Butler module and R S; C G . Hence, CG = CEnd G by Lemma 3.5.4 . Since G is a finite rank strongly indecomposable R-Butler module, QlEndRG / JQlEndRG = FEndRG/ J FEndRG is a division Falgebra. But QlEnd G = QlEndRG , since R S; CEnd G, and so QlEndG/ JQlEnd G is a division Q-algebra. Hence , G is strongly indecomposable, and QG has finite length as a QlEnd G-module. Since G has countably infinite rank, G is quasigeneric. PROOF.

For H E B(T)Q' define cdn H = (rank H, rank H(r) : rEST). Then cdn H = cdn V H, where V H = (Q H, QlH (r ) : rEST). The next lemma generalizes the construction of central-special quasi-generic groups given in Lemma 3.5.3.

Lemma 3.5.10 Suppose N = (No, N, : i E ST) is an indecomposable in rep(ST , Ql[x]) with EndQN = EndQ[x]N. There is a central-special quasigeneric T -group G such that V G = NA E rep(ST,Q(X)), End G = EndAG, and QEnd G/ JQEnd G is afinite-dimensional division Q(x)-algebra. Let A denote the principal ideal domain given in Lemma 3.5.2. Then EST) E rep(ST,Ql(X)) as QA = Q[x]A = Q(x) .Infact, N A is an indecomposable Ql(x)-representation, since N is an indecomposable

PROOF.

NA

= (NoA ,NiA:i

3.5 Quasi-Generic Groups

123

Q[x]-representation. This holds because EndQ(x)(N A) = Q(x)End!Q[xIN since Q(x) is the quotient field of Q[x] and N is finitely generated as a Q[x ]-module. Now, NA = Ua = (QG,QG(r) :r EST) for some finite rank A-Butler module G with EndQ(x)N A isomorphic to QEnd AG (Exercise 3.3.2), noting that QG = Q(x)G and Q(x) is the quotient field of the principal ideal domain A. Since N A is an indecomposable Q(x )-representation, QEnd AG has no nontrivial idempotents. This shows that G is a strongly indecomposable A-Butler module . By hypothesis, EndQN = EndQ[xIN, so that Q(x) £ CEndQN A = CEndQ Ua and A £ Ca = (CEndQUa)n(EndG).ItfollowsfromLemma3.5.9andTheorem 3.4.3 that G is a central-special quasi-generic T-group with End G = End-, G and QEnd G/JQEnd G a finite-dimensional division Q(x)-algebra. Given a finite lattice of types T , a countable torsion-free abelian group H is pregeneric if H is a torsion-free Z[x ]-module, H /(Z[x]X I + ... + Z[x ]Xn ) is bounded for pure rank-1 subgroups X; of H with type X; E T, and End H = Endz1xlH. A pregeneric group H is special , but a strongly indecomposable pregeneric group need not be quasi-generic. For instance, all the groups listed in Lemma 3.5.1 are pregeneric groups. As a consequence of the next theorem, if B(T)Q has tame representation type, then each strongly indecomposable group in B(T)Q can be constructed as a factor group of a pregeneric subgroup of a central-special quasi-generic T-group.

Theorem 3.5.11 The category B(T)Q has tame representation type ifand only iffor each sequence of nonnegative integers w = (wo, w r : tEST) there are finitely many strongly indecomposablepregenericgroups HI, .. . , Hn such that: (a) For each i, G; = H;A is a central-special quasi-generic T-group with EndAG; = End G;. (b) If H E B(T)Q is strongly indecomposablewith cdn H = w, then H is quasiisomorphic to H;A for some A = Z[x]/(f(xn and f(x) an irreducible polynomial in Z[x]. First assume that (a) and (b) hold. Fix wand define N; = (QH;, QH;(r) : rEST) E rep(ST, Q[x D. If U E rep(ST, Q) is an indecomposable representation with cdn U = w, then U = (QH, QH(r) : rEST) for some strongly indecomposable H in B(T)Q and cdn H = w by Theorem 3.3.2. By (b), H is quasi-isomorphic to K = H;A for some A = Z[x]/(f(xn, f(x) an irreducible polynomial in Z[x]. Hence , QA = Q[x]/(f(xy} with f(x) irreducible in Q[x]. Then U = (QH,QH(r) :r EST) is isomorphic to (QK ,QK(r) :r EST) = (QH;A, H;(r)Q[x]QA: rEST) = N;QA, noting that K(r) = H;(r)Z[x]A for each type r. This proves that B(T)Q has tame representation type. Conversely, assume that B(T)Q has tame representation type. Fix wand choose indecomposable N I , . .. , Nm E rep(ST, Q[xD with Q[x] £ CEnd N; such that if U E rep(ST, k) is indecomposable with cdn U = w, then U is isomorphic to N;B for some indecomposable cyclic Q[x]-module B. By Lemma 3.5.10, there is a central-special quasi-generic T -group G; such that (QG ;, QG; (r ) : rEST) =

PROOF.

124

3. Butler Group s

NjA E rep(ST, Q(x» , End G j = End"G j , and QE(Gj)/lQE(G j) is a finitedimensional Q(x)-algebra. In fact, G j is a finite rank strongly indecomposable A-Butler module, so there are pure rank-l subgroups X" ... , X m of G, such that G;/(AX, + ... + AXm ) is bounded. Hence, G, is a central-special quasi-generic T -group by Lemma 3.5.9. Define H, = Z [x ]X , +.. ·+Z[x]X n , a Z[x]-modulewithEnd H, = Endz[xlHj, since End G j = End" G j . Then H, is a pregeneric group with G i = H, ® Z[x I A. Moreover, H, is strongly indecomposable, since G, is strongly indecomposable. This proves (a). As for (b), let H be a strongly indecomposable in B(T)Q. Then UH = (QH, QH (r ) : r E ST) is an indecomposable in rep(ST, Q), and so UH = Nj®Q[xIB for some B = Q[x]/(g(x)') by the definition of tame representation type. Moreover, K = HjA is quasi-isomorphic to H for some A = Z[x]/(f(xY) with QA = B. This is so because UK = (QK,QK(r:):r: EST) = (QHjA, QHj(r:)A: r EST) = NjB = UH , whence K is quasi -isomorphic to H by Theorem 3.3.2. The proof of (b) is now complete. Open Question: Are there existence conditions on central-special quasi-gen-

eric T -groups, or their endomorphism rings, equivalent to tame representation type for B(T)Q? This question is the analogue of the open questions in Section 1.5 for k-representationsofa finite poset.

NOTES ON CHAPTER

3

Section 3.1 is an exposition of the traditional tools of the theory of torsion-free abelian groups of finite rank used in this book. Additional properties and example s of such groups are in [Fuchs 73] and [Arnold 82]. The notion of r -socle dates back to [Baer 37], but the first systematic development of r-radicals seems to be due to [Lady 79]. Finite rank Butler groups are called quasi-essential groups in [Koehler 65], purely finitely generated groups in [Bican 70, 78], R-groups in [Butler 65], and rr-diagrammatic groups in [Butler 87]. One ofthe attracti ve features of this category of groups is the number and variety of characterizations of these groups within the category of torsion-free abelian groups of finite rank. The work of [Koehler 65], as acknowledged by [Butler 87], is fundamental to Corollary 3.2.4, Theorem 3.2.5, and Theorem 3.3.2. Some other characterizations are given in [Arnold 81]. Results of [Bican 70, 78], including Proposition 3.2.15, give rise to a characterization of finite rank Butler groups in terms of balanced extensions, Theorem 3.4.6. This characterization provides both a natural definition for Butler groups of arbitrary rank and a host of associated problems for torsion-free abelian groups. The resolution of these problems involves sophisticated set theory and logic; see [Fuchs 94]. Included in [Fuchs, Metelli 92] is a development of properties of countable Butler groups using techniques potentially useful for Butler groups of arbitrary rank. An error in an argument of [Bican , Salce 83], repeated in [Arnold 86] as observed in [Mines, Vinsonhaler 92], is corrected in [Bican, Rangaswamy 95]. The corrected argument can be used to provide an alternative proof of Theorem 3.4.6. In another direction , the properties listed in Theorem 3.2.5(b) characterizing finite rank Butler groups have been considered for torsion-free abelian groups of infinite rank in [Mader, Mutzbauer, Rangaswamy 94].

3.5

Quasi-Generic Groups

125

In a series of papers, summarized in [Lady 83], the results of Section 3.2 are extended to a class of finite rank torsion -free modules over Dedekind domains called Butler modules . The setting is different from that of this manuscript, since the quasi-homomorphism category of finite rank Butler modules is subdivided into subcategories determined by a generalization of the notion of a splitting field. These subcategories are then related to categories of finitely generated modules over finite-dimensional algebras . Some partial results on representation type of these subcategories are obtained from classifications of representations of species [Ringel 76]. Finite rank Butler modules over l-dimensional Noetherian domains are considered in [Goeters 99]. Other generalizations include valuated vector spaces , [Richman 84], and groups with Murley groups playing the role of rank-I groups [Albrecht, Goeters 98]. Quasi-generic groups arose out of the notion of generic representations given in Section 1.5. There are functorial correspondences from categorie s of infinite-dimensional representations of posets to categories of countable Butler groups , [Duga s, Thome 91] and [Rangaswamy, Vinsonhaler 94], but the morphisms for groups are "local quasihomomorphisms," not quasi-homomorphisms as defined herein . This provides an obstacle to the determination of a Butler group analogue of generic representations. The point of Theorem 3.5.11 is that in the tame representation type case, strongly indecomposable Butler groups of finite rank are determined by quasi-generic Butler groups of infinite rank.

4 Representations over a Discrete Valuation Ring

4.1 Finite and Rank-Finite Representation Type There are several choices for extending the definition of representations of finite posets over fields to representations over discrete valuation rings [Plahotnik 76]. The choice made herein arises naturally from the connection between repre sentations of finite posets over discrete valuation rings and isomorphism at p categories of finite rank Butler groups; Section 4.3. Posets with finite representation type are characterized in Corollary 4.1.7. A discrete valuation ring is a principal ideal domain R with a unique prime ideal p R. If R is a discrete valuation ring, then R is a local ring, the Jacobson radical of R is p R , and R / p R is a field. An important example from abelian group theory is the discrete valuation ring Zp, p a prime of Z. Specifically, a torsion-free abelian group is a p -local group if and only if it is a torsion-free Zp-module. Observe that pZp is the unique prime ideal of Zp and Zp/ pZp = Z/ pZ is a field. Throughout this chapter, R denotes a discrete valuation ring. If M is a finitely generated free R-module and N is a pure submodule ofM, i.e., N n pM = pN, then M / N is a finitely generated torsion-free R-module. Since R is a discrete valuation ring, in particular a principal ideal domain , M / N is a finitely generated free R-module, and so N is a summand of M . Consequently, a submodule of a finitely generated free R-module is a summand if and only if it is a pure submodule. Let S be a finite poset and define rep(S, R) to be the category with objects V = (Vo, Vi : i E S) such that Vois a finitely generated free R-module, each Vi is a summand of Vo,and Vi is contained in U], provided that i :5 j in S. A morphism from V = (Vo, Vi : i E S) to V = (VO, Vi : i E S) is an R-homomorphism

4.1 Finite and Rank-Finite Representation Type

127

f: V o -+ Vo with f(V i) ~ Vi for each i in S. Since R is a discrete valuation ring, each Vi is a finitely generated free R-module, being a summand of a free R-module Vo. The rank of a representation V = (Vo, Vi : i E S) E rep(S , R) is the rank of Vo as a free R-module. Rank-I representations are easily described. Up to isomorphism, they are of the form V = tU«; Vi : i E S) with V o = R and Vi = or R, subject to the condition that if i ~ j in S and Vi = R, then U, = R. Given an integer j ~ 0, let rep(S, R, j) denote the full subcategory ofrep(S, R) with objects (Vo, Vi : i E S) such that pj V o ~ ~ {Vi : i E S} ~ Ue: In particular, rep(S, R , 0) consists ofrepresentations V = (Vo, Vi: i E S) with Vo = ~{Vi : i E S} and each Vi a summand of Vo. The category rep(S, R , 0) is a direct generalization of the category of elements V in rep(S, k) with no trivial summands, k a field. Define Ind(S, R, j) to be the set of isomorphism classes of indecomposable representations ofrep(S, R, j).

°

°

Example 4.1.1 If S is a finite poset with w(S) = 1 and j ~ is an integer, then S is a chain and rep(S, R, j) has finite representation type. Elements of Ind(S, R , j) are rank-l representations, and each indecomposable representation has endomorphism ring R. PROOF.

The argument of Lemma 1.3.3(a) for fields carries over to this setting.

For posets that are not chains, there are more indecomposable representations for discrete valuation rings than there are for fields . Example 4.1.2 The category rep(S2, R, j) has finite representation type for each j ~ 0. The elements of Ind(S2, R, j) are: (i) rank-l representations (R, 0, 0), (R, R, R), (R , R, 0), and (R, 0, R) with endomorphism ring R. (b) rank-2 representations (R EB R + (l + I)R(lj pn), R EB 0, EB R) with endomorphism ring {(r, s) E R x R : r - s E p" R} for each 1 ~ n ~ j .

°

Let V = (Vo, VI , V 2) E rep(S2, R, j) be an indecomposable representation with rank Vo > 1. Notice that each Vi is nonzero and VI n V 2 = 0, since otherwise V is a rank-I representation as in Example 1.1.3. Hence, p" Vo ~ VI EB V2 ~ Vo for some smallest positive integer n ~ i, observing that if n = 0, then U« = VI EB V2, and V would again be a rank-l representation. Now, V = (VI EB V2, VI, V2, pnvo) is an R-representation of S3 with pnvo a free R-module, being isomorphic to Us, and pnVo n Vi = pnVi, since Vi is a pure submodule of Ui; However, p" Vo need not a summand of VI EB V2. There is an isomorphism End V -+ End V given by sending f to the restriction of f to VI EB V2, recalling that End V is the ring of R-endomorphisms f of VI EB V2 with PROOF.

128

4. Representations over a DiscreteValuation Ring

f(Ui) £ u, and f(pnUo) £ ti; Then V is an indecomposable representation with rank greater than 1. The remainder of the proof is a matrix argument generalizing that of Example 1.1.5 for representations of S3 over a field. The setting is that U = (Uo, UI , U2) is indecomposable and V = CUI EB U2, UI, U2, pnUo) is an indecomposable representation with End V isomorphic to End U. Let BI = {XI, , x,} be an R-basis of UI, B2 = {YI, ... , Ys} an R-basis of U2, and B3 = {ZI, , z.l an R-basis of p'U«. For each i,

In particular, pnUo can be interpreted as the row space of a t x (r Mv

+ s) R-matrix

= (A I B),

where the rows of Mv are labeled by B3, A = (aij) is a t x r R-matrix with columns labeled by BJ, and B = (bij) is a t x s R-matrix with columns labeled by B2. The following invertible R-matrix operations on M» do not change V: (a) Elementary column operations within A (a basis change for VI); (b) Elementary column operations within B (a basis change for V2); (c) Elementary row operations on M (a basis change for pnvo). Write M v:::::: N if the matrix N can be obtained from Mv by a sequence of operations (a), (b), and (c). Since R is a principal ideal domain, the block matrix A can be reduced by a series of elementary invertible row and column operations to a diagonal matrix [Hungerford 74]. Since n is minimal and R is a discrete valuation ring, this diagonal matrix can be chosen to be

(01

0)

pD

with D a diagonal R-matrix and I an identity matrix, agreeing that a 0 x 0 matrix is empty. If X = UI EB U2 E pn Vo with u, E Vi, and p-height X :::: n, then p-height UI = p-height U2, since VI and V 2 are pure submodules of Vo. It follows that M» is equivalent to a matrix of the form

(:

0 pI 0

Bll pB21

B12 pB22

B2n ) pB"

pnl pnBnl

p nBn2

pnB nn

0 0

with each row of (l-I Bi!, pi-I Bi2, ... , pi-I Bin) containing an element of

4.1 Finite and Rank-Finite Representation Type

129

p-height i - 1. If E is an operation as in (c), then (l 0 . . . 0 I BII B12

Bin)

.. .

(E 0 . .. 01 EB II EB n . .. EBln) ~ (l = EE- I 0 .. . 0 I EB II EB I2 ... EB ln)

~

as an application of (c) and (a). Applying operations (b) and (c) to diagonalize the matrix (B II BI2 . .. Bin) yields M» ~ N =

(:

I pB 21

0 pI

0 0

0

pnl

pnBn1 p" Bn2

Now use (c) followed by (a) to see that N

0 pI

C

0

0 p B22

0 0

I 0

p" I 0

P~m

)

pnBnn

~

0 p C22

p nCn2

P~m

)

pnCnn

Since n is chosen to be minimal, I must be nonempty. As in Example 1.1.5, V = (Rx i EI1 RYt , RXI, RYI, Rtx, + Yt)), since V = (VI EI1 V2, VI, V 2, pnVo) is indecomposable. Then V = (Rx, EI1RYI + R(xI + Yd/ p", RXI, RYd is isomorphic to W = (R EI1 R + R(I + 1)/ p", R EI1 0, 0 EI1 R). A straightforward calculation shows that f E End W if and only if f = (r, s) E R x R with r - s E p" R. Hence, End V is isomorphic to {(r,s) E R x R : r - s E pnR}. For a finite poset S, define S* to be the disjoint union of S and an element * unrelated to any element of S. For example, If S = Sn is an antichain, then S* = Sn+I' Given n ::: 2, let Cn denote a chain with n elements. Then C~ is the poset (n, I). For example, q = (l , I) = S2 and q = (l, 2).

Proposition 4.1.3 Let n ::: 2 be an integer. (a) There is a fully faithful functor Fj : rep(C~_l' R, j) --+ rep(C~, R , 0) for each j ::: O. (b) If V E rep( C~ , R, 0) is indecomposable with rank> 1, then there is some integer j ::: 0 and indecomposable V E rep(C~_I ' R , j) with Fj(V) = V. (a) Write Cn- I = {I < 2 < ... < n - I} and let V = (V o, VI s:; ... s:; V n- I, V*) E rep(C~_I' R, j) . Define Fj(V) = (Vo, VI s:; ... s:; Vn-I s:; Vo, V*) E rep( C~, R, 0). It is routine to verify that with Fj(f) = f, Fj : rep( C~_l' R , j) --+

PROOF.

rep( C~ , R, 0) is a fully faithful functor. (b) Let V = (Vo, VI s:; .. . s:; Vn , V*) be an indecomposable representation in rep(C~ , R , 0) with rank> 1. Then Vo = Vn + V*, and so V* = WI EI1 (V* n Vn)

130

4. Representationsover a Discrete Valuation Ring

for some WI. This holds because V* n Vn is a pure submodule, hence a summand, of V*, since Vn is a pure submodule of Vo . Therefore,

and

is a representation direct sum. But WI is a free R-module. Since V is indecomposable with rank> I, it follows that WI = 0, Vo = Vn contains V*, and V =

(Vn, VI, "" Vn- I, Vn, V*) . Furthermore, p! Vn is contained in Vn- 1 + V* for some j. To see this, let W be the pure submodule of Vn generated by Vn-l + V* . Then Vn = W' ffi W for some

W' and

V =(W',O, .. . ,O, W',O)ffi(W , VI, . .. , Vn-I, W, V*). Again, W' = 0, since V is indecomposable of rank greater than 1, and so Vn W . Thus, there is some j with p! Vn contained in Vn-I + V*, since Vn is a finitely generated free R-module with rank Vn rank Vn- I + V* . Hence, V = (Vn, VI, .. " Vn-I, V*) E rep(C~_l' R , j) with V = Fj(V). As a consequence of (a), V must be indecomposable.

=

=

=

Corollary 4.1.4 If S (1,2), then rep(S, R , 0) has infinite representation type but each indecomposable has rank less than or equal to 2. Rank-2 elements of Ind(S, R, 0) are (Vo, 0 ffi R, R ffi 0 5; Vo)for j ~ 1, where Vo = R ffi R + R(1 + l)jp j.

=

=

=

PROOF. As mentioned above, q (1,2) and q (1,1) S2. By Theorem 4 .1.3, each indecomposable in rep«l, 2), R, 0) is of the form Fj(V) for some indecomposable V E rep(S2, R, j). Now apply Example 4.1.2.

Lemma4.1.5 If Sis a finite poset, S' is a subposet of S, and j' ::5 j, then there is a fully fa ithful functor F- : rep( S', R, j') ~ rep( S, R, j).

=

=

PROOF. Given V (Vo, U, : i E S') E rep(S', R, l'). define F-(V) (Vo, Vj : E S) E rep(S, 'lp, j) by Vo V o, Vj U, if i E S', Vj n{Vj: j E S', j > i} if i E S\S' and there is some j E S' with j > i, and Vj Vo otherwise. Each Vj is a pure submodule, hence a summand, of Vo, and pj Vo 5; :E j Vj, since j' ::5 j and pj Uo 5; :E j Ui. Then F- is a fully faithful functor, just as in Proposition 1.3.1 (b).

i

=

=

= =

The functor F+, as defined in Proposition 1.3.1 (a), is not directly relevant for representations over discrete valuation rings, since if U, and U, are pure submodules of Vo, then U, + U, need not be a pure submodule of Us;

4.1

Finite andRank-Finite Representation Type

131

Proposition 4.1.6 Let S be afinite poset ofwidth 2 that does not contain (1, 2) as a subposet. For each j :::: 0, rep(S, R, j) has finite representation type, and each indecomposable has rank ~ 2.

PROOF. The proof is by induction on lSI, the cardinality of S. If lSI = 2, then S = Sz, rep(S, R, j) has finite representation type, and each indecomposable has rank ~ 2 by Example 4.1.2. Now assume lSI:::: 3 and let a and b be incomparable elements of S. Since w(S) = 2 and S does not contain (1, 2) as a subposet, it follows that if s E S with s =f:. a and s =f:. b, then either s < a and s < b or else s > a and s > b. Hence, there is a partition S = AU B U C of S with C = Is E S : s < a, s < b}, A = Ia, b}, and B = {s E S: s > a, s > b}. If the pair (a, b) of incomparable elements of S is chosen to be minimal in the sense that w( C) = I, then C is a chain. In this case, the partition S = AU B U C is a splitting decomposition, as defined in Section 1.3, since s < t for each sEA and t E B. Use the proof of Proposition 1.3.5 to see that each indecomposable U in rep(S, R, j) is isomorphic to F-(V) for some indecomposable representation V in either rep(A U B , R, j) orrep(B UC, R, j), with F- defined as in Lemma 4.1.5. Both A U Band B U C are posets of width 2 that do not contain (1, 2) as a subposet, being subposets of S. By induction on lSI, AU Band B U C have finite represention type, and each indecomposable in rep( A U B, R, j) or rep( B U C, R, j) has rank less than or equal to 2. Consequently, rep(S, R, j) has finite representation type and each indecomposable has rank less than or equal to 2. A finite poset S is a garland if w(S) = 2 and S does not contain (1, 2) as a subposet. Garlands also arise for representations of finite posets over fields [Simson 92]. The terminology is motivated by the fact that if S is a garland, then S is a subposet of the poset Gn,adisjointunionoftwochains Cn = {l < 2 < .. . < n} and C~ = {I' < 2' < .. . < n'} subject to the additional relations that i < (i + 1)' and i' < i + 1 for each I ~ i ~ n - 1. Corollary 4.1.7 Let S be a finite poset. Then rep(S, R, j) has finite representation type if and only if w(S) = 1 or S is a garland.

PROOF. If w(S) = I or S is a garland, then rep(S , R, j) has finite representation type by Example 4.1.1 and Proposition 4.1.6. The converse follows from Corollary 4.1.4 and Lemma 4.1.5 in the case w(S) = 2 and from Example 4.2.4 and Lemma 4.1.5 if w(S) :::: 3. An additive category has rank-finite representation type if there is a bound for ranks of indecomposables in the category and rank-infinite representation type if there is no such bound. Finite representation type implies rank-finite representation type, but Corollary 4.1.4 illustrates that the converse is not true in general. The next proposition relates rep(Sn+z, R/pR) to rep(S, R , 2n + 1) for S = (1, 2). This provides a systematic construction of indecomposable representations

132

4. Representations over a Discrete Valuation Ring

in rep(S, R, j) for sufficiently large j from indecomposable representations of S; overthe field k = R] p R, If M is an R-matrix,then M(mod p) is the Rj pR-matrix obtained by reducing each element of M modulo p. Define E(n , k) to be the subcategory of rep(Sn+2, k) , consisting of representations V = (Vo, Vi: I ::: i ::: n + 2) such that Vo = VI E9 . .. E9 Vn , there is a fixed s with s = k-dimension Vi for each i ::: n, Vn+ 1 E9 Vn+2 S; Yo, and Vn + 1 = (1 + I + ... + l)k S is the image of the diagonal embedding of kS into Yo. Proposition 4.1.8 Let S = (1,2), n ::: 0, and k = Rj p R, There is a correspondence 4J : E(n, k) ~ rep(S, R, 2n + 1) such that if V E E(n, k) is an indecomposable representation, then 4J(V) is an indecomposable representation. PROOF.

Let k

= RjpR and V = (Vo, Vi)

E

E(n, k). Then Vo

Vn+1 E9 Vn+2 )j K, where K is the row space of a k-matrix

M'

= ( MI,I II, M2 I'"

= (VI

E9 .. . E9

I I I

... MI,n 0I 0I )

with each M: an s x s k-matrix, s the k-dimension of Vi for each i ::: n. The first row of M' gives the relations for Vn+ l , and the second row for those of Vn+2 . Choose s x s R-matrices M, with Mi(mod p) = M: for I ::: i ::: n. Let Wand X be free R -moduleswithrank 2s, and Y and Z free R-rnodules with rank(n + I)s . Define 4J(V) = V = (Vo, V), V 2 S; V 3) E rep(S, R, 2n + I) by

= W E9 X E9 Y ffiZ+ (1/p2n+I)(1 + 1 + 0+ O)R 2s , VI = W E9 Z, V3 = X E9 Y E9 Z, and V2 the row space of the matrix Mu = (E I, E2 , l) = Uo

pM I p 2M2

p 2I p 31

pnM n pn+1 I

pn+II 0

p 3I 0 0 pSI

0 0

0 0

0

0 p2n+1 I 0

0 0

I 0 0 I

0 0 0 0

0 0 0 p2n+31 0 0

I 0 0 I

a submodule of V3 = X E9 Y E9 Z. The columns of M u are indexed by X, Y , and Z, respectively, the l's stand for s x s identity matrices, the O's stand for s x s matrices of all zeros, and (1 + I + 0 + O)R 2s = {x E9 x E9 0 E9 0 E Vo : x E R2s }. Notice that V2 is a summand of V3. The strategy is to show that if f E End V with R-matrix M f relative to an R-basis of Vo, then Mf(mod p) = aI + N for some nilpotent k-matrix Nand some k-matrix a representing an endomorphism of V . In this case, it is verified that if f is idempotent,then f = 0 or I as a consequence of the assumption that V is indecomposable. Hence, 4J(V) = V must be an indecomposable representation of rank (6 + 2n)s.

4.1 Finite and Rank-Finite Representation Type

If I E End V, then I : VI = W E9 Z ~ VI = W E9 Z, I : V3 = X E9 Y E9 Z V 3 = X E9 Y E9 Z, I : Z ~ Z as VI n V3 = Z, I : V z ~ Uz, and I: (l

+ 1+ 0 + O)R 2s(mod pZn+l) ~

(l

133 ~

+ I + 0 + O)R 2s(mod pZn+I).

Then I can be represented as an endomorphism of W E9 X E9 YE9 Z with matrix

Mf

=

(

~I 0

o

o

0

hz hz

h3 133

o

0

where III : W ~ W, 114 : W ~ Z, [n : X ~ X, h3 : X ~ Y, h4 : X ~ Z, 132 : Y ~ X, 133 : Y ~ Y, 134: Y ~ Z, I44:Z ~ Z, and I: V z ~ UiSince I: (l + I + 0 + O)R zS ~ (l + 1+ 0 + O)R 2s + pZn+l Vo, it follows that III == hz(mod pZn+l), !J4 == h3 == h4 == 0 (mod pZn+I), and

o III hz

o

Also, I : U: ~ Ui. and so MUMf (Edll

0 0 133

0

== (0, E I, E z, l)Mf(mod pzn+l)

+ Ezhzl Ed33 I Ed34 + 144) == (g E I I g Ezi g l)(mod

pZn+l)

for some g. Equating block matrices yields

+ 144/ == g/(mod p2n+l), Edll + Ezhz == gEl == (Ed34 + I44l)E 1(mod p2n+l), Ed33 == g E; == (Ed34 + I44l)Ez(mod pZn+I). Ed34

Write matrices 133 = (aij)(n+l)x(n+l) representing an endomorphism of Y = (w)n+1 and 144 = (bij)(n+l)x(n+1) representing an endomorphism ofZ = (w)n+1 with each aij and bij an s x s R-matrix representingan endomorphismof W. Then Ed33

== (pZi+l aij)(mod p2n+l),

I44Ez

== (pzj+lbij)(mod pZn+I),

Ed33

==

Ezh4Ez

+ I44Ez(mod In+l).

Write 134 = (xij) with each xij an s x s R-matrix. Combine equations to see that pZi+l aij == pZj+2+Zi Xij + pZj+1bij(mod pZn+l) for 1 ::: i, j ::: n + 1. Then pZi+l aij == pZj+lbij(mod pmax(Zi+l ,Zj+I». If i < j, then aij == O(mod p), and if i > i. then bij == O(mod p). Consequently, h3(mod p) is lower triangular, I44(mod p) is upper triangular, and aii == bii(mod p) for each 1 ::: i ::: n + 1.

134

4. Representations over a Discrete Valuation Ring

Write the matrix fll = (Cij hX2 representing an endomorphism of X = (R s )2 with Cij an s x s R-matrixrepresentingan endomorphismof W. Recallfrom above that g E, == Edll + Ezf32 == (Ezf34 + f44l)E I(mod p2n+I). From thelast row of the latter equation, CII

== bn+1 ,n+l(mod p) and CI2 == O(mod p2).

The first row yields CII

== bll(mod

p)

== c22(mod p) .

From the first column, Micll == b.,Mj(mod p) for each i :::: n. Finally, the second column of this equation gives C22

== bii(mod p) for each i

:::: n.

Let a = clI(mod p), an s x s k-matrix. By the results of the preceding two paragraphs, a == c22(mod p) == bjj(mod p) == aii(mod p) for each i :::: n + 1, and a M, == M ja(mod p) for each i :::: n. In particular, aM' = M'a, recalling the definition of M' and the fact that M; = Mi(mod p) . Consequently, al(n+2)sx(n+2)s is a k-matrix representing an endomorphism of V E E(n, k) . This is so because a represents an endomorphism of VI E9 .. . E9 Vn+2 with a(K) ~ K, the row space ofM'. In summary, Mf(mod p) = al + N , where NI

N-

(

0 0

o

o

N2

132

o

0 0 N3

0

and N, = f ii(mod p) - al is of the form

A straightforwardcalculation shows that N is nilpotent. Finally, assume that V is indecomposable and f2 = f E End V . Then (al + N)2 = a 2I + aN + N a + N 2 = a I + N. Since N is a nilpotent matrix with zeros on the diagonal, it follows that a 2 = a. Thus, al(n+2)s x(n+2)s represents an idempotentendomorphism of V. But V is indecomposable, so a = 0 or 1.1f a = 0, then N is both nilpotent and idempotent. In this case, f E pEnd V ~ EndR(Vo) with f2 = f. Since Vo is a free R-module, f = O. On the other hand if a = 1, then 1 - f = 0 and f = 1. This shows that ¢(V) = V is an indecomposable representation. Corollary 4.1.9 If S tion type for j :::: 5.

= (1,2),

then rep(S, R, j) has rank-infinite representa-

4.2 Wild Modulo p Representation Type

135

PROOF. Let n = 2. Given a positive integer s, there is an indecomposable representation

in rep(S4, k) with A an indecomposable s x s k-matrix. In the notation of Proposition 4.1.8,

In fact, V E E(2, k), since Vo = VI EB V2, k-dimension VI = k-dimension V2 = s, and V3 EB V4 £; VI EB V2. By Proposition 4.1.8, ¢(V) is an indecomposable in rep(S, R, j). Since there are indecomposable k-matrices A withs arbitrarily large, it follows that rep(S, R, j) contains indecomposable representations of arbitrarily large finite rank for j 2': 5 = 2 . 2 + 1. Example 4.1.10 If S = (1,2), then rep(S, R, 4) has rank-infinite representation type.

PROOF. Exercise 4.1.1. Open Question : For S = (1, 2) and 1 ::: j ::: 3, does rep(S, R, j) have rankfinite or rank-infinite representation type? In this case, can indecomposable representations in rep(S, R, j) be classified up to isomorphism ?

EXERCISE

1. Find a matrix to show directly that if S = (1,2), then rep(S. R, 4) has rank-infinite representation type (Proposition 4.1.8 does notapply in this case).

4.2 Wild Modulo p Representation Type Includedin thissectionis a computationof therepresentation typeof rep(S, R, j) in terms of Sand j (Corollary4.2.5). The definition of wild modulo p representation typeis motivated by the definition of endowildforrep(S, k), k = R/ p R a field. The categoryrep(S, R, j) has wild modulo p representation type if for each R -algebra r that is finitely generatedand free as an R-module, there is U E rep(S, R, j) such that r / p r is a ring-homomorphic image of End U . It remains an open question as to whetherifrep(S, R, j) has wild modulo p representation type, then for each finite-dimensional k-algebra A there is U E rep(S, R, j) with A a homomorphic image of End U (see Open Question 2). If rep(S, R, j) has wild modulo p representation type, then it has infinite representation type. Moreover, in view of the diversity of endomorphism rings in

136

4. Representations overa Discrete Valuation Ring

this case (Exercise 4.2.1), it is not reasonable to expect to be able to classify the indecomposable representations. The first lemma gives a computational tool for determining wild modulo p representation type. The idea is that for a given I', there are R -matrices A and B with qA, B) = rand qA(mod p), B(mod p» = r/pr . The matrices A and B are then used to construct representations U with qA , B) S; U and a homomorphism End U --+ qA(mod p), B(mod p». It follows that there is an onto ring homomorphism End U --+ r / p r. The matrices A and B are somewhat special , since qA, B) --+ qA(mod p) , B(mod p» need not be onto for arbitrary matrices A and B (Exercise 4.2.3). Lemma 4.2.1 Suppose r is an R-algebra that is finitely generated andfree as an R-module. There are R-matrices A and B such that (a) qA, B) = r andqA(mod p), B(mod p» = r/pr, and (b) the ring homomorphism qA, B) --+ qA(mod p), B(mod p» given by M r+ M(mod p) is onto. (a) If {I, Y2, . .. , Yt} is an R-basis of I', then {I, X2, . . . , x t } is a k-basis for r / pi', where k = R/pR is a field and Xi = Yi + pr o Define (t + 2) x (t + 2) r-matrices

PROOF.

A=

0 0

1 0

0 1

0 0

0 0

0 0

0 0

0 0

0 0

I

, B=

0

0 1 Yl 0

0 0 I Y2

0

0

0 0 0 I

0 0 0 0 Yt

0 0 0 0 0

Then A and B are R-matrices with qA, B) isomorphic to Endj-Il") = r. The proof is exactly the same as for Example 1.1.7. Similarly, r / p r is isomorphic to qA(mod p) , B(mod p», as desired . Assertion (b) follows from (a). Corollary 4.2.2 If S = (1,2), then rep(S, R, j) has wild modulo p representation type for j ::: 7. PROOF. Let r be an R-algebra that is finitely generated and free as an R-module and, via Lemma 4.2.1, let A and B be s x s R-matrices with r = qA, B) and r / pr = qA(mod p), B(mod p» . Define V E rep(S3, k) by

V = (kS t1J k S t1J k", k S t1J 0 t1J 0, 0 t1J k' t1J 0, 0 t1J 0 t1J k S ,

(1 + I

+ 1)kS, (l + A' + B')kS)

for s x s k-matrices A' = A(mod p) and B' = B(mod p). Then End V = qA', B'), by Exercise 1.2.2(c), whence End V is isomorphic to r / pro It follows from the definitions that V E £(3 , k), as defined in Section 4.1. In the notation of

4.2 Wild Modulo p Representation Type

137

Proposition 4 .1.8,

M'

= ( ~ I ~, I ~, I~ I~ ) . =

Moreover, ¢(V ) = U = (Uo, UI, U2 £; U3) E rep(S, R , 7) is given by Uo WEB X EB Y EB Z + (1/ p7)(1 + 1 + 0 + O)R 2s , UI = WEB Z, U3 = X EB Y EB Z, and U2 , a submodule of X EB Y EB Z, the free R-module with basis the rows of M u = (E I , E2, I) =

p2I p3I p4 I

o

o o

p3I 0 0

p7 I

o

0

0 0 0 p9I

I

0

0 0 0

0 0

I

o

0)

o

I

I

0 0

.

Notice that C(A, B) £; EndU, since if b E C(A , B), then bMu = Mub, and so bI E EndU. As in Proposition 4.1.8 , if f E EndU, then Mf(modp) = aI + N for some unique aI E End V = C(A', B') and nilpotent k-matrix N . It follow s that there is a ring homomorphism x : End U ~ End V I' / pf given by x (f) = a I . Since C(A, B) £; End U, n is onto by Lemma 4.2.1. This shows that rep(S , R , 7) has wild modulo representation type. Now apply Lemma 4.1.5 to see that rep (S , R , j) has wild modulo p representation type for j ::: 7.

=

Submodules of finitel y generated free R-modules need not be summands, unless they are pure . Another category of representations of finite posets over discrete valuation ring s is defined to reflect this distinction. Let rep f (S , R) be the category of representations U = (Uo , U, : i E S) such that Uo is a finitely generated free R-module, each U, is a R-submodule of Ui ; U, £; U, if i ::::: j in S, and U« = b {U i : i E S}. The morphisms in this category are representation morphisms. Assertion (b) of the next lemma is a generalization of Theorem 1.2.3, while statement (a) for the special case that n = 2 was used in the proof of Example 4.1 .2. For an antichain Sn, define fl(Sn , R) to be the full subcategory of rep(Sn+1, R , 0) with objects of the form U (Uo, Ui, U* : i E S) such that Uo EB{ U, : i E S}, U, n U* = 0, and U, + U* is pure in Uo for each i E S« . Recall that the poset S* = S U {*} is the disjoint union of S and a point *.

=

=

Lemma 4.2.3 Suppose S is a finite poset. (a) There is afully faithful functor H' : rep(S, R , j) ~ rep f(S*, R). The image of H' consists ofall V (Vo, Vi, V* : i E S) with Vo b{Vi : i E S}, each Vi a pure submodule of Vo, and V* n Vi pi Vi for each i in S. (b) There is a category equivalence H: fl(Sn , R) ~ rep(Sn, R, 0), where Sn is

=

=

=

an antichain. PROOF. (a) Define H'(U) = V = (b {Ui : i E S}, Ui , pi Uo), where U = (Uo, U, : i E S ), and define H'(f) to be the restriction of f to b{Ui : i E S}. Then H' is

138

4. Representations over a Discrete Valuation Ring

a functor. For each i E S, pj Vo n Vi = p! Vi and Vi is a pure submodule of E i Vi, since Vi is a pure submodule of Vo. Each Vi is a summand of E {Vi: i E S}, but pj Vo need not be, which is why the image of H' is in rep reS, R) and not necessarily in rep(S, R) . The functor H' is fully faithful, since if g: H '(V) ~ H '(V') is a representation morphism, then g extends uniquely to f: Vo ~ V~ with H'(f) = g . To see that the image of H' is as described, let V = (Vo , Vi, V*: i E S) E rep [(S* , R) with Vo = E {Vi: i E S}, each Vi a pure submodule of Va, and V* n Vi = p! Vi for each i. Then pj Vo = E{pj Vi : i E S} S; V*. Define V = (Vo, Vi : i E S) with Vo = Vo + (11 pj)V* . Now, Vo is a finitely generated torsionfree, hence free, R-module, and pj Vo is contained in Va. To see that Vi is pure in Uo, suppose x E Vo with px E Vi. Then p! x E V* n p Vi = pHI Vi, and so x E Vi, as desired. This shows that V E rep(S , R , j) with H'(V) = V . (b) Given V = (Vo, Vi, V* : i E Sn) E D.(Sn, R) with Vo = EB{Vi : i E Sn}, define H: D.(Sn ,k) ~ rep(Sn, R,O) by H(V) = (Vo, Vi :i E Sn), where Vo = Vol V* and Vi = (Vi + V*)I V* for each i in S. Then Vo is a finitely generated free R-module, since V* is a pure submodule of the finitely generated free R-module Vo, and Vo = L{Vi : i E Sn} . Because Vi + V* is pure in Vo, Vo/(Vi + V*) is a torsion-free module, and so Vi is pure in Va . Thus, H(V) E rep(Sn, R, 0). The proof that H is a category equivalence is the same as that of Theorem 1.2.3. Example 4.2.4 The category rep(S3, R, 0) has wild modulo p representation type. Let r be an R -algebra that is finitely generated and free as an R -module. By Lemma 4.2.1, there are n x n R -matrices A and B with r = qA , B) and rip r = qA(mod p), B(mod p)) . Define V = (Vo, VI, V2, V3 , V*) E rep(S4, R, 0) by

PROOF.

Uo

= R7n EB R6n EB R5n, VI = R7n EB 0 EB 0, V2 = 0 EB R6n EB 0, V3 = 0 EB 0 EB R5n,

and V* the free R-module with the rows of the following matrix as a basis:

1000000 0100000 0010000 0001000 o 0 0 0 p2 1 0 0 o 0 0 0 0 p6I 0 o 0 0 0 0 0 p7I

o

000000010 0 0 0 0 I 0 0 0 p2I o p6I 0 0 0 0 I 0 0 p5B o 0 0 0 0 0 0 I 0 p6I o 010010000 o 001001000 0000/00100

p2I

A routine computation shows that V E D.(S3, R). If f E End V, then M [(mod p) = al + N for some nilpotent k-matrix N with zeros on the diagonal and al E qA(mod p), B(mod p)) (Exercise 4.2 .2). Then 1T : End V ~ qA(mod p), B(mod p)) = I defined by 1T(f) = al is a ring

r pr

4.2 WildModulo p Representation Type

139

homomorphism. Since C(A, B) embeds in End U (Exercise 4.2.2), n is onto by Lemma 4.2.1(b). This shows that rep(S3, R, 0) has wild modulo p representation type. Recall that if S is a finite poset with w(S) only if S contains (I, 2) as a subposet.

= 2, then S is not a garland if and

Corollary 4.2.5 The category rep(S, R, j) has: (a) wild modulo p representation type if either w(S) ::: 3 or w(S) = 2, S is not a garland, j ::: 7; (b) rank-infinite representation type ifw(S) = 2, S is not a garland, j ::: 4; (c) finite representation type if and only if either w(S) = 1 or S is a garland. PROOF. Statement (a) is a consequence of Corollary 4.2.2, Example 4.2.4, and Lemma 4.1.5 . Assertion (b) follows from Corollary 4.1.9, Example 4.1.10 and Lemma 4.1.5. Statement (c) is Corollary 4.1.7.

For antichains the characterizations are complete; there is only finite or wild modulo p representation type.

Corollary 4.2.6 The category rep( s.. R, j) has (a) wild modulo p representation type ifand only ifn ::: 3, and (b) finite representation type if and only if n ~ 2.

Open Questions 1. Suppose S is a finite poset with w(S) = 2 and S is not a garland. If 6 ::: j ::: 4, does rep(S, R, j) have wild modulo p representation type? In this case, can indecomposable representations in rep(S, R, j) be classified up to isomorphism? 2. For which finite-dimensional algebras I" over a field k = R / p R does there exist an R-algebra r that isfinitely generated and free as an R-module with r/pr = r '? EXERCISES

1. Let G be a finite group,not necessarily abelian, and kG the groupalgebra of G overthe field k = R/ p R. Provethat the group algebra RG of Gover R is an R-algebrathat is finitely generated and free as an R-module with RG/pRG = kG. 2. Complete the proofof Example 4.2.4.

?).

3. Let R be a discrete valuation ring, A = (A f) and B = (A Show that the natural homomorphism C(A, B) ~ C(A(mod p), B(mod p» is not onto. 4. Let rep/(S, R, j) be the full subcategory of rep/(S, R) with objects V = (Vo, Ut : i E S) such that pj Vo £; Ej Ui. Use the results of Sections4.1 and 4.2 to find bounds for the representation type of rep/(S, R, j) in terms of Sand j .

140

4. Representations over a Discrete Valuation Ring

4.3 Finite Rank Butler Groups and Isomorphism at p Computations of representation type for categories of representations of posets over the discrete valuation ring Zp can be transferred directly to isomorphism at p categories of Butler groups (Corollary 4.3.3). Recall that for a finite lattice T of types, B(T) is the category of finite rank Butler groups G with typeset G ~ T, and Sr is the oppos ite of the poset of join irreducible elements of T. For an integer j :::: 0, let B(T, j) denote the full subcategory of B(T) consisting of those groups G with p! G contained in :E{G(r) : r E Sr} . The isomorphism at p category of B(T, j) is denoted by B(T, j)p. Notice that indecomposable groups in B(T, j)p are indecomposable in B(T, j), i.e., indecomposable as groups . Up to rank-I summands and isomorphism at p, there is no loss of generality in considering B(T, j): Proposition 4.3.1 If G E B(T) with no rank-I summands and p is a prime, then G is isomorphic at p to a group in B(T, j)for some nonnegative integer j . Let IT(G) denote the inner type of G. Since G is a Butler group, IT(G) E typeset G by Lemma 3.1.12 and Theorem 3.2.5. Then :E{G(a):a E STl = G*(IT(G» . This holds because if x E G with type x = r > IT(G) and r E T\Sr is not join irreducible, then r > a for some join irreducible, a E Sr and x E G(r) ~ G(a). Since G is a Butler group, G = G(IT(G» = Grr(G) $ G#(IT(G» with Grr(G) an IT(G)-homogeneous completely decomposable group and G#(IT(G»jG*(IT(G» finite (Theorem 3.2.5). But G has no rank-I summands, so that Grr(G) = and GjG*(IT(G» is a finite group. Choose a subgroup H of G with H jG*(IT(G» the subgroup of G jG*(IT(G» consisting of elements with order a power of p. Then G j H is a finite abelian group with order prime to p. Hence, G is isomorphic to H at p. Moreover, for some i . pj(H jG*(IT(G» = 0, so that pj H ~ G*(IT(G» ~ H*(IT(G» ~ Hand H E B(T, j). PROOF.

°

Given a prime p, a finite lattice T of types is p-locally free if X p is isomorphic to Zp for each rank -I group X with type X E T . Restriction to the case that T is p -locally free amounts to the p-reduced case , in that if type X E T and X p is not isomorphic to Zp, then X p is isomorphic to Q (Exercise 2.1.3). A torsion-free abelian group G is p-locally free if G p is a free Zp-module. Corollary 4.3.2 [Richman 95] 1fT is a p-Iocally free finite lattice oftypes and j is a nonnegative integer, then there is a category equivalence F : B(T, j)p ~ rep(Sr, Zp, j) given by F(G) = (G p » G(r)p : r E Sr). PROOF. The proof is analogous to that of Theorem 3.3.2 , bearing in mind that integers prime to p are units in Zp.

4.3 FiniteRank ButlerGroups and Isomorphism at p

141

Corollary 4.3.3 If T is a finite p-locally free finite lattice of types, then B(T,

n, has:

(a) wild modulo p representation type if either W(ST) ~ 3 or W(ST ) = 2, ST is not a garland, j ~ 7; (b) rank-infinite representation type ifw(ST) = 2, ST is not a garland, j ~ 4; and (c) finite representation type ifand only ifeither W(ST ) = lor ST is a garland. PRO OF.

A consequence of Corollaries 4.3.2 and 4.2.5.

Following are illustrations of how representations over discrete valuation rings can be used to construct finite rank Butler groups . If G is a finite rank Butler group with linearly ordered typeset , then G is a completely decomposable group by Corollary 3.2.7(a) . This is, in the isomorphism at p category, a manifestation of the fact that representations of chains over a discrete valuation ring are direct sums of rank-l representations. Observe that a finite lattice of types T is a chain if and only if ST is a chain .

Example 4.3.4 If T is a finite lattice of types and ST is a chain, then rank- I groups are the only indecomposable groups in B(T, j) p. PROOF.

Corollary 4.3.2 and Example 4.1.1

Example 4.3.5 Let T be afinite lattice oftypes with ST = {type X I, type X z} = Szfor subgroups Xi ofQ with 1 E Xi and 1/ p rt. X;/or each i . Indecomposable groups in B(T , j)p are: (i) rank-I groups G = Xl n Xz, Xl, X Z, Xl + X z with (End G)p = Z p, and (ii) rank-2 groups G = Xl $ X z + (1 + I)Z(1/ pi ) with (End G)p = fer, s) E z p x z p: r - s E pi Z p} for each 1 ~ i ~ j .

Apply Corollary 4.3 .2 and Example 4.1.2, noticing that if G is as given in (ii), then (G p- G(r\)p , G(rz)p) = (Zp $ Zp + (1 + I)Zp(l/ pi), Zp $0, 0$ Zp) E rep(Sz , Zp, j). PROOF.

Example 4.3.6 Let p be a prime, 5

/ \

3

T

4

/ \ /

=1

2

\ /

°

142

4. Representations over a Discrete Valuation Ring

a p-Iocally free finite latticeoftypes. and Xi a subgroup ofG. with 1 E Xi, type Xi = i, and 1/ p r/. Xi. Then B(T, O)p has infinite representation type, but each indecomposable has rank less than or equal to 2. Indecomposable groups in B(T, O)p ofrank 2 areoftheform G = X4 EeX3 + X z(1, 1)/ pi for some j :::: 1. Moreover, B(T, j) p has indecomposable groups ofarbitrarily large finite rank for j :::: 4. PROOF.

Notice that Sr

= C~ = 2 I

4 Apply Corollaries 4.3.2 and 4.1.4 to see that indecomposable groups have rank less than orequalto 2. If G is as definedabove, then (G p' G(4)p, G(2)p, G(1)p) = (V o, Zp Ee 0 f Vo,O Ee Zp) with Vo = Zp Ee Zp + Zp(1, I)/pi, as required by Corollary 4.1.4. The last statement of the theoremis a consequenceof Corollaries4.3.2 and4.1.9 and Example 4.1.10. In particular, for r = 5,

G = Xf Ee Xf Ee X~s Ee X~s

+ X4M + (1/ p5)(1 + 1 + 0 + O)Z2s ,

where M is the row space of the matrix pZ[

p3[

p3[

0

0 p5[

000 and A is an s x s Z-matrix such that A(mod p) is an indecomposable Z/pZ-matrix. Notice that Tcr(G) = {type Xi; 1 ::: i ::: 4}. In view of Corollary 4.3.2, the following open question is a group-theoretic version of a special case of open questions in Sections 4.1 and 4.2. Open Question: What is the representation type of B(T, j)p if Sr = (1,2) and 1 ::: j ::: 3 or 6 :::: j :::: 4? Can the indecomposable groupsbe classified in these cases?

EXERCISES

J. Let T be a finite lattice of types with Sr = {type XI, type Xz, type X3}. an antichain, for subgroups Xi of Q with 1/ p ¢ Xi for each i. Provethat the following construction demonstrates that B(T3, 2)p has wild modulo p representation type. Given two n x n Z-matrices A and B, define a subgroup G of Q'1 $ Q'1 $ Q'1 $ Q'1 by G

= Xr $

Xr + (I + 0 + 0 + I)X3+ (0 + 1 + 0 + I)X~

+ (I/pz)(l +0 + A + l)zn + (l/p)(O+ I + B +O)zn.

2. Provide a proof for Corollary 4.3.2.

4.3 NarES ON CHAPTER

Finite Rank Butler Groups and Isomorphism at p

143

4

There is very little literature on representations of posets over discrete valuation rings . The contents of this chapter are distilled from [Arnold, Dugas 97, 99]. The methods , albeit ad hoc, are sufficient to determ ine the representation type of most of the isomorphism at p categories of finite rank Butler groups. In addition , indecomposable representations give rise to groups that are indecomposable in the isomorphism at p category, hence indecomposable as groups . The representation types of some more general categorie s of representations of posets over discrete valuation rings and their factor rings are given in [Plahotnik 76] and [Simson 96].

5 Almost Completely Decomposable Groups

5.1 Characterizations and Properties An almost completely decompo sable group is a torsion-free abelian group G of finite rank quasi-isomorphic to a completely decomposable group. An almost completely decomposable group is a Butler group, by Coroll ary 3.2.4, but any strongly indecomposable Butler group with finite rank greater than I is not almost completely decomposable. Almost completely decomposable groups are quite complicated, as a consequence of the variety of ways that a completely decomposable group can be embedded as a subgroup of bounded index in an almost completely decomposable group. Chapter 5 is a brief introducti on to this technic al subject. Properties of finite rank Butler groups given in Chapter 3 are assumed. In particular, if C is a subgroup of a finite rank Butler group G with G/ C bounded , then G/C is finite by Exercise 2.1.1. Moreover, Tcr(G) = [r : G(r)/G#(r) =f:. O} is the critical typeset of G, and a regulating subgroup of G is B = ~ {Gr : r E Tcr(G)} with G(r) = G r EB G#(r) and G, a r-homogeneous completely decomposable group. The first characterization of almost completely decomposable groups is in terms of the ranks of the G r 'so

Proposition 5.1.1 Afinite rank Butler group G is an almost completely decomposable group if and only if rank G = ~{rank G(r )/G#(r ): r E Tcr(G)}. In this case, each regulating subgroup of G is a completely decomposable group isomorphic to EB{G(r)/G#(r ) : r E Tcr(G)}.

5.1

Characterizations and Properties

145

PROOF. Let G be an almost completely decomposable group and C a completely decomposable subgroup of G with G I C finite. Then rank G = rank C and Ter(G) = Ter(C) . Moreover, G(r)/C(r) and G#(r)/C#(r) are finite, whence rank G(r)/G#(r) = rankC(r)/C#(r) for each type r E Ter(G). But rank C = E{rank C(r)/C#(r) : r E Ter(C)} by Example 3.1.5,so that rank G = E{rankG(r)1 G#(r): r E Ter(G)} . Conversely, assume that G is a finite rank Butler group with rank G = E {rank G(r)/G#(r) : r E Ter(G)}, G(r) = G, E9 G#(r), and B = E{G, : r E Ter(G)} a regulating subgroup of G . Then G I B is finite by Propo sition 3.2.12. Hence, Ter(G) = Ter(B), rank G, = rank G(r)/G#(r) = rank B(r)1 B#(r) for each type r E Ter(G), and E{rank B(r)1 B#(r): r E Ter(B)} = rank B. It follows that B = E9{ G, : r E Ter( G)} is completely decomposable. Since G I B is finite, G is almost completely decomposable. There is a criterion for deciding whether or not a completely decomposable subgroup B of an almost completely decomposable group G is a regulating subgroup of G. Define a positive integer i (G) to be the least element of {I G I CI : C completely decomposable, G I C finite}.

Proposition 5.1.2 [Lady 74B] Let G be an almost completely decomposable group and B a completely decomposable subgroup of G. (a) The group B is a regulating subgroup of G ifand only iflG I BI = i(G) . (b) IfC is a completely decomposable subgroup ofG with GIC finite, then

i(G) divides IG/CI. (c) The group G is completely decomposable ifand only ifi(G) = 1.

(a), (b) The first step is to show that if Band C are distinct regulating subgroups of G, then IGI BI = IGIC I. Write B = ${B, : r E Tcr(G)} and C = ${C, : r E Ter(G)} with G(r) = B, E9 G#(r) = C, E9 G#(r) and B, and C, r-homogeneous completely decomposable groups for each r E Tcr(G) . The proof that IG I BI = IGI CI is an induction on the cardinality of {a E Ter(G) : B" =I- C,,}. Since Band C are distinct, there is some a E Tcr(G) with B" =I- C". Define D = C" E9 (${B, :a =I- r E Ter(G)}) = E9{D, : r E Ter(G)}, another regulating subgroup of G with I{a E Tcr(G) : B" =I- D,,}I < I{a E Ter(G): B" =I- C,,}I. By induction, it is now sufficient to prove that IGIBI = IGIDI . To this end, define E = B + C" , Then E = B + E(a) = D + E(a), since B" and C" are subgroups of E(a). Consequently, lEI BI = jE(a)1 B n E(a)1 = IE(a)1 B(a)1 and, similarly, lEI DI = IE(a)1 D(a)l. Notice that B(a) = E9{B, : r ::: a} and D(a) = C" E9 (E9{B, : r > a}) are regulating subgroups of E(a), and B#(a) = D#(a) = E9{B, : r > a} is a regulating subgroup of E#(a) . Now,

PROOF.

IE(a)1 B(a)1 = I(B" E9 E#(a))/(B" E9 B#(a))1

= IE#(a)1 B#(a)1

146

5. Almost Completely Decomposable Groups

and

= I(Cu ED E#(a»/(Cu ED D#(a»1 = IE#(a)/ D#(a)1 with IE#(a)/ B#(a)1 = IE#(a)/ D#(a)l . Thus, IE(a)/ B(a)1 = IE(a)/ D(a)1 and IE(a)/ D(a)1

IG/BI

= IG/EIIE/BI = IG/EIIE(a)/B(a) 1 = IG/ E I IE(a )/ D(a )1= IG/EIIE/DI = IG/DI ,

as desired. It remains to prove that if B is a completely decomposable subgroup of G and B is not a regulating subgroup of G, then there is a completely decomposable subgroup C of G such that IG/C1 is a proper divisor of IG/ BI. In this case, if B is a completely decomposable subgroup of G with IG/ B I = i (G), then B must be a regulating subgroup of G. Conversely, if B is a regulating subgroup, then B = i(G). Otherwise, there is a completely decomposable subgroup C of G with IG/C1 = i(G) < IG/BI. Thus, C is a regulating subgroup of G with IG/CI:/; IG/BI ,acontradiction. Now assume that B is a completely decomposable subgroup of G and B is not a regulating subgroup of G. Choose a maximal in Tcr(G) such that B(a) = BuEDB#(a)butG(a):/; BuEDG#(a).DefineC = GuED(ED{Br :a:/; r E Tcr(G)}), where G(a) = Gu ED G#(a) and B(r) = B; ED B#(r) for each type r , Then C is a completely decomposable subgroup of G. By the choice of a, B#(a) = C#(a) is a regulating subgroup of G#(a). Define E = C + Bu = B + Gu. Then E = C + D(a) = B + D(a), and as above, IE/CI = IE(a)/C(a)1 and lEI BI = IE(a)1 B(a)l . Now,

IG(a)/C(a)1

= IG(a)/ E(a)IIE(a)/C(a)1

and IG(a)/B(a)1 = IG(a)/E(a)IIE(a)/B(a)l ·

However, IG(a)/CCa)1 = I(Gu ED G#(a»/(G u ED C#(a»1 = I(Gu ED G#(a»/ tG; ED B#(a»1 is a proper divisor of IG(a)/B(a)1 = I(Gu ED G#(a»/(Bu ED B#(a»1 by the choice of a. Hence, IE/CI = IE(a)/C(a)1 is a proper divisor of IE(a)/B(a)1 = IE/BI , and so IG/C1 = IG/EIIE/C1 is a proper divisor of IG/BI

= IG/EIIE/BI·

Assertion (c) is an immediateconsequence of (a).

If Band B' are tworegulating subgroupsof an almostcompletely decomposable

group G, then IG/BI = IG/B'I = i(G) by Proposition 5.I.2(a). However, the following example shows that G/ B need not be isomorphicto G/ B' . Example 5.1.3 [Lady 74B] Let P be a prime and choose rank-l subgroups X), X2, X3, X4 of Q such that 1 E Xi, lip ¢ Xi, type X), typeX 2, and type X3

5.1 Characterizations and Properties

147

are pairwise incomparable, and type X4 = type XI n X2 1: type X3. Define G = Xl EEl X2 EEl X3 EEl X4 + Z(1, 1,0, O)lp + Z(O, 0,1 ,1)1 p. Then (a) B = XIEElX 2EElX3EElX4 isa regulating subgroup ofG with G I B isomorphic to ZI pZ EEl ZI pZ, and (b) if Y = X 4(1, 1,0, p)1p, then C = XI EEl X2 EEl X3 EEl Y is also a regulating subgroup of G with G I C isomorphicto ZI p2Z. PROOF. (a) Observe that G is an almost completelydecomposablegroup. Let ri = type Xi . Then r3 and r4 are incomparable, since r4 1: r3 by assumption and r3 ~ r4 ~ rl is a contradiction. Hence, G(ri) = Xi with G#(ri) = for i = 1,2,3. Since r4 = rl n r2, G(r4) = X4 EEl G#(r4) with G#(r4) = (XI EEl X2 ) + Z(1, 1,0,0)1 p. This shows that B is a regulating subgroup of G. Clearly, G IBis isomorphic to ZI pZ EEl ZI pZ . (b) On the other hand,

°

G(r4) = (X 4 EEl XI EEl X2) + Z(1, 1,0, O)lp = YEEl«X I EEl X2) + Z (1, 1,0,0)lp)= YEElG#(r4),

observing that (1, 1,0, p)lp = (1, 1,0,0)lp + (0,0,0, 1) E G. Hence, C is a regulating subgroup of G. Finally, G I C = ZI p2Z, since G I C is generated by x = (0, 0, I, 1)1 p + C with order x = p2. This holds because px = (0,0, 0, 1) + C = -(1, 1,0 ,0)1 p + C with (0,0,0,1) rt C, and p2x = p«O, 0, 0, 1) + C) = (O,O ,O,p)EC. Corollary 5.1.4 [Lady 74B] Suppose G and H are almost completelydecom-

posable groups. (a) IfG is nearly isomorphicto H, then i(G) = i(H). (b) IfG = HEll K , then i(G) = i(H)i(K).

PROOF. (a) Choose a monomorphism f: G -+ H with IHlf(G)1 relativelyprime to i(H). By Proposition 5.1.2(b), i(H) divides HI f(B), for a regulating subgroup B ofG, since f(B) is a completelydecomposablesubgroupof H. But IHlf(B)1 = IHlf(G)llf(G)lf(B)I, so i(H) divides If(G)lf(B)1 = IGIBI = i(G). A symmetric argument shows that i(G) divides i(H). (b) If C = EEl{C r : r E Tcr(H)} is a regulating subgroup of Hand D = EEl{D r : r E Tcr(K)} is a regulating subgroup of K , then B = C EEl D is a regulating subgroup of G. This holds because G(r) = H(r) EEl K(r) = Cr EEl Dr EEl H#(r) EEl K#(r) = C, EEl Dr EEl G#(r) for each type r. By (a), i(G) = IGI BI = IHI CIIK I DI = i(H)i(K). The exponent of a bounded abelian group A, written e = exp A, is the least positive integer e with e A = 0. If G is a finite rank Butler group, then G I R( G) is finiteby Corollary 3.2.13(a), where R(G) is the intersectionof all of the regulating subgroups of G.

148

5. Almost Completely Decomposable Groups

Define eG = exp G/ R(G) and br(G) = exp G#(r)/ R(G#(r» for each type r, adoptingthe convention that s.t G) = 1if G#(r ) = O. Noticethat G#(r)/R(G#(r» is finite for each type r , since G#(r ) is a Butler group, beinga pure subgroupof the Butlergroup G. Moreover, br(G) is the least commonmultipleof{exp G#(r)/ D : D regulating subgroup of G#(r)}, since R(G#(r» = n{D : D regulating subgroup of G#(r)}. The br(G)'s are called the Burkhardt invariants of G. The following lemma gives a procedure for constructing R(G) from the Burkhardt invariants and a given regulating subgroup of an almost completely decomposable group G. Lemma 5.1.5 [Burkhardt 84] Let G be an almost completely decomposable group and B = EI7{B r : r E Tcr(G)} a regulating subgroupofG. Then R(G) = EI7{b r(G)Br : r E Tcr(G)}. Write br for br(G) and let C = EI7{ Cr : r E Tcr(G)} be an arbitraryregulating subgroup of G. Then G(r) = B, EI7 G#(r) = C, EI7 G#(r) for each r E Tcr(G). If r E Tcr(G), then EI7{CO': r < a} is a regulating subgroup of G#(r), since G(a) = G(r)(a) if r < a . Then b.B; ~ brG(r) = brC r EI7 brG#(r) ~ b.C, EI7 R(G#(r» ~ Cr EI7 (EI7{Ca : r < a}) = C(r) ~ C, using the fact that brG#(r) ~ R(G#(r» . Since C is an arbitrary regulating subgroup of G, EI7{b rBr : r E Tcr(G)} ~ R(G). Conversely, let x E R(G)\(EI7{b rBr : r E Tcr(G)}). Since R(G) ~ B, it suffices to assume that x = EI7{x r E B, \b r B; : rES} for some subset S of Tcr(G). In particular, b, =I- 1 for each rES. Let r be a maximal element of Sand write B r = AI EI7 ... EI7 An with each Ai a rank-I group of type t . Then X r = al EI7 . .. EI7 an for some a; E A i. Because x, rt brBr, there is some i, say i = 1, with a, rt b.B«. Since b, = exp G#(r)/ R(G#(r», there is some a E G#(r) with order (a + R(G#(r» = br =I- 1 in G#(r)/ R(G#(r» . Let D = EI7{D r : r E Tcr(G#(r»} be an arbitrary regulating subgroup of G#(r) and define C = EI7{CO' : a E Tcr(G)}, where Cr = A~ EI7 A2 EB··· EB An with A ~ the pure rank-I subgroup of G generated by al + a , CO' = DO' if a > r, and CO' = BO' if a t r. Then C is a regulating subgroup of G, so that x E R(G) ~ C, say x = EB{cO' E C u : a E Tcr(G)}. Now,O = x - x = I:{xu - CO' : a E Tcr(G)} , so that Xr - Cr = I:{cu - Xu : a =Ir] E G(r). Since r is maximal in S, X a = 0 if a > rand X r - Cr = I:{cu : a > r] E G#(r). Also, X r - Cr = (al EB · . . EBa n ) - (a~ El7a~ EI7 . .. EB a~) E Br EI7 Cr with a; E A 2 ~ B; for 2 ~ i anda; = ql(al+a)forsomeql E Q. Thenxr-cr+qla = (al -qla»EI7(EB{aj -a;:2 ~ i ~ n}) = I:{cu :a > r}+q la E B; nG#(r) = 0, since (al - qlal) EB (EI7{aj - a;: 2 ~ i ~ n}) E B; and a, cuG#(r). Consequently, al = qlaJ, ql = I, and a = I:{-cu :a > r} E D , recalling that CO' = DO' if a > r . Since D is an arbitrary regulating subgroup of G#(r), a E R(G#(r», a contradiction to the choice of a. PROOF.

Theorem 5.1.6 (a) [Burkhardt 84] If B = EB{G r : r E Tcr(G)} is a regulating subgroup of an almost completelydecomposable group G, then R(G) = I:{br(G)G(r):

5.1 Characterizations and Properties

149

r E Ter(G)} is a completely decomposable fully invariant subgroup

ofG. (b) [Mader, Vinsonhaler 94] If G and H are nearly isomorphic almost completelydecomposablegroups, then R( G) is isomorphic to R(H) and G/ R( G)

is isomorphic to H/ R(H). PROOF. (a) By Lemma 5.1.5, R(G) = ${brB r : r E Ter(G)}. Hence, R(G) is a completely decomposable subgroup of G contained in 'E{brG(r) : r E Ter(G)}. Conversely, in view of the definition of b, and Lemma 5.1.5, if r E Ter(G), then brG(r) = b.B; $ brG#(r) S; b.B; $ R(G#(r» S; R(G) as a consequence of Exercise 5.1.12(b). Consequently, 'E{brG(r): r E Ter(G)} S; R(G), as desired. If f E End G, then f(R(G» S; R(G) , since R(G) = 'E{brG(r): r E Ter(G)}, b r is an integer, and f(G(r» S; G(r) for each r. This shows that R(G) is a fully invariant subgroup of G. (b) Since G/ R(G) is finite, G is quasi-isomorphic to R(G). But G is nearly isomorphic to H, so that R(G) is quasi-isomorphic to R(H) and Ter(G) = Ter(H). By (a), R(G) and R(H) are completely decomposable groups, whence R(G) is isomorphic to R(H) by Theorem 3.1.7(b). In view of Theorem 2.2.2(a), it is sufficient to assume that there is an integer m relatively prime to ea = exp G/ R( G) with m H S; G S; H. Then m H (r) S; G(r) S; H(r) for each type rand br(G) = br(H) for each r E Ter(G). Hence, mR(H) S; R(G) = 'E{br(G)G(r) : r E Ter(G)} S; R(H) = 'E{br(H)H(r): r E Ter(G)}. Since eG = exp G/ R(G) is relatively prime to m, G/ R(G) is isomorphic to H/R(H).

If Hand K are almost completely decomposable groups, then H $ K is an almost completely decomposable group with R(H $ K) S; R(H) $ R(K) (Exercise 5.1.12(a» . However, the group constructed in Example 5.1.3 is an example of an almost completely decomposable group H$K with R(H$K) # R(H)$R(K). Example 5.1.3 (continued) Let G be as defined in Example 5.1.3. Then (c) G = H$K, where H = (X I$X2)+ Z(l, 1,0, O)/pand K =(X3 $ X 4) + Z(O, 0, I , I)/p; (d) R(H) = X I $ X 2 and R(K) = X 3 $ X4; and (e) R(G) = XI $ X2 $ X3 $ pX 4 • Statement (c) is immediate from the definition of G . As for (d), the critical typeset of H is {rl, r2} with rl and r2 incomparable and H(ri) = Xi. Then R(H) = XI $ X 2 . Alternatively, R#(H(ri» = 0, so that the Burkhardt invariants for H are b, = I for r = rio Hence , R(H) = XI $ X2 by Theorem 5.1.6(a). Similarly, R(K) = X 3 $ X 4 • (e) In this case, there is not a unique regulating subgroup of G , by Example 5.1.3(a) and (b). However, the Burkhardt invariants b, of G can still be computed. Now, Ter(G) = {rl , r2, r3, r4} with G#(ri) = for i = 1,2,3 and

PROOF.

°

150

5. Almost Completely Decomposable Groups

G#(r4) = (XI EB X 2 ) + Z(l , 1,0, O)/p. Hence, b r = 1 for r = ti, 1 ~ i ~ 3, and b r = p for r = r4. By Theorem 5.1.6(a), R(G) = I:{brG(r): r E Tcr(G)} =

XI EB X2 EB X3 EB pX4 •

The next examplecompletesthe computation of the integer n( G) for an almost completeIy decomposable group G mentioned just after the proofof Lemma2.2.7. Example5.1.7 If G is an almost completely decomposable group and eG is the exponent of G/ R(G), then eG = neG). PROOF. Since C = R(G) is a fully invariant completely decomposable subgroup of G, restriction induces a ring embedding ec End C £; End G £; End C by Theorem 5.1.6(a). Then eG(End C/ NEnd C) £; End G/ NEnd G £; End C/ NEnd C. By Proposition 3.1.8, End C/NEnd C is isomorphic to Matn(l)(R 1) x . .. x Matn(m)(R m) with each R, a subring of Q. Since a subring of Q is a principal ideal domain, each Matn(i)( R i ) is a maximalorder over Ri, as noted in Section2.2. Apply the definition of neG) to complete the proof. Parts (b), (c), and (d) of the nextcorollaryare consequences of Corollary2.2.11, the proof of which uses the fact that maximal orders are hereditary. A grouptheoretic proof for the special case of almost completely decomposable groups is a consequence of Exercise 5.104 and the proof of Corollary 2.2.11 . Corollary5.1.8 Suppose G and H are almost completely decomposable groups. (a) The groups G and H are nearly isomorphic ifand only ifthere is a completely decomposable group C with G EB C isomorphic to H EB C. In this case, C

may be chosen to be isomorphic to both R(G) and R(H). (b) The group H is a near summand of G if and only if G X EB X' for some X and X' with X nearly isomorphic to H . In particular, G is indecomposable if and only if and G are the only near summands of G. (c) IfG EB K is nearly isomorphic to H EB K for some torsion -free abelian group K offinite rank, then G is nearly isomorphic to H. (d) JfG is nearly isomorphic to GI EB · ·· EB G«. then G = HI EB·· · EB H; with each Gi nearly isomorphic to Hi.

°

=

PROOF. (a) SupposeG and H are nearlyisomorphic. Then C = R( G) is completely decomposable, C is isomorphic to R(H) by Theorem 5.1.6(b), and eG £; C with e = exp G/ R( G). By Theorem 2.2.2(a), there is f E Hom(G, H) and g E Hom(H, G) and a nonzero integer m prime to e such that gf = mIG and fg = mlH. Write 1 = rm + se for integers rand s and define ¢ : G ~ HEBe by ¢(x) = (fx, ex) and a: H EB C ~ G by a(x , y) = rgx + sy . Then a¢ = IG, so that H EB C is isomorphicto G EB D for some D. Since G and H are nearlyisomorphic, C is nearlyisomorphicto D by Corollary 2.2A(a). Thus, D is isomorphic to a summand of the completely decomposable

5.1 Characterizations and Properties

151

group C E9 C , by Corollary 2.2.6, hence completely decomposable by Theorem 3.1.7(a). Therefore, C is isomorphic to D, since quasi-isomorphic completely decomposable groups are isomorphic by Theorem 3.1.7(b). Conversely, if G E9 C is isomorphic to H E9 C , then G is nearly isomorphic to H by Corollary 2.2.4(a). Statement (b) is a consequence of Corollary 2.2.11, (c) is a special case of Corollary 2.2.4, and (d) follows from (b) and (c). There are examples of almost completely decomposable groups G and H with no rank-I summands such that G E9 H has a rank-I summand (Exercise 5.1.7). However, the next lemma demonstrates that an almost completely decomposable group contains a maximal completely decomposable summand that is unique up to isomorphism. Proposition 5.1.9 [Lady 74B] Assume that G is an almost completely decomposable group. (a) Then G = C E9 H , where C is completely decomposable and H has no

rank-l summands. (b) If G = C' E9 H ' is another such decomposition of G, then C is isomorphic to C' and H is nearly isomorphic to H '.

(a) Since G has finite rank, it follows immediately that G has such a decomposition. (b) First assume that C and C' are r-homogeneous completely decomposable. Then H n C' is a pure subgroup, hence a r-homogeneous completely decomposable summand, of C' by Lemma 3.1.6. Write C' = (H n C') E9 D . Then G = C' E9 H ' = (H n C') E9 D E9 H' , from which it follows that H n C' is a summand of H . But H has no rank-I summands, whence H n C' = o. Let f : G ~ C be a projection with ker f = H. Since H n C' = 0, f : C' ~ C must be a monomorphism. Similarly, there is a monomorphism g: C ~ C'. This shows that rank C = rank C'. Since C and C' are r-homogeneous completely decomposable groups of the same rank, they must be isomorphic. Moreover, H is nearly isomorphic to H ' by Corollary 5.1.8(c) . For the general case, let r be a maximal type in the union of the typeset of C and the typeset of C'. Then C(r) and C'(r) are r-homogeneous completely decomposable groups by the choice of r. Write C = C(r)E9 D and C' = C'(r) E9 D'. Then G = C(r) E9 (D E9 H) = C'(r) E9 (D' E9 H'), and the groups D E9 Hand D' E9 H' have no rank-I summands of type r. The argument of the preceding paragraph shows that C(r) is isomorphic to C'(r) and D E9 H is nearly isomorphic to D' E9 H' with D and D' completely decomposable. In view of Corollary 5.l.8(d), it suffices to assume that D E9 H = D' E9 H'. By induction on the rank of G, D is isomorphic to D' and H is nearly isomorphic to H ' . Thus , C = C(r) E9 D is isomorphic to C' = C'(r ) E9 D',

PROOF.

152

5. Almost Completely Decomposable Groups

There is a characterization,in terms of thejoin irreducibleelements ST, of those finite lattices T such that each finite rank Butler group with typeset contained in T is almost completely decomposable. Corollary 5.1.10 Let T be a finite lattice of types. Each group in B(T) is an almost completely decomposable group if and only if W(ST) :::: 2.

If W(ST) ::: 3, then there is a strongly indecomposable group G in B(T) with rank G = 2 by Example 3.3.3. In this case, G cannot be quasi-isomorphic to a completely decomposable group. Conversely, if W(ST) :::: 2, then each strongly indecomposable group in B(T) has rank 1, as a consequence of Theorem 3.3.2 and Example 1.1.3. Hence, each G in B(T) is quasi-isomorphic .to a completely decomposable group. PROOF.

The remainderof this sectionis devotedto thecase that T is a finitelatticeof types with ST = S2 = {r, a}. If G E B(T), then G is almost completely decomposable by Corollary5.1.10. If, in addition, G has no rank-l summandsand G(r)nG(a) = 0, then Tcr(G) = S2 and R(G) = G(r) $ G(a) is a unique regulating subgroup of G. It followsfromTheorem 3.4.5 that G = GI $ . . . $ Gn is a finitedirect sum of indecomposablegroups of rank 2. Moreover, R(G) = G(r) $ G(o;) = (GI(r) $ GI(a»$ '" $(Gn(r) $ Gn(a» = R(G I)$ '" $R(G n) with GjR(G) = Gti R(Gr) $ . .. $ Gnj R(G n) and each G;/ R(Gi) a finite cyclic group. The following proposition is the converse in that a decomposition of the finite group G j R( G) intoadirectsum of cyclicgroupsgivesrise to acorrespondingdirect sumdecomposition of G intoindecomposable groupsof rank lessthan or equalto 2. Proposition 5.1.11 [Mader, Vinsonhaler95] Suppose T is afinite latticeoftypes with ST = [r, a} . LetG E B(T) with no rank-l summandsandG(r)nG(a) = o and write G j R(G) = CI E9 .. . $ Cn with each C, a cyclic group. Then

G = G I E9 . . . $ G; for some indecomposable almost completely decomposable group G, of rank 2 with G i j R(G i) = C, for each 1 :::: i :::: n.

PROOF. Assume that C, is a cyclic group of order ai and let X and Y be subgroups of Q containing 1 with type X = r and type Y = a. Choose Xi E G(r) and Yi E G(a) with z, = (Xi + Yi)jai E G, z. + R(G) a generator of Ct. {XI, X2, ... ,X n, YI, , Yn} a Z-independent subset of G, and C = (XXI $ . .. $ Xx n) $ (YYI $ $ YYn) a pure subgroup, hence a summand, of R(G) = G(r) $ G(a). This can be done inductively, observing that G(r) and G(a) are non-zero homogeneous completely decomposable groups with G(r) n G(a) = 0 and by recalling that G has no rank-l summands. Define Gi = (XXi $ YYi) + ZZi, a subgroup of G. Then G;/ R(G i) = C, and (G I $ . . '$G n)+ R(G) = Gas a consequenceofthefactthat each Zi is a generator of Ci' Now, C = (XXI $ . .. $ Xx n) $ (YYI $ . . . $ YYn) = C(r) $ C(a) is a summand of R(G) = G(r) $ G(a), say R(G) = C $ D. Since GjC = Gj R(G), it follows that D is a summand of G. But G has no rank-l summands, whence

5.1

Characterizations and Properties

153

R(G) = C ~ G I EB . .. EB Gn . Thus, G = G I EB .. . EB G; with rank G, = 2 and G;/ R(G j ) = Ci, as desired.

Corollary 5.1.12 Suppose G is afinite rank Butler group with typeset ~ Tz = {r n a , r , a} and r and a incomparable. Then G is a finite direct sum of indecomposable groups with rank less than or equal to 2. PROOF. By Proposition 5.1.9, it is sufficient to assume that G has no rank-1 summands. In view of the hypotheses, G(r U a) = G(r) n G(a) = O. Since the finite group G I R(G) can be written as a direct sum of cyclic groups, Proposition 5.1.11 applies. Proposition 5.1.11 can be used to demonstrate that direct sum decompositions of almost completely decomposable groups are not unique up to near-isomorphism even if the critical typeset has 2 elements. As an illustration, suppose G satisfies the hypotheses of Proposition 5.1.11 with G I R(G) = Z/30Z . Then G I R(G) = Z/2Z EB Z/3Z EB Z/6Z = Z/6Z EB Z/5Z = Z/lOZ EB Z/3Z = Z/15Z EB Z/2Z. Each of these decompositions of G I R( G) into cyclic groups induces a direct sum decomposition of G into indecomposable groups of rank 2. These decompositions are not unique up to near-isomorphism by Theorem 5.1.6(b), since if X and X' are nearly isomorphic summands of G, then X I R(X) is isomorphic to X'I R(X'). On the other hand, if G I R(G) is a p-group for some prime p, necessarily finite, then any two decompositions of G I R( G) into cyclic groups are unique up to isomorphism and order. This uniqueness carries over to G , since it is shown in Corollary 5.4.3 that if G I R(G) is a p-group, then direct sum decompositions of an almost completely decomposable group G into indecomposable groups are unique up to isomorphism at p.

EXERCISES

1. Prove that if G is an indecomposable finite rank Butler group with critical typeset G ~ Tz = Iri . rz} and rank G = 2, then G is isomorphic to XI ffi Xz + Z(I, m)/n, where m and n are relatively prime integers with n > I and X, is a subgroup of iQ containing 1 with type Xj = rj. 2. Find a complete set of near isomorphism invariants for finite rank Butler groups G with typeset G ~ h 3. [Cruddis 70], [Amold, Dugas 93B] Let G be a finite rank Butler group with Tcr(G) ~ T3 = {rl, rz, r3} and rl + rz + r3 = type iQ. Prove that G is the direct sum of groups of rank less than or equal to 2. Find a complete set of near-isomorphism invariants for G. 4. Let G be an almost completely decomposable group with eGG ~ R(G) ~ G and K a near summand of G. Prove that if g: G --+ K, then there is h: g(G) --+ K with ghg = eag , Conclude that X = g(G) is a summand of G nearly isomorphic to K . 5. Let G be an almost completely decomposable group with eGG ~ R(G) ~ G and D the unique maximal subgroup of G with eGD = D. Write R(G) = ffi{A r : r E

154

5. Almost Completely Decomposable Groups Tcr(G)}, A r a r-homogeneous completely decomposable group . Prove that: (a) D E9{A, : r E Tcr(G), eGA, A,}; and

= =

=

D E9 E , where E is the pure subgroup of G generated by E9{A, : r E Tcr(G), eGA, =1= A r }.

(b) G

6. Assume that G is an almost completely decomposable group with Tcr(G) an inverted forest. Compute the Burkhardt invariants of G .

=

7. (a) [Jonsson 57] Give an example of an almost completely decomposable group G C E9 H = C ' E9 H ' such that C and C ' are completely decomposable groups , H and H' have no rank-I summands, but H and H ' are not isomorphic. (b) [Jonsson 57] Give an example of two almost completely decomposable groups G and H with no rank-I summands such that G E9 H has a rank-I summand. 8. Find almost completely decomposable groups G and Hand that f(R(G» is not contained in R(H).

f

E Hom(G , H) such

=

9. [Mader 95] Let G be an almost completely decomposable group and define G(O) G, G(I) = :E{G#(r) : r E Tcr(G)}, and, inductively, G(n+l) = (G(n»)O). (a) Prove that G(n) 0 for some n and that G(O) 2 G(1) 2 . .. 2 G(n) 0 is a descending chain of purely invariant subgroups of G. (b) Prove that G(IT(G»/G(1) is isomorphic to E9{G(r)/G#(r) :r is minimal in

=

=

Tcr(G)} .

(c) Prove that i(G) = IG/G(IT(G))i IG(1) /G(1)(IT(G(1»I · · · IG(n)/G(n)(IT(G(n»I. 10. [Burkhardt 84] Let G be an almost completely decomposable group and C a completely decomposable subgroup of G with G/ C finite. Prove that C is a regulating subgroup of G if and only if there is a decomposition C = E9{ C, : r E Tcr (G)} such that each for each r, C, 5; (expG#(r)/C#(r»G(r) 5; C(r). II . [Burkhardt 84] Let G be an almost completely decomposable group . Prove that if G 7l/ p n71 and G/ C (71/ p71)n, has regulating subgroups Band C such that G/ B then for each finite abelian group A with IAI = pn there is a regulating subgroup D ofG with G/D A.

=

=

=

12. Let G and H be almost completely decomposable group s. (a) Prove that R(G E9 H) 5; R(G) E9 R(H). (b) Prove that if r E Tcr(G), then R(G#(r» 5; R(G) .

5.2

Isomorphism at p and Representation Type

The representation type of the category of almost completely decomposable groups in B(T, j) is computed in Corollaries 5.2.4 and 5.2.9 by determining the representation type of a subcategory of rep(ST , Zp, j). Indecomposable representations in rep(ST, Zp, j) can be used to construct indecomposable almost completely decomposable groups in B(T, j). Let T be a finite lattice of types, j a nonnegative integer, and C(T, j) the full subcategory of B(T, j) consisting of almost completely decomposable groups. Specifically, the groups in CCT, j) are almost completely decomposable groups G with typeset G 5;; T and p! G 5;; b {G( r ) : rEST} 5;; G . Since ST consists

5.2 Isomorphism at p and Representation Type

155

of the join-irreducible elements of T, :E{G(r): r E Sr} = G *(IT(G». Henceforth , it is assumed that j ::: 1, since C(T, 0) consists only of almost completely decomposable groups (Exercise 5.2.2). For almost completely decomposable groups, there is little loss of generality, up to isomorphism at p, in assuming that T is a p -locally free lattice of types . By Proposition 5.1.9, an almost completely decomposable group G is the direct sum of a completely decomposable group C and an almost completely decomposable group H with no rank-l summands. Moreover, H is unique up to near isomorphism. As a consequence of Proposition 4.3 .1, if G is an almost completely decomposable group with no rank-l summands, the typeset of G is contained in T , and p is a prime, then G is isomorphic at p to some H E C(T, j) with en = exp H j R(H) a power of p . If D is the maximal p-divisible subgroup of H , then D is completely decomposable and H = DEBE for some E with typeset E a p-locally free set of types (Exercise 5.1.5). Hence, if H is indecomposable at p with rank H > 1, then H E C(T' , j) for some finite p-locally free lattice T ' . Recall from Corollary 4.3.2 that if T is a p -locally free finite lattice of types and j ::: 1 an integer, then there is a category equivalence Fr : B(T, j)p ~ rep(Sr, Zp, j) given by Fr(G) = (G p: G(r)p: r E Sr). Define Crep(Sr , Zp , j) to be the full subcategory ofrep(S, Zp, j) with objects of the form Fr(G) , G E C(T, j). Clearly, the categories C(T, j)p and Crep(Sr, Zp, j) are equivalent categories.

Lemma 5.2.1 1fT is afinite lattice, T' is a sublattice ofT, and j' :::: j, then there is a fully faithful functor F: Crep(Sr', R , j') ~ Crep(Sr , R, j). PROOF. A consequence of the definition of Crep(Sr, Zp , j) and the observation that C(T', j) is contained in C(T, j). Specifically, F = Fr(FT'r l • It is not all that easy to identify those representations in rep (Sr, Zp , j ) that are in Crep(Sr , Zp, j); see Exercise 5.2 .1. By Proposition 5.1.1 , G E B(T) is an almost completely decomposable group if and only if rank G = :E{rank G(r)j G#(r) : r E Tcr ( G)}. However, Tcr ( G) need not be contained in Sr, the join irreducible elements of T . For example, let T be a Boolean algebra with exactly 3 atoms. Then Sr consists of these three atoms . If G is an almost completely decomposable group with critical typeset consisting of three pairwise incomparable elements of T that are not atoms, then the critical typeset of G is clearly not contained in Sr . Define Ccrit(T, j) to be the category consisting of all almost completely decomposable groups in C(T, j) with critical typeset contained in Sr , the set of join irreducible elements of T. The image of the functor Fr on groups in Ccrit(T, j), denoted by CrePcrit(Sr , Zp, j), consists of those representations V = (Vo, Vi : i E Sr) E Crep(Sr, Zp, j) such that rank Vo = :E{rank Ui] Vr : i E Sr}, where Vr is the purification of :E{Vs : s < i in Sr} in Vi , as a consequence of Proposition 5.1 .1 and the definition of Fr. For each such representation, Vo is a finitely generated free Zp-module and each Vi is a summand of Vo.

156

5. Almost Completely Decomposable Groups

Notice that if Crepcrit(Sr , 7lp , j) has infinite or wild modulo p representation type, then Crep(Sr, 7lp , j) has, respectively, infinite or wild modulo p representation type by Lemma 5.2.1. This is because CrePcrit(Sr, 7lp , j) embeds as a full subcategory of Crep(Sr, 7lp , j). The representations constructed in the following examples are in CrePcrit(Sr, 7lp , j), hence automatically in Crep(Sr, 7lp , j).

Example 5.2.2 Let T be a p-locallyfree finite lattice oftypes. If either w(Sr) = 5, j ::: 1; w(Sr)::: 4 and j ::: 2; or w(Sr) = 3 and j ::: 3, then Crep(Sr, 7lp , j) has wild modulo p representation type. Let r be a 7lp-algebra that is finitely generated and free as an 7lp-module and A and B two n x n 7lp-matrices with C(A, B) = I', guaranteed by Lemma 4.2.I(a). (i) w(Sr) = 5 and j ::: 1. By Lemma 5.2.1, it is sufficientto prove that CrepeSs, 7lp , 1)haswildmoduloprepresentationtype.DefineU=(Uo, UI, UZ, U3, U4, Us) E Crep(Ss, 7lp , 1) by PROOF.

Uo = 7l; EI7 7l; EI7 7l; EI7 7l; EI7 7l;

+ (II p)(1 + 0 + 1 + A + B)71; ,

+ (llp)(O + 1 + 1 + 1 + 1)71;, UI = 7l; EI7 0 EI7 0 EI7 0 EI7 0, U2 = 0 EI7 7l; EI7 0 EI7 EI7 0, U3 = EI7 0 EI7 7l; EI7 EI7 0, U4 = EI7 EI7 EI7 7l; EI7 0,

°

°

Us = 0 EI7 0 EI7 0 EI7 0 EI7 7l; .

°°°

°

As a consequence of Lemma 4.2.3(a), End U is isomorphic to End V, with V = (Vo , UI, Uz, U3, U4, Us, pUo) E rep/(S6, 7lp ) and Vo = UI EI7 U2 EI7 U3 EI7 U4 EI7 Us = 7l~n . Moreover, pll« = pVo + W , where W is the free 7lp-module with the rows of the 7lp-matrix M = (~I~I~I~I~) as a basis and I an n x n identity matrix. To see that there is an onto ring homomorphism End V ~ C(A(mod p), B(modp)), let f E End V . Then f can be written as (f1,h,h,f4,/s) with each fi a 7lp-endomorphism of Ui, and f(pUo) S; pll«. In matrix terms, fM == M g(mod p) for some g. This equation implies that h == h == f4 == fs(mod p) (from the second row) and fl == h(mod p) with itA == Af4(mod p) and fIB == Bfs(modp) from the first row). Hence, f(modp) = (a,a,a,a,a), with a E C(A(mod p), B(mod p)). Then ¢ : End V ~ C(A(mod p) , B(mod p)), given by ¢(f) = a, is a ring homomorphism. Now, C(A, B) embeds in End V, viaa H- (a, a, a, a, a), whence ¢ is onto by Lemma 4.2.1. Since End U is isomorphic to End V, CrepeSS, 7lp' 1) has wild modulo p representation type. (ii) w(Sr) ::: 4 and j ::: 2. Onceagain,it is sufficient toprovethat Crep(S4, 7lp , 2) has wild modulo p representation type. Define U = (Uo, UI, U2, U3, U4) E

5.2 Isomorphism at p and Representation Type

V o = Z; ED Z; ED Z; ED Z;

157

+ (1/p2)(1 + I + A + B)Z;

+ (1/p)(O + I + I + 1)Z;, VI

= Z; ED 0 ED 0 ED 0, V 2 = 0 ED Z; ED 0 ED 0,

V 3 = 0 ED 0 ED Z; ED 0, and V 4 = 0 ED 0 ED 0 ED Z;. By Lemma 4.2.3(a), End V is isomorphic to End V, with V = (Vo, VI, V 2, V3, V 4, p2Vo) E repJ(Ss, Zp) and Vo = VI ED V 2 ED V3 ED V 4 = z~n. Then p2Vo = p2Vo + W, where W is the free Zp-module with the rows of the Zp-matrix

M_ I I pIAI pIB) - (I0 I pI as a basis. If I E End V , then I = (fl, h. h. 14) with each Ii a Zp-endomorphism of Vi and 1M == Mg(mod p2) for some g. This equation implies, from the first row of M, that II == h(mod p), f3A == A!I(mod p), and 14B == BII(mod p) . From the second row, h == 13 == 14(mod p). Hence, I(mod p) = (a, a, a, a) with a E C(A(mod p), B(mod p». Then ¢: End V --+ C(A(mod p), B(mod p) given by ¢(f) = a is a ring homomorphism and C(A, B) embeds in End V, via a r-+ (a , a , a , a). Thus, ¢ is onto by Lemma 4.2.1. Since End V is isomorphic to End V, rep(S4 , Z p, 2) has wild modulo p representation type, as desired. (iii) W(ST) = 3, j :::: 3. Define V E Crep(S3, Zp, 3) by Uo = z~n ED z~n ED z~n VI = z~n

+ (1/ p3)V*,

ED 0 ED 0, V2 = 0 ED z~n ED 0, V3 = 0 ED 0 ED z~n ,

and V* the free Zp-module with the rows ofthe Zp-matrix

M

I

=

0

(

o o

0 pI 0 0

0 pI 21 _p 0 pI o 0 o

Then End V is isomorphic to End V, where V = (Vo, VI, V2, V3, p3Vo) and Vo = VI ED V2 ED V3· If I E End V, then 1= (fl , h. 13) with Ii: Vi --+ Vi and 1M == Mg(mod p3) for some g. Write !I = (aij), h = (bij), and 13 = (cij) as 3 x 3 matrices with each aij, bij, and Cij an n x n Zp-matrix. The equation, 1M == M g(mod p3), yields, from the first row of M, all == b ll == c22(mod p) and a!2, an , C2h C23 all == 0 modulo p. Continuing with each row of M leads to the conclusion that a., == bu == cii(mod p) = a for each 1 ~ i ~ 3 with

158

a

E

5. Almost Completely Decomposable Groups

C(A(mod p), B(mod p)) and f(mod p)

N=

0 0 ( ail

= al + N with

0 0 0 b~2 0000 ai2 0 0 0

b~3 0 0 0) 0000 0 0 0 0

a nilpotent 7l/p71-matrix. Complete details of this computation are given in [Arnold, Dugas 97, Appendix] . The ring homomorphism if> : End V -;. C(A(mod p) , B(mod p)), defined by if>(f) = a, is onto by Lemma 4.2.1, since C(A , B) embeds in End V . Since End U is isomorphic to End V, Crep(S3 , 7lp , 3) has wild modulo p representation type. Finally, apply Lemma 5.2.1 to see that Crep(ST , 7lp , j) has wild modulo p representation type for W(ST) = 3 and j ::: 3.

Example 5.2.3 Let T be a p-locally free finite lattice of types. If W(ST) = 4 and j ::: 1 or w(Sr) ::: 3 and j ::: 2, then Crep(ST, 7lp , j) has rank-infinite representation type. Let A be an n x n 7lp-matrix with A(mod p) an indecomposable Z/ p71matrix. (i) W(ST) = 4 and j ::: 1. Define U = (Uo , UI, U2 , U3 , U4) E Crep(S4, 7lp , 1) by

PROOF.

Uo

UI

= 7l; EB 7l; EB 7l; EB71; + (1/p)(1 + 0+ 1+ l)71; + (1/p)(O + 1+ 1 + A)71; , = Z; EB 0 EB EB 0, U2 = EB 7l; EB 0 EB 0, U3 = 0 EB 0 EB Z; EB 0,

°

°

U4 = OEBOEBOEB71; .

As a consequence of Lemma 4.2.3(a) , End U is isomorphic to End V with V = (Vo , UI, U2, U3, U4, pUo) and Vo = UI EB U2 EB U3 EB U4. Then pUo = pVo + W, where W is the free R-module with the rows of the R-matrix M = (~I~I~I~) as a basis. An argument like that of Example 5.2.2 shows that if f E End V, then f(mod p) = aI, with a E C(A(mod p)). Iff is an idempotent, then a is an idempotent. Since A(mod p) is an indecomposable matrix, a = 0 or 1. If a = 0, then f2 = f E pEnd V . Since Vo is a finitely generated free Zp-module, f = O. On the other hand, if a = 1, then the same arguments show that 1 - f = 0 and f = 1. Hence, V, and consequently U is an indecomposable representation. Since there are indecomposable 7l/p71-matrices for each n, Crep(S4, 7l p , 1) has rank-infinite representation type. Now apply Lemma 5.2.1. (ii) W(ST) ::: 3 and j ::: 2. Define U = (Uo, UI, U2, U3) E Crep(S3, t.; 2) by

Uo = Z; EB Z; EB Z; + (1/p2)(1 + 1 + A)Z; + (1/p)(O + 1 + l)71;,

= Z; EB 0 EB 0, U2 = 0 EB 7l; EB 0, U3 = 0 EB 0 EB 7l;. Then End U is isomorphic to End V, with V = (VO, UI, U2 , U3, p2Uo) and Vo = UI

5.2 Isomorphism at p and Representation Type

159

VI ED V2 ED V3. In particular, p2Vo = p2Vo + W, where W is the free Zp-module with the rows of the Zp-matrix

M_(IIIIA) 0 pI pI as a basis . A by now familiar argument shows that if f E End V, then f(mod p) = al , with a E C(A(mod p» . As in the proof of (i), Crep(S3 , Zp , 2) has rank-infinite representation type . Again, apply Lemma 5.2.1. As a consequence of the preceding constructions, the representation type for C(T, j)p can be computed. The case that W(ST) = 3 and j = 1 is addressed in Corollary 5.2.9.

Corollary 5.2.4 Let T be a p-locally free finite lattice oftypes. Then C(T, j)p has: (a) wild modulo p representation type if either (i) WeST) ~ 5, j ~ 1; (ii) WeST) = 4, j ~ 2; (iii) W(ST) = 3, j ~ 3; or (iv) W(ST) = 2, ST is not a garland, and j ~ 7. (b) rank-infinite representation type if either (i) WeST) ~ 4, j ~ I; (ii) WeST) = 3, j ~ 2; or (iii) W(ST) = 2, ST is not a garland, and j ~ 4. (c) finite representation type if either (i) W(ST) = 1, j ~ 1 or (ii) W(ST) = 2, j ~ I, and ST is a garland. PROOF. As a consequence of Corollary 5.1. 10, C(T, j) = B(T, j) if W(ST) ~ 2. Thus, (a)(iv) , (b)(iii), and (c) follow from Corollary 4.3.3 . The remainder of the corollary is a consequence of the definition of Crep(ST , Zp , j) and Examples 5.2.2 and 5.2.3, wherein the representations constructed are in Crepcrit(ST, Zp, j).

An important consequence of Corollary 5.2.4 is the following corollary, demonstrating circumstances under which there are indecomposable almost completely decomposable groups of arbitrarily large finite rank . Representations constructed in this section can be used to give explicit descriptions of such groups ; see Exercise 5.2.3.

Corollary 5.2.5 Assume that T is a p-locallyfree finite lattice oftypes for some prime p. There are indecomposable almost completely decomposable groups ofarbitrarily large finite rank with typeset contained in T if either W(ST) ~ 4, j ~ 1; W(ST) = 3, j ~ 2; or W(ST) = 2, ST is not a garland, and j ~ 4.

160

5. Almost Completely Decomposable Groups

Use Corollary 5 .2.4 and the observation that if a group is indecomposable at p , then it is indecomposable.

PROOF.

=

=

Suppose X n {rl ' . .. , r n } is a p-locally free set of types with rj n rj ro for each 1 ~ i =P j ~ n. Define C(n) to be the category of almost completely U {roJ . Given j :::: 0, C(n , j) is decomposable groups with typeset G 5; the subcategory of groups G in CCn) with p jG 5; G*(ro) 5; G. In this case, G*(ro) = G(r))EB' . . EBG(rn ) = R(G). The category C(n , j)p is the isomorphism at p category of C(n, j). For example, X; could be the set ofatoms for a p-Iocally free Boolean algebra Tn. Then C(n, j) Ccrit(Tn , j), those almost completely decomposable groups in B(Tn , j) with critical typeset contained in X n , the join irreducible elements of

x,

=

r;

Corollary 5.2.6 The category C(n, j)p has: (a) wild modulo p representation type if either n :::: 5, j :::: 1; n = 4, j :::: 2; or n = 3, j :::: 3. (b) rank-infinite representation type if either n = 4, j :::: 1 or n = 3, j = 2. (c) finite representation type if and only if either n :s 2, j :::: lor n = 3, j = 1. A special case of Corollary 5.2.4, where T is the sublattice of the lattice of all types generated by X n, except for the case n = 3, j = 1. It is a consequence of Theorem 5.2.S(c) and Example 1.1.5 that CC3, l)p has finite representation type . PROOF.

Remark Let (S(n), j -1) be the posetdefined in Exercise 1.4.1. Observe that by comparison, the representation type of CCn, j)p is exactly that of rep«S(n), j - 1), k) , k a field, with the possible exception of n = 3, j = 2. A direct proof ofthis observation is not known .

:s

It remains to compute the representation type of CCT, j)p with W(ST) 3 and j = 1. Oddly enough, this apparently simpler case seems to require an additional hypothesis on ST. A finite poset S is a forest if it is a disjoint union of trees, equivalently, {s : s ~ i in S} is a chain for each i in S. For example, Sn is a forest. If ST is a forest, then CrePcrit(ST, 'Zp, j) has some special properties as given in (c) of the next lemma, the representation analogue of Theorem 3.2.5(b) for almost completely decomposable groups. The module EB{ V;* : i E ST} is analogous to a regulating subgroup for almost completely decomposable groups. If G is an almost completely decomposable group with critical typeset that is an inverted forest, then there is a unique regulating subgroup R(G) (Corollary 3.2.13). This property is reflected in CrePcrit(ST, 'Zp, j) in the case ST is a forest, remembering that ST is the poset of join irreducible elements of T with opposite order.

Lemma 5.2.7 Suppose T is a p-locally free finite lattice of types and V (Vo, U, : i E ST) E Crepcrit(ST, Zp, j).

=

5.2 Isomorphism at p and Representation Type

161

(a) For each i in ST, U, = V;* E9 Vr, where Vr is the pure submodule of U, generated by {Vi : j < i in ST}. (b) The Z p-module E9{U;*:i E Srl is a submodule of ~{Vj :i EST} with ~{Vj : i E Srl/(E9{V;* : i E ST}) and Ui] E9 {Vs* : s S i} bounded Zp modules for each i E ST. (c) /fST is aforest, then ~{Vj : i EST} = E9{V;* : i EST} and U, = E9{Vs* :s S i} for each i EST' (a) The pure submodule Vr is a summand of the finitely generated free Zp · module U: (b) If i is minimal in ST, then Vr = 0 and U, = V;*. Next assume that i is not minimal in ST and Vi/~{us*:s S j} is bounded for each j < i in ST. Then Vr 1~{Vi : j < i} is a bounded module, since Vr is a finitely generated free Zp. module and rank Vr = rank ~{Vi: j < i}. By induction, Ui] E9 {Vs* : s S i} is a bounded module for each i in ST. Hence, ~ {Vj : i E ST}I ~ {V;* : i E ST} must be a bounded module. Finally, ~ {V;* : i E Srl = E9{ V;* : i E Srl, since rank ~j V;* = rank ~ j Vi = rank V o = ~j rank Ui] Vr and Ui] Vr is isomorphic to V;* by (a). (c) In view of (b), it is sufficient to show that U, = ~ {V s* : s S i} for each i in ST. If i is minimal in ST, then U, = V;*. Now suppose i is not minimal in ST. Then C; = {s : s S i} is a chain, since ST is a forest. Thus, Vj# = VI for t the maximal element of Cj\{i}. By induction on the length of Ci, VI = ~{V;:s S t}. This shows that U, = ~{Vs* : s S i}.

PROO F.

There is a connection between CrePcrit(ST , Z p, j) and representations of S; over the finite ring Z I pi Z, recalling that S; is the disjoint union of ST and a point * unrelated to any element of ST. For a finite poset ST, let ~rep(ST , Z I pi Z) be the category of representations V = (Vo, U, : i E S;) such that Vo = ~ {Vi : i E ST} with each Vi a free Z I pi Z-module, Vi <; Vi if i S j in ST, and V* n U, = 0 for each i E S. Morphism sets are Zl piZ-homomorphisms f :Vo -+ V~ with f(V i) <; V; for each i E ST. Elements V of Mep(ST, Z I piZ) with Vo = E9{ Vi : i E ST} can be analyzed by solving matrix problems, as illustrated in Section 4.1 for rep(S, Zp, j). An additive functor F : X -+ X' between additive categories is a representation embedding if F preserves indecomposables and reflects isomorphisms. The functor F is a representation equivalence if F is a representation embedding that is dense, that is, for each Y in X' there is X in X with F(X) isomorphic to Y. A representation embedding, respectively equivalence, induces an injection, respectively a bijection, from Ind X to Ind X'. In the next theorem, (a) relates categories of Z p-representations of ST to categories of Z I pi Z representations and corresponding matrix problems, (b) generalizes the category equivalence of Lemma 4.2.3(b) for ST = Sn, and (c) and (d) are applications of (a) and (b) that relate the representation type of categories of almost completely decomposable groups to categories of representations of finite posets

162

5. Almost Completely Decomposable Groups

over the field k = Z/ pZ. A nonfunctorial partial generalization of (a), without the hypothesis that ST is a forest, is given in Theorem 5.4.2.

Theorem 5.2.8 Assume that T is a p-locallyfree finite lattice oftypes, ST is a forest, and j ::: 1. (a) There is afullfunctor F: CrePcrit(ST, Zp, j) ~ ~rep(ST, Z/pjZ) that is

a representation equivalence.

(b) There is a category equivalence H : Mep(ST, Z/pZ) ~ rep(Sr , Z/pZ). (c) The category of almost completely decomposable groups Ccrit(T, I)p has

finite representation type if and only if rep(ST, Z/ pZ) has finite representation type. (d) /frep(ST,Z/pZ) has wild representation type, then Ccrit(T, I) has wild modulo p representation type.

(a) For V = tU«, Vi : i E ST) E Crepcrit(ST, Zp, j), let hU = h {Vi: i E S} ~ Ue. Define F(V) = (Vo = hU/pjhU, Vi = (Vi + pjhU)/pjhU, V. = pjVO/pjhU : i E ST)and FU): F(V) ~ F(V), the morphism induced by amorphism f : V ~ V. Then F is a functor with F( V) E Mep(Sr, Z/ pj Z), noticing that f(hU) ~ hv . In particular, for each i, Vi is a pure submodule of Ui; Hence, Vi = (Vi + pj Vo)/ p! Vo is isomorphic to the free Z/ pjZ-module Vii p! Vi and

PROOF.

Vi

n V. = O.

To see that F is a full functor, let g: F(V) ~ F(V) be a representation morphism. By Lemma 5.2.7, Vi V;* ED V1, hU ED{V;*: i E ST}, and Vi = ED {V; : s ~ i} for each i E ST. It follows by induction that g lifts to an homomorphism f: hU ~ hV with f(Vi) ~ Vi for each i EST and f(pj Vo) ~ pjvo. Then f extends to h: Vo ~ Vo with F(h) = g, as desired. To confirm that F is a dense functorlet V = (Vo, Vi, V. : i E ST) E Mep (ST, Z/ p! Z) . Each Vi = vt ED Vi# , where Vi# is the injective envelope of h {VS : s < i} in Vi. This is so because Vi is a free Z/ pj Z-module and Z/ pj Z is self-injective. Then V/isafreeZ/pjZ-module,andjustasinLemma5.2.7, h{Vi :i EST} = ED{Vti E ST} and Vi = ED{Vs' :s ~ i} for each i EST. Choose a finitely generated free Zp-module Wo = ED{ V;* : i E ST} with Wo/pjwo = ED{Vt: i EST} , (V;* + pjwo)/pjwo = vt for each i EST, and a submodule K of Wo with K/ pj Wo = V•. Define Vi = ED{ Vs' : s ~ i) for each i in ST and Vo = Wo+(l/pj)K.Theneach Vi is pure in Vo,since KnVi ~ pjWo.Furthermore, (Vi + pjwo)/pjwo = Vi , pjvo/pjwo = K/pjWo = V., Wo = hU, and rank Vo = rank Wo = h{rank V;* : i E ST)' Then V = (Vo, Vi : i E ST) E Crepcrit(ST, Zp, j) with F(V) = (Vo, Vi, V* : i EST) = V, as desired. If F(V) is isomorphic to F(V), then, since F is a full functor, there are representation morphisms f: V ~ V and g: V ~ V with Iv - fg E pjEndV and I u - gf E pjEnd V. Since Vo is a free R-module and pZp = lZp, the Jacobson radical of Zp , it follows that pEnd V ~ lEnd V . Then gf = I - (I - gf) is a unit of End V, and similarly, f g = 1 - (1 - f g) is a unit of End V. Consequently, V is isomorphic to V.

=

=

5.2 Isomorphism at p and Representation Type

163

Finally, assume that F(V) = V' EEl V" is a representation direct sum. Since F is a dense functor, there are V' , V" E CrePcrit(Sr, 'lL p , j) with F(V') = V' and F(V") = V". Then F(V' EEl V") = F(V') EEl F(V") = F(V). But F reflects isomorphisms, so that V is isomorphic to V' EEl V" . If V is indecomposable, then V' or V" = O. Hence , V' or V" is 0, and so F(V) is indecomposable. This shows that F preserves indecomposables. (b) Let k = 'ILl p'IL, V = (Va, Vi, V*: i E Sr) E !:Hep(Sr , k), and define H: L\rep(Sr, k) --+ rep(Sr, k) by

H(V)

= «EEl{Vi: i E Sr})IV* , (Vi + V*)IV*: i E Sr).

Then H is a fully faithful functor . To see that H is dense, let V = (VA , Vi : i E Sr) E rep(Sr,k) with no trivial summands. Since Sr is a forest, Vi = V/ EEl Vi# and Vi = EEl{V/: j ~ i in Sr} for each i E Sr as a consequence of Lemma 5.2.7. Write Va = :E{Vi: i E Srl = (EEl{Vi : i E Sr})1 V* for some V*. Define Va = EEl{Vi:i E Sr}) and Vi = EEl{Vj:j ~ i}, where (V;* EEl V*)IV* = V/, for each i E Sr . Then V = (Va, Vi , V* : i E Sr) E !:Hep(Sr, k) with H(V) = V , as desired. (c) Observe that Ccrit(T, I)p has finite representation type ifand only ifrep(Sr, k) has finite representation type, by (a) and (b) and the fact that Ccrit(T, l)p is category equivalent to CrePcrit(Sr, 'lL p , 1). (d) Let I' be an 'lLp-algebra that is finitely generated and free as a 'lLp-module. Since rep(Sr, k) has wild representation type, rep(Sr, k) is endowild by Corollary 1.4.3. In view of the category equivalence (b), there is some V in !:Hep(Sr , 'ILl p'IL) with End V = I' I pi', Since F is dense and full, there is V E CrePcrit(Sr , 'lL p , 1) with F(V) = V and End V --+ End V = I' Ipf onto. This proves that CrePcrit(Sr, 'lL p , 1) has wild modulo p representation type. The next corollary completes the computation of the representation type of

C(T , l)p with w(Sr) = 3 for the case that Sr is a forest. Corollary 5.2.9 Suppose T is a p-locallyfree finite lattice oftypes and Sr is a forest with w(Sr) = 3. Then C(T , l)p has: (a) rank-infinite representation type if Sr contains (2, 2, 2), (1,3, 3), or (1,2 ,5)

as a subposet; and (b) wildmodulo p representation type ifSr contains (2, 2, 3), (1,3,4), or(l, 2, 6)

as a subposet. (a) By Theorem 5.2.8, Ccrit(T, l), has rank-infinite representation type if and only if rep(Sr, k) has infinite representation type. Now apply Theorem 1.3.6, bearing in mind that Sr is a forest with w(Sr) = 3, and that Ccrit(T, j)p is a full subcategory of C(T, j)p. Assertion (b) follows from Theorems 5.2.8 and 1.4.4.

PROOF.

164

5. Almost Completely Decomposable Groups

Open Questions 1. Suppose T is a finite p-locally free lattice oftypes with w(Sr) = 2 and Sr is not a garland. 1f6 :::: j :::: 4, does C(T , j)p have wild modulo p representation type? Ifl :s j :s 3, does C(T, j)p have rank-finite or rank-infinite representation type? In either case, can the indecomposable groups in C(T, j)p be classified? 2. What is the representation type of C(T, I)p if w(Sr) = 3 and Sr is not a forest ? In particular, what if Sr = (N , 4)? 3. Do the categories C(T, j)p and Ccrit(T , j)p have the same representation type? Ifw(Sr) = 3 and j = 2, does C(T, j)p have wild modulo p representation type? Does C(3 , 2)p have wild modulo p representation type? Can the indecomposable groups be classified ? 4. Is there a definition oftame representation type such that (i) if C(T, j)p has tame representation type, then indecomposables can be classified, and (ii) each C(T, j)p has exactly one offinite, tame, or wild modulo p representation type?

EXERCISES

1. Give a description ofrepresentations in Crep(Sr , Zp, j) . 2. Prove that each group in C(T, 0) is completely decomposable. 3. Give explicit constructions of indecomposable almo st completely decomposable group s corresponding to the indecomposable representations in Crep (Sr . 7lp • j ) given in Examples 5.2.2 and 5.2.3 .

5.3 Uniform Groups An almost completely decomposable group G is called uniform if there is a prime p such that G IG*(lT(G» is a free ZI pjZ-module, where IT(G) is the inner type of G. Define V(T, j) to be the subcategory of C(T, j) consisting of the uniform almost completely decomposable groups in C(T, j) and define Vcrit(T, j) to be the full subcategory of all uniform almost completely decomposable groups in Ccrit(T , j). Uniform almost completely decomposable groups arise naturally and are intimately related to more general finite rank Butler groups .

Example 5.3.1 Let T be a finite lattice oftypes. (a) Then V(T , I) = C(T, I) and Vcrit(T, I ) = Ccrit(T , I). (b) Let G be afinite rank Butler group in B(T) and write G = CI K for some completely decomposable group C E B(T). Then H = C + (11 p j)K is a uniform almost completely decomposable group in V(T , j).

5.3 Uniform Groups

165

(c) Conversely, suppose H = C + (11 pi)K is a uniform almost completely decomposable group in VeT, j) with C a completely decomposable group and K a pure subgroup of C. Then CI K is a finite rank Butler group in B(T). (a) Each 'I.I p'I.-module is free. (b) The group H is almost completely decomposable, pi H S; C, and HI C is isomorphic to pi HI pi C = (p i C + K)I pi C . But (p i C + K)I pi C is isomorphic to the free 'I.I pi 'I.-module K I pi K , since K is a pure subgroup of C. Furthermore, typeset H = typeset C S; T . (c) Since K is a pure subgroup of C , G = C I K is a finite rank Butler group. Moreover, G E B(T) by Lemma 3.3.1, since typeset C = typeset H S; T and T is a lattice.

PROOF.

The following three examples demonstrate that representation type changes from C(T, j)p to VeT, j)p; compare Corollary 5.2.4.

Example 5.3.2 The category Vcrit(T , j)p has finite representation type if ST = S3 and j :::: 2. In this case, each indecomposable in Vcrit(T, j)p has rank less than or equal to 9. [Arnold, Dugas 93A]. A complete and explicit list of indecomposables is given in [Mouser 93] .

PROOF.

Example 5.3.3 The category Vcrit(T, j)p hasfinite representation type if ST = (1, 2). Each indecomposable has rank less than or equal to 3. If G E Vcrit(T, j)p, then Fr(G) = (G p, G(T)p : TEST) = (Vo, VI, V2 ~ V3) = V is an indecomposable in Crep(ST, 'I.p, 2) with S = (1,2) and (End G)p

PROOF.

isomorphic to End V by Corollary 4.3.2. Then End V is isomorphic to End V for V = (VI EEl V3, VI, V2 S; V3, p2Vo) by Lemma 4.2.3(a). In particular, V is indecomposable. Assume that rank V > 1 and let A = (M I I M2 I M3) be a 'I.p-matrix with rows a basis of the free 'I.p-module p2Vo and the columns labeled by bases of VI, V2, and respectively. As demonstrated in Section 4.1, allowable operations that do not change the isomorphism class of V are invertible elementary 'I.p-column operations in M, for 1 :::: i :::: 3, invertible 'I.p-column operations from M2 to M3 (but not from M 3 to M2), and invertible 'I.p-row operations on M. Diagonalizing M I by invertible row and column operations yields A ~

V; ,

(LIMIN)= (I

o

I

o Mil pD M21

M12INII N 12) M22 N21 Nn

with p D a diagonal matrix having diagonal entries pi, i 2: 1. Since V is uniform,

166

5. AlmostCompletely Decomposable Groups

each linear combination of the rows of the matrix (pD I MZ1 Mzzi NZI Nn) must contain a unit in the (M zl M zz) block and a unit in the (N zI Nzz) block. Furthermore, (L I MIN) ~ (l 0 I Mz I M3). This can be seen by first diagonalizing (M Z1 Mzz) with elementary row and column operations. The result is a matrix (0 EpD I I 0 I NZI Nzz) with E a product of elementary row operations . Using column operations, this matrix may be assumed to be of the form (0 pD 1 I I 0 I NZI Nzz). Next, apply column operations from Mz to M3 to clear the M3 block. This yields the second row block of A equivalent to (0 pD I I 0 I 0), contradicting the assumption that V is uniform. In particular, the second-row block must be empty. A similar argument shows that A = (l 0 I Mzl M3) is equivalent to ( I

o

0II I 0

0 0 1 0) pD 0 I

with D a diagonal matrix. Since V is indecomposable with rank V > I, either rank V = 3 and, from the second-row block, V = (Zpx EB Zpy EB Zpz, Zpx, Zpy, Zpy EB Zpz, (I + pi + l)Zp), i < j, or else, from the first row block, rank V = 2 and V = (Rx EB Ry, Rx, Ry, Ry, (I + l)R) . Example 5.3.4 The category V(T, j)p has wild modulo p representation type if (a) W(ST) = 4, j ::: 2, or (b) W(ST) = 3, j ::: 3. (a) Suppose W(ST) = 4, j ::: 2. Let r be a Zp-algebra that is finitely generated and free as a Zp-module and let A, B be n x n Z p-matrices with r = C(A, B). Define V = (Vo, VI , V z, V 3, V 4) E rep(S4, Zp , 2) by

PROOF.

Vo = VI EB Ui EB V 3 EB V4

VI

= Z~ EB 0 EB 0 EB 0,

U3 = 0 EB 0 EB Z~ EB 0,

+ (II pZ)Vs, Ui = 0 EB z;n EB 0 EB 0, and

V4 = 0 EB 0 EB 0 EB

z;n,

where Us is the free R -rnodule with the rows of the following matrix as a basis:

I

o (o o

0 I 0 0

0 0 lOA 0 0 0 I 0 I 0 I 0 pI 0 I 0 I 0

0)

pB

I pI

'

for I an n x n identity matrix and 0 an n x n matrix ofo's. Notice that V = FdG) for some G E V(T', 2) ~ V(T, j) with ST' = S4, since V E CrePcrit(S4, Zp , j) with Vo/(VI + Ui + V3 + V4) a free ZI pZZ-module. Thus, (End G)p = End V. Now, End V is isomorphic to End V, where V = (VI EB U: EB V3 EB V 4, VI, Us. V 3, V 4, pZVo). A by now familiar computation yields an epimorphism End V ---+ C(A(modp), B(modp» and C(A, B) embeds in End V. It follows from

5.3 Uniform Groups

167

Lemma 4.2.1 that VeT , j)p has wild modulo p representation type in the case WeST) 2: 4, j 2: 2. (b) The case that w(S) = 3, j 2: 3 follows from Corollary 4.3 .2 and Example 5.2.2, since in that example VO/(UI + V z + V 3 ), being isomorphic to V*/ p 3V*, is a free Z/p3Z- module. Example 5.3.5 The category VeT , j)p has: (a) rank-infinite representation type if W(ST) = 2, j 2: 6, and ST contains either (1,3) or (2,2) as a subposet, and (b) wild modulo p representation type if W(ST) = 2, j 2: 11, and ST contains either (1,3) or (2,2) as a subposet. PROOF.

The associated representations are Exercises 5.3.3 and 5.3.4.

Corollary 5.3.6 Let T be a finite p-localiy free lattice oftypes and j a positive integer. Then VeT, j)p has: (a) wild modulo p representation type if either (i) W(ST) = 5, j 2: 1; (ii) W(ST) = 4, j 2: 2; (iii) W(ST) = 3, j 2: 3; or (iv) W(ST) = 2, j 2: 11, and ST contains (1,3) or (2, 2) as a subposet. (b) rank-infinite representation type if either (i) weST> = 4, j 2: 1; or (ii) W(ST) = 2, j 2: 6, and ST contains (1,3) or (2,2) as a subposet. (c) finite representation type if either W(ST) = 1; or W(ST) = 2, j 2: 1, and ST is a garland.

Moreover, Ucrit(T, j) has finite representation type if W(ST) = 2 and ST is a forest that does not contain (1, 3) or (2, 2) as a subposet. PROOF. Statements (a) and (b) follow from Examples 5.3.4 and 5.3.5 and Corollary 5.2.4, noticing that U(T , 1) = C(T, 1) by Example 5.3.1(a). As for (c), apply Corollary 4.3.3( c). For the last statement, if ST is a forest of width 2 that does not contain (2, 2) or 0 , 3), then ST = (1,2). Now apply Example 5.3.3.

Descriptions of representation type for V (T, R, j) p remain incomplete for those T with W(ST) = 2 or W(ST) = 3 and j :::: 2. This deficiency already arose in Section 5.2 for C(T, l)p with W(ST) = 3 and ST not a garland. Define V (n, j) to be the full subcategory of uniform almost completely decomposable groups in C(n, j) and Utn, j)p the isomorphism at p category of Utn , j). Corollary 5.3.7 The category V(n, j)pfor j 2: 1 has: (a) wild modulo p representation type j ?:. 3.

if either n 2: 5; n

= 4, j

2: 2; or n

= 3,

168

5. Almost Completely Decomposable Groups

(b) rank-infinite representation type ifn = 4, j = 1; and (c) finite representation type ifn ~ 2 or n = 3, j ~ 2; Corollary 5.3 .6 and Example 5.3 .2.

PROOF.

Open Questions 1. What is the representation type for the missing cases of Corollary 5.3.6 ? 2. Are there representation embeddings or equivalences from representations over fields that can explain the representation type of U(T , j)p? 3. Give a theoretical explanation of why Ccrit(S3, 2)p has rank-infinite representation type but Ucrit(S3, 2)p has finite representation type.

EXERCISES

1. Prove that there are indecomposable groups G in B(T3) of arbitrarily large finite rank with G = G*(lT(G)). 2. Prove that if ST = (1, 2) , then there is a bound on the ranks of indecomposable groups in B(T) with G = G*(IT(G)).

3. Assume that ST = (l, 3).

(a) Use the following representation to show that U(T, j)p, j ~ 11, has wild modulo p representation type: U = (Uo, UI, U2 ~ U3 ~ U4) with Uo = z~n 61 z~n 61 z~n 61 z~n + (l/pll)U* , UI = z~n 61 0 61 0 61 0, U2 = 0 61 z~n 61 0 61 061, U3 = 0 61 z~n 61 z~n 61 0, U4 = 0 61 z~n 61 z~n 61 z~n, and U* the row space of the matrix contained in UI 61 U2 61 Uj 61 U: : I

(

o o o o

0 I 0 0 0

0 0 0 0 0 0 I 0 0 0 I 0 0 0 I

p5 I p61 P7 I p8 I

0

p 61

I 0 P7 A 0 p 2 1 p8BOO

0

0 plO I 0

0 0

0 0

p 41

0 0

0 0 0

p 61

0

~ 0~ ~ 0~ 0~)

o o

I

0 0 I 0 p81 0 0 0 I

.

(b) Use the following representation to show that U(T, j)p , j ~ 6, has infinite representation type: U = (Uo, UI, U2 ~ U3 ~ U4) with Uo = z~n 61 z~n 61 z~n 61 z~n + (l/p6)U* , UI = z~n 610610610, U2 = 061Z;: 61061061, U3 = 061z~n 61 z~n 61061, U4 = 061z~n 61z~n 61Z;:, and U* the row space of the matrixcontained in UI 61 U2 61 U; 61 Ur I

(

o o

0 0 p 31 p4A I 0 I 0 p 41 p5 I 0 p2[ 0 [ p5 I 0 0 0

0 0

0 0) [ 0 p4 I 0 [ .

4. Assumethat ST = (2, 2). (a) Use the following representation to show that U(T, j)p , j ~ 10, has wild modulo p representation type: U = (Uo, UI ~ U2, U3 ~ U4) ~ with Uo = z~n 61 z~n 61 Z;: 61 z~n + (l / p IO)U*, U I = z~n 61 0 61 0 61 0, U2 = z~n 61 z~n 61 0 61 0,

5.4 Primary Regulating QuotientGroups

169

U3 = 0 $ 0 $ z~n $ 0, U4 = 0 $ 0 $ z~n $ z~n , and U* the row space of the matrixcontainedin U, $ Ui $ U3 $ U; :

o o

I 0 o 0 o 0 p 5I 0

0 I 0 0 0

0 0 I 0 0

0 0 0 I 0

0 0 0 0 I

4I p p5I p6I p7I 0

p5I I 0 p6A 0 I p 7 BOO 0 0 0 p9I 0 0

0 0 I 0 0

0 0 0 I 0

0) 0 0 . 0 I

(b) Use the following representation to show that U(T, })p, } :::: 6, has infiniterepresentation type: U = (Uo , U\ ~ U2 , U3 ~ U4) with Uo = z~n $ z~n $ z~n $

z~n +(I/p6)U* , U\ = z~n $0$0$0, U2 = z~n $z~n $0$0$0, U3 = 0$0$ z~n $ 0, U4 = 0 $ 0 $ z~n $ z~n , and U* the row space of the matrixcontained

in U, $ Ui $ U3 $ U; : I 0 0 p2I 0 I 0 p3I p 5I 0 0 I p4I

o o

p3A I 0 0 ) p4I 0 I 0 . 0 0 0 I

5.4 Primary Regulating Quotient Groups An almost completely decomposable group G is a p-primary regulating quotient group if p is a prime and G has a regulating subgroup B such that G j B is a pgroup, equivalently, i(G) is a power of p by Proposition 5.1.2. It follows that an almost completely decomposable group is a p-primary regulating quotient group if and only if eo = exp G j R(G) is a power of p. Notice that the class of pprimary regulating quotient groups is closed under finite direct sums , since if B is a regulating subgroup of Hand C is a regulating subgroup of K with IH j B I and IK j C I both powers of p , then D = B EB C is a regulating subgroup of G = H EB K with IGjDI a power of p . The p-primary regulating quotient groups are generalizations of groups in qT, j)p in the following sense . If Tcr(G) is an inverted forest and G is an almost completely decomposable group with no rank-I summands, then R( G) = G*(IT(G» is a unique regulating subgroup of G by Corollary 3.2.13(c). If, in addition, typeset G is contained in a finite lattice T of types, then G E qT, j)p if and only if G is a p-primary regulating quotient group . For p-primary regulating quotient groups, near-isomorphism is the same as isomorphism at p:

Lemma 5.4.1 Let G and H be p-primary regulating quotient groups. Then G and H are nearly isomorphic if and only if G and H are isomorphic at p. Moreover, G is indecomposable if and only if G is indecomposable at p. Since near-isomorphism is isomorphism at p for each prime p, it suffices to prove that if G and H are isomorphic at p , then they are nearly isomorphic. If G

PROOF.

170

5. Almost Completely Decomposable Groups

and H are isomorphic at p, then there is a monomorphism f : G ~ H such that HI f(G) is finite and relatively prime to p. Consequently, R(G) is isomorphic to R(H) as in Theorem 5.1.6(b). Since eG is a power of p , G is isomorphic to R(G) at q for each prime q -:;6 p . Similarly, H is isomorphic to R(H) at q for each prime q -:;6 p . Thus, G is isomorphic to H at q for each prime q , and so G is nearly isomorphic to H . The last statement follows from Corollary 5.1.8(b). The next theorem is a generalization of Theorem 5.2.8(a), wherein ST is assumed to be a forest. However, in contrast to that theorem , this construction need not be functorial , since there are almost completely decomposable groups G and H and a homomorphism f : G ~ H such that f(R(G)) is not contained in R(H); Exercise 5.1.8 . Theorem 5.4.2 [Mader, Vinsonhaler 97] Suppose G is a p-primary regulating quotient group with eG = pl . Define a 7l,lpj7l,-representation F(G) = (Uo, U T , U* : r E Tcr(G))by Uo = R(G)I p j R(G), U T = (R(G)(r)) + p! R(G))I p j R(G), and U* = pl Gf p! R(G). (a) There is a ring isomorphism ¢ : End G I pjEnd R(G) ~ End F(G). If f is an idempotent ofEnd F( G), then there is an idempotent g ofEnd G with ¢(g) = f . (c) The group G is indecomposable ifand only if(End G)p is a local ring.

(b)

PROOF. (a) Let ¢(/)EEndF(G) be the endomorphism induced by f, which is well-defined, since R(G) is a fully invariant subgroup of G by Theorem 5.1.6(a). Then ¢(/) = 0 if and only if f : R( G) ~ p j R( G). Equivalently, f E pj EndR(G) ~ EndG , since pjG ~ R(G). The proof that ¢ is onto is a direct analogue of the proof of Theorem 5.2.8(a). The crucial observations are that R(G) is a completely decomposable subgroup of G with p! G ~ R(G), and R(G) is a fully invariant subgroup of G . (b) Suppose f is an idempotent of End F(G) . By (a), there is h E End G with ¢(/) = hand h 2 - h E p jEnd R(G). Now End R(G)INEnd R(G) is a product of matrix rings over subrings of Q by Proposition 3.1.8. In fact, h + pjEnd R(G) lifts to an idempotentg of End R(G). This is so because idempotents of AIp! A lift to idempotents of A for A a matrix ring over a subring of Q, hence for A = End R(G)INEndR(G) . But idempotents of End R(G)I NEnd R(G) lift to idempotents of End R(G), since NEnd R(G) is nilpotent. Since R(G) is a fully invariant subgroup of G with G I R(G) finite, there is an idempotent g E End G with ¢(g) = f. (c) If (End G)p is a local ring, then 0 and 1 are the only idempotents of (End G)p. Hence, G is indecomposable at p and so must also be indecomposable as a group. Conversely, suppose that G is indecomposable. Then End G has only 0 and 1 as idempotents, and so End F(G) has only 0 and 1 as idempotents by (b). Since End F(G) is finite, it must be a local ring. By (a), (EndG)plp j(End R(G))p is a local ring , whence (End G) p is also a local ring , since p E J (End G) p -

5.4 PrimaryRegulating Quotient Groups

171

The near-isomorphism category of torsion-free abelian groups of finite rank has nonunique direct sum decompositions into indecomposables by Example 2.1.11. The near-isomorphism category of almost completely decomposable groups also has nonunique direct sum decompositions by Proposition 5.1.11. However, as a consequence of the next theorem, the near-isomorphism category of p -locally free almost completely decomposable groups G with ea a power of p is a KrullSchmidt category.

Corollary 5.4.3 [Faticoni, Schultz 96] Assume that G and Hare p-primary regulating quotient groups. If G = Xl ED . . . ED X n is nearly isomorphic to Y1 ED · ·· ED Yn with each Xi and Yi an indecomposable group, then m = nand there is a permutation C1 of {I, 2, . . . , n} with Xi nearly isomorph ic to YU(i) for each I ~ i ~ n. By Corollary 5.1.8(d), it suffices to assume that G = H. It is also sufficient to assume that both G and H are p-reduced, hence p -locally free; Exercise 5.1.5. Now apply Theorem 5.4.2(c) and Exercise 1.2.6 to see that there is a permutation C1 of {l , 2, . .. ,n} with Xi isomorphic at p to Yu (i) for each 1 ~ i ~ n. Since eo and en are powers of p, it follows from Lemma 5.4 .1 that Xi and Yu(i) are nearly isomorphic. PROOF.

There are examples of almost completely decomposable groups G and H with no rank-I summands such that G ED H has a rank-I summand; Exercise 5.1.7(b) . This cannot happen if G and H are p -primary regulating quotient groups:

Corollary 5.4.4 [Lady 74B] Suppose G and Hare p-primary regulating quotient groups. (a) IfG and H have no rank-Y summands, then GEDH has no rank-s summands. (b) Any two direct sum decompositions ofG into indecomposable groups have the same number of rank-t summands oftype r for each r.

This corollary is a special case of Corollary 5.4.3 and the observation that nearly isomorphic completely decomposable groups are isomorphic.

PROOF.

EXERCISE

I. Assume that G = A E9 H with A a completely decomposable group. Prove that if B is a regulating subgroupof G, then B E9 H is a regulating subgroup of Hand G/ B is isomorphic to H / H n B. Conclude that if G is a p-primaryregulating quotient group, then so is H . What can be said if A is not completely decomposable? NOTES ON CHAPTER

5

Finite rank completely decomposable groups are classified in [Baer 37]. Examples of pathological direct sum decompositions of almost completely decomposable groups were

172

5. Almost Completely Decomposable Groups

constructed in [Levi 17] and [Jonsson 57, 59]. For many years , almost completely decomposable groups were considered primarily as a good source of counterexamples; see [Fuchs 73]. Some positive results for almost completely decomposable groups , most of which are given in Section 5.1, first appeared in [Lady 74B]. An equivalence condition on almost completely decomposable groups G and H, G ffi C ::::::: H ffi C for some completely decomposable group C , led to the notion of near-isomorphism for torsion-free abelian groups of finite rank. Most of Section 5.2 is derived from [Arnold, Dugas 98]. As demonstrated in that section, representations of finite posets can be used to easily construct examples of indecomposable almost completely decomposable groups of arbitrarily large finite rank with types in a fixed typeset. The fact that categories of almost completely decomposable groups with typesets contained in a relatively small lattice of types can have wild modulo p representation type vividly demonstrates the complexity of almost completely decomposable groups. Almost completely decomposable groups are viewed as extensions of finite torsion groups by completely decomposable groups in [Mader 95]. Near-isomorphism is rediscovered in the sense that an equivalence, called type isomorphism, which arose naturally in this context , is the same as near-isomorphism. Mader 's approach, together with the results and techniques of [Burkhardt 84], [Krapf, Mutzbauer 84], and [Schultz 87], has rekindled interest in the subject of almost completely decomposable groups ; see [Mutzbauer 99], [Mader, Nonxga 99], [Ould-Beddi, Striingmann 99], and [Reid 99]. [Mader 00] includes a comprehensive, detailed exposition of the subject of almost completely decomposable groups and an extensive bibliography. Several classes of almost completely decomposable groups have been classified by invariants up to near isomorph ism: uniform block rigid groups [Dugas, Oxford 93]; groups divisible by all but three primes [Cruddis 70]; groups with critical typeset of two elements; and cyclic regulating quotient groups [Mader, Mutzbauer 93], [Mader, Vinsonhaler 95], [Burkhardt, Mutzbauer 96], and [Dittmann, Mader, Mutzbauer 97]. Examples given in these papers demonstrate that classification of almost completely decomposable groups up to isomorphism involves algebraic number-theoretic difficulties that can be avoided by classification up to near isomorphism. There is a substantial amount of literature on almost completely decomposable groups with small typeset; see [Arnold , Dugas 99]. Results of [Cruddis 70] are extended in [Arnold 73]. In the latter paper, some of the arguments are invalid. Corrected proofs and related results can be found in [Arnold, Dugas 95A], [Lewis 93], and [Mader, Vinsonhaler 94]. Finite rank Butler groups G such that each pure subgroup of G or each torsion-free quotient of G is almost completely decomposable are characterized in terms of the critical typeset of G in [Nonxga, Vinsonhaler 95]. Invariants for almost completely decomposable groups are discussed in [Arnold , Vinsonhaler 93] and [Vinsonhaler 95].

6 Representations over Fields and Exact Sequences

6.1 Projectives, Injectives, and Exact Sequences The terms exact sequence, kernel, and cokernel are defined for arbitrary categories [Mitchell 65]. In order to avoid lengthy arguments, these terms are defined herein in terms of their descriptions in rep(S, k) rather than derived. Let V = (Vo, Vi : i E S), V = (VO, Vi : i E S), and W = (Wo, Wi : i E S) be objects ofrep(S, k), I E Hom(V , V), and g E Hom(V, W). The sequence f

g

O-+V-+V-+W-+O

is an exact sequence if the induced sequence

o-+ Vi -+f Vi -+g Wi -+ 0 of vector spaces is exact for each i in S U {O}, i.e., for each i , I: U, -+ Vi is one-to-one, g : Vi -+ Wi is onto, and ker g = image I . A representation morphism I : V -+ V is a monomorphism if I :VO -+ Vo is l -to-l and an epimorphism if the k-linear transformation I : Vi -+ Wi is onto for each i E S U {O} . The kernel of I is Ker 1= (ker I, (ker f)

n Vi : i E S) E rep(S, k),

where ker I is the kernel of the k-linear transformation I: VO -+ Vo. If I is an epimorphism, then 0 -+ Ker I -+ V -+ V -+ 0 is an exact sequence of representations, since 0 -+ (ker f) n Vi -+ Vi -+ Vi -+ 0 is an exact sequence of vector spaces for each i E S U {O} .

174

6. Representations over Fieldsand ExactSequences

A representation morphism I: V -+ V is apure morphism and I(V) = (/(Vo), I(V;): i E S) is a pure subrepresentation of V if I(V;) = I(Vo) n V; for each i E S. For example, the kernel of a representation morphism is a pure subrepresentation. The terminology arises from the abelian group setting . A pure morphism is called a proper morphism in [Simson 92]. If I : V -+ V is a pure morphism, then the cokernel 01 I is Coker 1= (Voll(Vo), (V; + I(Vo»II(Vo): i E S) E rep(S, k) . The sequence 0 -+ V -+ V -+ Coker 1-+ 0 in rep(S , k) is exact if I is a pure monomorphism, since 0 -+ I(Vo)n V; -+ V; -+ (V; + I(Vo))//(Vo) -+ Ois an exact sequence of vector spaces for each i, A representation V E rep(S, k) is projective if whenever I : V -+ W is a representation epimorphism and g : V -+ W is a representation morphism with V, W E rep(S, k), then there is a morphism h: V -+ V with Ih = g. Dually, a representation V is injective if whenever I : V -+ W is a pure monomorphism and g : V -+ V with V, WE rep(S, k), then there is a morphism h: W -+ V with hi = g. Recall that V = (Vo, V; : i E S) E rep(S, k) is a trivial representation if Vo =P 0 but V; = 0 for each i in S. Up to isomorphism, the only trivial indecomposable representation is P(O) = (k, P(O); : i E S) with P(O); = 0 for each i in S. Call V E rep(S, k) cotrivial if V; = Vo for each i. The only cotrival indecomposable representation, up to isomorphism, is 1(0) = (k , 1(0); : i E S) with 1(0); = k for each i in S. Lemma 6.1.1 Let S be a finite poset, k afield, and a: rep(S , k) -+ rep(SOP, k) the contravariant duality given in Proposition 1.4.7. Then: (a) a sequence 0 -+ V -+ V -+ W -+ 0 is exact in rep(S, k) if and only if 0 -+ a(W) -+ a(V) -+ a(V) -+ 0 is exact in rep(SOP, k); (b) a representation V is projective in rep(S, k) if and only if a(V) is injective in rep(SOP, k); and (c) V is a trivial representation if and only if a(V) is a eotrivial representation. PROOF.

Exercise 6.1.1.

Let S be a finite poset and k a field. For t E S, define pet) = (k, pet); : i E S) E rep(S, k),

= k for each i>: tin Sand P(t)j = 0 if j t. tin S. Dually, define l(t) = (k , I(t);: i E S), where I(t)j = k for each j 1: tin Sand I(t)j = 0 if j ~ tin S. Observe that if pet) E rep(S, k), then a(P(t» = let) E rep(SOP, k) . For example, let S = {I,2 < 3}. Then P(O) = (k, 0, 0, 0), P(l) = (k , k, 0, 0), P(2) = (k, 0, k, k), and P(3) = (k, 0, 0, k). Hence, sop = {I,3 < 2} and a(P(O» = where P(t)j

6.1 Projectives, Injectives, and Exact Sequences

=

=

=

(k, k , k , k) 1(0), a(P(I» (k , 0, k , k) 1(1), a (P(2» and a(P(3» = (k , k , 0, k) = 1(3) E rep(SOP, k).

Lemma 6.1.2 Let S be a finite poset, k a field, and V rep (S , k ). Then:

175

= (k, k , 0, 0) = 1(2), = (Vo, Vi : i E S) E

(a) V is projective in rep(S , k ) if and only if V is isomorph ic to a finite direct sum of representations P (O) and P(t)for tin S; and (b) V is injective in rep (S, k) if and only if V is isomorphic to a finite direct sum of representations I (0) and I (t) for t in S. PROOF. The first step is to prove that P(t) is projective for each t E S U {O} . To this end, assume that f : V --+ W is an epimorphism and g : P(t) --+ W . Since P(t)o = k and f is onto, there is a k-linear transformation h : P(t)o --+ Vo with f h = g. Then h is a representation morphism, recalling that if t E S, then P(t)j = k if j ~ t and P(t)j = 0 otherwise. This shows that each P(t) is projective. It follows that finite direct sums of P(O) and P(t) are projective. Conversely, let V = (Vo , Vi : i E S) E rep(S, k) be a projective representation. There is an epimorphism f : P --+ V , where P is a finite direct sum of representations P(t ), t in S U {O}. This holds because for a given t , there is a finite direct sum V (t ) = (Vo, Vi : I E S) of copies of pet) and a representation morphism V (t ) --+ V such that V, is contained in the image of V,. Taking the direct sum of all such V (t)'s for t E S U {O} gives the desired P. Since 1u : V --+ V and V is projective, there is a morph ism h : V --+ P with f h = 1u- In particular, V is isomorphic to a representation summand of P . Since P is a finite direct sum of P(t)'s for t E S U {O} and each P(t ) is indecomposable, V is isomorphic to a finite direct sum of cop ies of P (O ) and P (t ) for t E S by Theorem 1.2.2 . (b) Apply (a), Lemma 6.1.1 (b), and the observation that a (P (t» = I (t) .

Example 6.1.3 Suppose S is a chain with n elements and k is a field. If V

E

rep(S , k), then V is both projective and injective.

PROOF . The representation V is a finite direct sum of indecomposable representations, and each indecomposable has dimension I by Lemma 1.3.3(a). If V E rep(S, k) has dimension 1, then V = P(s) = I(t) for some t < s E S or V = P(O) = I (n) . By Lemma 6.1.2, V is both projective and injective, as desired.

An epimorphism f: P --+ U in rep (S, k) , with P a projective representation, is a projecti ve cover of V if given g : P' --+ V an epimorphism with P' projective, then there is an epimorphism h : P' --+ P with f h = g. In this case , P is isomorphic to a summ and of P' , since P is projective. Notice that a representation U is projective if and only if the identity homomorphism U --+ V is a projective cover of V. A pure monomorphism f: V --+ I in rep(S, k) with I an injective representation is an injective envelope of V if for each pure monomorphism g : V --+ I' with I '

176

6. Representations over Fieldsand Exact Sequences

injective, there is a pure monomorphism h : I -+ I' with hf = g. In particular, I is a summand of I' . Moreover, V is injective if and only if the identity is an injective envelope of V. Projective covers and injective envelopes are unique up to isomorphism (Exercise 6.1.10) . Projective covers and injective envelopes of V E rep(S, k) can be constructed explicitly. Assume that V E rep(k, S) has no trivial summands. For each i in S, let Vr = 'E{V j : j < i in S}. Define P(V)

= (P(V)o , P(V)i) with P(V)o = E9{ Vi/Vr : i E sj and P(V)i =E9{Vj/Vrj~i}

for each i in S. If V is trivial, define P(V) = V. Then P(V) is a finite direct sum of copies of P(t), t E S U {O}, whence, by Lemma 6.1.2(a), P(V) is a projective representation in rep(S , k). If V E rep(S, k) has no cotrivial summands, let Vi = n{ U, : j > i in S} agreeing that Vi = Vo if there is no j in S with j > i , Define I(V)

= (/(V)o, I(V)i) with I(V)o = E9{V;j Vi : i E S} and I(V)i = E9{Vj/ u.. i i j}.

If V is cotrivial , define I(V) = V . By Lemma 6.1.2(b), I(V) is an injective representation in rep(S, k) for each V . Lemma 6.1.4 Let V

E

rep(S, k)for afinite poset S andfield k.

(a) There is a projective cover pu : P(V) -+ V and (b) an injective envelope qu : V -+ I(V).

PROOF. (a) It suffices to assume that V has no trivial summands. For each i, there is a k-linear transformation hi : Vi -+ Ui] Vr given by h(x) = x + Vr Since ker hi = Vr is a vector space summand of Vi, there is a k-linear transformation gi : Ui] Vr -+ Vi S; Vo with h ig, = 1. Define a k-linear transformation p : P(V)o = E9{ Ui] Vr: i E S} -+ Vo by p = 'E{gi: i E S}. Then p is a representation morphism, since P(P(V)i) = p(E9{Vj/ Vr j s i}) = 'L, {g/V j/ VJ): j ~ i} S; 'L, {Vi : j s i} S; Vi for each i E S. An induction argument, beginning with a minimal element of S, shows that P(P(V)i) = Vi for each i in S. In particular, if i is a minimal element of S, then Vr = 0, P(V)i = Vi, and P(P(V)i) = Vi . Now assume that i is not a minimal elementofS and p(P(V)j) = Vjforeachj
E

rep(S , k) is both projective and injective, then w(S)

=

= I.

6. Confirm the assertions of Lemma 6.1.4(b) and Example 6.1.5.

°

7. The Cartan matrix of a finite poset S is an lSI x lSI matrix C(S) = (cij), where Cij = 1 if i ~ j in Sand Cij = otherwise. Show that the ith column of C(S) is the transpose of the vector cdn P(i) and that the ith row of C(S) is cdn I(i). 8. Define the extended Cartan matrix to be an (lSI + 1) x (lSI + 1) matrix C*(S) consisting of C(S) together with last row (0, 0, . . . , 0, 1) and last column the transpose of (-1, -1, . . . , -1 ,1 ). Label the elements of S by {I, 2, . . . , n}. Given variables XI , , Xn+l , define the quadratic form as of S by qS(Xl , ... , Xn+l ) = (Xl , , Xn+1)C*(S )(XI, , xn+d ' . (a) Show that qS(XI, ,X n+() = L{x1:l~i~n+l}-L{xiXj:i<jinS}+ Xn+l (L{Xi : 1 s i s n}). (b) [Ringel 84] In each example below, confirm that qs (z ) = 0, where z = (zj , .. . , zn+() and S = {I, 2, . . . , n}.

178

6. Representations over Fields and Exact Sequences (i)z=(1 ,1 ,1 ,1,2),

(ii)

z = (1, 1, 1, 1, 1, 1, 3),

S=l

2

3

S= 2

4

6

I

(iii)

z = (2, 1, 1, 1, 1, 1, 1, 4),

I

3

5

S= 1

4

7

3

I

(iv) z = (3,2,2, 1, 1, 1, 1, 1, 6),

I

I

I

S= 1

I

6

I

2

5

3

8

I

2

4

I

7

I

6

I

5

I

4 (v) Z

= (1,2,2,1,1 ,1 ,1,1,5),

S=2

4

I\ I

1

3

8

I

7

I

6

I

5 (c) Provethat if qs(z) > 0 for each 0 =/: z = (ZI, .. . , Zn) with Zi a nonnegative integer for each i, thenS has finiterepresentation type(theconverseis alsotrue; [Drozd74] or [Simson92]).

9. Provethat U = (Uo, U, : i E S) E rep(S, k) is a sincere representation if and only if the projectivecover P(U) of U contains a copy of P(i) for each i E S. 10. Prove that projective covers and injective envelopes in rep(S, k) are unique up to isomorphism.

6.2 Coxeter Correspondences The construction of matrices to find indecomposable representations in rep(S , k) , as described in Section 1.2, is impractical for a general finite poset. In this section, correspondences are defined that can be used to generate indecomposables from a given indecomposable. In fact, ifrep(S, k) has finite representation type, then all indecomposables in rep(S, k) can be found by applying these correspondences to l-dimensional projective or injective representations (Theorem 6.2.6). Let V = (Vo, Vi : i E S) E rep(S, k) and PU : P(V) ~ V be a projective cover of Vas defined in Lemma 6.1.4(a) with P(V)o = tI7{V;/ Vr: i E S} and P(V)i =

6.2 Coxeter Correspondences

${ Vjl V J: j ~ i}. Define representations W (V ) = (Wo, Wi : i Wo = P(V)o

E

179

S) , with

= e] Vi/V!: i E S} , Wi = ${Vi/V;: i i

j}

and L (V ) = (L (V )o, L(V)i : i E S), with L(V)o = Vo, L(V)i

= 'E{Vj:i ij} .

Then

c+(V) = (ker pi), (ker Pu) n Wi: i

E

S)

is a kernel of the representation morphism pu : W (V) --+ L( V). Hence, 0--+ C+(V) --+ W(V) --+ L(V) --+ 0 is an exact sequence of representations with W(V) an injective representation by Lemma 6.1.2(b) . Let qu : V --+ I (V) be an injective envelope of V as defined in Lemma 6.1.4(b). Define representations

with

and K(U ) = (K (U )o, K (U )i ), with K(U)o = qu(Uo), K (U) i = qu(Uo) n W'(U)i. Then

is a cokernel of the pure monomorphism K(V) --+ W'(V). Hence, 0--+ K(V) --+ W'(V) --+ C- (V) --+ 0 is an exact sequence of representations with W' (V) a projective representation. Proposition 6.2.1 Let S be a finite poset and k a field. Then C+ and C- are correspondences from rep( S, k) to rep( S, k). If U E rep( S, k), then C+ V = 0 if and only if V is a projective representation, and C- V = 0 if and only if V is an injective representation. Moreov er, C+ and C- preserve finite direct sums.

180

6. Representations over Fields and Exact Sequences

The representation U is projective if and only if P(U) = U, equivalently, the kernel of pi): P( U) -+ U is O. Thus , C+ U = 0 if and only if U is projective. A similar argument shows that U is injective if and only if C- U = O. Both P(U)$P(V) and PCU$V) are projective covers of U$V . Hence P(U)$P(V) is isomorphic to P(U $ V) by Exercise 6.1.10 . Now, C+(U $ V) is constructed in terms of P (U $ V) , while C+ U $ C+ V is constructed from P(U) $ P(V). It follows that C+(U $ V) is isomorphic to C+ U $ C+ V. A similar argument, using injective envelopes, shows that C-CU $ V) is isomorphic to PROOF.

C-U $ C-V.

Example 6.1.5 (continued) (c) If U has no trivial summands, then

W(U)

= (U I $

.. . $ Un, Uz $ . . . $ Un '

UI $ U3 $ L(U) = (Uo , Uz +

$ Un,·· ·, UI $

+ Un "

' " UI +

$ Un-I),

+ Un -I>,

and C+U

= (ker pi) , (ker Pu) n CUz $

.. . $ Un) , .. . ,

(ker Pu) n CUI $ .. . $ Un-I» ·

(d)

If U has no cotrivial summands, then W'(V) = (Vo/V I E9 • . . E9 VO/V n, Vo/V 1 ,

••• ,

Vo/Vn),

K(U) = (qu(Uo), qu(Uo) n (Uo/ VI> , . . . , qu(Uo) n (Vo/ Un», and

C- U

= «Uo/ UI> $

. . . $ (Uo/ Un»/qu(Uo),

«Uo/Ui)+qu(Uo»/qu(Uo) : 1 ~i

~n).

=

(e) Suppose cdn U (uo, UI,"" un)' (i) If U is not a projective representation, then cdn C+U

= (UI + UI

(ii)

+

+ Un + Un -I

Uo , Uz +

... + Un -

Uo, . .. ,

- uo)·

If U is not an injective representation, then cdnC-U = «n -l)uo - UI - . . . - Un, Uo - UI ,.· · , Uo - Un)'

cv

For a positive integer i, define to be the composition of C+ repeated i times and define to be the composition of C- repeated i times. Define C+OU = C-oU = U. Call U in rep(S, k) preprojective if U is isomorphic to P for

c:'

c:'

6.2 CoxeterCorrespondences

181

some projective P and i ::: 0, and preinjective if U is isomorphic to C+ i I for some injective I and i ::: 0. Each projective is preprojective and each injective is preinjective. Example 6.2.2 Following is a listingofthe indecomposable preprojectives and preinjectivesfor rep(S3, k): (a) indecomposable preprojectives: P(O), P(1), P(2), P(3) ,

°

C- P(O) = (k EfJ k, k EfJ 0, EfJ k, (1 + l)k),

= 1(0), C- P(1) = 1(1), C- P(2) = 1(2), C- P(3) = 1(3).

C- 2 P(O)

(b) indecomposable preinjectives: 1(0),/(1),/(2),/(3)

C+ 1(0) = C- P(O), C+2 / (O) = P(O), C+ 1(1) = P(1) , C+ 1(2) = P(2), C+ 1(3) = P(3). Recall that P(O) = (k , 0, 0, 0), P(1) = (k, k, 0, 0), P(2) = (k, 0, k, 0), and P(3) = (k , 0, 0 , k) are the projectives for rep(S3, k) . Then cdn P(O) = (1,0,0,0), cdn P(1) = (I, 1,0,0), cdn P(2) = (1,0,1,0), and cdn P(3) = (1,0,0, I) . The injectives for rep(S3, k) are 1(0) = (k, k, k, k) , 1(1) = (k, 0, k, k) , 1(2) (k , k, 0 , k), 1(3) = (k , k, k , 0) with cdn 1(0) = (1,1 ,1,1), cdn 1(1) = (1,0,1 , I), cdn 1(2) = (1, 1,0, I), and cdn 1(3) = (I, I, 1,0). (a) By Example 6.1.5(e), cdn C- P(o) = (2, 1, I , I), cdn C- 2 P(o) = (l, I, I, I) cdnC-P(1)=(l,O, I, I), cdnC-P(2)=(1, 1,0, 1), and cdnC-P(3)=(l,I,

PROOF.

=

1, 0). Since l-dimensional representations are determined up to isomorphism by their coordinate vector, it remains only to determine C- P(O). By Example 6.1.5(d), W'(P(O» = (k EfJ k EfJ k, k EfJ EfJ 0 , EfJ k EfJ 0, EfJ EfJ k), K(P(O» = «1 + 1 + l)k , 0, 0, 0), and C- P(o) = (Vo, VI> V2 , V3) with

° °

°°

+ 1 + I)k, VI = «k EfJ EfJ 0) + (1 + I + I)k)/(1 + 1 + I)k, V2 = «0 EfJ k EfJ 0) + (l + I + I)k)/(1 + 1 + I)k, V3 = «0 EfJ EfJ k) + (1 + 1 + l)k)/(1 + 1 + l)k. It follows that C- P(O) is isomorphic to (k EfJ k, k EfJ 0, EfJ k, (1 + l)k) . Vo = (k EfJ k EfJ k)/(l

° °

°

(bj By Example S.l.Xej.cdn C't J'(O) = (2,1,1 , 1),cdnC+2/(O) = (1,0,0,0), cdn C+ 1(1) = (1, 1,0,0), cdn C+ 1(2) = (1,0, 1, 0) and cdn C+ 1(3) =(1,0,0, 1). Once again, only C+ I (0) needs to be computed. By Example 6.1.5(c),

182

6. Representations over Fields and Exact Sequences

W(l(O» = (k EB k EB k , 0 EB k EB k, k EB 0 EB k , 0 EB 0 EB k), L(l(O» and C+ 1(0) = (k EB k, k EB 0, 0 EB k, (l + 1)k), as desired.

= (k, k , k, k),

There are several intriguing observations about the list in Example 6.2 .2. The first is that C+ and C- applied to an indecomposable injective representation of 53, respectively an indecomposable projective representation, resulted in an indecomposable representation. This property is, in fact, true for each finite poset and each indecomposable representation by Corollary 6.2.5(d) . Secondly, each indecomposable in rep(53, k) is both preprojective and preinjective. This property holds for each 5 such that rep(5, k) has finite representation type by Theorem 6.2.6. Finally, if V E rep(53 , k) and V is indecomposable, then V is isomorphic to C- C+ V if C+ V =1= 0 and V is isomorphic to C+ C- V if C- V =1= O. This property holds for each finite poset 5 by Corollary 6.2.4(c) and (d). Fundamental properties of the correspondences C+ and C- are derived in Corollary 6.2.4 by showing that the duality a of Lemma 6.1.1 is a factor of both C+ and

C-.

Proposition 6.2.3 If V E rep(5, k), 5 a finite poset, and k afield, then (a) aC+(V)

=

C-a(V); = a C+ and V has no projective summands, then p2(V) is naturally isomorphic to V ; (c) 0 ~ C+V ~ W(V) is an injective envelope and (d) W' (V) ~ C- V ~ 0 is a projective cover of U, (b)

if p

«c-o, c-

PROOF. (a) There is a projective cover 0 ~ M (V) ~ P( V) ~ V ~ 0 of U in

rep(5, k) with P(U)o = EB{Ui/Ur i E 5}, and P(U)j = EB IUd UJ: j ::::: i}. Then o ~ C+(V) ~ W(V) ~ L(V) ~ 0 is an exact sequence of representations with

and L(V)o

= Vo, L(V)j = E{V j : i i

j).

Hence, 0 ~ a L(V) ~ aW(V) ~ aC+(V) ~ 0 is an exact sequence of representations with aW(U)o

= a P(V)o = EB{a(Vi/Vr) : i E 5}, aW(V)j = EB{a(Vi/V7): j::: q,

and a L(U)o

= aVo, a L(U)i = a(E{Vj : i i

j}).

On the other hand, 0 ~ a V ~ a P(V) ~ a M(V) ~ 0 is an injective envelope of a V . Then 0 ~ K (a V) ~ W'(a V) ~ C-(a V) ~ 0 is an exact sequence of

6.2 Coxeter Correspondences

183

representations with

= a P(U)o = EB{a(U;/Ur): i E S}, W'(aU); = EB{a(UdUJ): t s i},

W'(aU)o

and

= aUo , K(U); = a(~{Uj: i i j}) . Thus, aW(U) = W'(aU) and aL(U); = K(U); for each i E S. This shows that aC+(U) = C-aU, as desired. K(aU)o

(b) There is an exact sequence of representations

0---+ C+(U) ---+ W(U) ---+ L(U) ---+ 0,

recalling that W(U) is injective with W(U)o = P(U)o and L(U)o Applying a yields an exact sequence of representations 0---+ a L(U) ---+ a W(U) ---+ p(U)

= Uo.

= aC+(U) ---+ 0

with a W(U) projective. Now, P(p(U» is a projective cover of p(U), so that aW(U) = P(p(U» EB P for some projectiverepresentation P. The definition of C+(p(U» yields an exact sequence 0---+ C+(p(U» EB / ---+ W(p( U» EB / ---+ L(p(U» ---+ 0

for someinjective / with W(p(U»o EB /0 = P(p(U»o EB Po = a W(U)o Hence, 0---+ a L(p(U» ---+ a(W(p(U» EB I) ---+ aC+ p(U) EB a(l)

= a P(U)o.

= p2(U) EB a(l) ---+ 0

is an exact sequence of representations with a(W(p(U» EB I) projective and a(W(p(U» EB 1)0 a 2P(U)o. Thus, a 2 : P(U)o ---+ a(W(p(U»EB 1)0 is a vector space isomorphism that induces a representation isomorphismex : U ---+ p2(U) EB a(l) with a(l) projective. Since U has no projectivesummands, a(l) = 0, whence / = 0 and P = O. Hence, U is isomorphicto p2(U). It is straightforward to see that this isomorphismis nat-

=

ural. (c) As in the proof of (b), a(W(U» ---+ aC+U is a projective cover of aC+U . By Lemma 6.1.1, C+(U) ---+ W(U) must be an injective envelopeof C+ U. (d) Since, C-aU = aC+U, it follows that W'(U) ---+ C- U is a projectivecover ofC-U .

Corollary 6.2.4 Let p

= aC+ and U E rep(S, k). Then:

(a) ifU has no trivial summands, then C+U is isomorphic to ap(U); (b) ifU has no cotrivial summands, then C-U is isomorphic to pa(U); (c) if U has no projective summands, then C- C+ U is isomorphic to U ; (d) if U has no injective summands, then C+C- U is isomorphic to U;

184

6. Representations over Fields and ExactSequences

(e) V is preprojective if and only if C+i V = 0 for some i ~ I.. and (f) V is preinjective if and only if V = 0 for some i ~ I.

c:'

PROOF. (a) As a consequence of Lemma 6.1.1, a p( V) = a 2 C+V is isomorphic to C+V for each V E rep(S, k) . As for (b), pa(V) = aC+a(V) = C-(a 2(V» by Proposition 6.2.3(a), and C-(a 2(V) is isomorphic to C-(V) by Lemma 6.1.1. (c), (d) Notice that C-C+V = (C-a)(aC+)(V) = p2(V) is isomorphic to V as a consequence of Lemma 6.1.1 and Proposition 6.2.3(b) . Assertions (e) and (f) are consequences of Proposition 6.2.1 and (c) and (d), respectively.

Corollary 6.2.5 Suppose V, V

E

rep(S, k), Sis a finite poset, and k is afield.

(a) If V has no injective summands and V has no projective summands, then Hom(V, C+V) is isomorphic to Hom(C-V, V). (b) If V has no projective summands, then End V is isomorphic to End C+ V. (c) If V has no injective summands, then End V is isomorphic to End C- V. (d) If V is indecomposable, then C+ V and C- V are indecomposable. (e) If V is either preprojective or preinjective, then End V is isomorphic to k. PROOF. (a) It is sufficient to prove that Hom(V , V) is isomorphic to Hom(p(V), p(V». To see this, observe that Hom(V, C+V) = Hom(V , ap(V» is isomorphic to Hom(p(V), a(V» by Lemma 6.1.1. Then Hom(p(V) , a(V» is isomorphic to Hom(pa(U), p2(V» = Hom(C -U, V) via Proposition 6.2.3. Let Pu : P( U) -+ U -+ 0 be the projective cover of U given in Lemma 6.1.4 and f: U -+ V. By the construction of P(U), W(U),and L(U),f inducesg: W(U)-+ W(V) and h: L(U)-+ L(V) with hpu = Pv S» where pu : W(U)-+ L(U). Since C+ U = Ker Pu, f induces f': C+ U -+ C+ V. This yields a homomorphism a:Hom(V,V)-+Hom(C+U,C+V) given by a(f)=f'. By Lemma 6.1.1, a :Hom(C+V,C+V)-+Hom(aC+V,aC+V) is an isomorphism. Since p = o C", a a : Hom(V, V) -+ Hom(p(V), p(V» is a homomorphism. Similarly, there is a homomorphism Hom(p(V), p( V» -+ Hom(p2V, p2V). It follows from Proposition 6.2.3(b) that Hom(V, V) is isomorphic to Hom(p(V) , p(V» . (b), (c) Now, Hom(V, V) is isomorphic to Hom(V, C+C- V) by Corollary 6.2.4(d), and by (a), Hom(V, C+C-V) is isomorphic to Hom(C-V, C-V). On the other hand, Hom(V, V) is isomorphic to Hom(C-C+ V, V), by Corollary 6.2.4(c), which is isomorphic to Hom(C+V, C+V) by (a). Statement (d) is an immediate consequence of (b) and (c), since a representation V is indecomposable if and only if 0 and I are the only idempotents of End V . (e) Assume that V is an indecomposable preprojective representation. Then V = c:' pet) for some t E S U {O} . By (a), End U = End C+i c:' pet) = End pet). But End pet) = k, since pet) is a I-dimensional representation. Similarly, if V is an indecomposable preinjective representation, then End U = End I (t) = k for some tESU{O}.

6.2 CoxeterCorrespondences

185

Theorem 6.2.6 Let S be afinite poset and k afield. Thefollowing areequivalent: (a) rep(S, k) has finite representation type; (b) If U is an indecomposable in rep(S, k), then U is preinjective; (c) If U is an indecomposable in rep(S , k), then U is preprojective; (d) End U is isomorphic to kfor each indecomposable U in rep(S, k).

In this case, ifU is an indecomposable in rep(S , k), then U is determined up to isomorphism by cdn U. PROOF. (a)::::} (b) [Ringel, 80].

(b) ::::} (d) follows from Corollary 6.2.5(e) . (d) ::::} (a) If rep(S , k) has infinite represention type, then S contains one of the critical posets T = S4 , (2,2,2),0,3 ,3),0,2,5), or (N, 4) as a subposet by Theorem 1.3.6. Choose an indecomposable U in rep(T, k) with End U not isomorphic to k, as given in the proof of Theorem 1.3.6. By Proposition 1.3.1, U gives rise to an indecomposable V in rep(S, k) with End V not isomorphic to k. (c)::::} (d) is a consequence of Corollary 6.2.5(e). (a)::::} (c) follows from (a)::::} (b), Corollary 6.2.4, and Proposition 1.4.7.

Example 6.2.7 The categoryteot S«, k) has tame representation type. Following is a complete list ofelements of Ind(S4, k): (a) indecomposable preprojectives U; all have endomorphism ring k and are classified by cdn U: (i) cdn c:"P(O) = (2n + 1, n, n, n, n),

U(XI , " " x n)

= (0, XI ," "

V(XI , •• • ,xn)

= (XI, ...

X n, X n , ..• , XI)

, X n, X n , • • • , XI,

and

0).

(ii) cdn C- 2n PO) = (2n - 1, n, n - 1, n - 1, n - 1),

c- 2n PO) =

(kn E9 k n -

I,

k n E9 0, 0 E9 k n -

I,

u'(k n -

I ),

v'(k n -

I

», where

= (XI, . . . ,Xn-I, XI, X2, ••• , Xn-I , 0) and V'(XI , • •• , xn -d = (XI, . •. , Xn-I , 0, XI ,···, xn-d .

U'(XI , .•• , Xn-I)

(iii) cdn C- 2n P(2) = (2n - 1, n - 1, n, n - 1, n - 1),

(iv) cdnC- 2n P(3) = (2n -1 ,n - 1, n -1,n ,n -1),

C-2n P(3) = (kn E9 r:', 0 E9 r:', u'(k n -

I

),

r

E9 0, v'(k n -

I

».

186

6. Representations over Fields and Exact Sequences

(v) cdn C- 2n P(4) = (2n - 1, n - 1, n - 1, n - I, n), C- 2n P(4)

= (k n $ r:' , 0 $ r:', u'(kn- t ), v'(k n- t ), k n $

0).

(vi) cdn c-(2n-l) P(l) = (2n, n - 1, n, n , n),

C-2n- 1 P(l)

= (k"

$ k" , w(k n- I), k" $ 0,0 $ k", (l

W(XI,·· ·, Xn-l)

(vii) cdn c-(2n-l) P(2)

+ I)(k n»

,

= (0, Xl, ... , Xn- l , Xl, .··, Xn-l , 0).

= (2n , n, n -

C-(2n-l) P(2) = (k n $

1, n, n),

r, k n $

0, w(k n- I ) , 0,

r, (l + I)(kn».

= (2n, n, n, n - 1, n), C-(2n -l) P(3) = (k n $ k", k" $ 0, 0 $ k" , w(k n- l), (l + I)(k n».

(viii) cdn c-(2n-l) P(3)

(ix) cdn c-(2n-l) P(4) = (2n, n, n, n , n C-(2n-l) P(4) = (k n $

r, k n $

0,

0, 0 $

v, (l + 1)(k n) , w(k n- ». 1

(b) indecomposable preinjectives V; all have endomorphism ring k and are classified by cdn V. (i)-(ix) Apply a to the representations in (a).

(c) cdn VA

= (2n , n , n , n, n)

VA = (k n tIJkn,k n tIJ 0, 0 tIJ k n, (1 + l)kn,(l+A)kn),EndVA :::::k[x]j (g(x)") , A an n x n k-matrix with minimal polynomial g(xY, g(x) irreducible.

(d) cdn V = (2n

+ 1, n , n + 1, n + 1, n)

V = (k n $kn+l,k n $O,O$kn+l ,a(kn+l),b(kn»),EndV:::::k[x]/(xn), where a(xl, . . . , xn+t> = (Xl, . . . ,xn, Xl, ... , Xn+l) and btx«, . .. , x n) (Xl, ... , Xn, 0, Xl> . . . , Xn) andfive other similar representations obtained by permuting the subspaces, VI, V2, V3, V 4 corresponding to permutations ofcdn V.

=

PROOF . [Gelfand, Ponomarev 70] or [Brenner 74B].

EXERCISES

1. Prove that if S is a forest, then C+ , C" : rep(S , k) -? rep(S, k) are functors. What is the situation if S is not a forest?

2. Confirm the assertions of Example 6.1.5(c), (d), and (e). 3. Explicitly construct the representations in Example 6.2.7(b) and (d).

6.3 Almost Split Sequences

187

4. Suppose U E rep(S, k ) with cdn U = (uo, u; : i E S). (a) Prove that if cdn a (U ) = (vo, Vi : i E S), then Vo = uo and 'E{u j : j ~ i} + 'E{Vj :j > i} = Uo for each i E S. (b) Prove that if cdn p (U ) = (wo, Wi : i E S), then Wo = 'E{(Ui - uo) : i E S}and ui ; = ui for each i E S. 5. Let X, Y E rep(S, k) be l-dimensional representations. Prove that for each positive integer n : (a) C+n X is isomorphic to C+n Y if and only if cdn C+n X = cdn C+n Y. (b) C- n X is isomorphic to c » Y if and only if cdn c » X = cdn C- n Y.

6.3 Almost Split Sequences Almost split sequences are one of the most important tools for classification of finitely generated modules over finite-dimensional k-algebras with tame representation type. The most important theorem about almost split sequences in rep(S, k) is that they exist (Theorem 6.3.4). Even more, the left-hand term of an almost split sequence in rep(S, k) determines the right-hand term, and the right-hand term determines the left-hand term. Almost split sequences in rep(S, k) can be constructed explicitly (Corollary 6.3.5 ). A representation morphism f: V -+ V in rep(S, k) is a splitting monomorphism if there is g : V -+ V with gf = I u and a splitting epimorphism if there is g : V -+ V with f g = I v. An exact sequence

£ :0-+ V ~V";'W-+O of representations in rep (S, k) is split exact if there is h: V -+ V with hf = I u , the identity on V. Equivalently, there is h : W -+ V with gh = I w- Then E is an almost split sequence if: (i) E is not a split exact sequence; (ii) if h : Y -+ W is not a splitting epimorphism, then there is a : Y -+ V with ga = h; and (iii) if h : V -+ Y is not a splitting monomorphism, then there is a : V -+ Y with af =h. The sequence E is left almost split if (i) and (iii) hold and right almost split if (i) and (ii) hold. Given exact sequences

£ :0-+ V~V";'W -+ 0, J'

g'

£':0-+ V '-+ V '-+ W -+ 0, call (a , (3, y): E -+ E' a homomorphism of exact sequences if a: V -+ V' , {3 : V -+ V ' , and y : W -+ W ' are representation morphisms with {3f = f'a, and yg = g'{3 . The homomorphism (a , (3, y) is an isomorphism if a, {3, and y are representation isomorphisms.

188

6. Representations over Fields and Exact Sequences

As a consequence of the following lemma, almost split sequences are unique up to isomorphism. Lemma 6.3.1 Suppose Sis a finite poset, k is afield, and f

g

E :O~V~V~W~O

is an almost split sequence in rep(S, k) (a) Both End V and End Wa re local rings. In particular, V and Ware inde-

composable representations.

(b)

If

f'

g'

E':O~ V'~V'~W~O

is a right almost split sequence with V' indecomposable, then there is an isomorphism (a , a, 1) : E' ~ E.

(c)

If

f'

e'

E' :O~ V~V '~W'~O

is a left almost split sequence with W' indecomposable, then there is an isomorphism (1, a, a): E' ~ E.

(a) Suppose hand h' are nonunits in End W. Since End W is a finitedimensional k-algebra, h and h' are not splitting epimorphisms. Choose f3 , f3' : W ~ V with gf3 = hand gf3' = h'. Then g(f3 + f3') = h + h' = u is not a unit of End W ; otherwise, u- I g(f3 + f3') = 1, contradicting the assumption that E is not split exact. (b) Since E' is not split exact, g' is not a splitting epimorphism. Choose a : V' ~ V with ga = g'. This choice is possible, since E is almost split. But E' is right almost split, so that there is f3 : V ~ V' with g' f3 = g. Since g' f3a = g'l w. there is a homomorphism (y, f3a, 1) : E' ~ E' with Ba]' = I'r . If y is a unit of End V', then a: f'(V ') ~ f'(V') is a unit, since V ' is indecomposable. In this case, (a, a , 1): E' ~ E is the desired isomorphism. Now assume that y is not a unit of End V'. Since V'is indecomposable, y is nilpotent, say yn = 0 for some n. Then g'(f3a)n = g', an epimorphism from V' to W, and (f3a)n induces h': W ~ V', since yn = O. As g'(f3a)n = g'l w' and g' is onto , it follows that g'h' is a unit of End W, a contradiction to the fact that g' is not a splitting epimorphism. (c) The proof is dual to that of (b). PROOF.

Almost split sequences are minimal with respect to being a nonsplit exact sequence, provided that the left- and right-hand terms of the sequence are known.

6.3 Almost Split Sequences

189

Lemma 6.3.2 Assume that S is a finite poset, k is a field, and that there is an almost split sequence

Let

be a nonsplit exact sequence in rep( S, k) . The following statements are equivalent :

(a) E ' is a left almost split sequence. (b) If V' is a nonzero pure subrepresentation of V , then 0 --+ V / V ' --+ V' / !'(V') --+ W --+ 0 is a split exact sequence. (c) E' is an almost split sequence. (d) If W' # W is a pure subrepresentation of W , then the sequence 0 --+ V --+ (g')-I(W') --+ W' --+ 0 is split exact. (e) E ' is a right almost split sequence. PROOF . (a) =} (b) Since V' is nonzero, h : V --+ V / V ' is not a splitting monomorphism . By (a), there is a: V' --+ V / V ' with a]' = h. Now, af'(V') ~ ker h = V', so a : V' / f'(V') --+ V / V' with a]' = 1, as desired. (b) =} (c) Notice that g' : V' --+ W is not a splitting epimorphi sm, since E' is not a split exact sequence. Since E is almost split, there is a : V ' --+ V with ga = s'. If a' = f-Ia!' is a unit of End V, then (a ' , a, 1) : E' --+ E is an isomorphism, whence E' is an almost split sequence . If a ' is not a unit, then ker e ' is a pure nonzero subrepresentation of V . By (b), there is a split exact sequence 0--+ V /ker e" --+ V' / f '(kercx) --+ W --+ O. Since a sends V' /ker o' isomorphically to a subrepresentation of V and ga = s'. this split exact sequence induces a splitting of E', a contradiction. (c) =} (a) is clear. The equivalence of (c), (d), and (e) is proved by a dual argument.

Corollary 6.3.3 Suppose E: 0 --+ V

~V

s; W --+ 0

is an almost split sequence in rep(S, k) and either End V or End W is a division algebra. If I'

g'

E' : 0 --+ V --+ V' --+ W --+ 0 is an exact sequence that is not split exact, then E' is an almost split sequence. PROOF . Assume that End W is a division algebra. Then!, : V --+ V ' is not a splitting monomorphism, since E' is not a split exact sequence . Since E is almost

190

6. Representations over Fields and Exact Sequences

split, there is a: V ~ V' with af = f '. Then a induces f3 : W ~ W with f3g = g'«. Moreover, 0 =1= f3, because E is not split exact. Since End W is a division algebra, f3 is an automorphism of W. Thus, 0, a, f3) : E ~ E' is an isomorphism, and E' is an almost split sequence. The proof for the case that End V is a division algebra is dual. The proof of the following theorem is beyond the scope of this book .

Theorem 6.3.4 [Bautista, Martinez 79] Let S be a finite poset and k afield.

If W is an indecomposable in rep(S, k) that is not projective, then there is an almost split sequence 0 ~ C+ W ~ V ~ W ~ O. (b) If V is an indecomposable in rep(S, k) that is not injective, then there is an almost split sequence 0 ~ V ~ V ~ C- V ~ O. (a)

Given that almost split sequences exist, they can be constructed as follows :

Corollary 6.3.5 Suppose S is a finite poset, k is a field, and V E rep( S , k) is a nonprojective indecomposable representation. Let Pu : P(V) ~ V be a

projective cover of V and

o ~ C+V ~

W(V)~ L(V) ~ 0

the resulting injective envelope «c-u. (a) There is a morphism i : V ~ L(V) induced by inclusion of Vo in L(V)o. (b) Define B = (Bo, B, E S) with Bo = {(x, y) E L(V) EB W(V) :ip u(x) = iCY)} and B, = Bo n ((C+V) i EB l-t) for each i E S. The sequence 0 ~ C+V ~ B ~ V ~ 0 is an almost split sequence in rep(S, k) . PROOF. Exercise 6.3.4.

Let (XI, " " X n ) be an n-tuple of l-dimensional representations in rep(S , k). Choose a representation embedding f i : Xi ~ I (0) and define V (X I, .. . , X n) E rep(S, k) to be the kernel ofEBfi : EB{X i : i E S}~ 1(0). Representations V(X i , ... , X n ) are the representation analogue of bracket groups, as defined in Chapter 7.3 .

Example 6.3.6 Suppose X E rep(S, k) is nonprojectivewith dimension 1. Then C+ X = V(l(i): Xi =1= O)for I-dimensional injective representations I(i), and

the sequence

o~

-4 VeX, I(i) : Xi =1= 0) ~ X ~ 0 = (0, XI , • •. , X n ) , g(x , XI, •• • ,Xn ) = X is an almost split

V(l(i): Xi

with f(xI, . .. ,Xn ) sequence.

=1=

0)

6.4 A Torsion Theory andLocalizations PROOF. Up to isomorphism, it suffices to assume X cover PX : P(X)

= $(P(i) : X; =f. O} -+

~

191

/(0). There is a projective

X with Px($;Y;)

= ~;y;.

Then 0 -+ C+ X -+ W(X) -+ L(X) -+ 0 is an injective envelope of C+ U with W(X) = ${I(i) : X; =f. O} an injective representation and X ~ L(X) a l-dimensional representation. Hence, C+ X = U(l(i) : (X); =f. 0), being the representation kernel of PX : W(X) -+ L(X). Notice that U(X, /(i): (X); =f. 0) = B, as defined in Corollary 6.3.5.

EXERCISES I. A representation morphism f : U -+ V in rep(S, k) is irreducible if f is neither a split monomorphism nor a split epimorphism and if f = gh for some h :U -+ Wand g : W -+ V, then either h is a splitting monomorphism or g is a splitting epimorphism.

Prove that if 0-+

U~V~W-+O

is an almost split sequence in rep(S, k) and V = VI ED . . . ED Vn with each V; an indecomposable representation, injections i i - and projections tti - then it j f and gij are irreducible morphisms foreach j . 2. Prove thatif f : U -+ V is irreducible, then f =f. L,j h j g j whenever Zj is an indecomposable representation and g j : U -+ Zi - h j : Zj -+ V arenotisomorphisms. 3. Prove Lemma 6.3.I(c) andtheequivalence of (c), (d), and(e) in Lemma 6.3.2. 4. Provide a proofforCorollary 6.3.5.

6.4 A Torsion Theory and Localizations Given a finite poset S with unique maximal element 00 and discrete valuation ring R, there is an associated category of representations of a quiver over R and a torsion theory for which the poset representations of S are the torsion-free elements (Corollary 6.4.6). A poset S can be viewed as a quiver, a directed graph. In this case, vertices are elements of S, and there is a directed edge i -+ j if and only if i :::: j in S. Representations of quivers and finitely generated modules over finite-dimensional k-algebras, k a field, are investigated thoroughly in [Ringel 84] and [Simson 92]. This chapter is devoted to a brief discussion of the category of quiver representations of S over R with relations induced by the poset S and its connection with rep(S, R). Define a category Qrep(S, R) with objects M = (M; : fij : i :::: j) where each M; is a finitely generated free R-module, fij : M; -+ M, is an R-homomorphism with pure image whenever i :::: j in S, [u is the identity of M; for each i E S, and

192

6. Representations over Fields and Exact Sequences

/ik fij = hk whenever i ~ j ~ k in S. A morphism from M to M' is g = {gi : i E S} with each gi : M, ~ M; an R-homomorphism and gihj = fijgi whenever i ~ j in S. Alternatively, Qrep(S, R) can be thought of as a category of modules over the R-incidence algebra of S that are finitely generated and free as R-modules; Exercise 6.4.1.

Proposition 6.4.1 Let S be afinite poset with unique maximal element 00 and R a discrete valuation ring. Then Qrep(S, R) is an additive category closed under direct sums, idempotents split in Qrep(S, R), andrep(S, R) is afull subcategory of Qrep(S, R). PROOF . It is straightforward to confirm that Qrep(S, R) is a category and the set ofmorphisms from M to M' is an abelian group for each M, M' in Qrep(S, R). If M = (M j, f ij: i ~ j) and M' = (M;, f: j : i ~ j) arein Qrep(S, R), then M E9 M' = tM, E9 M;, fij E9 f: j : i ~ j) is a direct sum of M and M' in Qrep(S, R). If e = {ej: i E S} is an idempotent in End M , the endomorphism ring of M in Qrep(S, R), then el = e, for each i. Hence, M = K E9 L with K = (ei(Mi), ej h jei: i ~ j) and K E Qrep(S, R), since ei!ijei has pure image in ej(Mj) and ekf·ke· J J Je.·e· IJ I = ekJ"ke I I' whenever i -< J' < - kin S. Let V = (V oo , Ui : i E S) E rep(S, R). Then M = (V ;, hj : i ~ j) E Qrep(S, R), where fij : Vi ~ U, is inclusion for i ~ j in S. Each hi has pure image, since Vi is pure in V oo , hence in U], whenever i ~ j, and /ikhj = hk whenever i ~ j ~k in S. ThecorrespondencefromHom(V, V') in rep(S, R) to Hom(M , M') in Qrep(S, R) given by f ~ {h = f: i E S} is an isomorphism for each V, V ' E rep(S, R) . It follows that rep(S, R) is a full subcategory of Qrep(S, R) .

If N = tN], gij) and (M i, fij) are in Qrep(S, R), then N is a subrepresentation of M if N, is a submodule of M, for each i and each gij is the restriction of hj to Ni. Given M = (M i, fij : i ~ j) E Qrep(S, R), define T(M) = (T(M)i, gij : i ~ j) by T(M)oo = 0, T(M)i = ker hoo if i E S\{oo}, and gij : T(M)i ~ T(M)j the restriction of hj to T(M)i. Then gij is well-defined, and the image of each gij is pure, since /ioohi = f ioo and the image of f ioo is pure. Notice that T(M) is a subrepresentation of M and each T(M)i is a free R-module, being a submodule of the free R-module M) . Moreover, M;/T(M)j is a finitely generated torsion-free R -module, hence free .

Proposition 6.4.2 Let S be a finite poset with unique maximal element 00 , R a discrete valuation ring, and M , N E Qrep(S, R). Then T is a functor from Qrep(S, R) to Qrep(S, R) such that: (a) (b) (c) (d) (e)

T(M) is a subrepresentation of M that preserves monomorphisms; T(M E9 N) = T(M) E9 T(N); T(T(M» = T(M); T(M) = 0 ifand only if hoo is a monomorphism for each i E S; and T(M/T(M» = O.

6.4 A Torsion Theory and Localizations

193

If f: M -+ N, then T(f) = f : T(M) -+ T(N). Hence, T: Qrep(S, R) -+ Qrep(S, R) is a functor and T(f) is a monomorphism if f is a monomor-

PROOF. (a)

phism. (b) Apply the definitions of T(M) and direct sums . (c) By definition, T(M) = (T(M)j , gjj : i ~ j) with T(M)oo = 0, T(M)j = ker f;oo if i E S\{oo}, and gjj: T(M)j -+ T(M)j, the restriction of fij to T(M)j. Then 0 = gjoo : T(M)j = ker fjoo -+ T(M)oo = 0, whence ker gjoo = T(M)j for each i, This shows that T(T(M» = T(M). Statement (d) follows immediately from the definition of T(M). (e) Write M IT(M) = (M;/T(M)j , hij), where hij : M;/T(M)j -+ MjIT(M)j is induced by f;j . Since hjoo: M;/T(M)j -+ MooIT(M)oo = Moo is a monomorphism, ker h joo = 0 for each i, Now apply (d). Call M E Qrep(S, R) torsion if T(M) = M and torsion-free if T(M) = O. Then M = (M j, f;j) is torsion if and only if fjoo = 0 for each i E S. A nonempty subset P of S is an up-set in S if given i E P and i < j in S, then j E P. An up-set is called a filter by some authors. Let M = (Mj , fij) E Qrep(S, R) and define M p = «Mp)j,gij), where (Mp)j = Moo if i E P, (Mp)j = M, if i 1. P, gjj = 1· if i ~ j in S with i E P, and gjj = f;j if i ~ j in S with j 1. P. Since P is an up-set, M» is a well-defined element of Qrep(S, R) . Call M = (M j, fij) E Qrep(S, R) cotrivial if each M, = Moo and fij is the identity for each i ~ j in S. If S has a unique minimal element and M is cotrivial, then M is both injective and projective. The constructions of the next two propositions resemble localizations of modules.

Proposition 6.4.3 Suppose S is a finite poset with unique maximal element, R is a discrete valuation ring, M E Qrep(S, R), and P is an up-set in S. Then (a) M ~ M», (Mp)p = M», and T(Mp) = T(M)p; (b) if M is torsion-free, then M p is torsion-free, and if M is torsion, then Mp

is torsion; and (c) if P = S, then M p is cotrivial. PROOF . Immediate consequences of the definitions.

There is a localization "dual" to that of M r- If P is an up-set in Sand M E QRep(S, R), define M P = «MP)j, gij), where (MP)j = M, if i E P; (MP)j = 0 ifi 1. P;gjj = fijifi ~ j,iEP;andg jj =Oifi ~ j,j 1. P .SincePisan up-set, M P is a well-defined element of QRep(S, R).

Proposition 6.4.4 Suppose S is a finite poset with unique maximal element, R is a discrete valuation ring, M = (M j, f ij) E QRep(S, R), and P is an up-set in S. Then (a) M P ~ M, (MP)P

= M P, and T(M P) = T(M)P;

(b) (Mp)P = (MP)p = (Nj, gij) with N, = Moo for i E P, N, = gij = 1 for each i;

°

ifi 1.P, and

194 (c)

6. Representations overFields andExact Sequences

If M is torsion-free, then M P is torsion-free, and if M is torsion, then M P is torsion;

(d) M = I:{M P : P up-set}; and (e) there is a subset X ofthe set of up-sets in 5 with M = E9{M P: P E X}for

each torsion M in Qrep(5, R) ifand only if 5\{oo} is the disjoint union of {P\{oo}: P EX}. PROOF. Statements (a), (b), and (c) are immediate consequences of the definitions. As for (d), if i E 5, then i is an element of the up-set P, = {s E 5 : s ::: i}. Hence, if M = (M], fij) E Qrep(5, R), then M = I:{M P : P up-set}, since M, = (MP)j for P =Pj • Assertion (e) follows from the arguments in (d), observing that if M is torsion, then Moo = O. The next corollary is a reduction step in a characterization of those finite posets 5 with unique maximal element such that T(M) is a summand of M for each M E Qrep(5, R). The complete characterization is left as an exercise (Exercise 6.4.2).

Corollary 6.4.5 Assume that 5 is afinite poset with unique maximal element 00, R isa discretevaluation ring, and thereisa subset X ofthe set ofup-sets in 5 with 5\{00} the disjoint union of{P\{oo}: P E X}. If ME Qrep(5 , R), then T(M) is a summand of M ifand only ifT(M P) is a summand of M P for each P EX. PROOF. If T(M) is a summand of M and P is an up-set, then T(M)P is a summand of M P. But T(Ml = T(M P) by Proposition 6.4.4(a). Conversely, suppose that fP : M P ~ T(Ml is a splitting of the inclusion of T(M P) = T(Ml in M P for each P EX . By Proposition 6.4.4(e), T(M) = E9{T(M)P : P E X} and M, = E9{(M P)j : P E X} for each i E 5\{00}. Hence, /; = E9p M, ~ T(M) j is a splitting for each i E 5. Define foo = O. Then f = {lj : i E 5}: M ~ T(M) is a splitting of the inclusion of T(M) in M, since T(M)oo = O.

r:

For s E 5, pes) = (P(s)j , /;j) E Qrep(5, R) is defined by P(S)j = R if s ~ i, P (s), = 0 if s is not less than or equal to i, /;j = 1 if s ~ i, and /;j = 0 otherwise. Define P(O) = (P(O)j , /;j) E Qrep(5 , R) by letting P(O)j = Rand /;j = 1 for each i E 5 and i ~ j in 5. Observe that each pes) is torsion-free. The following result is given in [Butler 87], [Bautista, Martinez 79], and [Buenermannn 81, 83] for the case that R is a field.

Corollary 6.4.6 (a) Projectivesin Qrep(5 , R) are, up to isomorphism, summands ofdirectsums of representations ofthe form P(s) . (b) Torsion-free elements of Qrep(5, R) are, up to isomorphism, the elements of rep(5, R).

6.4 A Torsion Theory andLocalizations

195

(c) An ME Qrep(S, R) is torsion-free if and only if M is a subrepresentation ofa projective representation of Qrep(S, R). PROOF. (a) Each pes), hence the direct sum of P(s)'s, is readily seen to be projective. Moreover, if ME Qrep(S, R), then there is N E Qrep(S, R) with N isomorphic to a direct sum of P(s)'s and an onto morphism N -r M. Specifically, N = EB{N(s) :s E S}, where N(s) = (N(s);, f ij) with N(S)i = M, ifi :::: s, N (s), = 0 if its, fij = 1 if s ::: i ::: j, and fij = 0 otherwise. Notice that each N(s) is isomorphic to a direct sum of copies of pes). If M is projective, then M is a summand of N. (b) If V E rep(S , R), then fioo :Vi -r Voo is inclusion, so that T(V)i = ker fioo = 0 for each i. Thus, T(V) = O. Conversely, if M = (M] , f ij :i ::: j) E Qrep(S , R) is torsion-free , then each fioo is a monomorphism by Proposition 6.4.2(d). But hoofij = /;00 if i ::: j in S, whence each Iu is a monomorphism. Hence, M is isomorphic to V = (Moo, /; oo(Mi): i E S\{oo}) E rep(S , R), observing that each fioo has pure image in Moo . (c) In view of (a), Proposition 6.4.2(b), and the fact that each pes) is torsion-free, a projective representation is torsion-free . Since T preserves monomorphisms, a subrepresentation of a projective representation is also torsion-free . Conversely, if M = (Mo, Mi) E rep(S , R), then, by Proposition 6.4.3, M is a subrepresentation of a cotrivial projective representation.

EXERCISES I. Let S be a finite poset and R a discrete valuation ring. Define the incidence algebra RS to be the set of matrices M = (aij) such that aij E R if i ::::: j in Sand aij = 0 otherwise. (a) Prove that RS is a ring withidentity. (b) Prove thatthere is a fully faithful embedding Qrep(S, R) -+ mod RS, thecategory of finitely generated RS-modules. Describe the image of thisembedding. 2. Let S be a finite posetwith unique maximal element 00 and R a discrete valuation ring. Prove that T(M) is a summand of M for each M E Qrep(S , R) if andonly if S\{oo} is a forest.

NOTES ON CHAPTER 6 Descriptions of projectives, injectives, projective covers, andinjective envelopes in rep(S, k) aregiven in [Gabriel 73A). Coxeter correspondences on rep(S, k) arecommonly defined by regarding rep(S, k) as a subcategory of the category modkS of finitely generated modules over the incidence algebra kS of S and using the Auslander-Reiten correspondences DTr and TrD for these modules [Bautista, Martinez 79], [Drozd 74], and [Btinermann 81 , 83]. In this chapter, the definitionand development of Coxeter correspondences, in the spiritof [Gelfand, Ponomarev 70]forthecasethatS is an antichain, areentirely within thecategory

196

6. Representations over Fields and Exact Sequences

rep(S , k) and without reference to any module category. Hence, they are immediately available for quasi-homomorphism categories of finite rank Butler groups , as briefly discussed in Section 7.1. The existence of almost split sequences in rep(S, k) is derived in [Auslander, Smale 81], as well as [Bautista, Martinez 79] and [Biinermann 81, 83] using the torsion theory of Section 6.4, from the existence of almost split sequences in mod kS [Auslander, Reiten 75]. There does not seem to be a proof of their existence entirely within the category rep(S , k) . All indecomposable representations for the critical posets S4, (2, 2, 2), 0, 3, 3), (N, 4), and 0 , 2, 5) are given in [Ringel 84] via a description of the Auslander-Reiten quiver. The list is not quite as explicit as that of Example 6.2.8 for S4. An intriguing aspect that is not discussed herein is the characterization of the representation type of rep(S, k) in terms of an integral quadratic form qS, see Exercise 6.1.8. If rep(S, k) has finite representation type, then there is a one-to-one correspondence from the vectors cdn U, U E 1nd(S, k), to the roots of qS [Drozd 74]. Moreover, the critical posets can be associated with certain Dynkin diagrams . See [Ringel 84] and [Simson 92] for comprehensive discussions of the role of quadratic forms in representations of partially ordered sets and quivers and characterizations of the representation type of rep(S , k) in terms of properties of q s . Techniques from the theory of torsion-free abelian groups, particularly the analogue of rank-I groups and balanced exact sequences as discussed in Chapter 7, have been carried over to rep(S , k) for a finite poset S and field k in [Arnold , Vinsonhaler 92] and [Nonxga , Vinsonhaler 97]. The representation type of the category Qrep(S, k) for a finite poset S and a field k is completely described in [Loupias 75].

7 Finite Rank Butler Groups

7.1 Projectives, Injectives, and Exact Sequences Free groups are the projectives and divisible groups are the injectives for the category of torsion -free abelian group s [Hungerford 74]. Changing the category and the class of defining homomorphisms can change the projecti ves and injectives. Properties of Coxeter correspondences and almost split sequences are applied to categories of finite rank Butler groups in this section. A group G in B(T ) is projective in B (T ) if whenever f : H --+ K is an epimorphism of groups and g : G --+ K , then there is h : G --+ H with f h = g. The group G is injecti ve in B(T ) if for f: H --+ K a monomorphism of groups in B (T) with pure image and g : H --+ G , there is h : K --+ G with hf = g . Proposition 7.1.1 Let T be afinite lattic e of types. (a) The projectives in B(T) are .(D)-homogeneous completely decomposable groups, .(D) the small est element ofT. (b) The injectives in B(T) are .(oo)-homogeneous completely decomposable groups , .(00) the largest element ofT.

f: H --+ K is an epimorphism of group s in B(T) , X is a rank-l group with type .(D) , and D f. g : X --+ K. Let 0 f. x E X and 0 f. y E H with f( y ) = g(x). If Y is the pure rank-l subgroup of H generated by y , then Y E B(T), and so .(0) ~ type Y. Since type X = .(0), there is some h : X --+ H with f h = g. A standard argument using injections and projections for finite direct sums [Hungerford 74] shows that a . (D)-homogeneous completely decomposable group is projective in B(T).

PROOF. (a) Suppose

198

7. Finite Rank Butler Groups

Conversely, let G be projective in B(T). Since G E B(T), there is an epimorphism f : C -+ G , with C a completely decomposable group in B(T), by Lemma 3.3.1. Because G is projective, there is h : G -+ C with hf = IG . Therefore, G is a summand of C, hence completely decomposable by Theorem 3.1.7(a). It remains to show that if X is a rank-I summand of C with type X > r(O), then X is not projective in B(T). To see this, let Y be a rank-I group in B(T) with type Y = r(O) < type X. Choose 0 =I x E X and 0 =I y E Y. Since Y is reduced, there is an integer n such that na = x has no solution a E Y EI1 X. Define G to be the pure subgroup of Y EI1 X generated by ny + x . Then type G = type Y and type (Y EI1 X)/G = type X = r. Now assume, by way of contradiction, that X is projective in B(T). Since (Y EI1 X)/G is isomorphic to X, G is a summand of H = Y EI1 X, say H = G EI1 D. Then type D = r , from which it follows that H(r) = X EI1 D . But y tf. X EI1 D , since na = x has no solution a E Y EI1 X, a contradiction. (b) Let X bearank-l group with type X = r(oo), G and HE B(T), f : G -+ H a monomorphism with pure image, and 0 =I g : G -+ X. Then QX = Q @z X is an injective abelian group, so there is h : H -+ QX with hf = g. Now, type h(H) E cotypeset H ~ T, so that type h(H) ~ type X, the largest element of T . Consequently, there is h: H -+ Xwith hf = g . This shows that X, hence a r(oo)-homogeneous completely decomposable group, is injective in B(T). Conversely, suppose G E B(T) is injective in B(T). Then G is a pure subgroup of a completely decomposable group C in B(T) by Lemma 3.3.1. Since G is injective in B(T), G is a summand of C, hence completely decomposable by Theorem 3.1.7. It now suffices to show that if Y is a rank-l summand of G with type Y < type X = r(oo), then Y is not injective in B(T) . In this case, G must be a r(oo)homogeneous completely decomposable group. A construction like that of (a) shows that Y EI1 X has a pure rank-I subgroup with type equal to type Y that is not a summand. Hence, Y cannot be injective in B(T). A finite rank Butler group G is balanced projective if whenever f: H -+ K is a balanced epimorphism of finite rank Butler groups Hand K and g : G -+ K, then there is h : G -+ H with f h = g . The group G is cobalanced injective if for Hand K finite rank Butler groups , f : H -+ K a cobalanced monomorphism, and g : H -+ G , there is h : K -+ G with hf = g.

Proposition 7.1.2 For finite rank Butler groups, completely decomposable groups are the balanced projective groups and the cobalanced injective groups. PROOF. Suppose f

: H -+ K is a balanced epimorphism, X is a rank-I group with type X = r, and 0 i= g: X -+ K. There is a pure rank-I subgroup Y of H with type Y ~ rand f(Y) = g(X). But type Y ~ type f(Y) = type X = r , whence type Y = r = type X. It follows that there is an isomorphism h : X -+ H with f h = g, and so X is balanced projective. Consequently, completely decomposable groups are balanced projective.

7.1

Projectives, Injectives, andExactSequences

199

Conversely, if G is a balanced projective finite rank Butler group, then there is a completely decomposable group C and a balanced epimorphism f : C ~ G, by Lemma 3.2.3(a). Since G is balanced projective, G is isomorphic to a summand of C, hence completely decomposable. This shows that completely decomposable groups are the balanced projectives. To see that a completely decomposable group is cobalanced injective , assume that f : H ~ K is a cobalanced monomorphism and X is a rank-l group with type X = r. It suffices to prove that the induced homomorphism Hom(K, X) ~ Hom(H, X), given by h 1-+ hf, is onto. In this case, X, hence a completely decomposable group, is cobalanced injective . It is also sufficient to assume H [r] = K [r] = O. This is so because Hom( H, X) = Hom(HIH[r] , X) by the definition of G[r] and (HIH[r])[r] = 0 by Lemma 3.1.9(a). Similarly, Hom(K, X) = Hom(K I K[r], X) and (KI K[r])[r] = O. To complete the proof that X is cobalanced injective, let g : H ~ X and Y = g(H) with type Y = a ::: r . Then g: HI H[a] ~ Y is onto and f : HI H[a] ~ K I K [a] is a pure monomorphism, since f: H ~ K is cobalanced. Define G = (Y $ (KIK[a]))IM, M the pure subgroup of Y $ (KIK[a]) generated by {(g(a),-f(a)) :a E HIH[a]) . Then KIK[a] ~ G is onto, from which it follows that G[a] = 0, since (KIK[a])[a] = 0 by Lemma 3.1.9(a) . As Y is a pure subgroup of G with Yea) = Y and G[a] = 0, Y is a summand of G by Proposition 3.1.15(b) . Since Y is a summand of G, there is h': G ~ Y with the restriction of h' to Y the identity endomorphism of Y. Define h : K ~ Y by h(k) = h'«k + K[a]) + M) E h'(G) = Y. If a E H, then hf(a) = h'«f(a) + K[a]) + M) = h'(g(a) + M) = g(a), and so hf = g, as desired. Conversely, assume that G is cobalanced injective. Then G is a cobalanced subgroup of a completely decomposable group C by Lemma 3.2.3(b). Hence, G is a summand of C, so that G is completely decomposable, as desired. A sequence 0 ~ G ~ L ~ K ~ 0 of groups in B(T)Q is exact if f is a monomorphism, (ker g + image f)1(ker g n imagef) is finite, and (image g + K)/(image g) n K is finite. In this case, 0 ~ QG ~ QL ~ QK ~ 0 is an exact sequence of Q-vector spaces . Proposition 7.1.3 Let H : B(T)Q ~ rep(ST, Q) be the category equivalence (QG, QG(r) : rEST). Then 0 ~ G ~ given in Theorem 3.3.2 with H(G) L ~ K ~ 0 is an exact sequence of groups in B(T)Q if and only if 0 ~ H(G) ~ H(L) ~ H(K) ~ 0 is an exactsequence in rep(S, k).

=

PROOF. Assume that 0 ~ G ~ L ~ K ~ 0 is an exact sequence in B(T)Q. Then 0 ~ QG ~ QL ~ QK ~ 0 is an exact sequence of vector spaces. It is sufficent to prove that if rEST, then QG(r) ~ QK(r) ~ 0 is onto. In this case, QG(r) = QG n QL(r) and 0 ~ QG(r) ~ QL(r) ~ QK(r) ~ 0 is an exact sequence for each rEST , as desired . Let X be a pure rank-l subgroup of K with type X :::: rEST and A = g-l(X) the preimage of X in L . Then A E B(T), being quasi-isomorphic to a pure subgroup

200

7. Finite Rank ButlerGroups

of L E B(T). Hence, A = Y1 + ...+ Yn for pure rank-l subgroups Yi of A with g(Yi):f. 0, type Yi = a., and type X = U{ai : 1 ::: i ::: n} ::: r. Since the lattice of all types is distributive, r = U{ r nai : 1 ::: i ::: n}. But r is join irreducible, whence r r n a, and a, ::: r for some i, Thus, QX Qg(Yi) with Yi E L( r ), as desired. As for the converse, since H is a category equivalence , there is a category equivalence H' :rep(Sr, k) ~ B(T)Q with H H' naturally equivalent to the identity functor on rep(Sr, k) and H' H naturally equivalent to the identity functor on B(T)Q. It follows that if 0 ~ H(G) ~ H(L) ~ H(K) ~ Oisanexactsequence in rep(S, k), then 0 ~ G ~ L ~ K ~ 0 is an exact sequence in B(T)Q.

=

=

An element r of a finite lattice of types T is meet-irreducible if whenever r = 8 n a with 8, a E T , then r = 8 or r = a .

Corollary 7.1.4 Let T be afinite lattice oftypes. (a) Projectives in B(T)Q are direct sums of rank-l groups X with type X a join-irreducible type in T. (b) lnjectives in B(T)Q are direct sums of rank-l groups X with type X a meet-irreducible type in T . (c) Each G E B(T) has an projective cover PG : peG) ~ G in B(T)Q, where peG) = ${G(r)/G#(r) :r E SrI. (d) Each G E B(T)Q has an injective envelope iG : G ~ leG) in B(T)Q, where leG) = ${G*[r]/G[r] : r E Sr}.

PROOF. Statements (a) and (b) follow from Proposition 7.1.3 and Lemma 6.1.2, while (c) and (d) are consequences of Lemma 6.1.4 and Propo sition 3.2.8. The duality a of Lemma 6.1.1 induces a duality for categories of Butler groups .

Theorem 7.1.5 Let T' and T' befinite lattices oftypes and a : T'

~

T a lattice

anti-isomorphism. Then: (a) there is a contravariant category equivalence Da : B(T')Q ~ B(T)Q such

that Da-I D a is naturally equivalent to the identity functor on B(T')Q; (b) Da(G(r» is quasi-isomorphic to Da(G)/ Da(G)[a(r)]for each rET'; (c) Da sends balanced sequences in B(T')Q to cobalanced sequences in B(T)Q;and (d) Da sends cobalanced sequences in B(T')Q to balanced sequences in B(T)Q. PROOF. (a) Notice that a induces an anti-isomorphism a : Sr' ~ S", Sr the opposite of the poset of meet irreducible elements of T . Now, H : B(T')Q ~ rep( Sr', k) with H (G) = (QG , QG (r) : r E Sr) is an exact category equivalence by Proposition 7.1.3, and a: rep(Sr, k) ~ rep(Sr', k) is an exact contravariant category equivalence by Lemma 6.1.1.

7.1

Projectives, Injectives, and Exact Sequences

201

Define a correspondence F : B(T)Q -+ rep(ST, k) by F(G) = (QG, QG[T] : TEST). Then F is a functor, as a consequence of Lemma 3.1.9(b). Now, QG[T] = :E{QG(a) : a;fT} and QG(T) = n{QG[a] : atT} for each type T by Proposition 3.2.8. Since H is a category equivalence, a representation U in rep(ST, Q) is completely determined by (QG, QG(T): T E T). Hence, U is also completely determined by (QG , QG[T]: T E T). In fact, (QG, QG[T]: T E T) is completely determined by (QG , QG[T] : TEST), since QG[a n T] = QG[a] + QG[T] by Proposition 3.2.8(d). It now follows that F is an exact category equivalence, in particular, Hand F are manifestations of the same functor from B(T)Q to rep(T , Q). Define D cx by F-1a H: B(T')Q -+ B(T)Q . Then D cx is an exact contravariant equivalence. Since Hand F are manifestations of the same functor, D cx - 1 = H-1a F, whence D cx - 1Ds = (H-1a F)(F-1a H) = H- 1a 2 H is naturally equivalent to the identity functor. Assertions (b), (c), and (d) are left as exercises (Exercise 7.1.1). Corollary 7.1.6 Suppose Tis afinite lattice oftypes and G, H E B(T)Q. There are correspondences C+, C- : B(T)Q -+ B(T)Q such that: (a) C+ G = 0 if and only if G is projective in B(T)Q and C- G = 0 if and only if G is injective in B(T)Q. (b) If G has no injective summands and H has no projective summands, then QHom(G , C+ H) is isomorphic to QHom(C-G, H) . (c) IfG has no projective summands, then QEnd G is isomorphic toQEnd C+G , and if G has no injective summands, then QEnd G is isomorphic to QEnd C-G . (d) If G E B(T)Q is a strongly indecomposable group, then C+ G and C- G are strongly indecomposable groups. PROOF. Apply Proposition 7.1.3 and Corollary 6.2.5. Call G E B(T)Q preprojective if C+i G = 0 for some i ~ 1 and preinjective if C-iG = 0 for some i ~ 1. Let cdn G = {rank G, rank G(T): TEST}. Corollary 7.1.7 Let T be afinite lattice of types. Thefollowing statements are equivalent: (a) B(T)Q has finite representation type. (b) If G is a strongly indecomposable group in B(T)Q' then G is preinjective. (c) IfG is a strongly indecomposable group in B(T)Q, then G is preprojective. (d) The quasi-endomorphism ring of G is isomorphic to Q for each strongly indecomposable G E B(T)Q. In this case, if G is an strongly indecomposable group in B(T)Q, then G is determined up to isomorphism by cdn G.

202

7. Finite Rank Butler Groups

PROOF .

Theorem 6.2.6 and Proposition 7.1.3.

Corollary 7.1.8 Let T be a finite lattice oftypes. (a) If G E B(T)Q is a strongly indecomposable group in B(T)Q that is not projective, then there is an almost split sequence 0 --+ C+G --+ B --+ G --+

oin B(T)Q.

(b) IfG is a strongly indecomposablegroup in B(T)Q that is not injective, then there is an almost split sequence 0 --+ G --+ B --+ C-G --+ 0 in B(T)Q. PROOF .

Proposition 7.1.3 and Theorem 6.3.4

Open Question: Given a finite lattice T of types, is there a category X of abelian groupscontaining B(T)Qand a categoryequivalenceX --+ Qrep(Sr , Q) that restrictsto the categoryequivalence B(T) Q --+ rep(S, Q) of Theorem3.3.2 ? What is the situation for B(T)p and Qrep(Sr, Zp)?

EXERCISES

I. Complete the proof of Theorem 7.1.5.

2. Give explicit constructions for C+ G and C- G, G E B(T)Q' Give examples of preprojective groups that are not projective and preinjective groups that are not injective. 3. Given X a rank - I group in B(T k r X. (a) Compute C+ X and (b) Find an almost split sequence 0 -+ C+ X -+ B -+ X -+ O. (c) Find an almost split sequence 0 -+ X -+ B -+ C- X -+ O.

c:

4. For G E B(T)Q and M a subset of T , define G(M) to be the pure subgroup of G generated by {G(r): rEM} and G[M] = n{G[r] : rEM}. Let 0 -+ G -+ H -+ K -+ 0 be an almost split sequence in B(T)Q' Prove the following : (a) If G(M) =1= G, then 0 -+ G(M) -+ H(M) -+ K(M) -+ 0 is a split exact sequence in B(T)Q. (b) If G[M] =1= 0, then 0 -+ G/ G[M] -+ H/ H[M] -+ K/ K[M] -+ 0 is a split exact sequence in B(T)Q.

7.2

Endomorphism Rings

If R is a ring with additive group a reduced torsion-free abelian group, then R is the endomorphism ring of G for some reduced torsion-free abelian group G of finite rank by Theorem 2.4.6. However, the G constructed in that theorem is not, in general, a Butler group. Endomorphism rings of finite rank Butler groups are described in this section. Given a ring R, R+ denotes the additive group of R.

7.2 Endomorphism Rings

203

Proposition 7.2.1 If G is a finite rank Butler group, then (End G)+ is a finite rank Butler group. PROOF . By Corollary 3.2.4, there are completely decomposable groups C and

D such that G is a pure subgroup of D and an epimorphic image of C, say f: C -)0 G . Then 4> : End G -)0 Hom (C, D), defined by 4>(g) = gf, is a monomorphism with pure image. To see that 4> has pure image, suppose p is a prime, h E Hom (C, D) , and ph = gf for some g E End G. If x E C, then ph(x) = gf(x) E pD n G = pG . It follows that (g/p) E End G and h = is! p)f. But Hom(C, D) is a completely decomposable group, since C and D are completely decomposable groups. This shows that (End G)+ is a finite rank Butler group .

Proposition 7.2.2 Assume that R+ is a reduced torsion-free abelian group of finite rank. Then (a) R+ is a Butler group ifand only if(NR)+ is a Butler group and (R/NR)+ is

almost completely decomposable; and (b) if (R/NR)+ is a Butler group and QR/ JQR is a simple algebra, then (R/NR)+ is homogeneous completely decomposable. PROOF . (a) The ring R is a subring of QR, a finite-dimensional Q-algebra. Recall thatNR = R n JQR, JQR the Jacobson radical ofQR. First, assume that R+ is a Butler group . Then (R/NR)+ , a homomorphic image of R+, and (NR)+, a pure subgroup of R , are Butler groups by Corollary 3.2.4. As in Lemma 2.2.7, there is a nonzero integer n with nR* 5; R/NR 5; R* = R) x . .. x R m , each R; a finitely generated free S;-module, S; a Dedekind domain, is homogeneous, so that and QR; a simple Q-algebra. Since S; is a domain, Rt is homogeneous. But (R*)+, hence each Rt, is a Butler group, since (R*)+ is quasi-isomorphic to the Butler group (R/NR)+ . Furthermore, each (R;)+ is homogeneous coinpletely decomposable, because homogeneous Butler groups are completely decomposable by Corollary 3.2.7(a). This shows that (R/NR)+ is almost completely decomposable. Conversely, assume that (NR)+ is a Butler group and (R/NR)+ is almost completely decomposable. Then R+ is quasi-isomorphic to (NR)+ ED (R/NR)+ by Theorem 2.4.7. Thus , R+ is a Butler group by Corollary 3.2.4. (b) In this case , as in (a), (R/NR)+ is quasi-isomorphic to a homogeneous completely decomposable group, hence homogeneous completely decomposable.

st

Not every R with R+ a Butler group is the endomorphism ring of a Butler group. For example, if R is the ring of Hamiltonian quaternions over Zp, then R+ is a free Zp-module that is q-div isible for each prime q #- p, and QR is not a field, as required by (a) of the next corollary. Corollary 7.2.3 [Arnold, Vinsonhaler 89] Suppose QR is a division algebra and R is the endomorphism ring of a finite rank Butler group.

204

7. FiniteRank ButlerGroups

(a) If R+ is p-reducedfor at most 4 primes p of 7l. then QR is afield isomorphic

to Q[x]/ (g(x Y) for some irreducible g(x) E Q[x]. (b) If R+ is p-reducedfor at most 3 primes p of Z, then R is a subring of Q.

(a) Since R is a subring of a division algebra, NR = 0 and R+ is homogeneous completely decomposable. Hence, R+ is not p-reduced if and only if R+ is p-divisible. Thus, typeset R+ S; T4, the finite Boolean algebra of types generated by {type Zp : R is p-reduced}, since R+ is p-reduced for at most 4 primes , say PI,

PROOF.

P2, P3, P4·

Let G be a finite rank Butler group with End G = R. Then G is strongly indecomposable, since QR is a division algebra. Moreover, typeset G S; T4 • This is so because if q is a prime with q # Pi for each i , then q R = R, and so qG = G. In particular, G E B(T4 ) . By Theorem 3.3.2, B(T4 )Q is category equivalent to rep(S4, Q) . Moreover, each indecomposable in rep(S4, Q) has endomorphism ring Q[x]/(g(xY} for some irreducible g(x) E Q[x] by Example 6.2.7. Thus, QR = QEnd G must also be of this form . (b) In this case, just as in (a), if R = End G for some finite rank Butler group G, then G E B(T3) and G is strongly indecomposable. Thus, QR = QEnd G = Q by Example 3.3.3. Theorem 7.2.4 [Arnold, Vinsonhaler 87] Assume that R is a ring with additive group afinite rank Butler group. If R is p-reducedfor at leastfive primes p, then R is the endomorphism ring ofa finite rank Butler group. Let PI, . . . , PS be five primes for which R is p -reduced. For each 1 :::: i :::: 5, define Xi to be the subring of Q generated by {l/pjlj i= i} . By Example 1.1.7, there is

PROOF.

for some m x m Q-matrices A and B with QR = C(A , B) = End(U). Let H = (XIR+)m $ ... $ (XsR+)m, a finite rank Butler group, with QH = (QR)m $ . . . $ (QR)m. Define G = V n H, V the row space of the Q-matrix

Then G is a Butler group, being a pure subgroup of H . A routine matrix calculation shows that {(r, r): r E R} S; End G S; {(f, f) : f E C(A, B» = ({(r, r) :r E QR}. Notice that each XiR+ is a fully invariant subgroup of H, since R is Pireduced for each 1 :::: i :::: 5. It follows that if(r, r) E End G, then r E n{X i R : 1 :::: i :::: 5} = R. Hence, End G is isomorphic to R, as desired. Theorem 7.2.4 can be used to construct Butler groups with pathological direct sum decompositions just as for arbitrary torsion-free abelian groups of finite rank. For instance:

7.3 Bracket Groups

205

Example 7.2.5 (a) There is afinite rank Butlergroup G and a primep such that G is indecomposable at p but (End G)p is not a local ring. (b) If P ~ 5 is a prime, then there is a finite rank Butler group H with a nonunique decomposition into indecomposables in the isomorphism at p categoryoffinite rank Butler groups. (c) There is afinite rank stronglyindecomposable Butler group G with a subgroup H such that G is nearly isomorphic to H, H is a summandof G tB G, but G is not isomorphic to H. PROOF. (a) By Theorem 7.2.4, there is a finite rank Butler group G with End G = Z[i], the Gaussian integers, since Z[i] is a free abelian group of rank 2. The remainder of the proof is as in Example 2.1.5. (b) A consequence of Theorem 7.2.4 and Example 2.1.11, in which case R is also a ring with finitely generated free additive group. (c) By Theorem 7.2.4, there is a finite rank Butler group with End G = Z[(_5)1 /2] = R, since R+ is a free group of rank 2. Then R contains a nonprincipal ideal I with R/I finite, for example, I = (3, 1 + (_5)1 /2). Since R is a Dedekind domain, I and R are in the same genus class, as in Lemma 2.2.8. Hence, H = I G is a subgroup of G with G/ H finite and H nearly isomorphic to G by Theorem 2.4.3.

7.3 Bracket Groups This section consists of a brief introduction to a well-studied class of Butler groups. A bracketgroup, also called a B(I)-group [Fuchs, Metelli 91] or a Richman-Butler group [Hill, Megibben 93], is a group G = C/ X with C a finite rank completely decomposable group and X a pure rank-I subgroup of C. In this case, G is a finite rank Butler group with rank G = rank C - 1. Example 3.2.1 is an example of a bracket group, as is C+ Y for a rank-l group Y in B(T); Exercise 7.1.3. In the latter case, C+ Y = C/ X, where C is an injective group in B(T)Q and X is a rank-l group, as a consequence of Example 6.3.6 and Proposition 7.1.3. Bracket groups are intimately related to those almost completely decomposable groups containing a completely decomposable group of cyclic index. This relationship has yet to be investigated in any depth. Proposition 7.3.1 (a) If G = C/ X is a bracket group and n a nonzero integer, then H = C + (l / n)X is an almost completely decomposable group with H/ C a cyclic

group isomorphic to X/nX. (b) If G is an almost completely decomposable group and C is a completely decomposable subgroup of G with G/ C cyclic, then G is a bracketgroup. (c) If G = C/ X is a bracketgroup, then G is almost completely decomposable if and only if Tcr(C) has a uniqueminimal element.

206

7. FiniteRank ButlerGroups

PROOF . (a) The group H is almost completely decomposable. Moreover, H / C is isomorphic to nHInC = (nC + X)/ X = X/(X n nC) = X/nX, since X is a pure subgroup of C. (b) Let C = A I E9 . . . E9 An be a completely decomposable subgroup of G with rank Ai = I for each i and G = C + Zx. Write x = (I/m)(al , ... , an) with a; E Ai and m a nonzero integer. Then G = (C E9Zx)/ Y is a bracket group, where Y is the pure rank-I subgroup of G generated by (mx, -aI, .. . , -an)' (c) Exerc ise 7.3.2 .

Let A = (A I , , An) be an n-tuple of subgroups of iQl with I E Ai for each i E n+ = {I, 2, , n}. Define G[A] = (AI E9 . . . E9 An)/ X, where X is the pure rank-I subgroup of A I E9 . . . E9 An generated by (1, I, I , ... , I). For example, G[AI, A z] = Al + Az, since 0 -+ AI n Az -+ AI E9 A z -+ AI + A z -+ 0 is an exact sequence with I E A] n Az. Notice that type X = n{type Ai : I ~ i ~ n}, and if G[A] is an almost completely decomposable group, then G must be completely decomposable. Call A cotrimmed if each (Ai + X)/ X is a pure subgroup of G[A]. For example, G[A] , Az] is not cotrimmed unless AI = Az.

Lemma 7.3.2 Let G = G[A]for some A = (AI ," " An)' (a) Then A is cotrimmed if and only ifn{A j

(b)

= Ai +n{A j :j =/: G[A] = G[B].

If B,

i}, then B

:

j

=/: i} s:;; Ai for each i.

= (BI, . . . ,Bn) iscotrimmedwith

PROOF. Exercise 7.3.3.

Let A = (AI, . .. , An), E a subset of n+, and define rE = type n {A j : j E E} . Given a partition P = {P(1), . . . , P(m)} of n+, let r(P) = n{rp(i) U rp(j) : I ~ i < j ~ m}. For a type r , two elements i and j of n+ are t -equivalent if either i = j or there is a sequence i(1), ... , i(n) E n+ with i(1) = i, i(n) = j, and r 1:. type Ai(k) + Ai(k+l) for each I ~ k ~ m - 1. It can be readily continned that r-equivalence is an equivalence relation on n+ For instance, if r = type Ai for some i , then {i} is a r -equivalence class of n+, since type Ai ~ type Ai + A j for each j E n" .

Lemma 7.3.3 Let G be a bracket group. Then: (a) G is quasi-isomorphic to G[A] E9 D for some completely decomposable group D and cotrimmed A = (AI, . . . , An); (b) typeset G[A] = {r(P): P partition ofn+}; (c) G[A] is quasi-isomorphic to G[B(1)] E9 . .. E9 G[B(m)], where each B(i)

is a subtuple of Ai and (d) ifG = G[A] and r E typeset G, then G(r) is quasi-isomorphic to G[B], where B = (B I, . . . , Br ) , B, is a rank-Y group with type B, = rpi for a

7.3 Bracket Groups partition Pi

= {E(i) , E(i)C},

207

and E(1), ... , E(r) are the t-equivalence

classes of n" , In particular; rank G(r)

=r

-

1.

= C I Y, where C = XI EEl .. . EEl X; with rank Xi = I for each i and Y is a pure rank-I subgroup of G. It is sufficient to assume that o =1= Y = (XI, . . . , x n) E y with each Xi =1= O. Let Ai be a subgroup of Q containing I and Ii : Xi ~ Ai an isomorphism. Then f = fl EEl .. . EEl fn : C ~ C' = A I EEl . .. EEl An is an isomorphism inducing an isomorphism f : G ~ C' I f(y) · If 0 =1= y = (ql , .. . qn) E f(y), and X is the pure subgroup of C' generated by (1, I, . .. , I), then each qi =1= 0 and (ql-I, . .. , q; -I)f : G ~ G[A] is a quasi-isomorphism. By Lemma 7 .3.2, A may be assumed to be cotrimmed. (b) Let n : C A ~ G[A] be the canonical epimorphism with C A = Al EEl . .. EEl An and ker n = X = «I, .. . , I)) • . Recall from Section 3.2 that cosupport G[A] = {cosupportf: 0 =1= f E Hom(G[A], Q)}, where cosupport f = {i : frr(A i) = O}. It follows that the maximal elements of cosupport G[A] are the subsets I of n+ with III = n - 2. For each such I write n+\J = {i, There is an epimorphism !J : G[A] ~ G[Ai , Aj] = Ai + A j ~ Q induced by the projection of C A onto Ai EEl Aj. In particular, !J«al , '" ,an) + X) ai - aj ' There is a cobalanced monomorphism ¢: G[A] ~ EEl{!J(G): I E 5} EEl{A i + A j : I ::; i < j ::; n} given by ¢«al,'" ,an) + X) EEl{ai - aj : I::; i < j ::; n} (Lemma 3.2.3(b» . In particular, since the image of ¢ is pure, type (al, '" ,an) + X n{typeai -a j : I::; i < n} . Let 0 =1= y = (aI , . . . , an) + X with a, E Ai. There is an equivalence relation on n+ defined by i y -equivalentto j if a, = ai - Let P = {P(1), . .. , P(m)} be the partition of n+ consisting of y-equivalence classes. It follows from the definition of ¢ that type y type ¢(y) rep), as desired. (c) The proof is an induction on n. Let r = type A I and E(1) = {I}, E(2), .. . , E(r) be the r -equivalence classes of n+ . It suffices, by relabeling, to assume that E(2) = {2, 3, . .. , mI. Define B = (AI, A z, · · · , Am)' Now, G[B] is a summand of G[A]. To see this, write G[A] = CAl X with C A = Al EEl··· EEl An' X = «I , I, . .. , I))., and G[B] = CnlY with C n = AI EEl· ·· EEl Am and Y = «(1 , I, . . . , I)) • . A projection f: C A ~ Ce induces an onto homomorphism f : G[A] ~ G[B] . Next, define g : G[B] ~ G[A] by g«al, .. . ,am) + Y) = (aI, ... , am, aJ, ... , at) + X. Since f g is the identity on G[B], G[B] is a quasi-summand of G[A], provided that g is a quasi-homomorphism. To this end, let X E G[B]. By (b), type X = rep), where P = {P(1), ... , P(t)} is the partition of m+ given by the x-equivalence classes. The g(x)-equivalence classes in n+ are P' = {P '(l) , P(2) , . .. , pen)}, where P'(l) is the union of P(l) and {i: m + I ::; i ::; n} . If I ::; i ::; m and m + I ::; j ::; n, then type AI ::; type Ai + A j , since r = type A I and Ai and A j are in different r -equivalence classes. It follows that type x = rep) ::; type g(x) = reP '). Consequently, g is a quasi-homomorphism from G[B] to G[A] by Theorem 3.3.2 . In particular, G[A] is quasi-isomorphic to G[B] EEl ker f . However, ker f = G[C], where C = (Am+J, . . . , An)' An induction on n completes the proof. PROOF. (a) Write G

n.

=

ts

=

=

=

= =

208

7. Finite Rank Butler Groups

(d) Given Bi , define f; : B, -+ G[A] by f;(1) = (aJ, ... , an) + X, where = 0 if j ~ E(i) and aj = 1 if j E E(i). Then B, -+ G[A](r) is a quasitype B, :::: r . homomorphism as a consequence of (b), observing that type fi(1) Now, f fl EB .. • EB f,: BI EB ... EB B, -+ G[A](r), and f(l, 1, ... ,1) O. Hence, f induces f: G[B] -+ G[A](r). If y E G[A](r), then type y :::: r. It follows from (b) that y E image f, so that f is a quasi-epimorphism. Finally, image f has rank r - 1, since E(1), ... , E(r) is a partition of n+.

t. .

aj

=

=

=

Theorem 7.3.4 Let T be afinite lattice oftypes and G = G[A]for some n-tuple A = (AI , . . . , An) ofsubgroups ofQ. Thefollowing statements are equivalent: (a) G is strongly indecomposable; (b) rank G( r) 1 for each r typeA i ; (c) QEnd G Q.

= =

=

PROOF. (a) => (b) Ifn = 2, then rank G = 1, and so rank G(r) = 1 for each r = type Aj.Now suppose n :::: 3, r type Ai, and rank G(r) > l.ByLemma7.3.3(d), {r}, E(2), andE(3). there are at least three r-equivalence classes of n+, E(l) In view of Lemma 7.3 .3(c) , G[B] is a proper quasi-summand of G, where B (A }: i E E(1) U E(2». This is a contradiction to the assumption that G is strongly indecomposable. type Ai. Since (b) => (c) If f E End G, then f(G(ri)) ~ G(ri) for each ri rank G(ri) = 1, f: G(ri) -+ G(ri) is multiplication by some qi E Q . However, if 0:1= x = (al ,'" , an) E X , thenqlal + .. ·+qnan = f(x+X ) = f(O ) = 0 E G, and so qlal ED · · · ED qnQn = q(Q, ED··· ED an) E C for some q E Q . Thus, q = qi for each i and f = q E Q , as desired. (c) => (a) is a consequence of Proposition 2.1.1(a), since Q End G = Q has no nontrivial idempotents.

=

=

=

=

If G[A] and G[B] are strongly indecomposable, then they are isomorphic if and only if they are nearly isomorphic. This is a consequence of Theorem 2.4.3, since End G[A] is a subring of Q by Theorem 7.3.4, hence a principal ideal domain. If A (AI,"" An) and r E typeset G[A] with rank G(r) 1, then, by Lemma 7.3.3(d) and the distributivity ofthe lattice of all types, there is a subset E of n+ with r = type Dr, where Dr = n{A i: i E E} + n{A j : i ~ E}. Define 8(A) = (Dr: rank G(r) 1).

=

=

=

Theorem 7.3.5 Assume that T is afinite lattice oftypes, A = (A I, . . . , An) and B = (B I , •• • , Bm ) are tuples ofsubgroups ofQ, and G = G[A] and H = G[B] are strongly indecomposable groups in B(T). Then

=

(a) G and H are quasi-isomorphic if and only if rank E{G(r) :r E M} rank E {G (r) : rEM} for each nonempty subset M of T ; and (b) G and H are isomorphic if and only if they are quasi-isomorphic and there is 0:1= q E Q with q8(A) = 8(B).

7.3

Bracket Groups

209

PROOF. See [Arnold, Vinsonhaler 93] and references therein.

Theorem 7.3.6 [Goeters, Ullery, Vinsonhaler 94] Assume that Tis a.finite latticeoftypes, A = (A I,"" An) and B = (B I , .. . , Bm)are tuples ofsubgroups of Q, and G = G[A] and H = G[B] are strongly indecomposable groups in B(T). Then G and H are quasi-isomorphic ifandonly ifrank G(r) = rank H(r) for each rET.

EXERCISES

1. [Goeters, Ullery 95] Let A = (AI, ... , An) for n ~ 3. (a) Prove that G[A] is strongly indecomposable if and only if for each i E n+ and n rF. partition {{i}, E, F} of n'". type A; (b) Prove that iffor each 3-.element subset {i, i . k} of n +, type A; + A j, type A j + Ak, and type A j + Ak are pairwise incomparable, then G[A] is strongly indecomposable.

irE

2. [Fuchs, Metelli 91] Prove Proposition 7.3.1(c). 3. Prove Lemma 7.3.2. 4. Prove that if A = (A I , .. . , An) is an n-tuple of subgroups of Q, G = G[A] , T is a finite lattice of types, and D : B(T)Q -+ B(TOP)Q is the duality given in Theorem 7.1.5, then D (G[A]) = G(A ), the kernel of the homomorph ism Al E9 . . . E9 An -+ Q given by inclusion of the A;'s in Q.

5. State and prove the duals of the results of Section 7.3 for groups of the form G(A ), as defined in the previous exercise.

NOTES ON CHAPTER 7

Projectives and injectives in B(T) are characterized in [Butler 68] and [Lady 79]. Coxeter correspondences and almost split sequences for B(T)Q have yet to find significant application. An alternative development of Coxeter correspondences is given in [Lady 83] for Butler modules over Dedekind domains . There are scattered results on endomorphism rings of Butler groups of infinite rank; see [Arnold, Dugas 96], [Dugas, Gabel 97], and references therein. An interesting class of finite rank Butler groups, called Kravchenko groups and defined in terms of balanced projective resolution s, are investigated in [Nonxga, Vinsonhaler 96]. These classes of groups form a descending chain of subclasses of the class of finite rank Butler groups with intersection the class of finite rank completely decomposable groups . Representation analogues of Kravchenko groups are investigated in [Nonxga, Vinsonhaler 97]. There is an extensive body of literature on bracket groups, beginning with [Richman 83]. Although this is a restricted class of groups , it is rare in the theory of torsion-free abelian groups of finite rank to be able to classify groups by numerical invariants. A survey of this literature up to 1991, including a discussion of invariants and classification, can be found in [Arnold, Vinsonhaler 93]. More recent work includes [Hafting 93],

210

7. Finite Rank Butler Groups

[Goeters, Ullery 91, 92, 94], [Goeters, Ullery, Vinsonhaler 94], [DeVivo, Metelli 96, 99], and [Yom 94, 98]. Cyclic regulating quotient groups, those almost completely decompos able groups G containing a regulating subgroup B with G/ B cyclic, have been characterized , as noted in Notes on Chapter 5. In view of Proposition 7.3.1, cyclic regulating quotient groups are related to bracket groups. Implications of these characterizations of cyclic regulating quotient groups for the corresponding bracket groups , given in Proposition 7.3.1(a), remain unexplored.

8 Applications of Representations and Butler Groups

8.1 Torsion-Free Modules over Discrete Valuation Rings The theory of Butler groups has, a priori, little relevance for p-Iocal torsion-free groups. This is so because p-Iocal Butler groups are completely decomposable. Nevertheless, isomorphism at p classes of finite rank Butler groups can be interpreted as torsion-free Zp-modules of finite rank (Corollary 8.1.9). Rank-l groups correspond to purely indecomposable p-Iocal groups , and bracket groups correspond to copurely indecomposable p-Iocal groups . Let R be a discrete valuation ring with prime p and quotient field Q, R* the completion of R in the p-adic topology, and k = Rj pR a field. Then R* is a discrete valuation ring with prime p and R*jpR* = RjpR [Fuchs 70, 73]. For example, if R = Zp, then R* is the ring of p-adic number s with quotient field Q*, the field of p-adic rational numbers, and k = Z] pZ. A torsion-free R-module M is divisible if pM = M and reduced if M has no nonzero divisible submodules. There is a unique divisible submodule d(M) of M with M = d(M) ED Nand N a reduced module . The p-rank of M is the Rj pRdimension of M j pM, and M* is the completion of M in the p-adic topology. A pure submodule B of M is a basic submodule of M if B is a finitely generated free R-module with M / B divisible. Fundamental properties of completions [Fuchs 73] are listed in the following lemma. Lemma 8.1.1

If M is a reduced torsion-free R-module offinite rank, then:

(a) M is a pure R-submodule of M*, M* is a free R*-module with R*-rank M* = p-rank M, and M*j M is a divisible R-module;

212

8. Applications of Representations andButler Groups

(b) each R-endomorphism of M extends uniquely to an R*-endomorphism of

M*; (c) if B is a basic submodule of M, then B* = M*; and (d) if N is a reduced torsion-free R-module of finite rank, then each f HomR(M, N) extends to a unique R*-homomorphism f* : M* -+ N*.

E

PROOF. Exercise 8.1.1.

Lemma 8.1.2 Suppose M is a torsion-free R-module offinite rank. (a) If M is not divisible, then M has a basic submodule B with rank B = prank M s rank M. (b) If p-rank M = n and rank M = n + k, then M is free if and only if k = 0 and divisible if and only if n = O. (c) If 0 -+ K -+ M -+ N -+ Oisanexactsequenceoftorsion-free R-modules, then p-rank M = p-rank K + p-rank N. PROOF. (a) Notice that M/ pM is a nonzero finite-dimensional k-vector space. Let {x! + pM, . . . ,Xn+ pM} be a k-basis of M/ pM and define B = Rx, E9 ... E9 Rx n, a free submodule of M. Then M/ B is divisible as B + pM = M. To see that B is pure in M, let x E M and px = rlx! + ... + rnxn. Since the (Xj + pM)'s are k-linearly independent, p divides r, for each i . Thus, x E B, as desired. (b) Observe that M is free if and only if B = M, since if M is free, then M/ B is both free and divisible, hence zero. Thus, by (a), M is free if and only if p-rank M = rank M. Apply (a) to see that M is divisible if and only if M does not have a basic submodule, equivalently, p-rank M = O. Statement (c)follows from thefactthat 0 -+ K/ p K -+ M/ pM -+ N/ p N -+ 0 must be an exact sequence, since the image of K is a pure submodule of M. Let M be a finite rank torsion-free R-module. If M is reduced with p-rank M = I, then M is apurely indecomposable module. In this case, M* is isomorphic to R*, and so M is isomorphic to a pure submodule of R*. If p-rank M = I, then M = d(M) E9 N with N a purely indecomposable module. For an example of a purely indecomposable module, choose t E R*\R and let R, be the pure submodule of R* generated by {I, t }. Then R is a basic submodule of R R; = R*, p-rank R, = 1, and rank R, = 2. " Purely indecomposable R-modules have properties reminiscent of torsion-free abelian groups of rank 1 (Lemma 3.1.1):

Proposition 8.1.3 Suppose M and N are twopurelyindecomposable R-modules. (a) Thefollowing statements are equivalent:

(i) M and N are isomorphic. (ii) M is quasi-isomorphic to N. (iii) HomR(M, N) f. 0 and HomR(N, M) f. O. (iv) HomR(M, N) f. 0 and rank M = rank N.

8.1 Torsion-FreeModules over Discrete ValuationRings

213

(b) The endomorphism ring of M is a discrete valuation ring isomorphic to a

pure subring of R*. M and N are pure submodules of R*. The implications (i) =} (ii) =} (iii) are routine. (iii) =} (iv) Let 0 -:f= f E HomR(M, N) . Then f extends to an R*-endomorphism f* of R*, and so f is multiplication by some a E R*. Hence, f is a monomorphism, and rank M = rank f(M) ::: rank N . Similarly, rank N ::: rank M , since HomR(N, M) -:f= O. (iv) =} (i) Let 0 -:f= f E HomR(M, N). Since rank M = rank Nand f is a monomorphism, N /f(M) is bounded, say rN 5; f(M) with 0 -:f= r E R. Write r = pi u with p-height u = O. Then u is a unit of R, and pi N 5; f(M) 5; N. Since p -rank N = 1, N / piN is a cyclic R-module. It now follows that f(M) = pj N for some j , whence M and N are isomorphic. (b) Exercise 8.1.1.

PROOF. (a) It suffices to assume that

The next proposition gives analogues of the pure fully invariant subgroups G( r), for G a torsion-free abelian group and r a type. Here, an element t of R* plays the role of a type . Let M be a torsion-free module of finite rank with a basic submodule B and assume B 5; M 5; M*, a free R*-rnodule with an R -basis of B as an R*-basis of M*. Let t E R* and define M(t) = {m EM :tm EM}. Notice that if t E R , then M(t) = M .

Proposition 8.1.4 Assume that M and N are reduced torsion-free R-modules offinite rank and t E R*. Then: (a) M(t) is a pure submodule ofM , and iff : M --+ N is an Rshomomorphism, then f(M(t)) 5; N(t); (b) RrM(t) = Hom(R r, M)R r; (c) M(t) n M(s) 5; M(t + s); and (d) ifs E R*, then M(t)(s) = M(st) n M(t).

m E M and pm E M(t) . Then ptm E pM* n M = pM, so that tm E M. Thus, m E M(t) and M(t) is pure in M . The R-homomorphism f : M --+ N extends to an R*-hornomorphism f* : M* --+ N*. Hence, if m E M with tm E M , then f(tm) = f*(tm) = tf*(m) = tf(m) E N. This shows that f(M(t)) 5; N(t). (b) If r E R , then M(t) = M, R, = R, and RM = Hom(R, M)R. Now assume t E R*\R and let m E M(t). Then Rm 6' Rtm S; M , noting that Rm n Rtm = 0, sincet rf. R.Hence,Rrm 5; M,sinceR r = (R6'Rt)*andMispureinM* .Define f: R, --+ M by fey) = ym, an R-homomorphism with m = f(1) E image f. Thus, M(t) 5; Hom(R r, M)R r, and so RrM(t) S; Hom(R r, M)R r. Conversely, let 0 -:f= f : R, --+ M. Then f is a monomorphism, since R, is purely indecomposable. Hence, image f = Rsm 5; M for m = f(l) E M. In particular, m E M(t), and so image f 5; RrM(t), as desired.

PROOF. (a) Let

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(c) If m EM and tm, sm are in M, then (t + s)m is in M. (d) Let m E M(t)(s). Then m E M(t) with sm E M(t), and so tsm E M. This shows that M(t)(s) s:; M(st) n M(t). Conversely, if m E M(st) n M(t), then st m E M and tm EM. Hence, m E M(t)(s). Let TF R denote the category of reduced torsion-free R-modules of finite rank . As a consequence of the next two theorems, there is a variety of embeddings of rep(Sn, R, j) into TF R • Theorem 8.1.5 Let A = {tJ, . . . , tn} be a subset of R*\R such that {I, tj , t;tj: I ::s j ::s n} is a Q-independent set for each i, (a) Foreach integer j 2: 0, there is afully faithfulfunctor FA : rep(Sn, R, j) -+ TF R • (b) The image of FA consists ofall those M E TF R such that (i) M has a basic submodule B with Mj(B + R" M(t,) + ... + R,.M(tn)) torsion; (ii) B is afully invariant submodule of M; and (iii) p! B s:; M(tl) + ...+ M(tn) s:; B. PROOF. Let V = (V o, V; : i E Sn) E rep(Sn, R , j) and define M = FA(V) to be the pure submodule of Va generated by V o + Uit, + ... + Vntn. Given a representation morphism f: V -+ V, define FA(f) = [": FA(V) -+ FA(V), noticing that !*(u;t;) = f(u; )t; E Viti for each i. Then FA is a faithful functor. Observe that Vo is a basic submodule of M , since Vo is a free R-module, Vo is a pure submodule of Va' hence of M, and M j U» is divisible , since M is a pure submodule of Va and j Vo is divisible. Morever, M (t;) = V; for each i, To see this, first observe that V; is contained in M(t;) . Conversely, let m E M(t;). Then m = rouo + rltlu, + + rntnun for some r, E Q and U; E V;. Moreover, tim = rot;uo + rlt;t,ul + + rnt;tnun = qovo + q,vltl + ...+ qnvntn E M for some qi E Q, V; E V;. Fix a basis of Vo, express each U; and V; as a linear combination of the elements of this basis , equate coefficients, and use the hypothesis that {I. tj. t;tj : I ::s j ::s n} is Q-independent for each i to see that m E V;. Now, M j(Vo + RtlM(td+' " + R,kM(tn)) is torsion, since Viti s:; R,;M(t;) = R,;Vi. Moreover, pjvo s:; M(td + ... + M(tn) s:; Vo, since V; = M(t;) and V E rep(Sn. R, j) . Each M(t;) is fully invariant in M by Proposition 8.1.4(a) . Thus, if f E End M , then pj f( V o) s:; Vo n pj M = pj V o, since Vo is a pure submodule of M . This shows that Vo is a fully invariant basic submodule of M. To see that F is a full functor, let g : FA(V) -+ FA(V) and f the restriction of g to Vo, a basic submodule of FA(V). Now, f: Vo -+ Vo with !(V;) = !(FA(V)(t;)) s:; FA(V)(t;) = V; for each i. Thus, f : V -+ V is a representation morphism with FA(f) = g, as desired. Finally, let M be in TF R satisfying the above conditions (i)-(iii). Then V = (B, M(t;) : i E Sn) E rep(Sn, R, j), since M(t;) is a pure submodule, hence a

o;

8.1 Torsion-FreeModules over Discrete Valuation Rings

summand, of the finitely generated free R-module Band p! B S; M(td M (tn) S; B. Moreover,

215

+ .. . +

As a consequence of (b) of the next corollary, FA sends rank-I representations to purely indecomposable modules. In view of (a), not all modules in TF R are in the image of FA . For example, if p-rank M = n ~ 2 and rank M = n + 1, then M is not in the image of FA. Corollary8.1.6 Let FA: rep(Sn, R, j)

~

TF R be thefunctorofTheorem8.1.5.

If FA(V) = M, with V = (Vo, U, : i

E Sn), then p-rank M = rank Vo and rank M = rank Vo + rank VI + ...+ rank U; ~ 2(p-rank M). (b) If V = (Vo, Ui : i E Sn) with rank Vo = I, then M = FA(V) is purely indecomposable and End M = R.

(a)

PROOF. Assertion (a) followsfrom the definitionof FA (V), since Vo+ Viti +...+ Vktk = Vo $ Viti $ . .. $ Vktk by the independencecondition on A and p-rank M = rank Vo S; rank VI + ...+ rank Uv, Statement(b) is a consequenceof (a) and the fact that R = End V is isomorphic

to End FA(V), since FA is a fully faithful functor.

Recall from Section 4.2 that !!..(Sn, R) is the full subcategoryof rep(Sn+l, R , 0) with objects V = (Vo, VI, .. . , Vn, V*) such that Vo = VI $ . . . $ Vn, V* is a pure , tn) to be submodule of Yo, and V* n Vj = 0 for each 1 :5 i :5 n. Define V*(tl , the set of elements (tl VI , . . . , tnvn) E V; = $ .. . $ Vn*, with (VI , , vn) E V* S; Vo = VI $ . . . $ Vk . Let M E TF R with basic submodule B = B I $ $ Bn. Then M is a pure submoduleof B* = Bj $ .. . $ B~ . Given (t) = {tl' , tn}, a subset of R*, define M(I) = {(VI , .. . , Vk) EM: (tl vI , . . . , tkVk) EM} . There is another uncountable family of embeddings ofrep(S, R, 0) into TF R •

vt

Theorem 8.1.7 Let A = {tl, " " tn} be a subset of R*\R such that {I, tj , tjtj : 1 :5 j :5 n} is a Q-independent set for each i. There is a faithful embedding G A :rep(Sn, R, 0)

~

TF R

that is full for all those V E rep(Sn , R, 0) with no rank-Y summands. The image of G A consists ofall those M such that:

»

+ M(1)(tI ' . . . , tn is torsion, where B = B I $ .. . $ Bn is a fully invariant basic submodule of M with each B, fully invariant in M,. and (ii) M(1) is contained in B. (i) M I(B

PROOF. In view of Lemma 4.2.3(b), it suffices to define G A: !!..(Sn , R) ~ TF R • = (Vo, Vj, V* : i E Sn) E !!..(Sn , R), recalling that Vo = VI $ . . . $ Vn

Let V

216

8. Applications of Representations and ButlerGroups

«», o

and V* is a summand of Vo. Define M = G A(V) = (Vo + V*(tl.·.· , ~ Va. Given a representation morphism f : V ~ V' , extend f: Vo ~ V to an R*-homomorphism f* : Vo* --+ VQ* , and let G A (f) : G A(V) ~ G A (V') be the restriction of f* to G A (V). It follows that G A is a faithful functor. Now, Vo is a basic submodule of M . since Vo is free, Vo is pure in Vo*' hence in M, and M/ Vo is divisible, since M is pure in Va and Va/ Vo is divisible. Moreover, M(t) ~ B. as a consequence of the independence conditions on A and the fact that

Va =

vt $

... $ Vn* ·

To see that G A is full, assume that V and V' have no proper rank-l summands and let g : M = G A(V) --+ M' = G A(V') be an R-homomorphism. Extend g to Then g : M(t ) ~ M(t): an R*-homomorphism g* : M* = Va ~ (M ')* = (V with M(t) = V*. Since V has no proper rank-l summands, each V;/rri(V*) must be torsion; otherwise. there is a cyclic summand of Vi that gives rise to a representation summand of V . It now follows that g : Vi ~ V;' for each i, since g : V* ~ V~. Now. Vo = VI $ .. . $ Vb so letting f = g gives a representation morphism V ~ V' with G A (f) = g. as desired. Finally, let M be as described in the statement of the theorem. Then V = (B, Bit M(t) : i E Sn) E /::}.(Sn, R). noting that M(t) is pure in Band M, n B, = 0 for each i , Hence. G A(V) = (B + M(t )(tl, .. . , tn))* = M. as desired.

o)*·

A torsion-free R-module of finite rank with no free summands is copurely indecomposable if rank M = p-rank M + 1. Observe that if M is a finite rank torsion-free module with rank M = p-rank M + 1, then M = L $ F, where F is a free R-module and L is copurely indecomposable.

Corollary8.1.8 LetG A :rep(Sn, R .O) ~ TF R be thef unctor of Theorem 8.1.7. = G A (U), then p-rank M = rank UI + ... + rank U; and rank M = 2 (p-rank M) - rank Uo ~ 2(p-rank M). (b) LetU = (Uo.Ui:i E Sn) E rep(Sn,R.O) with n :::: 2, Uo = R n- I, UI = R $ 0 $ . . . $ 0, . . . , Un-I = 0 $ 0 $ .. . $ R, and Un = R(l, 1•...• 1). Then M = G A (U) is copurely indecomposable with End M = R.

(a)

If M

Assertion (a) is a consequence of the definition of GA. (b) Apply the functor of Lemma 4.2.3(b) to see that U ~ V, where V = (R" = R$ · ··$R. R$O$ · ··$0, . . . • 0$ ·· ·$R, R(I, ...• 1)) E ~(Sn, R). By (a), M = GA(V) = tR", R(tl, . . . , tn))* has p-rankn, rankn + 1, and endomorphism ring R. In particular, M is copurely indecomposable. PROOF.

Corollary 8.1.9 Let T be a finite lattice of types with Sr = Sn and A = {tl' ... , td a subset of 7l; \71 p such that {I, tj . tits- : 1 ~ j ~ k} is a Q-independent set for each i. (a) There is a fully faithful embedding FA : B(T. j)p

FA(G) = (Gp + G(rl)ptl

~

TF given by

+ ...+ G(rk)ptk)* 5; G; .

8.1

Torsion-Free Modules over Discrete Valuation Rings

217

(b) There is a faithful embedding G A : B(T, 0) p ~ TF that is full for groups H with no proper rank-I summands given by GA(H)

= (H(rt>p E9 ~ H(rt>;

E9

E9 H(rk)p

+ H*p(t) , ... , tk»)*

E9 H(rk);,

where H* is the kernel of H(rl) E9 . . . E9 H(rd

~

H.

PROOF. A consequence of Corollary 4.3.2, Theorem 8.1.7, and Corollary 8.1.8. Indecomposable modules in the image of FA and G A have wild modulo p representation type if IAI ~ 3, since rep(S3, R, 0) has wild modulo p representation type by Example 4.2.4. Following is a complete list of indecomposable modules in the image of FA and the image of G A for the case that IA I = 2. This list arises from a complete list of indecomposables in rep(Sz, Zp, j); Example 4.1.2.

= {r), rz} = Si and A = {t), tz} a subset ofZ; \Zp such that {I, tl, tztI. ti, tll is a Q-independent set for each i. Corollary 8.1.10 Let T be afinite lattice of types with Sr

(a) IfG to: (i) (ii) (iii) (iv)

E

B(T, j) and M

= FA(G) is indecomposable, then M is isomorphic

tl p if M has p-rank I and rank I ; (71 p, 7l pt))* or (71 p, 7lptz)* if M has p-rank 1 and rank 2; (Zp , Zpt), 7lptz)* ~ Z; if M has p -rank 1 and rank 3; or (B, Zpt) E9 0, 0 E9 Zptz)* ~ B* if M has p-rank 2 and rank 4, where E9 + 7lp (l , 1)/ pj 5; Q EEl Q . B =

z, z,

(b) If H E B(T, 0) and M = G A(H) is indecomposable, then M is isomorphic to: (i) tl p if M has p-rank I and rank 1; or (ii) (71 p E9 + 7lp(t) , tz»)* ~ 7l; E9 7l; if M has p-rank 2 and rank 3.

z,

PROOF. Exercise 8.1.2. The next corollary demonstrates a relationship between rank-1 groups in B(T) and purely indecomposable p-local groups and between bracket groups and copurely indecomposable p-local groups .

Corollary 8.1.11 Assume that T is a finite lattice of types with Sr = S; = {r), .. . ,rn } . (a) IfG E B(T, j) has rank 1, then FA(G) is purely indecomposable. = G[X), . . . , X n ] isa bracket group, where each Xi isa rank-I group with type Xi = Ti, then G A (H) is copurely indecomposable.

(b) If H

218

8. Applications of Representations and Butler Groups

PROOF. Exercise 8.1.2.

OpenQuestion: Canthe resultsofthissectionbe extendedtofunctorial embeddings ofrep(5, R , j)for an arbitraryfinite poset 5 ? Can each reduced p-local torsion-freegroup offinite rank be so represented? EXERCISES I. Prove Lemma 8. I.l and Proposition 8.1.3(b).

2. Confirm Corollaries 8.1.10 and 8.1.11. 3. (a) Show thatif M is a purelyindecomposable R-moduleof rank2, then M is isomorphic to M, for some t E R*. (b) Compute the endomorphism ring of Mr . (c) [Richman 68] Find necessary and sufficient conditions in terms of sand t for M, and M, to be isomorphic. 4. [Arnold 72] Let M be a copurely indecomposable module with p-rank n and rank n + I. (a) Prove that QEnd M is a subfield of the p-adic completion of the quotient field Q of R. (b) Give an example to demonstrate that quasi-isomorphic copurely indecomposable modules need not be isomorphic. Why must such an example have rank greater than or equal to 3? (c) Prove that if N is another copurely indecomposable module, then M is quasiisomorphic to N if and only if the nth exterior power of M is isomorphic to the nth exterior power of N .

8.2 Finite Valuated Groups There are correspondences between finite valuated p -groups and representations of finite po sets over the field Z/ pZ that can be used to determine the representation type of some categories of finite valuated groups. These correspondences also give rise to correspondences between indecomposable finite rank Butler groups and indecomposable finite valuated p-groups with torsion-free groups of rank 1 corresponding to cyclic valuated p-groups. Let p be a prime. An abelian p-group G is a valuated p-group if there is a mapping v from G to the ordinals together with 00 such that v(O) 00 , v(px) > v(x) , v(rx) = v(x) if r is an integer prime to p , and v(x + y) ::: min{v(x) , v(y)} for each x , y E G . Here, it is agreed that 00 > 00. As an example of a valuated p-group, let G be a subgroup of a p-group Hand define v(x) to be the p -height of x in H for each x E G. Then G is a valuated p-group, as a consequence of properties of the p-height function on H . In fact, this is the only example, since if G is a valuated p-group with valuation v, there is a p-group H such that G is a subgroup of Hand v(x) is the p-height of x in H for each x E G [Richman, Walker 79].

=

8.2 Finite Valuated Groups

219

The class of finite valuated p-groups forms a category. Morphisms, called valuated homomorphisms, are group homomorphisms f: G -4 H with v(x) ::: v(f(x» for each x E G. This category is an additive category; the direct sum of G and H is the group direct sum G EB H with v(x, y ) = min{ v(x) , v( y)} for x E G, y E H [Richman, Walker 79]. As will be seen , indecomposable valuated groups need not be indecomposable as groups. Let G be a finite valuated p-group and x E G. The set T(x , G) = {y E G : p" y = x , some n ~ O} is a finite poset, where y ::: y' if y = r" y' for some m ~ O. Then T(x , G) is a tree with unique minimal element x , called a root of T . If y , yl E T(x, G) with y < v', then V(yl) < v(y). Hence, T(x , G) is an example of afinite valuated tree, a rooted tree T with each vertex y of T labeled by v(y), an ordinal, or 00, such that if y , y' E T with y < v', then v( y') < v(y). Valuated trees play the same role for valuated p-groups as height sequences do for torsion-free groups. The point is that divisibility by integers is unique for a torsion-free group G, so that there is no need to record those y for which ny = x . However, divisibility by p is not unique for p-groups. Thus, vertices of the finite valuated tree T(x, G) record the elements y of G for which p" y = x ; in addition, these vertices are labeled by their values . For finite valuated trees Sand T, define S ::: T if there is a function f : S -4 T such that if y < y' in S, then f( y) < f( y') in T and v( y) ::: v(f (y)) for each y E S. Two finite valuated trees Sand T are equivalent if S ::: T and T ::: S. If S is a finite valuated tree and y E S, then length y is the length of a chain of elements from the root of S to s.

Lemma 8.2.1 The set of equivalence classes [T] offinite valuated trees forms a lattice. Specifically: (a) [S] ::: [T] if there is an SI equivalent to S and a T ' equivalent to T with SI s T ' . (b) [S] n [T) = [U], where U = {(s, t) E S x T: length(s) = length(t)} isafinite valuated tree with root U = (root S, root T) and v(s, t) = min{ v(s), v(t)}. (c) [S] U [T) = [V] , where V, the disjoint union of Sand T with the roots of S and T identified, is afinite valuated tree with v(root V) = max{v(root S) , v(root T)} and the values ofall other elements unchanged. PROOF. Exercise

8.2.1.

If x is an element of a finite valuated group G, then u-height x is the equivalence class of the valuated tree T(x, G). For an equivalence class [T] of valuated trees, define G(T) = {x E G : v-height x ::: [T]}.

Lemma 8.2.2 Suppose G and Hare finite valuated p-groups. (a) If f: G -4 H is a valuated homomorphism, then f(G(T)) ~ H(T) for each finite valuated tree T. (b) If[S] and [T] are equivalence classes offinite valuated trees with [S] ::: [T], then G(T) ~ G(S).

220

8. Applications of Representations and Butler Groups

PROOF. Statement (a) follows from the fact that v(x) :::: v(f(x» for each x and (b) is clear.

E

G,

The machinery is now in place for the finite valuated p-group analogue of the category of finite rank Butler groups with typeset contained in a finite lattice of types. Let L be a finite sublattice of the lattice of equivalence classes of valuated trees. An L-group is a finite valuated p-group G with v-height x E L for each x E G . Each finite valuated p-group G is an L -group for L the finite lattice generated by {v-heightx : x E G} . Define Ji. to be the poset ofjoin irreducible elements of L with reverse order. For a p-group G, and integer j ::: 1, G[pj) = {x E G: p j x = OJ. Notice that if f : G ~ H is a group homomorphism, then f(G[p j)) is contained in H[pj) .

Theorem 8.2.3 Let L be afinite sublattice ofthe lattice of equivalence classes offinite valuated trees and j ::: 1 an integer.

Z/ pjZ) given by Fj(G) (G[p j) , G(T)[pj) : [T) E h). (b) If U = (U o, U, : i E h) E rep(h, Z/ pZ), then there is an L-group G with Ft(G) = U . (a) There isafunctor Fj from thecategoryofL-groups torep(h,

=

(a) For L -groups G and Hand f : G ~ H a valuated homomorphism, define Fj(f) : G[pj] ~ H[p j). Then Fj(f) is a representation morphism by Lemma 8.2.2(b). It now follows that Fj is a functor. (b) Let n = lSI and m the Z] pZ-dimension of Us; Define G to be the direct sum of m copies of Z/pnz with Uo = G[p) . Write S = {I, 2, ... , n}, so that U (Uo, UI> . . . , Un)' There is a valuation v of G given by v(O) 00 , v(x) 2(n - k) + 1 if order x pk with k ::: 1 and pk-I x E Ui, and v(x) 2(n - k) if order x pk with k ::: 1 and pk-I x fj; Ui, To see that v is a valuation, notice that if x , y E G with order y < order x , then v(y) > v(x) . Thus, v(px) > v(x), since order px < order x. Moreover, v(x + y) ::: min{ v(x) , v(y)}, since order x+ y :::: max{order x , order y}. If order x + y order x order y k, then v(x) , v(y), and v(x + y) are either 2(n - k) or 2(n - k) + 1. If v(x) = v(y) = 2(n - k), then pk-I x and pk-I y are 2(n - k) . The other elements of Ui; whence pk-I(x + y) E Uk and v(x + y) cases are routine. Finally, if gcd(r, p) = I, then v(rx) = v(x), since pk-I x E U« if and only if p' . . . > tn} with v(t;) = 2i - 2 if Uk n Un-i+1 0 and v(t;) 2i - I if Uk n Un-i+I =1= O. It remains to prove that U, G(Tk)[p) for each k, in which case F(G) U . First let 0 =1= x E Ui, Then order x p so that v(x) 2n - 2 if x fj; U I and v(x) = 2n - 1 if x E UI. Thus, v(tn) :::: v(x) . Similarly, T(x , G) ;2 Ti, so that v-height x ::: Tk and x E G(Tk)[p). Therefore, Uk ~ G(Tk)[P) . Conversely, let x E G(Tk)[p) . Since T(x , G) ;2 Ti; there is y E G with pk-I y = x and v(y) ::: v(tn-k+)). Then x E Ui; otherwise, v(y) = 2(n - k) < PROOF.

=

=

=

=

=

=

=

=

=

=

=

=

=

=

=

=

8.2 Finite Valuated Groups

221

V(tn-k+I) = 2(n - k + I) - I = 2(n - k)+ I, since Uk n Un-(n-k+I)+1 = Uk # 0 . This shows that G(Tk([p]) is contained in Ui , as desired.

If U is an indecomposable representation and G is an L-group with Fj(G) = U, then G must be an indecomposable as a valuated group. This observation allows an explicit construction of indecomposable valuated L-groups from a given representation . Example 8.2.4 Assume J: = S3. The only indecomposable L -groups constructed in Theorem 8.2.3(b) are: (i) G = (Zl p3Z)x with (v(x) ,v(px),v(p2x» = (0,2,4), (1,2,4), (0,3,4), (0,2,5), (1,3 ,4), (1,2,5), (0,3 ,5), or (1 , 3,5); and (ii) G (Zl p 3Z)a El7 (Zl p3Z)b with (v(a), v(pa), v(p2 a»= (1,2,4), (v(b), v(pb), v(p 2b» (0,3,4) and (v(a + b), v(p(a + b), v(p2(a + b» = (0,2,5).

=

=

k), l-dimensional representations and a 2-dimensional representation. The groups in (i) correspond to the l-dimensional representations (k, 0, 0, 0), (k, k, 0, 0), (k , 0, k, 0), (k, 0 , 0, k),(k, k, 0), (k , k, 0, k), (k , 0, k, k), and (k, k, k, k), respectively, Similarly, the group in (ii) corresponds to the 2-dimensional representation (k El7 k, k El7 0, El7 k, (1 + I)k).

PROOF. Via Example 1.1.5, the only indecomposable representations in rep(S3,

k=

'Ill pZ, are the

°

=

If G is an L-group with FI (G) = U (U«, U, : i E S), as constructed in Example 8.2.4, then p" G 0, where n is the number of elements of S. Moreover, v(x) ::: 2n + 1 for each x E G .

=

Example 8.2.5 There are infinitely many isomo rphism classes of indecomposable p4-boundedfinite valuated groups G with v(x) ::: 9 for each x E G. PROOF. Apply Theorem 8.2.3 and the fact that rep(S4,

'Ill pZ)

has infinite repre-

sentation type; Example 1.1.6.

Example 8.2.6 The category of p5-bounded finite valuated groups G with v(x) ::: 11 for each x E G has wild representation type. PROOF. A consequence of Theorem 8.2.3 and the fact that rep(S5,

'Ill pZ) has wild

representation type ; Example 1.4.1 . The connection between finite rank Butler groups and finite valuated p-groups is intriguing. A group G E B(T)p is sent to UG (G p» G(r)p : r E Sr) E rep(Sr , Zp). If L is a finite lattice of equivalence classes of finite valuated trees Sr as posets, then G' (Gf p G , (G(r) + pG)lpG: r E Sr) E with Jt.

=

=

=

222

8. Applications of Representations and Butler Groups

rep(h, Z/pZ) . Moreover, there is an L-group H with Ft(H) = G' by Theorem 8.2.3. Consequently, G -* H is a correspondence from B(T) to L-groups such that if G is indecomposable at p, then H is indecomposable as a valuated group . Moreover, rank G = 1 if and only if H is a cyclic group.

Open Questions: 1. Do the functors of Theorem 8.2.3 and Proposition 8.2.4 reflect isomorphisms?preserve indecomposables? 2. Is Theorem 8.2.3(b)truefor all j ::: I?

EXERCISES 1. Prove Lemma 8.2.1. 2. Prove that there are infinitely many isomorphism classes of finite valuated p-groups G that are p3-bounded with v(x) :::: 13 for each x E G.

NOTES ON CHAPTER 8 There is a duality on the quasi -homomorphism category of finite rank torsion-free modules over a discrete valuation ring that sends purely indecomposable modules to copurely indecomposable modules [Arnold 72]. A more abstract version of this duality is given in [Lady 83] for Butler modules over a Dedekind domain R. The representation type for subcategories of TF R, defined in terms of the Q-dimension of a splitting field of a finite rank torsion-free R-module, is partially computed in [Lady 77B, 80A]. These computations employ represent ations of species [Ringel 76] instead of representations of partially ordered sets. Other papers on torsion -free modules of finite rank over discrete valuation rings include [Turgi 77] and [Franzen 83]. There is an extensive and evolving body of literature on torsion-free modules over valuation domains ; see [Fuchs, Salce 85] and references therein . Section 8.2 is a very elementary introduction to valuated groups . The seminal paper on the subject is [Richman, Walker 79]. Other embeddings from categories of representations of finite posets over 'Ill pZ into the category of finite valuated p-groups are given in [Arnold, Richman, Vinsonhaler 91]. These embeddings are functorial and send rank-l groups to simply presented valuated groups , but the functors are not additive functors . It is evident in [Beers, Hunter, Walker 83] that classification of finite valuated p-groups is intimately related to the solution of matrix problems for representations of finite partially ordered sets over 'Ill pi Z. A classification of valuated p-groups with values less than or equal to 4 is given in [Richman , Walker 99]. Each such group is the valuated direct sum of a short list of indecomposable finite valuated p-groups. This parallels the fact that if rep(S , k) has finite representation type, then a representation of arbitrary dimension is a direct sum of finite-dimensional indecomposable representations. On the other hand, there are arbitrarily large valuated p-groups with values less than or equal to 5. The question of whether or not the category of finite valuated p-groups with values less than or equal to 5 has wild representation type remains unresolved . Just as for representations of finite posets over discrete valuation rings, a suitable definition of tame representation type remains elusive.

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List of Symbols

A=(A1, . . ·,A n )

a<1

A(m, A)

Ap

AQ

b'> bf(G)

B(T) B(T, j) B(T, j)p B(T)Q C(A) C(A, B) cdn H cdn U

CG

CCn

-

C(n)

-

coker f Coker f == (mod p)

-

n-tuple of subgroups of Q subset of a poset S with a E S Jordan block matrix isomorphism at p category of torsion -free abelian groups quasi-homomorphism category of torsion-free abelian groups subset of poset S with b E S Burkhardt invariant of group G category of finite rank Butler groups, T a lattice of types subcategory of B(T), j ::: 0 isomorphism at p category of B(T, j) quasi-homomorphism category of B(T) centralizer of matrix A centralizer of two matrices A and B coordinate vector of a torsion-free abelian group H coordinate vector of representation U subring of the center of the center of End UG, UG E rep(ST , Q) Coxeter correspondence chain with n elements a category of almost completely decomposable groups with typeset contained in an antichain the cokemel of an R-module homomorphism the cokemel of a representation morphism congruence of integers modulo a prime p

236

List of Symbols

c(f)

C+ CR Crep(Sr,

CTcr(G)

CCT, j) CCT, n.

c-sr. j)

8(A) 8(a,b)S

d(G)

z; j)

-

content of polynomial I(x) E 'll[x] Coxeter correspondence center of ring R the subcategory of rep( Sr , 7l p » j) corresponding to

CCT, n,

- the subcategory of rep( Sr , 7lp » j) corresponding to Ccrit(T, j) - critical cotypeset of G - almost completely decomposable groups in B(T, j) - isomorphism at p category of CCT, j) - almost completely decomposable groups in B(T, j) with critical typeset S;; Sr - n-tuple of subgroups of Q - derivative of poset S with respect to suitable pair (a, b) - maximal divisible subgroup of torsion-free abelian group

G d(M)

dimU

EB eG E

~ EndU EndR(M) E(n, k)

Ig

r:

frA frAlp

frA p frAQ

G[A]

gcd GIH

Gn

Gp G[pi]

- maximal divisible submodule of torsion-free module over a discrete valuation ring - a subcategory of rep(Sr, 711 pi'll) of representations in matrix problem form - a subcategory of rep(Sn+l, R) of representations in matrix problem form over a ring R - dimension of a representation U - direct sum - exponent of G I R( G) for a finite rank Butler group G - element of - not element of - ring of endomorphisms of representation U - R-endomorphism ring of R-module M - a subcategory ofrep(Sn+2, k) - composition of morph isms I and g defined by Ig(x) = I(g(x» - composition of a morphism I repeated n times - category of torsion- free abelian groups of finite rank - mod p category of torsion-free abelian groups of finite rank - isomorphism at p category of torsion-free abelian groups of finite rank - quasi-homomorphism category of frA - bracket group - greatest common divisor - group of cosets of a subgroup H of an abelian group G - direct sum of n copies of abelian group G - localization of torsion-free abelian group at a prime p - pi -socle of an abelian group

Listof Symbols

G(T)

- localizationof a torsion-free abelian group at a set of primes S - socle of a valuated abelian group for a tree T

Gr

-

Gs

G[r] G*[r] G(r) G*(r) G#(r)

hex) hp(x)

H+K

Hom(A , B)

Hom(G, H) Hom(G, H)p HomR(M, N) (i)

i(G) 1M

In lnd(mod R) lnd(n, k)

Ind(S, k) Ind(S, R , j) I (t)

IT(G) iu I(U)

n

J/(T)

it

JR

ker f Ker f kn

k[x] k(x) k(x, y) k(U)

237

- r-radical

subgroup containing r -radical r -socle of torsion-free abelian group G subgroup of G generated by {x : type X > r ] pure subgroup generated by G*( r ) r-homogeneous completely decomposablesubgroup of Butler group G with G(r) = G r E9 G#(r) - height sequence of an element x in an abelian group - p-height of an element x in an abelian group - subgroup of group G generated by Hand K - morphism group of objects A and B in additivecategory - group of homomorphisms of abelian groups G and H - localization of Hom(G, H) at a prime p - group of R-homomorphisms of R-modules M and N - chain of length i - regulating index of an almost completely decomposable group G - submoduleof R-module M generated by {xm: x E I , m E M} for ideal I of R - nth power of ideal I - isomorphismclasses of indecomposables in mod R - isomorphismclasses of indecomposable k-representations of an antichain - isomorphismclasses of indecomposable representations of poset S - isomorphismclasses of indecomposables in rep(S, R , j) - l-dimensional injective in rep(S, k) - inner type of G - injection for direct sum - injectiveenvelopeof U E rep(S, k) - intersectionof sets, greatest lower bound of types - join-irreducible elements of a poset T - opposite of the poset of a lattice of v-heights - Jacobson radical of ring R - the kernel of an R -homomorphism f - the kernel of a representation morphism f - n-dimensional k-vector space for field k - polynomial ring in indeterminate x with coefficients in k - quotient field of k [x] - polynomial ring in two noncommuting variables x and y - a representation determined by U E rep(S, k)

238

List of Symbols

km L(U) Matm(k) M~N

M- 1

MIN modR

Mp

M(t) MOl

JLr

Mu

=(Ail, . · IA n )

n(G)

n+

NR OT(G)

n

P(A)

Pf(R) P(n, j)

pWG

peR) P(t) Pu

P(U)

o Q

qG

iQG Qrep(S, R) qu

R

R* rep(S, k)

-

least common multiple representation determined by U E rep(S, K) ring of all m x m matrices equivalence of matrices M and N under row operations and block column operations - inverse of square matrix M - module of cosets of submodule N of module M - category of finitely generated R-modules - a localization of M E Qrep(S, R) at an up-set P containing M - a localization of M E Qrep(S, R) at an up-set P contained in M - a module over R, t E R* - a module over R, (t) a subset of R* - right multiplication by an element r of a ring - matrix of representation U with block matrices Ai , - an integer determined by the endomorphism ring of a torsion-free abelian group of finite rank - {1,2, . .. ,n} - nil radical of ring R with finite rank torsion-free additive group - outer type of G - set of primes of the integers - category of summands of direct sums of copies of an abelian group A - category of summands of finite direct sums of copies of an abelian group A - category of finitely generated projective right R -modules - a poset - p-divisible subgroup of abelian group G - category of projective right R -modules - I-dimensional projective in rep(S, k), in Qrep(S, R) - projective cover P(U) ~ U E rep(S , k) - projective representation determined by U E rep(S , k) - field of rational numbers - quotient field of discrete valuation ring - {qx: x E G} for q a rational and G a torsion-free abelian group - {qx: q E iQ, x E G} for torsion-free abelian group G - category of quiver representations of a finite poset S over a principal ideal domain R - injective envelope U ~ leU) E rep(S, k) - ring, associative, with identity - completion of a ring R - category of finite-dimensional k-representations of S

List of Symbols rep(S, R)

rep(S, R , j) Rep(S, k) Rep(S, R)

R1

X

R(G)

R2

R+

RT(G)

Ru lSI

S* a

(SI," ·, Sn) Sn S(n , j) sop

I:{Vi:iES} I:{G i :iES} Sr (S) (S)*

T(M) r(P) T(x , G)

VI +V2 V = (Vo, Vi: Vcrit(T, j)

239

- category of finite rank free representations of poset S over ring R, over discrete valuation ring - category of finitely generated representations containing rep(S, R) - subcategory ofrep(S, R) for R a discrete valuation ring - category of countable representations of poset S over field k - category of countable representations of poset S over ring R - ring product of rings R1 and R2 - regulator subgroup of a finite rank Butler group G - additive group of a ring R - Richman type of G - endomorphism ring of matrix M u - cardinality of set S - disjoint union of a poset S and a point - dualityfromrep(S ,k)torep(SOP ,k) - disjo int union of posets S, - antichain with n elements - poset - oppos ite of poset S - subspace of Vo generated by subspaces {Vi : i E S} of Vo E rep(S, k) - subgroup of G generated by subgroups G, of G - opposite of poset of join irreducible elements of T - subgroup of an abelian group generated by a subset S - pure subgroup of an abelian group generated by a subset S - critical typeset of G - tensor product - meet of a set of types with indices in E - category of torsion-free modules of finite rank over a discrete valuation ring R - torsion-subrepresentation of M E Qrep(S, R) - type of a partition of n+ - finite valuated tree for x E G, a valuated abelian p-group - subspace of Vo generated by VI and V2 i E S) - representation of poset S - subcategory of Ccrit(T, j) of uniform almost completely decomposable groups - subspace of Vi generated by {V j : j < i} - n{Vj : j > i} for V = tU«, Vi) E rep(S, k) - direct sum of n copies of representation V - union of sets, least upper bound of types

*

240

List of Symbols

VeT, j) V

VA w(S) W(V) W'(V)

Z Z[ljn]

- subcategory of qT, j) of uniform almost completely decomposable groups - valuation of an abelian p-group - k[x]-module defined by matrix A - width of poset S - injective envelope of C+V for V E rep(S, k) - projective cover of C- V for V E rep(S, k) - ring of integers - subring of rationals generated by 1 and I j n for nonzero integer n - 'Zj i'Z if i integer and 'Zp/'Z if i = 00 - localization of integers at a prime p

Index of Terms

abelian group almost completely decomposable, 51 balanced projective , 198 B(I) -,205 bounded ,48 bracket, 205 Butler, 91, 105 central special, 115 cobalanced injective, 198 completely decomposable, 51, 105 divisible, 56 exponent of finite, 147 finitely Butler, 105 homogeneous , 79 indecompo sable at p , 51 injective in B(T), 197 L- ,220 locally completely decomposable, 99 p-local,56 p-primary regulating quotient, 169 pregeneric, 123 preprojective , 20 I preinjective , 201 p-reduced, 52 projective in B(T), 197 quasi-generic, 113 reduced ,56 Richman-Butler, 205 self-cancellation, 68 self-small, 69 semilocal, 58 special, 115

strongly indecompo sable, 50

T- , 116 r-homogeneous,79 torsion -free, 47 uniform almost completely decomposable, 164 valuated p-, 218 algebraically independent elements , 73 antichain, 10 Boolean algebra, 84 Burkhardt invariants, 148 cancellation self,68 Cartan matrix, 177 category additive, 10 definition , 10 equivalence , 69 categories of abelian groups almost completely decomposable, 154, 155, 160 Butler, 100, 140 generically wild, 121 isomorphi sm at p , 51 Krull-Schmidt, 50 modulo p , 53 quasi-homomorphism, 47 torsion-free , 47 uniform almost completely decomposable, 164

242

Index of Terms

categories of representations of posets genericalIy trival,41 genericalIy wild, 43 over a discretevaluation ring, 126, 127 over a field, 1, 11,36 over a ring, 31 centralizer matrix,7 pair of matrices, 8 chain, 20 completion p-adic,211 l-adic,72 composition series of a module, 39 coordinate vector Butlergroup, 122 representation, 25 Comer's theorem,72 cosupport homomorphism, 90 Butlergroup, 90 cotrimmed n-tuple of groups,206 cotypeset critical, 83 definition, 82 Coxetercorrespondences, 179,201 dimension of representation, 21 direct sums abelian groupsat p, 51 objects in a category, 11 representations, 2, 12 quasi-,49 duality, 38 endolength of a module, 40 endomorphism ring representation of antichain, 1 representation of poser, 11 endowild, 33 equivalence relationson n+ r-equivalence,206 y-equivalence, 207 exact sequences abelian groupsin quasi-homomorphism category, 199 almostsplit, 187 balanced, 109 homomorphism, 187 isomorphism, 187 left almostsplit, 187 representations, 173 right almostsplit, 187 filter, 193 functor additive, 31 contravariant, 38 dense,69 faithful; 31

full,31 naturalequivalence, 69 preserves indecomposables, 31 reflects isomorphisms, 31 genusclass of a module, 60 Grothendieck group,67 height sequence, 77 equivalence, 77 p-,77 idempotents, I lift in a ring, 40 split in a category, 19 injective envelope of representation, 175 isomorphism of abelian groups at p,51 near-,56 quasi-,47 join irreducible elementsof a lattice, 100 k-algebra finite-dimensional, 8 Kleiner'sTheorem,23 KruII-Schmidt-Azumaya theorem for additive categories, 19 Krull-Schmidtcategory, 50 lengthof a module, 39 localization group at a set of primes,98 integers at p , 50 representation, 193 matrix centralizer, 7 equivalence, 9 indecomposable, 15 problems, 15, 128 similarity, 10 simultaneous equivalence of pairs, 10 simultaneous similarityof pairs, 10 meet-irreducible elementsof a latttice,200 module Butler, 122 copurelyindecomposable, 216 divisible, 211 endolength, 40 length,39 purelyindecomposable, 212 rank,71 reduced, 211 R-Iattice,60 simple,39 stronglyindecomposable, 122 torsion-free, 71

Index of Terms morphisms of abelian groups balanced epi-, 89 cobalanced mono-, 89 endo -,47 epi-,47 iso- at p, 51 modulo p, 53 mono-,41 quasi iso-, 49 valuated ,219 morphisms of representations definition , I, 11 endo-, I epi-, 173 injection s for direct sum, 12 irreducible, 191 iso-, 173 mono-, 173 projections for direct sum, 12 proper, 174 pure, 174 splitting epi-, 187 splitting rnono-, 187 Nakayama's lemma , 40 Nazarova's theorem, 33 nilpotent ideal, 40 p-Iocally free groups , 140 lattice of types, 140 polynomials content, 114 primitive, 114 Pontryagin 's lemma, 72 poset chain, 20 definition, 10 derivative, 27 forest, 160 garland, 131 opposite of, 38 sincere, 25 splitting decomposition, 23 sub,19 suitable pair, 27 tree, 219 V-free, 98 width,20 p-rank of module, 211 projective cover of representation, 175 quadratic form of poset, 177 quiver, 191 quotient field of polynomial ring, 41 radical Jacobson, 40 nil,59 T-,82

243

rank module, 71 representation, 127 representation of poset cotrivial, 174 definition, I , 11,31,36, 127 generic, 40 indecomposable, 2, II injective, 174 preinjective , 181 preprojective, 180 projective, 174 sincere, 25 trivial,2 representation of quiver, 191 representation type endowild,33 finite, 19 fully wild, 32 infinite, 19 rank-finite, 131 rank-infinite , 131 strongly wild, 37 tame, 37 wild,32 wild modulo p, 135 ring artinian, 40 discrete valuation, 126 incidence algebra of poset, 195 local, 11 maximal S-order, 59 one in the stable range, 63 Ore domain, 120 polynomi al in two noncommuting variables , 31 S-order,59 semiperfect , 40 semisimple, 40 totally definite quaternion, 66 two in the stable range, 65 root of tree, 219 socle T-,78 subcategory full,13 subgroup of an abelian group pure, 52 regulating, 97 regulator, 97 T -radical , 82 T-socle,78 submodule basic, 211 pure, 71, 126 substitution property definition , 63 one-,63 two-, 64

244

Index of Terms

summandof an abelian group at p,51 near, 56 quasi-,50 support element,90 Butler group, 90

inner,84 lattice, 76 outer,84 Richman, 88 typeset critical, 79 definition, 77

topology p-adic, 211 Z-adic,72 tree, 219 type definition, 76

up-set, 113 valuated tree definition, 219 equivalence, 219 v-height, 219