Banach Algebras and Automatic Continuity (London Mathematical Society Monographs New Series)

obtained from two formulae of { ponens. Two sentences are equivalent in a theory if each can be deduced from the other in the theory. A theory T is conszstent if, for each sentence
4

Algebrau; foundatwns

We shall frequently apply AC in the form of an equivalent axiom, Zorn's lemma. Before stating this. wp give some definitions about ordered sets. Definition 1.1.1 Let S be a non-empty set. A preorder on S is a binary relation :::; such that:

(i) ifr:::; sand s:::; tin S. then (ii) 8:::; 8 for each 8 E S.

7':::;

t;

The relatwn :::; is a partial orcier, and (S,:::;) is a partially ordered set if, f1LrtheT:

(iii) if l'

:::;

sand

8 :::; l'

in S, then

l'

= 8.

The relation:::; is a total order and (S,:::;) zs a totally ordered set if, f1LrtheT:

(iv) for each

1',8

E

S, ezther

7' :::; ,<;

or s :::; r.

A s1Lbset T of a partially ordered set (S,:::;) TxT is a total order.

·t8

a chain if the rest-rzction of:::; to

Let:::; be a preorder on S, and set l' '" 8 for T. S E S if r :::; 8 and 8 :::; r. Then '" is an equivalence relation on S: the equivalence class containing s is denoted by [8]. Define [r] :::; [s] in S/'" if r :::; 8. Then:::; is well-defined. and (S/'" , :::;) is a partially ordered set. For example, let S be a non-empty set, and let T be a non-empty subset of P(S). Then the relation C on T is a partial order; T is ordered by inclusion. We shall often refer to a chain of s'llbsets of S to mean a chain in (S, C). Let A be a non-empty set, let {So: a E A} be a family of partially ordered sets, and take T C TIoEA. So. For f,g E T, we set f :::; g if f(o:) :::; g(o) in So for each a E A. Then:::; is a partial order on T, called the p7'Oduct partwl order. ,The product partial order on IRA is called the standard order. There is an equivalent approach to partially ordered sets. A strict partial or'der on a non-empty set S is a binary relation < such that: (i) if r < sand s < t in S. then 7' < t; (ii) 8 1- s for each .5 E S. Let:::; be a partial order on S, and set l' < S if T :::; 8 and r -=f:. s. Then < is a strict partial order on S. Oil the other hand, let < be a strict partial order on S, and set r :::; s if r < s or r = s. Then:::; is a partial order on S. Thus partial orders and strict partial orders correspond in this natural way, and we shall move freely between the notations :::; and <. Let (S, :::;) be a totally ordered set. \Ve use standard notat.ion for intervals in S: for example, for Sl, 82 E S, we write

[81,S2]={SES:S1:::;S:::;S2},

and

(Sj.s2)={sES:81<S<S2}.

Let (S,:::;) be a partially ordered set, and let t E S, Then t is: (i) max'imal if t 1- 8 (8 E S); (ii) minimal if 81- t (8 E S); (iii) a maximum if 8:::; t (8 E S); (iv) a minimum if t :::; 8 (8 E S). Let T be a subset of S. Then 8 E S is: (v) an upper bound of T if t :::; 8 (t E T) ; (vi) a lower bound of T if s :::; t (t E T); (vii) a supremum of T if t is a minimum in the set of upper bounds of T; (viii) an infimum of T if t is a maximum in the set of lower bounds of T. The subset T is: (ix) cofinal in S if, for each s E S, there exists t E T with s :::; tj (x) coinitial in S if, for each s E S, there exists t E T with t :::; s.

Foundatwns and ordered sets

5

Maxima and minima, suprema and infima are necessarily unique if they exist; we write maxT, min T, sup T, and infT for the maximum, minimum, supremum. and infimum. respectively, of a set T. Definition 1.1.2 Let (S,:::;) and (T,:::;) be partzally ordered sets. A map S ~ T is order-preserving ~f 'l,0(sd :::; ~)(S2) zn T whenever 81 :::; 82 in S. and isotonic [anti-isotonic] 2f '~j(sd < 1/)(S2) ['~)(8d > '1)(S2)] 'in T whenever 81 < 82 in S. The sets Sand T are order-isomorphic if there tS a bijedwn

'Ii' :

1/; : S

---->

T such that

1.)

and

1/)-1

are isotomc.

Definition 1.1.3 Let (S.:::;) be a partially ordeTed set. Then (S.:::;) is a lattice

tf {.'I, t} has a supTemum and an mfimum for' each Let (S, :::;) be a lattice, and let

8.

t

5,

t

E

S.

E S. Then we set

s V t = sup{s, t}.

.~ 1\

t = inf{s, t}.

For example, let T be a Ilon-empty set. Then (P(T), C) is a lattice: for subsets T1 and T2 of T. T1 V T2 = T1 U T2 and T1 1\ T2 = T1 n T 2 . Similarly, the family of all finite sllh.,ets of T is a lattice. Again. (JR A ,:::;) is a lattice: for j, 9 E JR A . we have

(f V g)(n) = f(ex) V g(o.)

and

(f 1\ g) (0:) = f(a) 1\ g(n)

for

ex EA.

Definition 1.1.4 Let (A,:::;) be a paTfzally oTdered set. Then A tS a directed set tf, fOT each Q. (3 E A. there ensts "Y E A wtth n :::; 1 and [j :::; "y; a tail fa S' if; a subset of the form {Q E A : Q 2:: Qo} for some Qo E A. A netm a set X 'i8 a

function from a dtlw:ted set into X. For example. N is a directed set for the obvious ordering; more generally. each lattice is a directed set. Often a net f : A ~ X is denoted by (,cO' : n E A). or just (,Tt», where the directed set is left unspecified. A sequence (xn) is a net indexed by N. The product of a family of directed sds is a directed set with respect to the product partial order. Let A be a directed set, and let f : A ~ X be a net in a set X. A sub71ct of ~ X. where B is a directed set and l/J : B ~ A is an order-preserving map such that 'Ij!(B) is cofinal in A' We now consider the notion of a projective systenl.

f is a map of the form fOil' : B

Definition 1.1.5 Let (A,:::;) be a d2rccted set, and let {Xc> : 0: E A} be a farmly of non-empty sets. Suppose that, for each Q. (1 E A 'With 0 :::; (3. there 2S a map 'IT"r3 : Xfj ~ X" 81Lch that 'IT"" Z8 the identtty on X"a and 'lTO'd 0 'lTih = 'lTt>;, whenever 0:::; {j :::; "Y. Then {Xl>; 'lTai;; A} is a projective system. Defin(. X

== lim proj{Xa; 1l'a,e; A}

=

{ext»~ E II Xc< : 1l'o.,e(x,e) =

Xc< for all n, j3 E A with

O'EA

Then X is the projective limit of the system {XO'; 1l'a{3; A}.

Q;

S

fJ}

Algebraic foundations

6

Clearly 1ra = 1raf3 0 1rf3 on X whenever (X S [3. In the special case where (with the usual order), we refer to a project we sequence {Xnj7l'mn}. Let (Xn) be a sequence of non-empty sets, and let en : X n+1 ---> Xn be a map for each n E N. For n > m, set 1rmn = ()m 0 ()m+l 0 . . . 0 en-I: Xn ---> X m , and let 1rmrn be the identity on X rn · Then {Xn;1rmn} is a projective sequence. denoted by {Xn; en}; its projective limit is

A

=N

X

== limproj{Xnj ()n}

=

{(xn ) E I1nENXn : ()n(xn+J)

= Xn

(n

E

N)}. (1.1.1)

Definition 1.1.6 Let 5 be a non-empty set. Let'R be a family of subsets of 5 such that Rl n R2 E whenever R J , R2 E 'R. An 'R-filter on 5 25 a non-empty subset F ofP(5) such that: 0 tJ. F; if E, FE F, then En FE F; if FE F and R E 'R with FeR, then R E:F. An 'R-ultrafilter on 5 is an 'R-filter' F on 5 such that either E E For' FE F whenever E, FE 'R wdh E U F = 5. A filter on 5 'i.~ a P(5)-filter; an ultrafilter on 5 is a P(5)-ultrafilter.

n

Let. 'R be as in the above definition. The family of all 'R-filters on 5 is a partially ordered set with respect to inclusion; an 'R-filter is an ultrafilter if and only ifit is maximal in this family. For example, let s E 5. Then {T C 5 : SET} is an ultrafilter OIl 5; these are the fixed ultrafilters on 5, and ultrafilters which are not fixed are free. Suppose that (A, S) is a directed set, and let F be the family of all subsets T of A which contain a tail. Then F is a filter on 5, called the order filter, and each ultrafilter containing F is a free ultrafilter on A. In the case where A = N, the order filter consists of all subsets F of N such that N \ F is finite; it is sometimes called the Frt£chet filter. Let F be a filter on 5, and let f,g E ~s. Then we say that f SF 9 if {8 E 5 : f(8) S g(s)} E:F. Clearly SF is a preorder on ~s. We say that f ""F 9 if f SF 9 and 9 SF f. and so f "'F 9 if and only if {s E 5 : f(s) = g(s)} E :F. In this way we obtain a partially ordered set which we denote by (~S / F, SF)' In the case where F is an ultrafilter, this set is totally ordered. Definition 1.1.7 Let (5, S) be a totally ordered set. Then (5, S) is well-ordered if each non-empty 8'ubsft of 5 conta'ins a minzmum element. Let (sn : n E N) be a sequence in a partially ordered set (5, S). Then (sn) is lnc'1'ea8ing [strictly zncreasing] if 8 n S 8 n +l [sn < Sn+l] (n EN). We define decreasing and 8trzctly decreasing sequences similarly. The sequence (8 n ) has a property eventually if there exists no E N such that (sn : n 2: no) has the property; we also say that 'sn has the property eventually'. In ZFC, a set (5, S) is well-ordered if and only if each decreaHing sequence in 5 is eventually constant. Let 5 be a well-ordered set, let T be a totally ordered set, and take f, gETS with f -I- g. Define

6(1,g) = min{s

E

-I- g(s)}. f(6(1,g)) -I- g(6(1, g));

5 : f(8)

Since 5 iH well-ordered, 6(1,g) exists and we set f -< 9 if f(6(1,g)) < g(o(1, g)) in T. For f,g E TS, we set f :::S 9 if f -< 9 or f = g. Then :::S is a total order on T S : it is called the lexicographic order. In particular, (1'1, .•• , r n) -< (Sl' ... , sn) in the lexicographic order on ]Rn if and only if 1'1 < Sl or there exists k E {2, ... ,n} such that ri = 8i (i E Nk-d and rk < 8k.

Foundatwns and order'ed sets

7

\Ve now give two sentence::; which are equivalent to AC in ZF.

Zorn's lemma is the sentence: each partially ordered set ill which each chain has an upper bound has Ii maximal element. The 'Well-oTdeTing pnnciple is the sentence: on each non-empty set 5, there is a binary relation:::::; such t.hat (5, :::::;) is a well-ordered set.

Theorem 1.1.8 The following sentences aTe eq'uiualeut in ZF: (a) the axiom of choice, AC,. (b) Zorn's lemma; (c) the well-ordering pr'inczple. 0 Thus, as we shall be working in ZFC, we shall be able to apply Zorn's lemma and the well-ordering principl('. Among the many facts that hold in ZFC and which we shall usc in this book, but which cannot be proved ill ZF, are th(, following: (i) each lin('ar space has a basis; (ii) in a unital algebra. each proper ideal is contained in a maximal ideal; (iii) each field is contained in an algebraically closed field; (iv) each R-filter is contained in an R-ultrafilter; (v) each continuous linear functional on a linear subspace of a normed space has a norm-preserving extension to the whole space. A famous model 9J1 of the theory ZF has been constructed by Solovay. In this model, the axiom of choice is false. but a weaker axiom, the 'axiom of dependent choice (DC)'. is true. and it is proved that every subset of each complete. separable metric space has the Bairc property (see Definition A.5.13). It follows rather easily from this (see A.5.24) that, in 9J1, every linear map from an (F)-space into a locally convex space is automatically continuous. Thus one cannot provE' in the theory ZF that discontinuous linear maps exi::;t, and so ZF is not a sufficiently rich theory in which to study the questions of automatic continuity that interest us. Although somewhat weaker axioms would suffice, it seems reasonable for us to assume AC itself, and thus to work in the theory ZFC. This 'We shall do j7'Om. this powt on'WaTds. \Ve now discuss ordinal and cardinal nnmbers.

Definition 1.1.9 Let 5 be a set. Then 5 1,8 transitive if Fach element of 5 is a subset of 5, and 5 ~s an ordinal if 5 2S transitive and (5, E) t8 a totally ordered set, Tegar-dmg 'E' as a stnct paTtwl onteT. Specifically. let 5 be transitive. and, for n.!J E 5. set 0: < (J if 0: E (3. Then 5 is an ordinal if (5,:$) is totally ordered. Suppose that wc are given a property that a set Illay have. The collection of all sets which have this property is the class determined by the propert.y. Certainly, every set is a class, but consideration of Ru::;sell's paradox shows that, for example, the cla::;s of all set::; is not a set. We denote the class of all ordinals by Ord; Ord is not a. set, but it is convenient to extend to Ord the notions of ordering that were previously defined for sets. Thus, let 0, (3 E Ord. Then it can be easily seen that 0 E (3 if and only if 0 <;; fl. Each non-zero ordinal a is exactly the interval [0,0) in Ord.

Algebraic foundations

8

Theorem 1.1.10 (i) The class (Ord,:S) is well-or·dered.

(ii) Each well-ordered set is onier-zsomorphic to a unique ordinal.

0

Let S be a set of ordinals. Then U{ a : a E S} is an ordinal, and it is the supremum of S in (Ord. :S). For each ordinal 0, a U {a} is an ordinal; it is the successor of n, written (t + l. If a = (3 + 1 for Home ordinal (3, then a iH a successor ordznal; if a i= 0 and a is not a successor ordinal, then 0: iH a limzt oTdinal, and in this case 0: = sup{,B: f:J < a}. The minimum ordinal is 0, denoted by 0, and the minimum limit ordinal is denoted by w. The ordinals a with n < ware the finite ordinals, and each is identified with an element of Z+, so that Z+ is order-isomorphic to w. The ordinals (t with (t ;::: ware the znfinzte ordinals. Let 0: be an ordinal, and let 5 be a non-empty set. An clement s of Sa. iH a sequence in 5 of length £(8) = 0'. If 5 iH a totally ordered set, then (50:, j) is also totally ordered. \Ve define

5
: (j

< (t};

for example, 5<w is the set of finite sequences in 5. Definition 1.1.11 An ordinal a is a cardinal zf, for each ol'dznal (3 wzth (3 < a, and ,B are not equt,potent. The class of all cardznal n'umbers is denoted by

0:

Card. Let 5 be a non-empty set. By the well-ordering principle and 1.1.10(ii), there is an ordinal equipotent to 5, and so there is a minimum ordinal which is equipotent to 5. This ordinal is a cardinal, and it is denoted by 151. Clearly, two sets have the same cardinal if and only if they are equipotent. For each cardinal r;" we write 1\;+ for the minimum cardinal A with r;, < A. Suppose that 5 is a set of cardinals. Then sup 5 is clearly abo a cardinal. The class Card is well-ordered by the relation :S on Ord, given above. The finite ordinals are cardinals: t.hey are the finite cardinals. The ordinal w is the minimum infinite cardinal. and aH a cardinal number w is denoted by ~o. A set 5 is countable if 151 :S ~o. and uncmmtable if 151 > ~o. A standard theorem of Cantor asserts (in ZF) that 151 < IP(5)1 for each set 5, and so ~o = INI < IP(N)I. In particular, there is an uncountable set. The minimum uncountable ordinal is called Wl; WI is a cardinal, and as such iH denoted by ~l' It is clear that a subset of WI is cofinal if and only if it is uncountable; we shall several times use the fact that, if (an) iH a sequence of ordinals with an < WI (n EN), then there is an ordinal 0' with an < a < Wi (n EN). Let 5 be a partially ordered set. The cofinality [coinitiality] of S is the minimum cardinal r;, such that 5 has a cofinal [coinitial] chain of cardinality K,; we write cof Sand coi S , respectively, for these cardinals. Clearly cof S :S cof r;, = r;,; otherwise K, is singular.

lSI.

A cardinal

r;,

is regular if

Foundatwns and ordered sets

9

We i:!hall use the principles of transfinite inductwn and of definitwn by transfinite recur.szon in the following forms.

Theorem 1.1.12 (Trani:!finite induction) LP.t a be a non-zero ordinal, and let 8 be a subset of a. Suppose that: (i) E 8; (ii) for each 13 < a, it follows fmm the hypotheszs that [0,13) c S that 13 E 8. Then S = a. 0

°

Theorem 1.1.13 (Trani:!finite recursion) (i) Let a be a non-zero ordinal. Suppose that f(O) is defined, and that, for each 13 < a, f((3) ~s defined whenever f is defined on [0, (3). Then f is defined on a. (ii) Suppose that f(O) is defined, and that, for each ordznal j3, f(j3) zs defined whenever f zs defined on [0. ,8). Then f(a) zs defined for each ordinal a. 0 We now define a map a ~ Net from the cla.'is Ord onto the class of infinite cardinals by trani:!finite recuri:!ion.

Definition 1.1.14 Let 0: be a non-zero ordmal. If fl' = (3 + 1 for an ordinal/3, definF Nn = Nt· n is a non-zero lzmit ordinal, define N" = sup{Ne : (3 < (t}.

rr

At a i:!mall number of points in this book. we shall require some elementary facts of cardinal arithmetic; we collect these facts here.

Definition 1.1.15 Let K, and A be non-zero cardmals, and let A and B be disJoint sets wzth IAI = K, and IBI = A. Then:

K,+A=/AUBI;

K,A=IAxBI;

K,A=/A B

/.

The definitions of the cardinal numbers K:, + A. /'i,A, and K,A are independent of the choices of the sets A and B. The facts that we shall Ui:!e are the following. Let K" A, and It be non-zero cardinali:!. Then: (i) if A 'S K, and K is infinite, then K + A = K,A = K,; (ii) (K,A)11 = K,AIl; (iii) if {As: s E S} is a family of sets with IAsl 'S K, (s E S), then / U{As : s E 8}/ 'S K, 181. For each non-zero cardinal K" we see that 2K = IP(K,)I. The real line JR( has cardinality 2 No , and, for this rea.son, 2N o is said to be the cardinal of the continuum; it is often denoted by c. Cantor's theorem implies (in ZFC) that K,+ 'S 2K for each cardinal K,. In particular, Nl 'S 2No and Na + 1 'S 2N« for each ordinal a. The contZn1wm hypothesis (CH) and the generalzzed continuum hypothesis (GCH) are the sentences: Nl = 2 No

and

Nn+l = 2N "

for each ordinal a,

respectively. It is known that both CH and GCH are independent of the theory ZFC. The key fact that is true in the theory ZFC + CH and that we shall use is the following: each set of cardinality c can be well-ordered in such a way that each element has only count ably many predecessors in the ordering. We shall use the continuum hypothesis in the proof of some results of §4.2, §4.8, and §5.7. When our proof does appeal to CH, we shall state this explicitly. The role of CH in our constructions of discontinuous homomorphisms from Banach algebras will be discussed in §5.7.

Algebraic foundatzons

10

vVe shall also require some less elementary facts about ordered S()ts: they arc summarized below. Let (5,~) be a partially ordered set with subsets 51 and 52. Then 51 « 52 if 81 < S2 whenever 81 E 51 and 82 E 52. We write 8 « 51 for {s} « 51, etc. Definition 1.1.16 Let

(5,~)

be a partwlly or'dered set. Then 5 is:

(i) an 0:1 -set 'if each non-em,pty subset of 5 has a countable cojinal and coinitial 8ub8et; (ii) a .81-set if 5 = U{SII : v < wIl, where {5" : v < wd in (P(5), C);

tS

a cham of Ctrsets

(iii) an Til-set if; fur each pair {51, 52} of countable s'ubsets of 5 such that 51 u52 is totally or-riered and 51 «52, there eX"l8ts s E 5 wI,th 51 « s « ,'h: (iv) a Semi-7]I-set if, for' each strIctly increasing sequence (8 n ) and 8triCtly decrea8ing sequence (t n ) in 5 such that Sn < tn (71 EN), ther'e exi8ts s E 5 with 8 m < 8 < tn (m, n E N). For example, JR. is an nl-set and a semi-Til-set, but it is not an T/l-set: there is no x E JR. with x > 0 and x < l/n for each n E N. By taking 51 or 52. respectively, to be 0 in (iii), above, we see that coi 5 > ~o and cof 5 > ~o for an 7]l-set 5; no set is both an aI-set and an 7]l-set. Let U be a free ultrafilter on N. Then it is easy to see that the totally ordered set (JR.N /U, ~u) is an 7]rset; this set will be discussed further later. Let S = {O, 1 }Wl . so that S is the collection of dyadu: sequences of length WI: S is a totally ordered set with respect to the lexicographic order :5. Definition 1.1.17 The set Q is the subset of elements a of S such that {T < WI : aT = I} is non-empty and has a maximum. For each IJ < WI. Qa is the set of elements of Q for which this maximum is less than IJ. Proposition 1.1.18 (i) For each (ii) Q

= U{Qa : IJ < wd

i8

IJ

<

WI,

Qa

i8

an

al

-8et.

a totally ordered fh -Tl1-set, and IQI

= c.

0

Theorem 1.1.19 (i) Let 5 be a totally oTdered .81 -8et, and let T be an 7]1 -set. Then there i8 an isotonic map from 5 into T. In paritcular, there zs an isotonic map from 5 into Q and from Q into T. (ii) Each totally ordered .81 -7]1 -set is order-z80morphic to Q.

o

Clause (i) of the above result implies that Q is univer8al in the class of totally ordered .81-sets: Q contains a copy of each such set. Let 5 be a totally ordered .81-set. Since 5 embeds into Q and IQI = c, necessarily 151 ~ c. On the other hand, it is clear that each partially ordered set of cardinality ~1 is a /3l-set, and so, with CH, a totally ordered set 5 is a .8l-set if and only if 151 ~ ~1· It may be said that the .81-condition replaces a cardinality condition on a partially ordered set in the case where CH fails.

Semigronps

11

Notes 1.1.20 For expositions of axiomatic set theory, see (Enderton 1977). (Jech 1978), or (Kunen 1980), for example. Proofs of the results that we have stated in this section are given in these sources. An account of some topics that are particularly related to the work of this book is given in (Dales and Woodin 1987); in this work, the independence of AC from ZF and of CH from ZFC is proved. The language of set theory that we have in mind is, strictly. 'first-order language, with equality'. There are many versions of the axiom of choice, and some other forms would probably be preferred by logicians. Theorem 1.1.8 is proved in (Hewitt and Stromberg 1965, 3.12) and (Jech 1978, §5). A full discussion of the axiom of choice is given in (.Jech 1973); the examples following the theorem are taken from this book. For an interesting historical account of this axiom, see (Moore 1982). Solovay's model ~ is from (1970); the model is discussed in (Jech 1973). The discussion of ordinal and cardinal numbers is a standard one, essentially due to von Neumann. See (Jech 1978, §2.3) and (Kunen 1980, I, §6), for example. The discussion of a]-, /31-, and 1]l-sets is taken from (Dales and Woodin 1996. Chapter 1), where 1.1.18, 1.1.19, and other related results are proved and full attributions are given; the set Q is called Sierpinski's set. Note however that. in (ibid.), a semi-1]I-set is required to be totally ordered.

1.2

SEMIGROUPS

Our aim in this nection in to introduce some elementary results about semigroups, concentrating on abelian semigroups. In 1.2.10, we shall identify a cancellative, abelian semigroup as a subsemigroup of a group; later, in §1.3, a similar construction will be used in the theory of localizations of commutative algebras. In 1.2.18 and 1.2.19, we nhall show that certain abelian groups and semigroups can be embedded in totally ordered rational linear spaces, and we shall discuss semigroups over Q+-, obtaining results which will be used when we dincuss rational semigroupn in Banach algebras in §4.9 and in our constructions in §5.7. Finally we shall nummarize some remIts about (31 -groupn, 771-groups, and Hahn groupn, and we shall give two explicit examples involving some ordered cones that will be important later. Definition 1.2.1 A semi group 'is a non-empty set S with an aS50ciative binaTY operatwn, S x S -+ S. A semigro1Lp tS abelian tf the operatwn is comm1Ltatwe. A subHemigroup of S is a subset of S whtch is a semigT01Lp fOT the gwen binary operation. We "hall initially denote the operation in a Hemigroup by juxtaposition. having in mind the multiplicative "emigroup of an algebra. Later, however, we shall often u"e additive notation when we are considering abelian semigroups. Let S be a semigroup. Then sop is the nemigroup formed by reversing the order of the product in 8; sop is the opposite semig7'Oup to 8. Let T be a subsemigroup of S, and let s E S. We denote by Ls and Rs the maps (1.2.1) Ls : t f--+ st, Rs: t f--+ ts, T ---> 8. A subset T of a semigroup S generates S if each element of S is a finite product of elements of T.

12

Algebraic foundations Let 8 1 and 8 2 be non-empty o;ubsets of a o;emigroup 8. Then

81

.

82 =

{81S2 : 81 E

81,

S2 E

8d,

and, if 8 is written additively, then

8 1 +82 =

{81 +S2: S1 E

81,

S2 E

8 2 }.

We write s . 8 for {s} ·8 = Ls(8) and s + 8 for {s} + 8, etc. Let 8 be a semigroup. An element e of 8 is a left [right] identity of 8 if es=s

[se=s]

(sE8).

An element e is an identity of 8 if it is both a left identity and a right identity; such an element of 8 is unique. A semigroup which has an identity is a umtal o;emigroup; we o;hall often denote the identity of 8 byes or bye. A semigroup without an identity is non-unital; in this case, take e O}; Q+., iR.+., and II are each subsemigroups of the semigroup (e, + ). Let 8 be a semigroup. An element s E 8 is an idempotent if S2 = 09; left identities and right identities of a semigroup are each idempotents. Let 8 be a unital semigroup, and let s E 8. Then 09 is left [rzght] invertible if there exists t E 8 with ts = es [st = es]. The element s is invertible if it is both left and right invertible. In this case, there is a unique element t E 8 with ts = st = es; t is the mverse of s. written S-1. The set of invertible elements of 8 is denoted by Inv 8, and 8 io; a group if Inv 8 = 8. The set Inv 8 is a group with respect to the operations of 8, and (8t)-1 = C 1 S- l (s, t E Inv 8). A subset T of a unital semi group 8 is a subgroup of 8 if T is a group for the operation of 8 and the identity of T is es. Let G be a group. For 8 c G, we define 8- 1 = {S-1 : s E 8}; 8 is symmetric if 8- 1 = 8. For f E e G , set ](09) = f(8- 1 ) (s E G); f is symmetric if ] = f. Let a E G. Then the left o;hift of f by a il:l defined by (8a f)(s) = f(a- 1 s)

(s

E G).

(1.2.2)

The product [t:>EA 8 a of a family {8a : Q E A} of o;emigroups is a semigroup with respect to the operation (sa)(t a,) = (Sata) (sa, ta E 8 a ). Suppose that 8 is a unital o;emigroup. Then we identify 8<w with 8<w = {(sn) E 8 w : Sn

= es eventually};

clearly 8<w is a subsemigroup of 8 W •

Examples 1.2.2 (i) Let 8 be a non-empty set. The set of bijections from 8 to itself is a group with respect to composition. In the case where 8 = Nn , this group io; the group of permutations on n letters; it is the symmetric group of order n, denoted by en. The group en is a finite group consisting of n! elements. A word with letters from the set 8 is a finite formal product S~l . . . s;;;, where mEN, S1,'" ,Sm E 8 and CI,'" ,Cm E {-I, I}; we also take the formal product with no factors to be a word (denoted bye). A word is reduced if it is e or if

Semigroups

13

whenever Sk+l = Sk· Each word is equivalent to a reduced word in an obvious way. The free group on S, denoted by IF'(S), consists of all reduced words; the product of S~1 •.• s;;; and t~1 ••• t~n iH the reduced word equivalent to S~1 ..• s;;;' t~1 •.• t~n. The identity of IF'( S) is e, and the inverse of s1 1 •.. s;~' is

Ck+l = Ck

s;;,,"m .. '8;-"1

Suppose that lSI = n. Then IF'(S) is denoted by IF'n: it is the free gr01tp on n generators. In particular, IF'2 is the free group on two generators. Abo, if lSI = ~o, then IF'(S), denoted by IF'N o , is the free group on countably many generators.

(ii) Let n E N. Then zn, Qn, ]Rn, and en are groups with reHpect to coordinatewise addition. Similarly, Z<w and ZW, etc., are groups. The former group contains (Z+)<w as a subsemigroup; (Z+)<w is the free abelian semzgroup on countably many generators. For j E N, set X J = (8;,j : i EN). Then each element of (Z+) <w can be written as Xr'1 ... X;;k , where ml, ... ,mk E Z+ and mi + ... +mk > O. (iii) Let n E N. Then §n = N;;:w, the set of finite sequences of elements of N n . The set §n is a semigroup with respect to juxtaposition of sequences; it is the free semigroup on n generators. By replacing N n by N, we obtain §N o , the free semzgroup on countably many generators. In fact, Het aj = (j) for J E N n · Then the generators of §n are a}, ... , an, and each element of §n has the form a'J:.k, where jI,'" ,jk E N n and mI,"" mk EN. Clearly §n is a su bsemigroup of IF' n' 0

aJ:1 ...

Definition 1.2.3 Let S be a semzgroup. Then S zs:

(i) cancellative zf rs

i= rt

and sr

i= tr

i= t; with t n =

whenever r, s, t E Sand s

(ii) divisible if, for each s E Sand n E N, there eX'tsts t E S

s;

(iii) torsion-free if r = s whenever r, s E Sand rn = srt for some n E N; (iv) a cone if S is cancellative, non-umtal, and abelian. Thus S is cancellative if and only if Lr and Rr are injective on S for each rES. Note that an abelian group is torsion-free if r = eo whenever rn = eo for some n E N. Also, if S is a divisible, torsion-free semigroup, then, for each s E S and each n E N, there exists a unique element t E S such that t n = s; this element t is denoted by si/n (or by sin if we are using additive notation). Suppose that S if, a cone and that s E S. Then sm = sn for m, n E N only if

e

m=n. It is clear that a subset T of an abelian group (e, +) is a cone (with respect to the operation of e) if and only if T + T C T and 0 ¢. T. Definition 1.2.4 Let Sand T be semigroups. A map'ljJ : S --> T is a morphism if'ljJ(rs) = 'ljJ(r)'ljJ(s) (r, s E S); 'l/J is an isomorphism if it is also bijective. Definition 1.2.5 Let S be a semigroup, and let K be a subsemigroup of (e, + ) with 1 E K. A K-semigroup in S is the image 'l/J(K) of a morphism 'ljJ ; K --> S; the semigroup 'l/J(K) is non-zero if'l/J(K) c S·.

Algebraic foundations

14

A K-semigroup 'lj;(K) is usually denoted by (s< : ( E K), where 'lj;(1) = s. We refer to rational and real semigroups when K is Q+. or jR+., respectively. Let 5 be a semigroup, let", be an equivalence relation on the set 5, and let 1r: 5 -- 51'" be the quotient map onto the set of equivalence classes of "'. The relation", is compatible with 5 if rs '" rt and sr '" tr whenever 1', s, t E 5 and s '" t. In this case, set ?r(r)1r(s) = ?r(rs) (1', s E 5); we obtain a well-defined operation on 5/", with respect to which 5/", is a semigroup. The quotient map 1r is a morphism. Let G be a group, and let H be a subgroup of G. Then H is normal if s· H = H· s (s E G), and in this case G/ H, the set of cosets of H, is a group. Let '" be an equivalence relation compatible with G, and let H = {s E G : s '" eo}. Then H is a normal subgroup of G, and G I H = G / '" . Let G and H be groups, and let IjJ : G -- H be a morphism. Then the kernel ker~ of '1jJ, defined to be {s E G: ~(s) = eH}, is a normal subgroup of G, and the map s· ker'lj; f--7 'lj;(s). G/ker'lj; -- 'lj;(G), is a bijective morphism.

Definition 1.2.6 Let 5 be a cone, and let G be an abelian group. operates on 5 if there zs a map (a, s) f--7 as, G x 5 -- 5, such that;

Then G

(i) a(st) = (as)t (a E G, s, t E 5); (ii) (a(3)s = a«(3s) (a, (3 E G, s E 5). Gs

In this case, it follows from (i) and (ii) that eos = {as: a E G} for s E 5.

=

s (s E 5). We define

Proposition 1.2.7 Let 5 and T be cones, and let G be a divisible, abelian group that operates on T. 5uppose that 80 E 5 and that X : 5 -- T is such that X(st) E GX(s)X(t) (s, t E 5). Then there zs a morphz8m ~ : 5 -- T such that 'lj;(so) = X(80) and'lj;(s) E GX(8) (s E 5). Proof Let Uo = {sa: n EN}, and define 'Po : sa -- X(so)n, Uo -- T, so that 'Po is a morphism with 'Po(so) = X(so) and 'Po(s) E GX(s) (s E Uo). Denote by F the family of pairs (V,
'lj;(t)

E

GX(t) = Gx(r)x(s) = Gx(r)'lj;(s) ,

and so 'lj;(t) = x'lj;(s) for some x E Gx(r). Set 'Ij)(r) = x. Suppose that also s', t' E U with rs' = t' and that 'lj;(t') = x''lj;(s') for some x' E T. Then st' = ts' because 5 is cancellative, and so X'lj;(8)'lj;(S') = x''lj;(s)'lj;(s'). Since T is cancellative, x = x'. Thus 'lj;( r) is well-defined as the unique element x E T such that 'lj;(t) = x'lj;(s) whenever rs = t with s, t E U and 'lj;(r) E Gx(r). It follows easily that 'lj; : V ....... T is a morphism extending 'lj;, and so (V, 'lj;) >.:= (U, 'lj;) in :F. By the maximality of (U, 'lj;), we have V = U, and so r E U whenever there exist s, t E U with rs = t.

15

Semigroups

For l' E S, set Vr = U u U{ I'm . U : mEN}, so that Vr is a subsemigroup of S with Vr :J U. Assume that there exist l' E S \ U and n :2:: 2 with 1',1'2, . .. ,rn - l ¢. U and rn E U. Then 1/)(1''') E Gx(r n ) = Gx(r)n. Since G is divisible, there exists 0' E G with '1/;(1''') = anx(r)n. Set 'IjJ(rmu) = O'mx(r)m'IjJ(u) for mEN and u E U. Suppose that rmlul = rm2u2 for some mI, m2 E N with mi > m2 and some UI. U2 E U. Then rml-m2111 = 'U2, and so r m, - m2 E U. This shows that m'l = m2 + np for some pEN. We have

'I/;(U2) = 'tf;(rnY'tf;(ud = xnpx(r)"P'Ij)(ut) , and so ami x(r)m l 1j;(ud = O' Tn2 x(r)1n 21/)(U2) , Thus 'I/; is well-defined on the set U{rm . U : mEN}. This extension of'IjJ to \1;. is a morphism such that 'IjJ(v) E GX(v) (v E V). But this contradicts the assumed maximality of (U, '1/). Hence l' E U whenever rn E U for some n E N. Finally, assume that there exists l' E S \ U. Suppose that r1nlul = rm2U2 for some ml,rn2 E N with mi :2:: m2 and some UI, U2 E U. If mi > m2, then rTnl-1n2uI = 112, and so r m, - m2 E U and l' E U, a contradiction. Thus mi = m2. and the map 1/) : I'm f---t x(r)m'IjJ(u), Vr ----- T, is a morphism extending 1/J with 1/J(s) E GX(s) (s E Vr ). Again Vr = U, and so l' E U, a contradiction. Thus U = S, and the result is proved. 0 \Ve now give a construction that will show that each cancellative, abelian semigroup can be embedded in a group. Let S be an abelian semigroup, and let T he a subsemigroup of S. We say that (SI' tt) '" (S2' t2) in S x T if there exists l' E T such that r8 l t 2 = rS2tl' Routine verifications show that the relation", is an equivalence relation which is compatible with S x T.

= (S x T)/"" and let 7r : S x T ----- S[T- I ] be T, set 1/J(s) = 7r«(8t, t») (8 E S).

Definition 1.2.8 Define S[T- I ]

the quotient map. For t

E

The definition of 1/J( s) is independent of the choice of t ill T because it is clear that (Stl,tt) '" (st2,t2) in S x T whenever t I ,t2 E T.

Proposition 1.2.9 Let S be an abelian semigroup with a subsemigroup T.

(i) The semigroup S[T- I ] zs an abelian, unital semigroup; '4): S a morphism: and 1jJ(T) C Inv S[T-I]. (ii) S[T- I ] = {1/J(s)'tf;(t)-1 : S E S, t E T}. (iii) In the case where S is canceliatwc, 'IjJ is an mjection.

f---t

S[T- I ] is

Proof (i) Certainly S[T- 1 ] is an abelian semigrollp. Define e = 7r(t, t» for t E T. Then e is independent of the choice of t in T, and e is the identity of the semi group S[T-I]. The map 1/J is a morphism, and 1/J(T) C Inv S[T- I ] because 7r«rt, t)(t, rt» = e (1', t E T). (ii) Let x E S[T-I]. Then x = 7r(8, t» = 'IjJ(8)'¢(t) -1 for some s E Sand t E T. (iii) Let 81,82 E S with '¢(8d = '1/;(82)' and take t E T. Then there exists l' E T with r81t2 = r82t2. Since S is cancellative, necessarily 81 = 82, and so '¢ ~~~w.

0

Algebraic foundations

16 We now apply 1.2.9 in the case where T notation.

=

S; we continue to use the above

Proposition 1.2.10 Let S be an abelian semigroup.

(i) The semzgroup S[S-1] is an abelian group. (ii) Suppose that S is cancellative. Then 1j; : S

-+

S[S-1] is an mjective

morphism. (iii) Suppose that S is cancellatzvc and tOri,ion-free. Then the gr'oup S[S-1] is torsion-fTee, and IS[S-1]1 = lSI. Proof Set G

=

S[S-l].

(i) Since 1l'((s. t)(t, s)) = eu (8, t E S), each element of G is invertible. (ii) This is immediate. (iii) Suppose that x E G with xn = ee, say x = 1j;(r)ljJ(s)-l. Then rn = 8" in S, and so r = .., because S is torsion-free. Thus x = cu, and G is torsion-free. If S is a singleton, then lSI = IGI = 1. If S is not a singleton, there is an element s E S which is not an identity of S. Since 'ljJ is injective, 1/;(8) i ee. Since G is a torsion-free group, the map n f--+ 1j;(8)", N -+ G, is an injection. Thus lSI = 1'4)(S)1 ;::: No, and so IS x 81 = lSI. Hence IGI ~ IS x SI = lSI ~ IGI, and so IGI = lSI, as required. 0 The group S[S-1] is the group of the scmigTOup S. Henceforth we shall identify S with ljJ(S); each cancellative, abelian semigroup is a subsemigroup of a group. Definition 1.2.11 Let S be a 8emigroup, and let ~ be a binary relation on S. Then (S,~) is an ordered semigroup zf S is a partial order on S and if Ls and Rs are isotonic for each s E S; (S,~) is totally ordered or well-ordered if the underlymg ordered set zs totally oTdered or well-ordered, respectively. Note that a totally ordered semigroup is cancellative and torsion-free, and that each abelian semi group is an ordered semigroup with respect to the binary relation =. A group [cone] which is an ordered semigroup is an ordered group [ordered cone]. For example, let n E N. Then (zn,::5) is a totally ordered group with respect to coordinatewise addition, where ::5 is the lexicographic order. Similarly, (Z<w,::5) and (zw,::5) are totally ordered groups. Now set s --< t in §n if either £(8) < £(t) or £(8) = £(t) and 8 --< t in the lexicographic order on Nr~(s). Then the semigroup (§n,::5) is well-ordered. We now change to additive notation for semigroups, as we wish to identify certain abelian semigroups with subsets of linear spaces. An identity of a semigroup is now denoted by 0, and the inverse of an element 8 is -8. Let (S, +) be an ordered, abelian semigroup, and set G = S[S-1]. Take Xl, X2 E G. By 1.2.9(ii), there exist 81,82, tl, t2 E S with Xl = S1 - t1 and X2 = 82 - t 2 · Define Xl ~ X2 if 81 + t2 ~ 82 + t 1. Then ~ is well-defined on G, and (G, +,:::;) is an ordered group; G is totally ordered if S is totally ordered.

Semzgroups

17

Definition 1.2.12 Let (G,

+,::;)

be an ordered abelzan group. Then

G+={sEG:O::;S},

G-={SEG:8::;O}.

Thus G+· = {s E G : 0 < s} and G-· = {s E G : 8 < O}. In the C8...'le where G 1= {O}, the semigrollps (G+·, +) and (G-·, +) are cones; they are the positive and negative cones of G, respectively. Definition 1.2.13 Let (G, For s E G, define s+ =

8

+,::;)

V 0,

be an ordered abelzan group which is a lattice.

8-

= 81\0,

lsi

=

8

V (-05).

For example, let 5 be a non-empty set. Then (IRS". +. ::;) is an ordered, abelian group which is a lattice, and the definitions of f+, f- and Ifl for f E IRS" coincide with the usual pointwise definitions; clearly we have f = f+ + f- and If I = f+ - f-· Definition 1.2.14 Let T be a cone whzch is a subsemzgroup of an abelzan semigroup 5, and set r < .<; zn 5 if s E r + T. Then::; is the T-order on 5. The T-order on 5 is denoted by ::;T; ::;T is always a preorder, and it is a partial order and (S'::;T) is an ordered semi group in the case where 5 is cancellative. If 5 is unitaL then «T. In the case where 5 is a group, (5, ::;T) is an ordered group, and 5+· = T; the order ::;T is a total order on 5 if and only if 5 = T U ( - T) U {O}.

°

Definition 1.2.15 An ordered abelian semzgroup 5 is a difference semigroup if s E r + 8 whenever r < 05 in 5. Thus an ordered abelian semigroup 5 is a difference semigroup if and only if the giVen order coincides with the 5-order. Definition 1.2.16 Let (8, +) be an abelian semigroup, and let K be a subset of C 'Whzch is a subsermgroup of both (C, +) and (C, . ), with 1 E K. Then S, together 'With a map (a, s) f--+ as, K x 5 ---+ S. is a semi group over K zf: (i) (a + jJ)s = as + fJs (a,lJ E K, s E 5); (ii) a(r + 05) = ar + as (a E K, r,s E 5); (iii) a(jJs) = (a(J)o5 (a,jJ E K, 05 E 5);

(iv) Is

= 05 (s

E 5).

In the case where (5,

+ ) is

a gr07lp, a semigroup over K is a K-linear space.

We refer to mtional-, real-, and complex-linear spaces when K is Q, IR, and C, respectively. Let IK be Q or IR, and let 5 be a cancellative semigroup over IK+·. Take G to be the group of S, and take a E IK and x E G, say :r = r - 05, where r,s

E

5. Define

ax = ar -

0'.8

if a 2:: 0,

ax = (-n)s - (-a)r

if

(t

< O.

(1.2.3)

By 1.2.16(ii), ax is well-defined. The map (a,x) f--+ ax, IK x G ---+ G, extends the given map, and routine verifications show that G is a lK-linear space.

Algebrazc foundations

18

Definition 1.2.17 Let E be a lK-linear space, where lK 105 Q or IR. Then E is an ordered lK-linear space zf (E,:::;) is an ordered gr'oup and o:x E E+- whenever :1' E E+- and n E lK+-. For example, (IRs, +, :::;) is an ordered real-linear space for each non-empty set S. Clearly each divisible, totally ordered abelian group can be regarded as an ordered rational-linear space. Let S be a cone which is a semi group over lK, and let e be the group of S, so that e is an ordered group for the S-order. Since S = e+-, necessarily e is an ordered lK-linear space.

Theorem 1.2.18 Let (e,:::;) be a totally order'ed, abelwn group. Then there tS a totally ordered rational-linear space (H,:::;) with IHI = lei and an injective, zsotonic morphzsm 1/J : e ---> H such that H = {1/{.,) / n : .'i E e, n E N}. Proof Set (Sl,PI/qr) ~ (S2,P2/q2) in e x Q if Plq2S1 = P2Q182. Since e is torsion-free, the relation rv is an equivalence relation on G x Q, and rv is compatible with G x Q. Let H = (G x Q)/rv, let 7f : G x Q ---> H be the quotient map, and define 7f«SI,PI/q1» + 7r«S2,P2/q2) (p/Q)7f«Sl,pI/Ql»

= =

7f«PIQ2 S1 + P2Q1 S2. 1/QlQ2), 7r(Sl,ppI/QQl».

These operations are well-defined, and H is a rational-linear space with respect to the operations. The map 1/J : S f-+ 7f((8, 1»). G ---> H, is an injective morphism, and H = {1jJ(s)/n : s E G, n EN}, so that IHI = IGI. Set 7f«s,p/Q» > 0 in H if s > 0 and p/Q > 0 or if s < 0 and p/q < O. Then < is well-defined on H, and (H. :::;) is a totally ordered rational-linear space. The map '1/; : (G,:::;) ---> (H,:::;) is isotonic. 0

Theorem 1.2.19 Let S be a totally ordered conr.. Then there 2S a totally ordered rational-linear space H wzth IHI = lSI and an mjectwe morphism '1/) : S ---> H with 1/'(S) c H+-. Proof Let G be the group of S. Since S is torsion-free, G is torsion-free and IGI = lSI by 1.2.1O(iii). Also, G is an ordered group with respect to the S-order on G. The result now follows from 1.2.18. 0 Let S be an abelian sernigroup. For n E N. set Sn 1/Jn : S

f-+

(n

+ l)s,

S,,+1

--->

S, and define

=

Sn.

Then {Sn; 1/J,,} is a projective sequence of semigroups and morphism;". Define S = lim proj{ Sn; 1/J,,}' so that S is the set

S = {s

=

(sn)

E

SW : Sn

=

(n

+ 1)8n+1

(n E N)}.

In general S may be empty; in the case where S is non-empty, S is clearly a subsemigroup of SW. Note that, in the next theorem, the semigroup S is not assumed to be either cancellative or torsion-free; this will be important for our applications of the result in §4.9 and §5.7.

]9

Semigroups

Theorem 1.2.20 (Esterle) Lft S be an abelian serru,group, and let S be as alioue. Assume that S =I- 0. For each (\' = plq E Qt-. and.'l = (lin) E S, dfjine p((n q )!) ) as = ( q(n!) S,,'!: n EN. Then S, togetheT 71nth thf'. map (n. s)

f--->

0'8, lS

(1.2.4)

a semzgroup 01leT Q-t-•.

Proof Let S = (sn) E Sand () = plq E Q+ •. Since (nq)!/q(n!) E N (n EN). the right-hand side of (l.2.4) belongs to SW. We show that 0'8 is well-defined. First note that. if rn, n E N with In :5 11. then 8 m = (n!/m!)sn because.9 E S. Suppose that jlk = plq in Q, with k ? if in N, say. Since Snq = ((nk)!/(nq)!)8 n b we have

p((nq)!) p((nk)!) j((nk)!) q ( n.I) Snq = q (') n. Snk = k( n.') Snk, and so (plq).'> = (jlk);:;, as required. Also, have

(

n+]

0;8

E

S because, for each n E N, we

) _ p((nq + q)!)((nq)!) _ p((nq)!) . ) ( p((nq + q)!) . )1)( n.') Snq+q ' ) Snq· q (( rz + 1)') . 8 nq + q q ((,nq. q (n.

We nov.. check that the conditions of l.2.16 are satisfied. Clauses (ii) and (iv) clearly hold. Let s = (;;n). and let plq.jlk E Q-t-•. Then

q + 'kj)

(p

.'I

=

((Pk+ qj ) ((nqk)!) qk (n!) .'inqk: n

E

N

)

=

p j qS + 'k o5

and

!!. (Ls) q

k

=

(p((nq)!)j((nqk)!) 8 n k : n EN) = PJ 8. q(n!)k((nq)!) q qk

o

so that clauses (i) and (iii) also hold.

Note that, in multiplicative notation, the image of (0:, s) under the ab<\{'e map is 8(~ = (s~~(nq)!)/q(nl): n EN). In this notation, set T = 71'1 (S), and, f~~' t

= 71'1(8)

E

T, define

. (1.2.5 )

Then T is a semigroup over Q-t-. with respect to this operation. A cone S is an T]1-cone if (S, ::;s) is an T]1-set. In §5.7, we shall be particularly concerned with T]l-cones over Q+.; we shall require the following minor result. Proposition 1.2.21 Let S be a cone over Q+. s'uch that (S, ::;s) is a semz-T}l -set and cof S > No.

(i) Suppose that coi S > No. Then S is an (ii) Let 80 E S, and set T over Q-t-•.

= n{n8o + S

7/1 -cone.

: n EN}. Then (T, -S:T) 'ts an 7/l-cone

Proof (i) Let A and B be countable subsets of S such that A « B. If A = 0 and B has a minimum b, then b/2 E Sand b/2 « B. If B = 0 and A has

AlgebrQ:l,c fonndatwns

20

a maximum a, then 2a E S and A < < 2a. If A has a maximum a and B has a minimum b, then (a + b)/2 E S and A « (a + b)/2 « B . If A has no maximum and B has no minimum, then either A = 0 or A containb a cofinal, strictly increasing sequence, and either B = 0 or B contains a coillitial, strictly decreasing sequence. In each ccu.;c, it follows from the hypotheses on S that there exists.5 E S with A « .'; « B. If A h3...'> a maximuIIl a and B has no minimum, then there exists ,<; E S with ,<; « B - (J" and then A « a + ,.; « B, If B has a minimum b and A has no maximum, thcn there exists .5 E S with ,<; < < b - A, and then A « b - s «B. In each case, we have found an element ,<; E S with A « s « B, and RO (S, ~8) is an 17l-Set. (ii) Clearly T is a cone ovcr Ql+., The identity map (T. ~T) -> (S.::;s) is isotonic. and S + T c T. and so r No. Let (tn) be a strictly decreasing sequence in T. Then nso <8 til (71. EN), and so there exists 8 E S with 7L<;0 <8 .9 <8 fn+1 No. By (i), T is an 17I-cone. 0 Definition 1.2.22 Let (C, +,~) be a totally order'ed abelwn group, and let :/:, y E C. Define:

(i) x = o(y) if n Ixl

~

Iyl for each n EN; 71. Iyl for some 71. E N;

(ii) x = O(y) zf Ixl ~ (iii) x rv y zf both x = O(y) and y = O(x). Clearly

rv

is an equivalence relation on C.

Definition 1.2.23 Let C be a totally ordered abehan gnmp. The set

r

=

rc =

C- /rv

of equivalence classes zs the value set of C, and the elements of r are the arc:himedean classes of C. The qlLotient map from Canto r is the archimedean valuation on C, denoted by Ve; or'L'. Set vex) ~ u(y) for :E, y E c· ~f y = O(.T). It is easily checked that set, and that

vex

~

is well-defined on

r,

that (L,

~)

is a totally ordered

+ y) 2:

min{ vex), v(y)}

(.T,y E C),

=

min{v(x),v(y)}

if v(x)-j.l.'(y).

'II(x+y)

We set v(O) = 00, and adopt the convention that.., < 00 for each s E Let S be a non-empty set, and let f E
r.

suppj = {s E S: j(8) -j. O}. Definition 1.2.24 Let S be a totally ordered set, and let lK be lR or
=

{f E J(lK, S): supp f is countable} ;

S) = {J E OC s : supp j is finite} .

Sem~groups

21

We write J(S) for J(C, S), etc. It is clear that each set that we have defined is a IK-linear subspace of IK s . Define 1'(f)

= inf supp f =

min supp f

(f E J(K

st) .

Then v is the Hahn valuatwn on J(IK, S). In the case where IK = R set f > 0 if f(v(f)) > O. Then (J(IK, S). +.:::;) is a totally ordered group, called the Hahn gr'oup of S. Clearly f "" 9 in J(IK, S) if and only if v(f) = v(g), and we can identify S as the value set of J(IK, S) and of J(l) (IK, S): the Hahn valuation coincides with the archimedean valuation in this case. In the case where S is an (Xl-set, J(IK, S) = J(1)(K S) because each wellordered subset of an (Xl-set is countable. Definition 1.2.25 Let (G,

+.:::;)

be an ordered abelian gr·oup. Then G ts:

(i) an Ctj-group [rl1-groupj tf (G.:::;) zs an (Xl-set [1'/I-setj; (ii) a {il -group ~f G = (Xl -subgTOupS of G. In the theory ZFC

U{ GI/

+ CH,

:

1/

<

a group

WI},

where {G,/ : //

< :.vt} is a cham of

G is a ;1l-groUP if and only if IGI :::; Nl .

Proposition 1.2.26 Let S be a totally ordered set. (i) SupposethatS is ann1-set [rl1-setj. ThenJ(RS) andJ(1)(J~,S) are both (Xl-gTO'UPS [Til -gTO'I1PSj. (ii) Suppose thatS zs a,Brset. Then JU)(RS) 'is a (3l-groUP, Definition 1.2.27 For each and define G = J(1)(R Q).

(J"

< WI, define G a =

J(~,Qa) (w1th Go

\Ve regard each Go- as a subgroup of G; we write P for the positive coneI'. of G and G er , respectively.

= G+·

and Per

Proposition 1.2.28 (i) For each a < WI. G er 1,S a totally ordered (ii)

G 'is

0

a totally ordered, dwis'ible (31-rl1-group, and

IGI =

c.

0'1

= {O}),

=

Gt·

-gTOUp. 0

Theorem 1.2.29 (i) Let G be a totally ordered /31 -group, and let H be a diviszblc 771 -group. Let So E G+· and to E H+·. Then then: is an isotonu; morphism 'Ii' from G into H with '~)(so) = to. (ii) Each totally ordered, div'isible (31 -T{I-gTOuP 1,S zsoionically 1,somorphic to

G.

0

Clause (i) of the above result implies that G is 'I1mversal in the class of totally ordered (31-groUPS: G contains a copy of each such group. Definition 1.2.30 Let S be a. cone. Then S is universal if, for each Xo E P and So E S, there is a morphism '!j; : P -> S with '!j;(xo) = So. The morphism '!j; of the above definition is necessarily an injection, and so

lSI 2':

c.

Algebrmc foundations

22

Lemma 1.2.31 Let (S. +) be an 'f/l-cone, and let G be the group of S. Then G is an 'fIl -group. Proof \Ve have explained that G is an ordered group. Let A and B he countable suhsf'ts of G with A U B totally ordered and A « B. First, suppose that A =/:. 0, and take an E A. Set

A' = {s - ao : sEA, B - ao. Then A' B ' , whence A « Then' exists 8 E S with -B « follows.

«

Theorem 1.2.32 Let S be an

'fII -cone OVf'f'

alld B' =

A'

«

.9

«

8

8,

.9

> ao}

B' in S, and so there exists s E S with B. Second, suppose that A = 0. and then -8 E G with - 8 « B. The result

+ Uo «

0

Q+-. Then S

l8

a unwersal cone.

Proof Let G be the group of S. Then (G, :S;s) is a divisihle, totally ordered group; by 1.2.31, G is an fII -group. Take :co E P and 80 E S. By 1.2.29(i), there is an isotonic morphism '~' : G ----+ G with 'Ij!(xo) = so; clearly we haW' l;!(P) C G+- = S.

0

We conclude this section with some examples of 'Til-cones over Q+-; the examples will be important in §5.7. Example 1.2.33 (Esterle) The set (lR'.+-)f''' of functions

f = (.f(n)) : N ----+ lR+is a cone over lR'.+- with respect to adclition, and NN is a subcone; particular elem{'nts of NN are 1 : n f--> 1 and Z : n f--> n. Let f, g E (lR'.+_)N. Then: if g(n) > f(n) eventually; if g(n) - f(n) ----+ 00 aH n ----+

00.

Cleally
= {.f E

(lR+-/>I : 1

«F

f and flp

E

co},

N(p)

= lR'.(p) n NN.

(1.2.6)

In particular, wc define lR'.(Z) and N (z). Thus lR'.(p) is a Hub set of the set of functions whoHe graph is 'eventually under' the graph of p. Clearly (lR(p), +) is a cone over lR+ and (N(p), +) is a cone. Take S to be either lR(p) or N(p). Then the S-order on S is exactly the strong Frechet order « F . The cone N (p) is not a cone over Q+-; we describe a small modification of N(p) that is such a cone.

Sem1,gToup8

23

First, define NIp] =

n {J

E N(p)

:

f

E

(1.2.7)

kN eventually} .

kEN

Thus, for f E NN, we have f E NIp] if and only if, for each kEN. there exists nk E N with kf(n) < pen) and fen) E kN for each 71 ~ nk. Choose a strictly increasing sequence (mk) in N such that pen) ~ (k + I)! (n ~ md, and define fen) = k! (n E {m,k, ... , mk+l - I}). Then f E Nip], and so N[p] -I 0. Clearly (N[p] , +) is a cone. The N[p]-order coincides with <
+ 2k ::; min{gl (n), ... , gk(n)} .

Define

fen) = max{h (n), ... ,Jk(n)}

+k

(n E {nk, .. . , nk+l - I}) .

Then f E N[p], and [lk]F s h, then f > s h. We clazm that (5, ::;s) is an 'T7l-COne over Q+ •. This iH a similar argument to that in 1.2.33. For example, let (In) be a strictly increasing sequence in S. and take a strictly increasing sequence (t n ) in lR such that tn ---> ex; as 71 ---> 00 and

fn+l(t)

~

fn(t)

(t

~ t n ),

1

00

Ifn(t)1 d/t < 2- n

(n

E

N).

t"

Set f(t) = 0 (t < td and f(t) = In(t) (t In <s I (71 EN), so cof 5 > No· By 1.2.32, 5 is a universal cone.

E

(tn' tn+1))' Then I E 5 and 0

24

Algfbmic fo'undations

Notes 1.2.35 The construction of S[T- I ] in 1.2.9 is an adaptation to semigroups of a standard COlH;truction (P. ;"'1 Cohn 19H9). The group G of a cancellativt', abelian semigroup S iH sonH'tirrH's called the GT'Othendieck g1'O'Up of S; it is the HOlution of a universal proLlem, in the sense t.hat, if H is a group and X : S ..... H is 11 morphiHm. then there is a group morphism {} : G ..... H such that X = {} 01/) Theorem 1.220 is baRed on part of Theorem I) 1 of (Esteri{' 19S:3a). Ordered groups. including the Hahn group J(K S), are diHcussed in (FuchH 196:3), (Priess-Crampe 1(83). and, especially, (Dales and \Voodin 1996. Chapter 1). On each torsion-free, abelian group G. there is a relation :::5 such that (G.:::5) is a totally ordered group. Propositions 1.2.26. 1.2.28, and 1.2.29 are proved in (DaleR and Woodin 1990, Chapter 1), Proposition 1.2.7 and the example Nil>!/F in 1.2.:~3 are taken from (Esterle 19961', 1(98).

1.:3

ALGEBRAS, IDEALS, AND FIELDS

\Ve present in this section a review of some essentially standard material about. algebras and their ideals; the section concludes with a brief discussion of fields, including ordered fields. Our account is. of course, prejudiced in favour of topics that will form the algel)) aic background to our later work; results are included because they will be required later. ancl our aim is to be efficient. In particular. we shall introduce notation that will be standard throughout the book. We begin hy Iecalling some basic' facts about linear spaces over a tieldlK; it will be assumed throughout that lK is either JR or C. Let E be a linear space over K As in §1.2, we write E· for E \ {a}. Let S be a non-empty subset of E. Then lKS = {(Xx: (l' E lK ..r E S}; we write lKx for lK {.1;}. The l'meaT .'ipan of 8 is lin 8 = lin K8 =

{t

(ti·'Ci :

nl· ... ,(tn

ElK,

'Cl, ... ·.E n

ES, n EN} .

For example, let 8 be a BOil-empty set.. Then coo(8) = lind Ob : S E S} C ([s, so that Coo = coo(N). In the case where 8 is a totally ordered set. coo(8) = ~(())(8). Let E be a linear space over R and set F = Ex E. Then F is a linear space over ([ for the operations:

(Xj·yd + (J'2.Y2) = (Xl +.r2, Yl +Y2) (.(1,:r2,Yl,Y2 E E); (n + i;3)(x, Y) = (nx - j3y, av + j3x) (0, (j E JR, x. Y E E). The space F is the eomplexijieation of E. Let {E-y : '"Y E r} be a family of linear suhspaces of a linear spac(' E. Then the sum L-YEI' E-y = lin(U E-y) is a d'iTect sum if E6 fllin(U-y1'O E-y) = {a} for each 8 E r. In this case, we write O-yEI' E-y for lin(U E-y); an element of (0 E-y)· has the form L~=l X-Yil where X-Yi E E;i (i E N n ) and '"Yi i= '"Yj if i i= j. Let {E-y : '"Y E r} be a family of linear spaces over lK. The Cartesian product TI-YEI' E-y is a linear space with respect to the coordinatewise operations; it is the dZTect pmduct of the family. We identify Eo with {(x-y) : x-y = 0 if ~f i= 8}, and then the direct sum O-YErE-y, also denoted by coo(r,E-y) (ef. A.3.74), is the subspace {(X-y) E TI-YEr E-y : XI'

= 0 for

all but finitely many'Y E

r}.

25

Algebra8, zdeals, and fields

Let E be a linear space. A basis for E is a linearly independent subset 8 of E such that lin 5' = E. A maximal linearly independent subset of E is a hasiti, and each linearly independent suhset is contaiued in a Imf:'is. The cardinality of bases is an invariant of E; it is the dirnension, dim E, of E. The space E is finzte-dzrnrnswnal if dim E is finite, and in.jinde-dzrnenslOnal if dim E is infinite. Let F be a linear subspace of E. and let E / F })(' the quotient space. Then F has codirnen.9wn n in E if dim( E / F) = n, F has Jinzte codirnenswn in E if dim( E / F) is finite, and F has znjinzte corhrnenswn in E if dirn(E / F) is infinite. Let E a.nd F be linear spaces over lK. A map T : E ---7 F is lK-lmea.r (or, usually, lmear) if

T(n:r

+ By)

= nTJ;

+ /3Ty

(0, (J

E lK, :r,

y E E) ,

and T is conJngate-lmear if

T(n:£

+ ~3y) = liT:r + STy

(0,3 E lK, x,y E E).

We often refer to linear maps from E to F as operators. A bijective linear map T : E --> F is a linear ?sornuTplnsrn; in this case, E and Fare lznearly isornorphzc, and we write E ::::: F. The set of linear maps from E into F is denoted by £'(E, F); it is itself a linear space over lK for the usual operations. Let G be a third linear space, and let S E £'(E, F) and T E £.(F. G). Then the composition T 0 S of Sand T belongs to £'(E, G); it is often denoted just by TS. \\Te write £'(E) for £'(E, E). The identity operator on E is often denoted by h. Let T E £'(E). Then z E C is an eigenva17te of T if zh; - T is not injective'. We write EX for £'(E, lK); EX is the algebmzc dual spacr of E. and clements of EX are linear j1J:nctionals on E. \Ve often write (:1',).) for ).(:r), where x E E and), E EX. There is a natural embedding /, : E --> EX x; here I(.l')().) = ).(x). Let E and F be linear spaces, and let T E £'(E, F). For)' E F X, we define TX), by the formula (x, TX),) = (Tx,).) (:1: E E) . (1.3.1) Then TX), E EX and T X E £'(FX, EX); T X is the algebraic dual map to T. The map T X : (Fx. a(FX, F)) ---7 (EX, a(EX, E)) is clearly continuous. An operator T E £( E, F) is rank-one if dim T( E) = 1 and jinite-mnk if dim T(E) is finite; each finite-rank operator is the sum of finitely many rankOlle operators; the ::;et of finite-rank operators is a linear suhspace, denott'd ·by F£(E,F), of £'(E, F). We write F£'(E) for F£'(E,E). Let F be a linear subspace of a linear space E. An element P E £.(E) with P(E) = F and Px = x (.T E F) is a projection onto F. Let (En: n E :2:) be a sequence of linear spaces (each possibly 0), and let Tn E £'(En+l' E,,) (n E :2:). The beqllence '"""' . T,,+l L .... -->

=

Tn

---+

En

n-1 T----> ...

(1.3.2)

0 Tn = 0 (n E :2:). The sequence (1.3.2) is e.Tact at rz E :2: if kerTn - 1; it is exact if it is exact at each n E:2:. Thus a sequence

is a cornplex if T,,-1

Tn(En·n)

E n+1

26

Algebr'aic foundatwns

is a shori exact sequence if S is injective, T is surjective, and S(E) = ker T; in this case, there exists Q E £(G, F) with To Q = la, and F = S(E) (;) Q(G). In the case where 2: is not exact. we measure 'the failure of exactness' by certain 'homology groups'.

2:

Definition 1.3.1 Let

Hn

be a r-07nplex, as in (1.3.2). The linear spaces

(2:)

(n

= kerTrt_I/im T"

E

Z)

are the homology groups of the complex.

Corresponding to a complex denoted by ""' L.....t:

(where we have Tn+!

0

...

E ,,+1 ~ 1~, E ,,~ T n - 1 ...

Tn+l ~

T" = 0 (n E Z)), we obtain the cohomology gT'O'IlpS

H"

(2:) =

(n E Z).

kerT,,+I/imTn

The sequence

L.....t

""'

x

: ...

TX n-1 ---+

x EX Tn EX ,,---+ ,,+1

T

X

n+1 ---+ ...

(1.3.3)

corresponding to the above complex (1.3.2) is also a complex, for clearly T;:: 0 T:_ 1 = (Tn - 1 0 Tn)X for each n E Z; 2: x is called the algebmzc dual complex to 2:. The algebraic dual of a sequence exact at n is also exact at n. Let E J , ••• , En, F be linear spaces over IK. A map T : I1:'=1 Ei - 7 F is 1/,lmeaT if, for each j E N" and each Xi E Ei (i = 1, ... ,j - 1,j + 1, ... , n), the map X f--+ T(Xl, ... , X)-I, X, Xj+l, ... , Xn), E j - 7 F, is linear. The set of n-linear maps from

n:';,,1 Ei into F is denoted by

£"(El"'" E,,; F),

and we write £1'(E, F) in the special case where El = ... = En = E: again. £n(E1 , . .. ,En; F) is itself a linear space over lK. A 2-linear map is said to be bilineaT.

The tensor pT'Oduct of linear spaces E 1 , ... , En is a pair (~, F), where F is a linear space and ~ E £1' (E1' ... , En; F), snch that. the following universal property holds: for each linear space G and each S E £n(E1 • ... ,En: G). there is a unique linear map T : F - 7 G such that S = To I. Such a tensor product always exists, and it is unique (up to linear isomorphism). We write n

® Ei

or

E1 0 ... 0 En

for

L(Xll' .. ,

i=1 for the space F and

Xl

0· .. 0

Xn

x n ), and we regard ®~=J Ei as

the tensor product of E 1 , ... , En; we identify (®~=1 Ei) 0 (®~=k+l Ei) with ®~=1 E i . Each element z of ®7=1 Ei can be written in the form m

Z = LXl,j 0 X2,j 0··· 0 xn.j j=1

Algebras, ideals, and fields

27

for some rn E N, where Xi,} E Ei (i E N n , j E N m ); in the case where z -I- 0, we may suppose that each of the sets {Xi,j : j E Nm } is linearly independent. Let E and F be linear spaces over lK. Then the following identities hold in E®F:

+ y) ® z = X ® z + y ® z r8' (y + z) = X ® y + x ® z

(X X

a(x ® y)

=

(ax) ® y

=x

® (ay)

(x, Y E E, Z E F) ; } (x E E, y, Z E F) ; (x E E, y E F, a E lK) .

(1.3.4)

For each bilinear map S : Ex F --t G, there is a unique linear map T : E®F such that T(x ® y) = Sex, y) (x E E. y E F), and so

--t

G

(E®F)X ",:C(E,F;C) ",:C(E,FX). For example, let Sand T be non-empty sets. Then coo(S) ® coo(T) "': coo(S x T);

here (f ® g)(s, t) = J(s)g(t) (5 E S, t E T) for J E coo(S) and 9 E coo(T). Let E be a linear space, and let ®P E for p E Z+ denote the tensor product of p copies of E (taking ®o E = C and ®l E = E). We define the linear space Q9E=O{®PE:pEZ+} , and denote a generic element of ® E by U = (up), where up E ®P E (p E Z+). Let U E ®P E and l' E ®q E, where p, q E Z+. Then U ® v E ®p+q E. For each a E 6 p, there is a linear map (j : ®P E --t ®P E such that (j(Xl ® ... ® .Tp) = Xa(l) ® ... ® Xa(p)

(Xl, ... , Xp

E E).

(1.3.5)

An element U E ®P E is symmetric if (j(u) = u (a E 6 p); we write VP E for the linear subspace of ®P E consisting of the symmetric elements. The linear map 1 Sp : 11!---> I ' 2)a(u) : a E 6 p.

(1.3.6)

p}

is a projection from ®P E onto VP E; Sp is called the symmet'r'iz'ing map. We define VE=O{VPE:PEZ+} . Note that Sp+q(u ® v) = Sp+q(Spll ® Sqv) (u E ®P E, v E

® qE).

Definition 1.3.2 An algebra over lK is a lK-linear space (A. +) which has an additwnal binary operatwn . such that (A, .) is a semzgroup, the di8tributwe law8 hold, and a(a· b) = (aa) . b = a· (ab) (a E lK, a, bE A). The algebra A i8 commutative zJ (A, . ) is abelian. Henceforth, we shall usually write ab for a . b, but sometimes we shall retain the '; (A, . ) is the multiplicative 8ernigroup of A. Note that the map rnA : (a, b)

!--->

ab,

A x A

--t

A,

is a bilinear map, called the product map. There is a linear map 7rA : A ® A --t A such that 7rA(a 0 b) = ab (a, bE A); this map 7rA is called the induced product map.

28

Algebraic fo'undations

Let K be a subsemigroup of (c, + ) with 1 E K. Then a K -semigroup in A is a K-semigroup (ar: : ( E K) in (A, . ). An algebra over C is a complex algebra. These algebras are the main objects of study in this book, and in Chapters 2-5 a complex algebra will just be called an algebra. A complex algebra is also a real algebra when scalar multiplication is restricted to R The set {O} is an algebra: it is the zero algebra, and it is denoted by O. Let A be an algebra. Then AOP is the algebra formed by reversing the order of the product in A; AOP is the opposite algebra to A. Note that, even when we are considering AOP, ab always denotes the product of a and b in the original algebra A. A subset of an algebra A is a .mbalgebra of A if it is an algebra with respect to the operations of A. The intersection of the subalgebras of A containing a subset S of A is the subalgebra of A generated by S; it is denoted by algAS

or

Let A be an algebra over JR, and let B containing A described above. Set (Xl, Yl) (X2' Y2)

=

alg S.

=A

(XIX2 - YIY2, :CIY2

x A be the complex linear space

+ X2yd

(Xl, :1'2, Yl, Y2 E A) .

Then B is an algebra over C containing A as a (real) subalgebra; B is the complexification of A. An element eA of an algebra A is an identdy of A if fA i- 0 and eA is the identity of the semigroup (A, . ); similarly, we define left and right identities of A. An algebra A with an identity is a unital algebra, and in this case we identify the field lK with lKCA. A subalgebra B of a unital algebra A is a unital subalgebra if eA E B. We sometimes write (eA - a)A and A(eA - a) for {b - ab : b E A} and {b - ba : b E A}, respectively, even when A does not have an identity. Let A be a unital algebra. An element of A is left mvertible [nght mvertible], [invertl,ble] if it is left invertible [right invertible], [invertible] in the semigroup (A, .). The set of invertible elements in A is denoted by Inv A. A unital subalgebra B of A is inverse-closed if B n IllV A = Inv B. Let B be a unital subalgebra of a commutative, unital algebra A. Clearly {be -

J :

b E B, c E B

n Inv A}

is a unital subalgebra of A, and it is the smallest inverse-closed subalgebra of A containing B. It is called the znver8c-clos'ure of B in A. Definition 1.3.3 Let A be an algebra over lK. Then AD i8 the linear space lK x A together wzth the product (a, a)({3, b) = (a{3, ab +;Ja + ab) Define A #

=

A if A is umtal and A #

=

((X, (3

E

lK, a, bE A).

AD if A is not unital.

Certainly A' is a unital algebra over lK; the element (1,0) is the identity of A', and we regard A as a subalgebra of AI>. Note that, if A is unital, then the element (1,0) is different from the identity eA of A. The algebra A# is the algebra formed by adjoining an identity to A; we shall usually be concerned with

Algebras, ideals, and fields

29

this 'conditional unitization'. We often denote the identity of A# by eA, even when A is non-unital; thus, in the non-unital case, we write O:eA + a for the element (a,a) of A# = AO. However, eO always denotes the identity of AO.

Definition 1.3.4 Ld A be an algebra, and let a, bE A. The quasi-product of a and b i8 a <> b, where a <> b = a + b - ab . An element a of A is left [right] quasi-invertible if there exists b E A such that b <> a = 0 [a <> b = 0]. The element a 1,1, quasi-invertible tf it is both left and right quasl,-mverttble, and a subset of A 1,8 left quasi-invertible if each of tts elements is left quasi-inver·tible, etc. The quasi-product is an associative operation with identity O. Suppose that a <> b = (' <> a = O. Then clearly b = c, and so, if a is quasi-invertible, there is a unique dement b E A such that a <> b = b <> a = 0; b is the quasi-mverse of a, and is denoted bya'l. \Ve write q-Inv A for the set of quasi-invertible elements of A. Clearly 0, <> b = 0 if and only if (eA - a)(cA - b) = eA in A#, and so Inv A# = eA + q-Inv A. Let A be an algebra. and let Sand T be non-empty subsets of A. As in §1.2, S . T = {ab : a E S, Ii E T}. We define ST = lin S . T, so that ST =

{t

aiaibi : 0'1, ... , an E

1K, aI, .... an E S, b], ... ,bn E T, n E N} ;

t=1

we write aT for {a}T, etc. Note that aA = a . A, for example, but that, in general, A . a . A i= AaA. Also, A#T, calculated in A#, is a subset of A. There is a special notation which we introduce because we wish to distinguish between two sets: for a nOll-empty subset S of A and n E N, set s[n]

= {al ... an

: ai, ... ,an E S},

sn

= lin s[n]

(whereas s(n) denotes the n-fold Cartesian product of S). The set S is mulhplicative if S[2] c S.

Definition 1.3.5 Let A be an algebra. Then: an element 0, E A factors if a E A[2], and A factors if A = A[21; A factors weakly tf A = A2; a pazr {al,a2} in A has a common factor b E A if {ai, 0'2} C bA, and A has factorization of' pairs zf each pair in A has a common factor. Of course, a unital algebra factors and has factorization of pairs; later we shall be concerned with non-unital algebras that either factor or factor weakly.

Definition 1.3.6 Let A be an algebra. A linear functional T on A is a trace if T(ab) = T(ba) (a, bE A). Examples 1.3.7 (i) The linear space IK n is a complex, unital, commutative algebra over IK with respect to the coordinatewise operations; IKn is taken to have this product, unless otherwise stated. More generally, let A')' be algebras for 'Y E r. Then I1-yH A')' is an algebra with respect to operations defined

Algebraic foundatwns

30

coordinatewise, and O-YEr AI' is a subalgebra thereof. (ii) Let S be a non-empty set. The pomtwise opemtzons on S are defined as follows. For f, 9 E OC S and a E OC, define

(af)(s) = af(s), (f (fg)(s) = f(s)g(s),

+ g)(8) = 1(8) =

f(s) 1,

+ g(S)'}

(s E S).

(1.3.7)

Then OC S is a commutative algebra with identity 1, and coo(S) is a subalgebra of CS. For f E OC S , we also define

(3tf)(s) = 3t(f(s)),

(:sf)(s) = :s(f(s)),

If I (8) = If(s)1

(s

E

S).

(1.3.8)

Let A be a subalgebra of OC S , and let U be a non-empty subset of S. Then A I U is an algebra; it is the restrictzon algebra of A to U. (iii) Let S be a semigroup; the linear space J(O)(S) = coo(S) was defined in 1.2.24. Clearly there is a unique product * on coo(S) such that

8..

* 8t = 8st

(s,t E S);

the operation * is called the convolutwn product on S, and (c 00 (S), *) is the algebrazc semigroup algebra. In the cru:;e where S hru:; an identity e, 8e is the identity of (coo(S), *). We refer to the algebraic gmup algebra when S is a group. Of course, properties of the algebra coo(S) correspond to properties of S; for example, coo(S) is commutative if and only if S is abelian. Now take S = +) X (Q+, Then the non-unital algebra coo(S) factors weakly. but it does not factor. (iv) Let qX] be the algebra of polynomials in one indeterminate, with coefficients from Co (See §1.6, below.) Then qX] is a commutative, unital algebra. We shall use the fact that each p E qXJ- can be written as

+»-.

w:.)!+,

p = no(X - al)" . (X - on), where aI, ... ,an E C and aD E C-. This algebra will be studied in §1.6. (v) Let E, F, and G be linear spaces, and let T : E x F ~ G be a bilinear map. Set A = E x F x G, as a linear space. The map

«.rl, Yl,

zd,

(X2' Y2, Z2»

f->

(0,0, T(Xl' Y2)),

A x A ~ A,

is bilinear and defines a product with respect to which A is an algebra with A:3 = O. Clearly A2 = {O} X {O} x lin T(E, F); in general, A[2j =I- A2. (vi) Let E be a linear space over K Then £(E) is a unital algebra over OC (with composition as product), and T E £(E) is invertible if and only if T is a bijection. Suppose that E is a finite-dimensional space. Then the following are equivalent: (a) T E Inv £(E); (b) T is injective; (c) T is surjective; (d) det T =I- 0, where det T is the determinant of T. Let E and F be linear spaces, and take Yo E F and Ao E EX. We define

Yo ® Ao : x

f-+

(x, Ao)Yo,

E

~ F.

The map Yo ® AO is either a rank-one operator on E or 0, and we may identify F ® EX with F£(E, F). Note that, given Xo E E- and Yo E E, there exists AO E EX with (xo, Ao) = 1, and then Txo = Yo, where T = Yo ® Ao.

Algebras, ideals, and fields In particular, E product

(9

EX

(Xl

(9

31

= F£(E); AI)

the product in F£(E) corresponds to the

(X2 (9 A2) = (X2'

0

Al)Xl 0

(1.3.9)

A2,

and the product::! of Xo 0 Ao E F£(E) with T E £(E) are

To (.1:0 ® Ao) = Txo ® AD,

(TO 0 AD) 0 T = Xo ® T X(AD) ;

(1.3.10)

we have (xo 0 AoV = AD ® ~(XO) E F£(EX). There is a unique linear functional Tr : E 0 EX --> C such that

Tr(x0A)

=

(x, A)

(x

E

E, A E EX).

(1.3.11 )

This map is called the trace map; clearly Tr is a trace on E 0 EX. Indeed it follows from (1.3.9) that each trace on E 0 EX is a scalar multiple of Tr. (vii) Let A

= C4

with the product given by

(Zl, Z2, Z3, Z4)( WI, W2, W3, W4)

=

(Zl WI, Zl W2, Z3W3, Z4W3)

for Zl, ... , Z4, WI, ... , W4 E C. Then A is a non-commutative algebra which factors, but the pair {CO, 1,0, 1), (1,0, I,D)} does not have a common factor. We shall note in 2.2.51, 2.9,45, and 2.9,48 that there are various Banach algebras that factor weakly, but do not factor, and in 4.3.19(i) that there is a commutative Banach algebra that factors, but does not have factorization of pairs. 0 Definition 1.3.8 Let E be a linear space, and let m, n E No Then MIm.n(E) zs the linear space of m x n matrices with coefficients mE.

An element of MIrn.n(E) is denoted generically by

(Xij) = (Xij : i E N m , j E Nn ). The transpose of (xi]) E MIm,n(E) is the matrix (Xij)t = (Xji) E MIn,m(E). An element (Xij) E MIm.n(E) can be identified in the mmal way with an element T E £ (E(rn),E(n»): for (Xl, ... ,Xn ) E E(n), T(XI"" ,Xn ) = (YI,'" ,Yrn), where Yi = L7=1 XijX) (i E N m ). Let n E N. We write MIn(E) for MIn,n(E). A matrix (Xi) E MIn (E) is upper-trzangular if Xij = (i > j); strictly upper-trzangular if Xij = 0 (i ::::: j); lower-triangular if Xij = (i < j); strzctly lower-triangular if Xlj = (z ::::: j); . and dzagonal if Xi) = (i =I- j). Let A be an algebra. Then MIn(A) is an algebra for the product given by (ai])(bij ) = (L~=l aikbkj): it is the full matrix algebra of order n over A. We write MIn for MIn(C), so that MIn is a complex algebra of dimension 11. 2 . \Ve also write G L( 11.) for Inv MIn, the general {meaT group of order n. Let A be a commutative algebra, and take (ai]) E MIn(A). Then the determinant of (aij) is

° ° °

det(aij) =

2) (-1)10'Ia1,O'(l) ... an.O'(n) : IY E Sn} ,

where IIYI is the sign of IY. The following result is standard.

°

Algebrazc fo'undatwns

32

Proposition 1.3.9 Let A be a commutative algebra, and let n EN,

(i) deteST) = (det S)(det T) (S, T E Mn(A», (ii) Let T E Mn(A). Then T E Inv M,,(A)

'tf and

only 'if det T E IllV A.

(iii) Suppose that (ai]) E Mn(A) and (b 1 , ... ,b,,) E A(n) are such that I:~=l aijbi = 0 (j E N n ). Then (det(aij»bi = 0 (i E Nn ). 0 Definition 1.3.10 Let A be an algebm. A set (eij : i,j E N n ) of elements of A is a system of matrix units (of size n) for A if CijekC=6j.kCi£

(i,j,k,fEN n );

in the case where A is umtal, the system

28 unital

2f, further, I:~1 eii =

CA.

In particular, let n E N, and consider the full matrix algebra Mn. \Ve sometimes write En for the identity matrix (6i,j) ill Mn. For i, j E N n , take Eij to be the matrix with 1 in the (i,])th position and 0 elsewhere. Then (Ei] : i,j E N n ) is a unital system of matrix units for Mn; it iH the standani system of matrix units. Proposition 1.3.11 Let AI, .... An be algebras. and let A be the linear- space ®:~1 Ai' Then there is a unique pmduct on A wzth respect to which A is an algebm and such that (0.10'" ® a n )(lJ1 @ ... 0 b7l ) = a l b1 @ ... 0 anb n

(a,. bi E Ai, i E N n ).

If A!, ... ,An are commutatwe, then A 2S commutative; if Ai has the zdentity (i E N n ), then C1 ® ... 0 en zs the 2dentity of A.

ei

Proof Take

(!

= (al,"" an) E

rr=l Ai'

The IIlap n

(b i

, ... ,

bn )

1--7

a) bl 0 ... 0 an bn .

II Ai

is n-linear, and so there is a unique linear map /La : A /La(b 1 0 ... 0 bn )

= a 1 bl

-->

/,=1

-->

A,

A such that

(hi E Ai, i E N n ).

® ... 0 anbn

The map a 1--7 f.La, I1:~ 1 Ai --> £: (A), is n-linear. and so there is a unique linear map f.L: A --> C(A) such that f.L(aI0·· ·®an ) = f.La (a = (al, ... ,an) E I1~~I Ai). The map (x, y) 1--7 fL(X)(Y), A x A --> A, is the required product. 0 The algebra A is the tensor- product of the algebras AI, ... ,An. For example, the space A# 0 A#op is an algebra for a product which satisfies (1.3.12)

This algebra is the algebmic enveloping algebra of A. Let E be a linear space. For u u0v=(up)0(Vq)=

=

(up) and v

(L

p+q=r

=

(vq) in

® E,

Up0vq:rEZ+).

we define (1.3.13)

33

Algebras, ideals. and fields

Then ® E is an algebra with respect to the product (u, v) >----+ u C>9 v; the sequence (1,0,0, ... ) is the identity of ® E. In the case where dim E > 1, ® E is noncommutative. For u E VP E and v E Vq E, define

uVv

=

Sp+q (U C>9 v) E

where Sp+q is the symmetrizing map. For U

Vv =

(lIp)

U

L

V (v q ) = (

Vp+q E ,

= (up) and v = (uq) in IIp

V 11q : r E Z+) .

(1.3.14)

VE. define (1.3.15)

p+q=r If 1I E Ep and v E Eq, then U C>9 v = a(v C>9 u) for a certain a E 6 p+q, and so U V v = 11 V 11. Thus V E is a commutative, unital algebra with respect to the product (u, v) >----+ U V 11. (However, VE is not a subalgebra of the algebra ® E.) Definition 1.3.12 Let E be a linear space. Then (®E,C>9) and (V E. V) are the tensor algebra and the symmetric algebra over E, respectzvely. Definition 1.3.13 Let A be an algebm. A subset S of A zs commutative if ab = ba (a, bE S). The commutant of S is

Se

=

{b E A : ab

= ba

(a E S)} .

The centre of A zs A e, and zt zs denoted by 3(A).

For each subset S of A, S e is a subalgebra of A, and S (' is a unital subalgebra in the case where A is a unital algebra. Clearly S iH commutative if and only if SeSe. The serond commutant, sec, of S is defined to be (se)c; always S c sec and se = secc. Let S be a commutative subset of A. Then it follows from Zorn 'H lemma that S is contained in a maximal (with respect to inclusion) subset AI of A. Let a E Ale. Then AI U {a} is commutative, and so a E AI. Thus AI = AIC and AI iH a subalgebra. Since se ~ MC and see c Mce = lvI, the set sec is a commutative subalgebra. The centre 3(,4) of A is a commutative subalgebra of A. Definition 1.3.14 Let A be an algebra. A subset M of A which is maximal (with respect to inclusion) in the fam'ily of commutatwe s'ubsets of A is a maximal commutative Hubalgebra of A.

For example, 3(Mn(lK») = IKEn. For let T = (O~rB) E 3(Mn(IK»). For each = TEij , and so ari = 0 (r i= i), a JB = 0 (s i= j), and (tii = Q jj. Thus T E IKEn- Also. we ::;ee ::;imilarly that lin {E 11 • ... , Enn} is a maximal commutative subalgebra of Mn(IK).

i,j EN", we have EijT

Definition 1.3.15 Let A be an algebra. Then a E A is nilpotent zf an = 0 for some n E N; a set SeA is nil if each a E S zs nilpotent, and S is nilpotent if s[n] = {O} for some n E N.

We write SJ1(A) for the set of nilpotent element::; of A, so that A is nil if and only if A = IJ1(A). The index of a E IJ1(A) is the minimum n E N such that

34

Algebrazc foundatwns

an = 0, and the mdex of a nilpotent set S is the minimum n E N such that s[n] = {O}. Let a, b E ')'teA). Then in general neither a + b nor ab belong to ')'teA), but both a + b E ')'teA) and ab E ')'teA) in the ca.<;e where ab = ba. Definition 1.3.16 Let A be an algebra. An element p of A is an idempotent if p2 = p. An idempotent in 3(A) 1,S a central idempotent. The set of idempotents of A is denoted by J(A). and

<E(A)

= lin J(A) .

Let p E J(A). Then p is the identity of the subalgebra pAp of A. Proposition 1.3.17 Let A be a non-unital algebra. Then <E(A#) = lKeAC:)<E(A). Proof Let b = o:eA + a E J(A#). Then 0: 2 = 0:. If 0: = 0, then a2 = a and b E J(A). If 0: = 1, then a 2 = -a, -a E J(A), and b E OCeA + <E(A). Thus J(A#) C lKe.l\ + <E(A), and so <E(A#) C OCeA + <E(A). The result follows. 0 Definition 1.3.18 Let A be an algebra. Then:

(i) elements a, b E A are (mutually) orthogonal in A, wrztten a 1. b, zf ab = ba = 0; a subset S of A zs orthogonal if a 1. b whenever a and baTe dzstinct elements of S; a sequence (an) zs orthogonal if am 1. an whenever m # n; (ii) p ~ q in J(A) ifpq = qp = p; (iii) p'" q in J(A) if there e:rist a, b E A such that p = ab and q = ba. For example, if A is unital and p E J(A), then p and eA - P are mutually orthogonal idempotents. An orthogonal set of non-zero idempotents is linearly independent. Clearly (J(A), ~) is a partially ordered set. A minimal clement of J(A) \ {O} is a minzmal zdempotent. Suppose that p is an idempotent which is not minimal, and take q E J(A)\{O} with q -
35

AIgebms. ideals. and jields

= O~=l Mn (lK),

where nl, .... nk E N; this is a.n

algebra of dimension nI + ... + n~. For j E

Nk , let (E~{) : r. s E Nn )) be the

Consider the algebra A

j

standard system of matrix units in Mn) (lK). Then E~~ 1.. E[l,l whenever i -j. j. Also, {E,\P : r' E Nn),.J E Nd is an orthogonal set of minimal idernpotents in A; it is the standard set of minimal idempotents in A. In the definition of p rv q. we may suppose that a = po, = aq = paq and b = qb = bp = qbp: to see this, replace the original a and b of the definition by paq and qbp, respectively. Proposition 1.3.20 Let A be an algebm. Then:

(i)

rv

is an equivalence relatwn on J(A);

(ii) ApA

=

AqA whenever p

rv

q in J(A);

(iii) for p, q E J(A), p + q E J(A) if and only if p 1.. q. Proof (i) Suppose that p rv q and q rv r in J(A). Then there exist a, b, c, d E A with p = ab, q = ba = cd, and r = dc. We have (ac)(db)

and so p

cv

= abab = p2 = P and (db)(ac) = dcdc = r2 =

r. It is now clear that

rv

is an equivalence relation.

(ii) Suppose that p = ab and q = ba. Then p q E ApA. It follows that ApA = AqA. pq

r',

= abab = aqb E AqA. Similarly,

(iii) Suppose that p + q E J(A). Then pq + qp = 0, and so, successively, + pqp = pqp + qp = O. pq = qp, and p 1.. q. The converse is immedia.te. 0

Definition 1.3.21 Let A be an algebm, and let p E J(A). Then p zs finite zf p = q whenever q E J(A) and p rv q ::5 p; otherwise, p zs infinite; p is properly infinite if there e:rist q, r E J(A) such that q ::5 p, r ::5 p. q 1.. r, and q rv r rv p. A unital algebm A is finite [infinite], [properly infinite] if eA zs a jinite [injinzte], [properly injinzte] idempotent in A.

Clearly a properly infinite idempotent is infinite. Proposition 1.3.22 Let A be a properly injimte, unital algebm. Then:

(i) there is an orthogonal sequence (Pn) zn J(A) such that Pn (ii) the only jinite centmll,dempotent in A is O.

rv

eA

Cn EN);

Proof Take p. q E J(A) such that q 1.. rand p rv q cv CA. There exist a, bE A with ab = q and ba = eA. (i) Define Pn = o,1!pbn (n EN). For n E N, we have = a1!pb n o,"pbn = Pn, bn(anp) = p, and (o,np)b" = Pn, and so Pn E J(A) and P1! r v P r v eA. For rn < n in N, we have PmPn = ampbTnanpb n = a m pa n - m pb1! = o,mpqan-mpbn = 0. Similarly PnPm = 0, and so (Pn) is an orthogonal sequence. (ii) Suppose that r E J(A) n3(A) and r is finite. Then qr E J(A) and qr ~ r. Since qr = a(br) and r = rba = (br)a, we have qr rv r. Thus r = qr because r is finite. Similarly r = rp. But now r = (rp)(qr) = 0, as required. 0

p;,

36

Algebrmc f01tndatwns

Proposition 1.3.23 Let A be a cornnmtat'ive algebra. Then: (i) \E(A)

Z8

a subalgebra of A;

(ii) for each finite subset T of J(A), there is a fimte orthogonal 87tbsd S of J(A) .mch that lin T c lin S; (iii) in the case where \E(A) is jinite-d,trnensumal, thf're is an or'tlwgonal subset {PI, .. ' ,Pn} ofJ(A) \ {O} wlnch,ts a basis for \E(A) and 1,$ s'uch that PI + ... +Pn is the zdentity of\E(A). Proof (i) Since A is cornrnutative, J(A)[2J of A.

c J(A), and so \E(A) is a subalgebra

(ii) The proof is by induction on the cardinality of the set T. The result trivially holds for subsets of cardinality 1. Assume that the result holds for all suh.,ets of J(A) of cardinality at most n. and let T = {p,Pl,'" ,p,,} be a subset of J(A). There is a finite orthogonal set S = {q], ... ,qm} in J(A) snch that lin {Pl' ... ,p,,} c linS. Set q = q] + .. '+qm' Then q E J(A) and qqj = qj (j E NmJ and so

U = {p - pq.pq], ... ,pqm' ql - pql,"" qm - pq",} is an orthogonal set of idempotents. Clearly Selin U and P E lin U. and so lin T c lin U. Thm; the resnlt holds for all subsets of cardinality at most n + l. (iii) This follows from (ii).

0

Definition 1.3.24 Let A be a unital algebra. A continued bisection of the identity for A is (], pazr {(p,,), (qn)} of seq71ences zn J(A) sneh that C.'l = PI + ql and, fOT each n E N, Pn = P,,+1 + (j,,+1 and APnA = Aq"A.

We note that, in this case, ]In ..1 q" (n E N) by l.3.20(iii) and that the set {qn : n E N} is orthogonal. Indeed, fix n E N. and assume that q"+J(j" = 0 for j E N k . Then

and similarly q,,(jn+k+] = O. Tlw result is proved by induction. It follows that PmP" = PmVn (rn, n EN). Definition 1.3.25 Let A and B be algebras. A homomorphism [anti-homomorphism] fmm A into B zs a linear map A --+ B snch that

e:

e(ab)

=

e(a)e(b)

[e(ab) = e(b)e(a)]

(a, bE A).

Let A and B be unital algebras. A homomorphism e : A --+ B is unital if e( eA) = eB. Let A be a non-unital algebra, let B be an algebra, and let e : A --+ B be a homomorphism. Then the map

e# : aeA

+ a f--t

aeB

+ 8(a),

A#

is a unital homomorphism. Let 8 : A --+ B be a homomorphism. Then 8(J(A)) and 8(3(A)) c 3(8(A)).

--+

B# ,

c J(B), 8(\E(A)) c \E(B),

37

Algebras, ideals, and fields

Let A and B be algebras. A homomorphism () : A --4 B is a rnonornO'rphzsm if it is injective, and in this case 0 is an ernbrdding of A into B. The homomorphism () is an epzrnoTph'lsrn if it is surjective. and an 1,8orrwTphz,ml if it is bijective. If there is an isomorphism 0 : A --4 B. then A and Bare isornorphl,c, and we write A ~ B. A homomorphism 0 : A --4 A is an endomorph1srn of A, and an endomorphism which is an isomorphism is an a1LtOTnorphisrn. Examples 1.3.26 (i) Suppose that A is a commutative algebra. Then the induced product map 7r A : A (9 A --4 A is a homomorphism.

(ii) The algebras qX] and coo(Z+) are isomorphic by an isomorphism that identifies X" and 871 for n E Z+. Let Sand T be non-empty sets. Then we have coo(S) (9 coo(T) ~ cuu(S x T); in particular, for each n E N, we can identify C n (9 C n with C n2 by setting (f (9 g)(i.j) = f(i)g(j) (i,j E N n ). (ii) Let n E N, and take a = (al, .... an) E C n and A = (Al .... , An) E (C7!) X. Then the rank-oIle operator a 0A in £(C n ) iH specified by the matrix (aiAj). and Mn ~ (9 (Cny ~ £(C n ). The trace map Tr : M" -+ C satisfies the condition that Tr(a cg A) = (n. A) = L.~1=1 ajA}. and HO

cn

n

Tr(A)

= Ln.jj

(A

= (nij)

E Mn).

(1.3.16)

j=l

The trace of an element in £( cn) is independent of the choice of the isomorphism with Mn. We see directly that Tr(AB) = TI'(BA) (A. BE Mrl)' 0 Let A l , A 2 , B l . B2 he algebras, and let 0 1 : Al ----> Bl and O2 : A z -+ B2 be homomorphiHHls. There iH a unique homomorphism 01 (9 ()2 : Al (9 A2 --4 Bl (9 B2 such that (0 1 ;g, ()2)(al (9 (2) = ()l (ad @ ()Z(a2) (a) E AI, a2 E Az). We have Mn(A) ~ M,,(IK) @A when A is a IK-algebra. Let A and B be algebras, and let A --4 B be a homomorphism. Then we have a homomorphism

o:

(1.3.17) where I,,, is the identity map on Mn(IK). Let E be a linear space. It is easily checked that the tensor algebra (® E, @) is characterized by the following universal property. Suppose that U is a unital algebra for which there is a linear map [ : E --4 U such that, for each unital algebra A and each linear map 1] : E -+ A. there is a unique unital hOIllomorphism U --4 A with 1] = 00 L Then U ~ ® E. Similarly, the symmetric algebra (V E, V) is characterilled as follows. Suppose that U is a unital algebra for which there is a linear IlIap I : E --4 U such that, for each unital algebra A and each linear map 1] : E ---; A with 17( x) I](Y) = r](y )1]( x) (T. Y E E). there is a unique unital homomorphism (j : U --4 A with 7] = () 0 /" Then U ~ VE.

o:

Definition 1.3.27 Let A be an algebra over lK. A character on A is a non-zem hornomorphisrn fmm A mto IK. The set of characters on A is the character space of A, denoted by

We regard
Algebrmc foundations

38

Note that ]\;[0 = A. Suppose that A is unital and that cp E CPA. Then cp(eA) = 1 and cp(a) #- 0 (a E Inv A). For each cp,1/; E CPA with cp #- 1/;, there exists a E A with '1'(a) = 0 and 1/;(a) = 1. It follows that CPA is a linearly independent set in A x .

Definition 1.3.28 Let A be a non-unital algebra. Set

+a

CPoo: (YeA

f-+

n,

A#

--+]K.

Clearly '1'x is a character on A # and ker'1'oe = A. For each cp E CPA, the map (WA + a f-+ 0: + cp(a) is a character on A#, and the map

'1'

f-+

'1' I A,

CPM

--+

CPA U {O},

is a homeomorphism; we regard cP A# as a subset of AX via this identification.

Proposition 1.3.29 Let A be a umtal algebra, and let>. E AX be such that A(eA) = 1 and A(a 2) = >.(a)2 (a E A). Then A E CPA. Proof Let x,y E A. Since >.((x +y)2) = (A(:r)

+ A(y))2,

we have

A(:ry + yx) = 2A(X)A(y). Further, (xy - yx)2

(1.3.18)

+ (xy + YT)2 = 2(x(yxy) + (yxy)x), and so, by (1.3.18),

(A(XY - YJ.:))2

+ 4A(X)2 A(y)2 = 4A(X)>'(Yxy).

(1.3.19)

Take a, b E A. and apply (1.3.19) with x = a - A(a)eA and y = b to see that >.(ab) = A(ba). It now follows from (1.3.18) that A(ab) = >.(a)>.(b), and so AECPA. D

Definition 1.3.30 Let A be an algebra for whzch CPA#- 0. For a E A, define 0:('1') = cp(a) (cp E CPA), and define

9 :a

f-+

0:, A

--+ ]K

•

Then 0: is the Gel'fand transform of a, and Q is the Gel'fand transform of A. We set A = {o:: a E A} = Q(A). Certainly Q : A --+ ]K

Definition 1.3.31 Let A be an algebra. A lmear subspace I of A I.S: a left ideal if A . I c I,. a right ideal if I . A c I,. and an ideal zf A . I + I . A c I. Thus, for us, an ideal is always 'two-sided'. The algebra A is an ideal in AD. The trivial ideals in an algebra A are 0 and A; a [left or right] ideal I is proper if I #- A. A linear subspace I of A such that AI = I A = 0 is an ideal in A; these ideals are the annihilator ideals of A. Let {Iv} be a family of left ideals in an algebra A. Then nIv and lin UIv are left ideals in A. Suppose that I and J are ideals in A and that n E N. Then I + J, I J, and In are ideals in A. Let p E J(A). Then clearly pIp = I n pAp.

Algebras, ideals, and fields

39

Let S be a subset of an algebra A. The left 'Ideal generated by S is the intersection of the left ideals in A which contain S; clearly this left ideal is equal to A # S = lin S + AS. A left ideal generated by a finite set {a 1 •.... an} is a finitely generated left ~deal; it is 'L,7=1 A#aj. A left ideal generated by a single element is It principal left zdeal. In the case where A is commutative, we speak of finitely generated ideals and of principal 1,deals. A commutative algebra in which every ideal hoi principal is a przncipal ideal algebra. An algebra is left Noetherzan if the family of left ideals satisfies the ascending chain condition; this is the case if and only if each left ideal is finitely generated. A commutative left Noetherian algebra is Noetherian. The ideal generated by a suhset S of an algebra A is the intersection of the ideals containing S; it is equal to A#SA#. Let A be a commutative algebra, and let a E A. Later, we shall often be concerned with the following specific ideal:

Ia = n{anA : n

E

N}.

( 1.3.20)

The ideal I" is the intersection of a chain of principal ideals. Let I he an ideal in an algebra A. The quotient space A/lis an algebra with respect to the product defined by the formula (a + 1) (b + I) = ab + I (a. b E A); the algebra A/lis the quotient algebra. We shall frequently use without specific mention the following standard isomorphism theorems for algebras. a

(i) Let A and B be algebras, and let e : A -----7 B be a homomorphism. Then -----7 (} ( A ), is an isomorphism.

+ ker e t-+ () ( a ) , A / ker e

(ii) Let A be an algebra, let B be a subalgebra of A, and let I be an ideal in A. Then b + I t-+ b + B n I, (B + 1)/1-----7 B/(B n 1), is an isomorphism. (iii) Let J be a [left] ideal in A with I c J. Then J / I is a [left] ideal in AI I. The map J t-+ J I I is a bijection from the set of [left] ideals of A which contain I onto the set of [left] ideals of AI I. Proposition 1.3.32 Let I be a left 'Ideal in an algebra A, and s'uppose that eEl is a left ident1,ty for I. Then the map (} : a t-+ !Le, A -----7 I, is a homomorphism. Let B be a 8ubalgebra of A such that b = 0 whenever b E Band bI = O. Then I B : B -----7 I is an embedding.

e

Proof Let a, bE A. Then be = ebe because be E I and e is a left identity for I, and now O(ab) = abe = aebe = O(a)O(b). The remainder is immediate. 0

Suppose that A is a commutative algebra. Then !)'leA) is an ideal in A, called the mlradical of A. A nil ideal in A is not necessarily nilpotent, but we do have the following result. Theorem 1.3.33 (Nagata-Higman) Let A be an algebra, and let n pose that an = 0 (a E A). Then A2 n - l = O.

E

N. Sup-

Proof The proof is by induction on ni the result holds trivially for n Assume that the result holds for n = k, and that ak+1 = 0 (a E A).

1.

40

Algebmic foundations

I:7=0 aiba"-i

First, set v(a, b) =

for a, bE A. For each t E IR-. we have

o = ~(a + tb)k+l

k-1

=

'11'(0., b)

+L

tjCj

)=1

for some C1,"" Ck-l E A, where we note that a k + 1 = bk + 1 = O. It follows that 1f;(a.b) = O. Now take a,b,c E A. Then akeb k = 0 because k

(k

+ l)a k cb k = akc L j=O

=

t i=O

aie

k

bJb k -

(t

j

k-1

= akc L fJJb k - J + L j=O

fJJa k - 1 bk -

aie'lj)(b, a k - 1 )

;=0

j)

j=O

k

= Llf'(a.cfJJ)b k - j = O. j=O

Set I = lin{A#akA#: a E A}. Then IAI = 0 and xk = 0 for each x E A/I. By the inductive hypothesis, A 2k - 1 C I, and so A 2k + 1 - 1 C IAI = O. The induction continues. 0

Proposition 1.3.34 Let A be a unital algebm with a continued blsectwn of the identlty. Then A has no proper- ldeal of fimte cod'tmens1On m A. Proof Let {(Pn), (qn)} be a continued bisection of the identity, and let I be an ideal of finite codimension in A. Since {qn : n E N} is linearly independent, there exist n E Nand a1 ..... an E C with an =f:. 0 such that I:~ GJqj E I. But then anqn = qn (I:~ a)qj) E I, and so qn E I. Since ApjA = AqjA and Pj-l = Pj +qj for j E N, successively p", Pn-1, q,,-l, .... P1, f' A E I, and so I = A. 0 A proper left ideal I of an algebra A is modular- if there exists 'U E A such that A(eA - 1L) C I; in this case,u is a right modular identity for I. An ideal I is modular- if there exists 71 E A such that A(eA - u) + (eA - u)A C I, and :';0 a proper ideal I is modular if and only if A/lis a unital algebra. Note that, if 1.1, is a right modular identity for a modular left ideal I, then u is not left quasi-invertible, For assume that there exist:,; a E A with 0.+ u - all = O. Then 'It = -(a - au) E I, and so A = I. which is a contradiction.

Definition 1.3.35 Let A be an algebm. A maximal [modular] [left] ideal in A is a maximal element in the famdy of pTOper- [modular-] [left] ideals in A. A minimal [left] ideal in A is a minimal element m the family of non-zeTO [left] ideals of A. Since each ideal containing a modular ideal is also a modular ideal, a maximal modular ideal is a maximal ideal. Let I be a left ideal with right modular identity '11" It is immediate from Zorn's lemma that the set of left ideals J such that I C J and 'It f}. J has a maximal member, say M. Clearly M is a maximal modular left ideal, and so

41

Algebras, zdeals, and fields

each proper, modular left ideal is contained in a maximal modular left ideal. Similarly, each proper, modular ideal is contained in a maximal modular ideal. Let I be an ideal in A. The map J f---> J I I is a bijection from the set of maximal modular [left] ideals of A which contain I onto the set of maximal modular [left] ideals of AI I. Let 1\1 be a maximal ideal in a unital algebra A, and let n E N. Then lH is the unique maximal ideal of A containing Afn. For take I to be an ideal in A with M n c I and I ~ AI. Then A = M + 1= (M + I)n C M n + I c I, and so

I=A. Example 1.3.36 Let E be a non-;!;ero linear space. It follows from (1.3.10) that FC(E) is an ideal in C(E). We claim that FC(E) is the minimum ideal in C(E). Indeed, let J be a non-zero ideal in C(E). Take S E r, take Xo,!lo E E· with S:r:o = !lo, and choose AD E EX with (yo. Ao) = 1. Consider T = :r:] ® Al E FC(E), and set Rl = Xl ® Ao and R2 = Xo ® AI. Then R l , R2 E C(E) and

(x

E

E),

so that T = R 1 SR 2 E J. Thus FC(E) c J. This establishes the claim. We see that, for each X E E, {T E C(E) : T:r = O} is a maximal left ideal in C(E) of codimension 1. and that every maximal left ideal has this form. D

Proposition 1.3.37 Let A be an algebra over'IK.

(i) The map cp f---> AI'f' is a b'\Jection from
(ii) A proper ideal I zn A# IKeA +1.

1,S

the kernel of a character on A# if and only zf

c

(iii) Suppose that M is an zdeal of codirnension one in A. M

Then elther

=

Proof (i) Let cp E
=

1.

(iii) There exists u E A with A = IKu 8 M. If u 2 E M, then A2 C M. If 71 2 E au + 1\1, where a E C·, then a-1u is a modular identity for M, and M = M

Algebraic f01L1ulatwns

42

Definition 1.3.38 A proper, modular ideal m a commutatzve algebra is primary 11 'it zs contained m a unique maximal modular ideal. Definition 1.3.39 A local algebra is a commutative, unital algebra which has a umque maximal ideal. The unique maximal ideal of a local algebra A is usually denoted by AlA. Clearly a commutative, unital algebra A iH local if and only if A \ IIlV A is an ideal in A, and in this case Inv A = A \ AIA, Note that we do not require that a local algebra be Noetherian. Let 1 be a primary ideal in a commutative, unital algebra A. Then AI I is clearly a local algebra with AIAIl = 1'11/1, where M if; the unique maximal modular ideal containing 1, For example. let AJ be a maximal ideal in A, and let n E N. Then Al n is a primary ideal, and AlAIn is a local algebra with maximal ideal MIMn. Definition 1.3.40 Let A be an algebra. A subset.6. of A Z8 a Mittag-Leffler set zf, for each sequence (an) in .6., there exists a E A such that n

2:: a1 ... ak E a1 ,., anA

(n E N), (1.3.21 ) k=1 The algebra A is a Mittag-Leffler algebra if A zs local and a Mzttag-LeJfier set. a-

Note that A is a Mittag-Leffler algebra if AlA if; a Ivfittag-Leffler set, and that, if A is a Mittag-Leffler algebra and (an) C 11'(1, then the element a E A which satisfies (1.3,21) is f;uch that n

a

-2::al"

·ak E a1" 'a n +1 .

IIlV A

(n EN).

(1.3.22)

k=1

for we have n a k=1

-2:: al'"

n+2

ak

-2:: a1'"

ak + a1'" an+1(eA + an+2) k=1 C a1 ... an+l . (CA + an+2A) Cal" . a n+1 . Inv A.

=a

For example, suppose that A iH a local algebra and that MA is nilpotent, say M;r+1 = O. Then a = L~l al ... ak satisfies (1.3.21) for (an) C MAl and so A is a Mittag-Leffler algebra. Further examples of Mittag-Leffler algebras will be given in 1.7.7(ii) and 2.2.18. Definition 1.3.41 Let A be an algebra. An zdeal P of A is prime zf it is a proper ideal and if either I c P or J C P whenever 1 and J are ideals in A with I J C P. An ideal Q of A is a semiprime ideal zf it is a proper ideal and 1 C Q whenever 1 is an ideal in A with [2 c Q. The algebra A is a prime algebra [semiprime algebra] if 0 is a prime [semiprime] ideal in A. Thus our convention is that A is not a prime ideal in itself. Suppose that P is a proper ideal in an algebra A such that either a E P or b E P whenever

Algebras, tdeals, and fields

43

ab E P. Then clearly P is a prime ideal in A. Let I be an ideal in A. Then the map P f--+ PI I is a bijection from the set of prime ideals in A which contain I onto the set of prime ideals in AII. For example, every algebra A which has a minimum, non-zero ideal I with /2 -lOis a prime algebra; by 1.3.36. this covers the case where A = £(E) for a non-zero linear space E.

Proposition 1.3.42 Let P be a proper ideal in A. Then the followzng condttions on P are equivalent: (a) P is a przme ideal; (b) either a E P or b E P whenever aAb C P; (c) either I C P or J C P whenever I and J are both left ideals in A wtth IJc P. Proof (a)*(b) Suppose that aAb C P. Then (AaA)(AbA) C P, and so either AaA C P or AbA C P, say AaA C P. But then, successively, (A#aA#)3 C P. A#aA# C P, and a E P. (b)*(c) Suppose that I and J are left ideals with IJ C P and J ct. P, and take bE J \ P. For each a E I, aAb C IJ c p, and::;o a E P. Thus I c P. (c)*(a) This is trivial.

D

Essentially the same argument as the above gives the following result.

Proposition 1.3.43 Let Q be a proper ideal in A. Then the following conditions on Q are equivalent: (a) Q is a semiprime tdeal; (b) a E Q whenever aAa C Q; (c) Ie Q whenever I is a left tdeal in A with 12 c Q. D We note the following trivial fact. Let I be a non-zero ideal in a semi prime algebra A, and take a E Ie. Then, by the above result, there exist::; b E A with aba =I O. and so (aI) n (fa) =I O. Let A be an algebra. An ideal in A is a mimmal prime ideal if it is minimal in the family of prime ideals in A. Let C be a chain of prime ideals in (A, c), and let 1= n{p : P E C}. Suppose that aAb C I. Then, for each P E C, either a E P or b E P, and so either a E I or bEl because C is a chain. By 1.3.42, I is a prime ideal. Thus it follows from Zorn's lemma that each prime ideal in A contains a minimal prime ideal. A non-empty subset 11;[ of an algebra A is an rn-system if, for each a, bE lvI, there exists x E A with axb E M. A multiplicative set is clearly an rn-system. By 1.3.42, A \ P is an rn-system whenever P is a prime ideal in A.

Proposition 1.3.44 Let I be an ideal in an algebra A. (i) Suppose that M is an m-system with In 11;[ = 0. Then there is a prime ideal P in A such that I C P and P n 11.1 = 0. (ii) Suppose that a E A and an f/:. I (n EN). Then there is a prime ideal P in A such that I C P and a f/:. P.

44

Algebrazc foundatwns

Proof (i) The set of ideals J in A such that J :=; I and JnM = 0 is non-empty, and so, by Zorn's lemma, it has a maximal element, say P. We prove that P is a prime ideal. Assume that a, b E A \ P with aAb c P. Then P + A#aA# and P+A#bA# are ideals in A strictly containing P, and so, by the maximality of P, there exist c, d E M with c E P + A#aA# and d E P + A#bA#. Take x E A with cxd E ]1.1. Then cxd E (P + A#aA#)A(P + A#bA#) c P, and so P n M i- 0, which is a contradiction. Thus P is a prime ideal in A. (ii) The set {an: n E N} is an m-system. 0 Corollary 1.3.45 Each maximal modular ideal in an algebra is a prime ideal. Proof Let !v! be a maximal modular ideal in an algebra A; take '/l to be a modular identity for M. Then un rt M (n EN), and so, by 1.3.44(ii), there is a prime ideal P in A with M c P. Since M is a maximal ideaL M = P. 0 Proposition 1.3.46 Let A be a commutatwe algebra. (i) A proper ideal P is a pnme ideal zf and only if ezther a E P or b E P whenever ab E P. (ii) dim(<E(A)j(<E(A) n P)) :::; 1 for each prime ideal P of A. (iii) A p'r'/,me ideal P in A is a mzmmal pnme ideal if and only zf, for each a E P, there eX'tst bE A \ P and n E N with anb = O. Proof (i) This is immediate from 1.3.42. (ii) First note that, if p E J(A) \ P, then. for each a E A, (a - ap)p = O. and so a - ap E P. Suppose that <E(A) r:t. P. and take Po E J(A) \ P. Then, for each p E J(A), either pEP or p - Po = (p - PPo) - (Po - Pop) E P, and so J(A) cPo + P. Thus <E(A) c OCpo 8 P and dim(<E(A)j(<E(A) n P)) = 1. (iii) Suppose that P satisfies the given condition, and let Q be a prime ideal in A with Q c P. Let a E P, and take b E A \P and n E N with anb = O. Since b rt Q, we have an E Q, and so a E Q. Thus Q = P, and P is a minimal prime. Conversely, suppose that P does not satisfy the given condition, and take a E P such that 0 rt S, where S = {aTib : b E A \ P, n E Z+}. Since S is a multiplicative set, by 1.3.44(i) there is a prime ideal Q in A with Q n S = 0. Since A \ PeS and Q C A \ S, necessarily Q C P. Take b E A \ P. Then ab E P \ Q, and so Q i- P. Thus P is not a minimal prime. 0 Definition 1.3.47 An algebra A is a domain if Ai- 0 and either a = 0 or b = 0 whenever ab = 0 in A. A commutative algebra which is a domain is an integral domain.

A complex algebra which is an integral domain is a complex integral domain, and a principal ideal algebra which is a unital integral domain is a principal ideal domam. Let A be a non-zero, commutative algebra. By 1.3.46(i), A is an integral domain if and only if A is a prime algebra. Let A be a local algebra which is an integral domain, and set S = (MA' .). Then S is a cone. The S-order on S is called the divisibility order, and is denoted by -<. Thus, for a, bEMA' we have a -< b if and only if bE aMA.

45

Algebras, ideals, and fields

Proposition 1.3.48 Let A be a complex integral domain, and suppose that a tS divisible in (A-, . ). Then there is a rational sermgroup (a Q : 0 E Q+-) in A w1,th a 1 = a. Proof First, suppose that x, yEA with xn = yn for some n E N, and let ( = e2i7r / n . Then (x - y)(x - (y) ... (x - (n-l y ) = 0, and so :1: = ryy for some 17 E 1[' because A is an integral domain. Now take n E N. There exist b, c E A with bn ' = c(n+l)! = a. We have cn+1 = (b for some ( E 1l', and so b = (7]c)n+l for some ry E 1l'. Thus we can inductively construct a sequence (an) in A with 1 = d. and an = a~t~ (71, EN). Define a P/ q = a~((q-l) I) for plq E Q+-. As in 1.2.20, a P/ q is well-defined, and (a P/ q : plq E Q+.) is a rational semigroup in A with a 1 = a. 0

a

Definition 1.3.49 A unital algebra in which each non-zero element is mvertible is a division algebra. A commutatwe diviswn algebra tS a field; it ts a real field or a complex field tf the algebra is real or complex, respectively. The identity of a field is often denoted by 1. Let A be a commutative algebra, and let I be a proper ideal in A. Then I is a prime ideal if and only if AI I is an integral domain, and I is a maximal modular ideal if and only if AI I is a field. For example, if A is a local algebra, then the quotient algebra. AlMA is a field; it is called the residue field of A. Definition 1.3.50 Let A be an algebra. Then A is a simple algebra tf A2 0 and A are the only ideals in A.

cI

0

liiUl if

Proposition 1.3.51 For each n EN, the algebra Mn(lK) is a stmple algebra. Proof Suppose that I is a non-zero ideal in Mn(lK), and take T = (Oij) E r, say 0rs cI 0, where r, s E N n . Since orsErs = ErrT Ess E I, necessarily E rs E I. Thus En = L~=1 EirErsEsi E I, whence I = A. 0 Proposition 1.3.52 (i) Suppose that A is a stmple algebra. Then aAb cI 0 whenever a, b E A-, and A is a prime algebra. (ii) A stmple, commutatwe algebra ts a field. (iii) Let M be a maxtmal ideal m a commutatwe algebra A. Then either M is modular or A 2 C AI. A maxtmal, pnme ideal in A is modular. Proof (i) Set I = {a E A : AaA = O}. Then I is an ideal in A, and I cI A because A2 cI A. Thus I = O. For a E A-, we have AaA cI 0, and so AaA = A. For a, b E A-, A2 = (AaA)(AbA), and so aAb cI O. By 1.3.42, A is a prime algebra. (ii) Let A be a simple, commutative algebra. By (i), aA cI (a EN), and so aA = A (a E A·). Fix a E A·, and choose e E A with ae = a. For each b E A, abe = ab, and so e is the identity for A. For each a E A-, there exists b E A with ab = e, and so A is a field. (iii) Clearly, either A2 C M or AIM is a simple algebra. By (ii), AIM is unital in the latter case, and so M is modular. 0

°

46

Algebraic foundations

Proposition 1.3.53 Let A be a commutative algebra, let I be an ideal in A, let P be a pnme ideal of the algebra f. and set

P= Then

P zs the

{a E A: af

c P}.

umque prime ideal Q in A such that I n Q = P.

Proof Certainly P is an ideal in A, and Pel n P. Take a E In P and ~ E 1\ P. Then ab E P, and so a E P. Thus In P = P, and, in particular, Pi=A. Let a, bE A \ P. Then there exist x, y E I with ax, by E 1\ P. Since P is a prime ideal in f, abxy E 1\ P, and so ab tf. P. Thus P is a prime ideal in A. Suppose that Q1 and Q2 are prime ideals in A with I n Q1 = I n Q2 = P, and take a E Q1 and bE f \ P. Then ab E In Q1 = Pc Q2 and b tf. Q2, and so 0 a E Q2. Thus Ql c Q2. It follows that P is uniquely specified. Corollary 1.3.54 Let A be a non-umtal. commutative algebra, let P be a pnme ideal in A, and set Q = {a E A# : aA C Pl. Then:

(i) Q is the unique prime ideal in A # such that Q n A = P; (ii) Q is a maximal zdeal in A# if and only if P is a maximal ideal m A. Proof (i) This is immediate from the proposition. (ii) Suppose that P is a maximal ideal in A. By 1.3.52(iii), P has a modular identity, say u. Since (eA - u)A C P, we have eA - u E Q, and so Q A. Let M be a maximal ideal in A# with M ~ Q. By 1.3.45, M is a prime ideal, and u tf. M, for otherwise eA E AI. So P c M n A i= A, P = M n A, and M = Q, showing that Q is a maximal ideal. Conversely, suppose that Q is a maximal ideal in A #, and let !vI be a maximal ideal in A with M ~ P. Set M = {a E A# : aA eM}. Then M is an ideal in A# with Q c M <; A#, and so Q = M. Thus M c !vInA = QnA = P, and hence P is a maximal ideal. 0

ct

Corollary 1.3.55 Let A be an integral domain. Then A# is an integral domain.

o Proposition 1.3.56 Let A be a non-zero, finite-dimensional complex algebra which 'is a domain. Then A has an identzty e, and A = Ce. Proof Let a EA·. The map b >---+ ab, A ---+ A, is injective hecause A is a domain, and it is bijective because A is finite-dimensional, and so there exists e E A· with ae = a. Since ae 2 = ae, we have e 2 = e, and so

e(eb-b)=(be-b)e=O

(bEA).

Thus eb = be = b (b E A), and e is the identity of A. Let a E A. Since A is finite-dimensional, there is a non-zero polynomial p over C with pea) = 0, say p = ao(X - ad'" (X - an), where a1,"" an E C and ao E C·. Then (a - ale)··· (a - ane) = 0, and so a = aJe for some j E N n . Thus A = Ceo 0

47

Algebras, tdeals, and fields

Corollary 1.3.57 Let A be a commntative complex algebra, and let P be a prime ideal with finite codimension in A. Then P is a maximal modnlar ideal of codimension one. Proof The quotient A/Pis a non-zero, finite-dimensiollal complex algebra which is a domain. 0

Let A be a commutative algebra over OC, and let S be a multiplicative set in A, so that S is a subsemigroup of (A, .). Let rv be the equivalence relation on A x S introduced in §1.2, and let 7r : A x S -+ A[S-l] and 'Ib : A -+ A[S-l] be as in Definition 1.2.8, so that A[S-l] is a unital, abelian semigroup, and 'I/J is a

morphism. 'Ve now set (a, s)

-

+ (b, t)

= (at

+ bs, ."It)

(a, b E A,

.9,

(1.3.23)

t E S) .

It is easily checked that + is well-defined and that (A x S, +) is an abelian semigroup. Suppose that (aI, sd rv (a2, S2) in A x S -say, ralS2 = ra2S1 for some rES-and that (b, t) E A x S. Then r(alt + bsds2t = r'(a2t + bS 2 )Slt, and so (aI, sd + (b, t) rv (a2, S2) + (b, t). This shows that rv is compatible with (A x S, + ), and so the formula 7r((a, 09))

+ 7r((b, t))

= 7r((at

+ bs, st))

defines an operation on A[S-I] with respect to which A[S-I] is a scmigroup. The map 'Ij;: (A,+) -+ (A[S-I],+) is a morphism. Let e = 7r((09, s)), as in 1.2.9(i). Suppose that 0 ¢ S. Then rs2 1= 0 for each rES, and so e 1= O. Thus, in this case, e is the identity of A[S-l]. Set ct:7r((a,s)) = 7r((aa, 05)) for (a,s) E A x S and a E K It is easily checked that this product is well-defined and that A[S-l] is an algebra over OC. The following result is an immediate consequence of 1.2.9. Proposition 1.3.58 Let A be a commntative algebra over' OC, and let S be a mnltiphcative s'ubset of A·. Then:

(i) (A[S-I]. +, -) is a commutative, unital algebra over K, 'Ij; : A is a homomorphzsm, and '¢(S) c Inv A[S-l]; (ii) for each x E A[S-I], there exists a E A and s E S wzth x = (iii)

'ljJ

-4

A[S'-I]

'Ij;(a)1/-{~)-I;

is an injection in the case when: A is an integral domain.

0

The element 7r((a, 09)) of A[S-l] is traditionally written as a/so In this notation, the algebra operations of A[S-I] are a

:; +

b

at

+ bs

t = -s-t-'

a

b

s

t = st'

ab

a

a:; =

aa --:;.

(1.3.24)

An important application of the above construction is to the case where Sis the complement of a prime ideal in the algebra A. Definition 1.3.59 Let A be a commntative algebra, and let P be a prime ideal in A. Then the algebra A[(A \ P)-I] is the localization of A at P.

48

Algebraic foundations

Let B = A[(A \ p)-I], and set P[(A \ p)-I] = {als E B : a E Pl. Then P[(A \P)-I] is an ideal in B, and, by 1.3.58(i), B\P[(A \p)-I] c Inv B. Thus B is a local algebra with maximal ideal P[(A \ P)-I]. Let A and B be commutative, unital algebras, let P be a prime ideal in A, and let () : A --+ B be a unital homomorphism with ()(A \ P) c Inv B. Then it is clear that there is a unique homomorphism cp: A[(A \p)-I] --+ B with cpo'ljJ = (); the formula for cp is cp(als)

= ()(a)()(s)-I

(a E A, sEA \ P).

(1.3.25)

A particularly important example of a localization is the construction of a quotient field. Let A be an integral domain over IK, so that ais a prime ideal in A. Then the localization A[(N)-I] is a field over IK, and the map 'ljJ : A --+ A[(N)-I] is an embedding. Definition 1.3.60 Let A be an mtegral domain. Then the field A[(Ae)-I] zs the quotient field of A. The quotient field of A is also called the field of fractions of A. Let A be a subalgebra of a field K. Then A is an integral domain, and the quotient field of A is equal to the subfield {alb: a E A, bE Ae} of K. Definition 1.3.61 Let K = (K, +, .) be a real field with a bmary relation ::;. Then K = (K, +, .,::;) is an ordered field if (K, + , ::;) is a totally ordered group and if ab E K+ whe~ever a, b E K+. Let K be an ordered field, and let a E K. Then a zs: (i) an infinitesimal if a = 0(1); (ii) finite zf a = 0(1); (iii) infinitely large if it zs not finite. Thus a is infinitely large if and only if lal > n (n EN). The sets of infinitesimals and of finite elements in K are denoted by

KO

and

KO#,

respectively. Clearly KO and K O # are subalgebras of K, and KO is an ideal in KO#. Let f K be the value set of the ordered group (K, +, ::;), with the archimedean valuation v, as in 1.2.23, and define + on fK by the formula v(a)

+ v(b) = v(ab) (a, bE

Ke).

+ is well-defined and that (f K, group; it is called the value group of K. We have v(l)

It is easy to check that

KO = {a E K : v(a) > a}, so that KO#

=

R1

+, ::;) is a totally ordered = a and

KO# = {a E K: v(a) ~ O},

+ KO.

Definition 1.3.62 Let K be an ordered field. Then K is: (i) an Cil-fieid [1]l-field] if (K, :s:) is an Cil-set [1]l-set]; (ii) a i3I-field if K Cil -subfields of K.

= U{Kv :

/.I

<

wd,

where {Kv :

/.I

<

wd

is a chain of

Algebras,

~deals,

49

and fields

Let U be a free ultrafilter on N. As in §1.1, define f "'U 9 for f, 9 E eN if {n EN: fen) = g(n)} E U, so that "'U is an equivalence relation on eN; the class containing f is denoted by [flu or [fl. For f, 9 E eN and a E e, define

[f]

+ [g] =

[f

+ g],

a[l] = [af],

[I][g] = [lg].

Then eN ju is a complex algebra with identity [1] for these operations; we regard ]RN ju as a real sub algebra of eN ju. It is directly clear that eN ju is a complex field and that ]RN ju is a real field; further, (]RN ju, S:.u) is an 1Jl-fie1d. The fields ]RN ju and eN ju are the (real and complex) ultrapowers by U of JR and e, respectively. We present some important examples of ordered fields. Definition 1.3.63 Let G be a totally ordered abelian group. For f, 9 E ;J(lK, G), define (f * g)(t) = '2)f(r)g(s) : r, s E G, r + s = t} (t E G). In fact, (f * g)(t) is given by a finite sum, and f * 9 belongs to ;J(lK, G). It is easy to verify that (;J(lK, G), *) is a commutative, unital algebra, and that ;J(l) (lK, G) and ;J(O) (lK, G) are unital subalgebras of ;J(lK, G). For example, let s, t E G. Then 88 *8t = 8s + t . The algebra ;J(G) is defined to be the formal power series algebra or Hahn algebra over G, and ;J(O)(G) = coo(G) is the algebraic group algebra on G. Theorem 1.3.64 Let G be a totally ordered abelian group.

(i) ;J(JR, G) is an ordered field, and ;J(1) (JR, G) lS a subfield of ;JC'R, G). (ii) The ordered group G is the value group of both ;J(JR, G) and ;J(1) (JR, G). (iii) Suppose that G is a f31 -group. Then ;J(1) (JR, G) is a f31-field. (iv) ;J(G) is a complex field; it is the complex~fication of ;J(JR, G). 0 Let S be a subsemigroup of G. Then we regard ;J(lK, S) and ;J(1)(lK, S) as subalgebras of ;J(lK, G). We identify the algebra of finite elements in the field K = ;J(JR, G) as

K O # = ;J(JR, G+) = {J E K : v(f) ~ O} . By analogy, we define ~(G)O# = ;J(G+), but we note that, in the case of this complex field, there is no intrinsic way of defining this subalgebra. The following fields will playa very important role later; the groups GO' and G were defined in 1.2.27. Definition 1.3.65 For each a < WI, define RO" = ;J(IR, GO') and CO' = ;J(GO") (with Ro = Co = {O}), and define R = ;J(I)(IR, G) and C = ~(I)(G). The maximal ideals Co and C~ of CO# and C~# are denoted by M and MO", r'espectively, and we write M# and Mff' for CO# and C~#, respectwely. Clearly we have R = U{RO" : a < wr} and C = U{CO" : a < wr}, and hence M = U{MO' : a < wr}. The field C is the complexification of R.

Algebraic foundations

50 Theorem 1.3.66 (i) For each a <

WI,

Ra is an aI-field.

(ii) The field R is a (31 -TIl -field. (iii) The algebra M contains a non-zero, real semigroup.

Proof (i) This is immediate from 1.2.26(i) and 1.2.28(i). (ii) Since R is an TIl-field.

= U{Ra : a < WI}, R is a (31-field; by 1.2.26(i) and 1.2.28(ii), R

(iii) Take 8 E G +•. Then the semigroup {as: a E JR +.} is a subset of G +. , and {bas: a E JR+.} is a non-zero, real semigroup in M. D Notes 1.3.61 Almost all the results of this section can be found in the following standard texts: (Hungerford 1980), (Jacobson 1974, 1980), (McCoy 1973), (Palmer 1994), (Pierce 1982). For 1.3.9, see (Hungerford 1980, VII.3). Our 'algebra' is strictly an associative algebra; there is a theory of non-associative algebras. For an account of n-linear maps and of tensor products, including the tensor algebra and the symmetric algebra, see (Greub 1978); for 1.3.11. sec (Pierce 1982, 9.2a). The algebraic group algebra of a group G is often denoted by IKG in the algebraic literature. The definition of a Mittag-Leffier algebra is new; the name is suggested by a connection, to be explained later, with the Mittag-Leffier theorems of Appendix 1. The ideal P of 1.3.53 is minimal if and only if P is minimal; the ideal Q of 1.3.54 is equal to P if and only if P is not a modular ideal in A. The localization of a commutative algebra at a prime ideal is discussed in (Hungerford 1980, IlIA), where the details that we have omitted are given. For an extensive discussion of ordered fields, including the fields ~(lR, G) and ~(J)(lR,G), see (Dales and Woodin 1996, Chapter 2); 1.3.64 follows from (ibid., 2.15). Two fields related to R are fully discussed in (ibid.): these are R = ~(IR, G) and

R=

{J

R : supp f n (-00, xl

E

is countable for each x E G} .

Each of these fields is a real-closed Til-field which has G as its value group. Let w denote the weight of an ordered field, as in (ibid.), so that w(S) is the minimum cardinality of an order-dense subset of a totally ordered sct (S, S). Then IRI = w(R) and w(R) = c, and IRI I

I

=

w(R) = 2NJ. The field

R

=

c.liil = 2NJ

is the 'Cauchy completion' of

R; it is claimed in (ibid., p. 66) that Ii is the natural analogue of the real line lR 'one cardinality higher'. No two of the three fields R, Ii and R are mutually orderisomorphic. 1.4

MODULES AND PRIMITIVE IDEALS

We now turn to the theory of modules over an algebra A. We shall define left and right A-modules, and also A-bimodules. The theory of left A-modules is equivalent to the theory of representations of A on linear spaces; we shall usually use the language of modules. The main theorems to be proved are 1.4.31 and Jacobson's density theorem 1.4.32; these results are the classical foundations of the subject. The primitive ideals of an algebra are the kernels of its 'simple' representations; we shall describe the hull-kernel topology on the set of primitive ideals.

Mod1Lles and primitwe Ideals

51

Definition 1.4.1 Let A be an algebm over lK. A left A-module is a linear SP']('C E over lK together with a Inlmear map (a, x) ....... a . x, A x E -+ E. such that

a . (b· x) = o,b . x

(a,b E A, x E E).

A right A-module zs a lmear space E over lK togethcT WIth a bilinear map (a,x) ....... x . a, A x E -+ E, snch that (x . a) . b = x . o,b

(a, bE A, x E E).

An A-bimodule is a linear space E together with maps with respect to which it is a left A -module and a right A -modnle and which are .mch that

a . (x . b) = (a . :1:') . b (a, bE A, x E E) .

( 1.4.1)

Trivially a lK-linear "pace is a lK-bimodule. The space {O} is an A-himodnl(;, denoted by O. For example, an algebra A is itself an A-bimodule with respect to the map!'> given by the product in A, and a left ideal in A is a left A-module. If A is a subalgebra of an algebra B, then B is an A-bimodule with respect to the product in B. Let <.p, tf; E A U {O}. Then lK is an A-bimodule for the operations a . z = <.p(o,)z,

z·

0,=

'lj!(o,)z

(a E A, z E lK);

(1.4.2)

this module is denoted by lKcp,,p, and we write lKcp for lK-{'.cp' Each one-dimen"ional A-bimodule has the form lKcp,1/I for some <.p,'ljJ E A U {O}. Let A be an algebra, and let E be a left A-module. Then E is a right AOP_ module for the same module operation. vVe shall give some definitions and results for left A-modules; similar definitions and results apply to right A-modules and to A-bimodules. Let A be a unital algebra. A left A-module E is 'unital if eA . x = x (x E E), and an A-bimodule E is unital if eA . x = x . eA = x (x E E). Let A be an algebra, and let E be a left A-module. \Ve make the convention that aO • x = x (a E A, x E E). We set

(a, a) . x

= ax + a

. x

(0: E lK, a E A, x E E);

(1.4.3)

E is a now unital left AI> -module. Let A be an algebra, and let E he an A-bimodule. Then E is symmetric (or commutati11C) if a . x = x . a (a E A. x E E) . For example, the A-himodule lKcp,,p is symmetric if and only if <.p = 'l/!. Definition 1.4.2 Let A be a commutative algebm. An A-module A-bimodule.

IS

a symmetnc

Suppose that A is commutative and E is a left A-module, and define x . a = a . x (a E A, x E E). Then E is an A-module. Let A be an algebra, let E be a left A-module, and let Sand T be non-empty subsets of A# and E, respectively. Then S . T = {a .

X :

a

E

S, x

E

T},

ST

= lin S

. T.

52

Algebraic fov,ndations

We write a . S and as for {a} . Sand {a}S, respectively. This notation is consistent with our previous notation in the ca..<;e where E is a left ideal in A. A linear subspace F of a left A-module E is a submodule of E if A . F c F. For example, A . x = Ax iH a Hubmodule of E for each x E E. Let S be a subHet of E. Then the interHection of the submodules of E containing S is the submodule generated by S; it is equal to A # S. A submodule F of E is finitely generated if there is a finite subset S of E such that F = A# S. Let F be a submodulc of E. Then ElF is a left A-module for the operation a . (x

+ F)

= a . x

+F

(a E A, x E E);

ElF is the quotient module. In particular, AI I is a left A-module for each left ideal I in A. Let I be an ideal in A. and let F be a submodulc of E such that lEe F. Then ElF iH a left (AI I)-module for the operation (a

+ I)

+ F)

. (x

= a . x

+F

(a E A, x E E) .

Let {E, : 'Y E r} be a family of left A-modules. Then A-module for the operation a . (x/)

= (a . x,) (a

E

TI,Er' E,

A, x, E E/),

and O,Er E/ is a submodule of TI E/. For example, for each n E N, left A-module; modules of this form are the free left A-modules.

..

is a left

A(n)

is a

Definition 1.4.3 Let A be an algebra. A left A-module E is neo-unital A . E = E.. and an A-bimodule E is neo-unital ~f A . E . A = E.

~f

Thus A itHelf is neo-unital if and only if it factors. Certainly a unital module over a unital algebra is neo-unital, but we shall see later that there are important non-unital examples. Definition 1.4.4 Let A be an algebra, and let E be a left [right] A-module. The annihilators of a subset S of E are s.L

= {a

E

A :a . S

= O} and

ST

= {a

E

A :S . a

= O},

respectively. We often write x.L for {x}.L. For each SeE, S.L is a left ideal in A; if S is a submodule of E, then S.L is an ideal in A. In the case where E.L = A, E is a (left) annihilator module; we say that A acts trivzally on the left. We may take E to be an ideal I of A in the above definition; we then have

= O}) IT = {a E A : I a = O}) = {b E A : ab = O} (a E A) , for example, and so A is a domain if and only if A -# 0 and a.L = 0 (a I.L

a.L

= {b

= {a

E A : ba

E A : aI

= O},

aT

E

Ae).

Definition 1.4.5 Let A be an ideal in an algebra B. Then B is left faithful over A if {b E B : bA = O} = 0, right faithful over A if {b E B : Ab = O} = 0, and faithful over A if {b E B : bA = Ab = O} = O. An algebra A is [left], [right] faithful if it is [left], [right] faithful over itself.

Modules and primitive ideals

53

Thus A is left faithful if and only if A~ = 0, and A is faithful if and only if A~ nAT = 0 (so that A is faithful if it is left or right faithful). For example, a semiprime algebra is left and right faithful. Let E be an A-bimodule. We shall write (I - eA) . E = {x - eA . x : x E E}, etc. Then

E = eA' E· eA 8 (I -eA) . E· eA 8 eA' E· (I -eA) 8 (I -eA) . E· (I -eA); (1.4.4) here eA . E . eA is a unital A-bimodule, and A acts trivially on either the left or the right on the other three bimodules. Definition 1.4.6 Let A be an mtegral domain, let E be a left A-module, and set E t = {x E E : x~ =j:. a}. Then E t zs the torsion submodule of E; E zs a torsion module if E t = E, and E is torsion-free if E t = O.

Clearly the torsion submodule E t is a submodule of E. Definition 1.4.7 Let A be an algebra, and let E be a linear space. A representation of A on E is a homomorphism p : A ---> £(E). The representatzon is: faithful if P is a monomorphism; of dimension n if dim E = n; finite-dimensional if E is Jinzte-dimensional.

Let E be a left A-module, and set p(a)(x) = a . x

(a E A, x E E).

(1.4.5)

Then P is a representation of A on E. Conversely, if P is a representation of A on a linear space E, then equation (1.4.5) defines a module operation with respect to which E is a left A-module. In this situation, we shall change freely between speaking of modules and of the corresponding representations. The situation for right modules is the same, save that, if p(a) (x) = x . a, then p is an anti-homomorphism from A and a representation of AOP on E. Let I be a left ideal in an algebra A. Then the corresponding representation on AI I is the left regular representation on AI I. Let E be an A-bimodule, and set pe(a)(x) = a . x and Pr(a)(x) = X • a for a E A and x E E. The 'representation form' of (1.4.1) is (1.4.6)

In the case where A is a commutative algebra, S

= alg {IE, pe(A), Pr(A)}

(1.4.7)

is a commutative, unital subalgebra of £(E).

Examples 1.4.8 (i) Let E be a linear space. Then E is a left £(E)-module for the map (T,x) f---> Tx, £(E) x E ---> E. (ii) Let A and B be algebras, and let () : A ---> B be a homomorphism. Then B is an A-bimodule for the maps defined by a . b = ()(a)b,

b· a = b()(a)

(a E A, bE B) .

o

5-1

Algebraic foundatwns

Examples 1.4.9 Let A be an algebra.

(i) Let E be a left A-module or a right A-module. Define A . a and a . A in EX for a E A and A E EX in the two cm;es by

(:1'. A . a) = (a . :1'. A),

(x, a . A)

= (x . a, A)

(x E E).

(1.4.8)

Then EX is, respectively, a right A-module for the map (a, A) I--> A . a and a left A-module for the map (0, A) I--> a . A. In the case where E is an A-bimodule, EX is also an A-bimodulc for the given maps. The space EX with these maps is the alyelrraic dual module of E. (ii) Let E and F be left and right A-modules. respectively. For each a E A. the maps (x,y) I--> (a· x)0y and (x,y) I--> X0(y· a) from Ex F into E0F are bilinear, and so there exist pt(a), Pr(O) E £(E0F) with Pe(a)(x0y) = (a· x)<'5J)y and p,(0)(x0y) = x0 (y. a) for x E E and y E F. The maps a

are

It

f--+

pe(a),

a

I-->

Pr(a),

A ~ £(E 0 F),

homomorphism and anti-homomorphism, respectively, and (1.4.6) holds.

It followH that E 0 F is an A-bimodule for the products specified by

a· (x0y)

=

(a· x)0y, (x0y)· a

=

x0(y· a)

(a E A, x E E, Y E F). (1.4.9)

The module E 0 F is the tensor product of E and F. We write a . x Q9 y for a . (x 0 y), etc. In particular, A Q9 A is an A-bimodule; for a E A and u E A 0 A, the module products a . u and u . a are the products (a Q9 CA)U and u(eA Q9 a), respectively, in the algebra A Q9 A. (iii) Let E be a right and F a left A-module, respectively, and set N = lin {x . a 0 y - x 0 a . y : a E A, x E E, y E F}.

Then E0A F denotes the quotient space (E0F)/N. Now suppose that E is an A-bimodule. Then N iH a submodule of the left A-module E Q9 F, and so E Q9A F is a left A-module. (iv) Let E be an A-bimodule. For a,b E A#, set P(a.b) (x) = a· X· b (x E E). Then P(a,b) E £(E), and (a, b) I--> P(u,b), A# x A# ~ £(E), is bilinear, and HO there is a linear map p: A# Q9 A# ---> £(E) with p(a0b)(x)=a·x·b

(a,bEA#.XEE).

The map P is a homomorphism from the algebraic enveloping algebra A# Q9A#op into £(E), and so E is a left A# Q9 A#oP-module for a product that satisfies the condition that (a Q9 b) . x = a . x . b (a, b E A #, x E E). In the case where A is unital and E is a unital A-bimodule, E becomes a unital left A Q9 AOP-moduie. 0 Definition 1.4.10 Let A be an algebra, and let E and F be left [nght] Amodules. A map T E £(E, F) is a left [right] A-module homomorphism if T(a . x) = a . Tx

[T(x· a) = Tx . a]

(a E A, x E E).

(1.4.10)

Let E and F be A-bimodules. A map T E C(E, F) is an A-bimodule homomorphism if it is both a left and right A-module homomorphism.

55

Modules and przmitwe zdeals

The linear spaces of left and right A-module and A-bimodule homomorphisms are denoted by AL(E, F).

LA(E, F),

and

ALA (E. F),

respectively: we write AL(E) for AL(E, E), etc. Note that ALA (E) is a unital subalgebra of L(E). Suppose that A is non-unital and that T E AL(E, F). Then E and Fare unital left A#-modules and T E A#L(E, F). Let E, F, and G be left A-modules. For T E AL(E, F). there is a linear map LT : S,....... SoT,

AL(F, G)

--->

/IL(E. G) .

(1.4.11 )

The terms monomorphism. isomorphism, etc., are defined in the obvious way for module homomorphisms. Two left A-modules [A-bimodulesJ E and Fare isomorphzc if there is a left A-module [A-bimoduleJ isomorphism from E onto F; we write E 9:' F. We shall use the standard isomorphism theorems for modules which are analogous to those for algebras; for example, if T E AL(E, F), then kerT is a left A-module and E/kerT 9:' T(E). Let E and F be A-bimodules, and set B = A ® AOP. Take T E ALA(E, F). Then T E BL(E, F). The converse holds in the case where E is neo-unital. Example 1.4.11 Let A be an algebra, so that A ® A is an A-bimodule. The induced product map ?fA : A®A ---> A is an A-bimodule homomorphism, and so ker ?fA is a su bmodule of A ® A. In fact, ker?fA is a left ideal in A® AOP: if ?fACE?=l aj ®bj ) = 0 and a, bE A, then ?fA ((a ® b)(E]=l aj ® bJ )) = a(E]=l ajbj)b = O. The ideal ker?fA is called the diagonal ideal; ker?f A is an ideal in A ® A when A is commutative. Let S be a non-empty set. Then we have coo(S) ® coo(S) = con(S x S), and the diagonal ideal is {F E coo(S x S) : F(s, s) = 0 (8 E S)}. 0 Definition 1.4.12 Let A be an algebra, and let E and F be left A-modules. For a E A and T E L(E, F), define (a . T)(x)

=a

. Tx,

(T x a)(x)

= T(a

. x)

(x E E).

(1.4.12)

Now let E and F be right A-modules. For a E A and T E L(E, F), define (T . a)(x)

= Tx

. a,

(a x T)(x)

= T(x

. a)

(x E E).

(1.4.13)

Certainly a . T, T x a, T . a, and a x T are linear maps. Suppose that E and F are left A-modules. Then T is a left A-module homomorphism if and only if a· T = T x a (a E A). Also, L(E, F) is an A-bimodule with respect to the maps (a, T) ,....... a . T,

(a, T) ,....... T x a,

A x L(E, F)

--->

L(E, F).

(1.4.14)

Similarly, if E and F are right A-modules, then L(E, F) is an A-bimodule with respect to the maps (a, T) ,....... a x T and (a, T) ,....... T . a. Suppose that E and F are left A-modules. In general, a . T ~ AL(E, F) for a E A and T E AL(E, F). However, we do have the following result. Proposition 1.4.13 Let A be a commutative algebra, and let E and F be Amodules. Then AL(E, F) is an A-module for the map (a, T) ,....... a . T. 0

56

Algebrazc foundations

Proposition 1.4.14 Let A be an algebra.

(i) Let E be a left A-module, and let F be a rzght A-module. (E 0 FY ~ C(E, FX) as A-bzmodules.

Then

(ii) Let E be a right A-module, and let F be a left A-mod'ule. (E0A F)X c::: CA(E,F X ) as lzneaT spaces.

Then

Proof (i) For

T

E (E 0 F)x. define TT : E

(y, TTX) = (x 0 y. T)

---+

F X by

(x E E. y E F) .

The map '1: T f---+ TT' (E0 F)X ---+ C(E,FX), is a linear isomorphism. Now take a E A and T E (E 0 F)x. For each x E E and y E F, we have

(y, TaTX)

= (J:0y, a·

T) = (x0y· a. T) = (y. a, TTX) = (y.(a· TT)(X)).

and so Ta T = a . T r . Similarly TT a = TT X n. and so 11 is an A-bimodule homomorphism. (ii) The map 17 : T f---+ Tn (E 0A F)X ---+ LA(E,F X ), is now a linear isomorphism. D Definition 1.4.15 Let A be an algebra, let (En: n E Z) be a seq1tenCe of left A-mod'ules, and let Tn E AL(En+l, En) fOT n E Z. The sequence

(1.4.15) is a complex of left A-modules zf it zs a complex; L splits if it is exact and if, faT each nEZ, there is a submodule Fn of En such that En = Tn(En+d 0 F". It is immediately checked that the complex splits if and only if, for each nEZ, there exists a map Qn E AC(En , En+d with Qn-1 0 Tn- 1 + Tn 0 Qn = h: n • Similar definitions apply to sequences of right A-modules and of A-bimodules, Consider a short exact sequence of left A-modules: '~ "

:0

----+

E

S ----+

F

T ----+

C

---+

°.

It is easy to check that the following conditions are equivalent: (a) the sequence L splits: (b) there is a submodule H of F such that F = SeE) 0 H; (c) there exists Q E AL(C, F) with To Q = Ie (T has a right inverse): (d) there exists R E AL(F, E) with R 0 S = Is (S has a left inverse); (e) there exists P E AL(F, F) with p 2 = P and P(F) = kerT = See). For example, suppose that (e) holds. Given y E C, take x E F with Tx = y, and set Qy = x - Px. If also Tx' = y, then T(x' - Px') = y = T(x - Px) because TP = 0, and so (x' - Px') - (x - Px) = pz for some z E F. But then pz = p 2 z = and so x' - Px' = x - Px. This shows that Q is well-defined. Clearly Q E AL(C, F) and To Q = Ie, and so (c) holds. The algebraic dual complex to the complex I: of (1.4.15) was defined in (1.3.3); it is a complex ofright A-modules with T;: E LA(K:;,E::+ 1 ). It is easy to check that LX splits if L splits.

°

Modules and primztive zdea18

57

Definition 1.4.16 Let A be an algebm, and let E be a left A-module. Let a E A. Then E Z8 a-divisible if a . E = E. Let U be a subset of A. Then E zs U -divisible if a . E = E for each a E U \ {o}. The mod'ule E zs divisible tf it zs A-divi8ible and injective if, for each left A-modnle G and each 8nbmodule F of G, each

S

E AC(F,E) has an e.T-tension to T E AC(G,E).

Suppose that E, F. and G are left A-modules with E = F 8 G such that E and G are divisible. Then F is divisible. Suppose that E is an injective left A-module. Then clearly every short exact sequence 0 --> E --> F --> G - t 0 of left A-modules splits, and so there is a submodule H of F such that F = E 8 H. Let A be an integral domain, and suppose that E is a divisible left A-module. Then it is clear that the torsion submodule E t is also divisible. Lemma 1.4.17 Let A be an algebm, and let E be a left A-modnle. For each U c A, E contmns a maxzmnm U -diV'tsible snbmodule.

Proof The family F of all U-divisible submodule:o of E is non-empty because it contains {a}. Set F = lin{G : G E F}. Then F is aU-divisible submodule. and F ::> G (G E F). Thus F is the maximum member of F. 0 Let A be an algebra, let E be a left A-module, and let (an) be a sequence in A. Then

lima1" +--

.a

n .

E

=

{x

E

E: t~~e exists (X1~ in E ,such that }. (1.4.16) ], Xl and Xn - an . Xn +1 (n E N)

Thw;, in the the notation of §l.l,

~

a] ... an . E is

1l'1 (lim

proj{E; en}). where

n:=l

en(x) = an . X (x E E). Note that ~al" ·an . E = a1' ··an . E in the case where E is torsion-free. Let a E A. Then clearly ~ an . E is the maximum a-divisible submodule of E. Let A be a unital integral domain, and let E be a unital A-module. If E is injective, then E is divisible. For let a E A·, and take x E E. Define T : ab I---> b . x, aA --> E. Then T E AC(aA, E), and so T hru; an extension T E AC(A, E). We have x = T(aeA) = a . T(eA) Ea· E, and so E = a . E, as required. We shall show that the converse of this result is true in the case where A is a principal ideal domain. Lemma 1.4.18 Let A be a unital algebm, and let E be a unztallejt A-modnle. Suppose that each A-module homomorphism from a left zdeal of A mto E has an extension to an A-modnle homomorphi8m from A into E. Then E i8 an injective modnle.

Proof Let G be a left A-module, let Fo be a submodule of G, and let To E AC(Fo, E). Let F be the family of pairs (F, T), where F is a submodule of G with F ::> Fo , T E A£(F, E), and T I Fo = To. Set (Fl. Td ::5 (F2' T 2 ) in F if F1 c F2 and T2 I F1 = T 1. Then (F,::5) is a partially ordered set, and each chain in (F,::5) has an upper bound. It follows from Zorn's lemma that (F,::5) . has a maximal element, say (F, T).

58

Algebraic foundatzons

Assume towards a contradiction that F i= G, and take :1: E G \ I = {a E A: a . x E F}, so that I is a left ideal in A and define ():a ..... T(a.x),

F.

Set

I-+E.

Then () E A£(I, E), and so, by hypothesis, (I has an extension B E A£(A, E). Define H = F + A . x and S : y + a . x ~ Ty + a . B(eA)' H -+ E. If YI + al . x = Y2 + (L2 . x in H, then T(YI - Y2) = {}(a2 - ad = (a2 - al) . B(eA), so that S is well-defi~ed. Clearly S E A£(H, E) and (H, S) ~ (F, T) in F, a o contradiction. Thus F = G, and E is injective. Theorem 1.4.19 Let A be a prmcipal ideal domam, and let E be a umtal left A-module. Then E is injective 2f and only 2f it is dwisible. Proof We must show that E is injective in the case where it is divisible. Let I be a non-zero ideal in A, and take T E A£(I, E). Since A is a principal ideal domain, there exists ao E A· such that I = aoA. Since E is ao-divisible, there exists x E E with ao . x = Tao. Define T : a ~ a . x, A -+ E. Then T E A£(A, E) and T I I = T. It follows from 1.4.18 that E is injective. 0 Corollary 1.4.20 Let A be a princ2pal ideal domam, and let E be a umtal left A-module. Suppose that E 2S dwisible. Then there is a torsion-free, divisible, unital left A-module F such that E = F 8 E t . Proof Since E is divisible, E t is divisible, and so, by 1.4.19, E t is injective. Thus there is a left A-module F such that E = F8Et . Clearly F is torsion-free, divisible, and unital. 0 Definition 1.4.21 Let E be a lmear space, and let T E £(E). For each subset W of C, the algebraic spectral space ET(~V) is the maximum linear ,mbspace F of E such that (zIe - T)(F) = F for each z E C \ W. As in 1.4.8, E is a left £(E)-module, and so the existence of ET(W) is a special case of 1.4.17. Clearly ET(W1 ) C ET(WZ ) whenever WI C W 2 in C. Proposition 1.4.22 Let E be a linear space, and let T E £(E).

(i) Suppose that F is a linear subspare of E such that T(F) = F and such that, for each x E F, there exists n E N with Tnx = O. Then F C ET(0). (ii) Suppose that W C C, z E W, and x E E with (zIE - T)x E Br(W). Then x E ET(W),

n

(iii) For each W C C, ET(W) = {ET(C \ {z}) : Z E C \ W}. (iv) For each family {Wv } of subsets ofC, ET Wv) = ET(Wv ),

(nv

Proof (i) Take x E F and z E C·. There exists n S = 2.:::~=1 z-kT k - 1 and y = Sx. Then y E F and (zIE - T)(F) = F for each z E
nv

N with Tn x = 0; set (zIe - T)y = x. Thus

E

there exists y E ET(W) = ((IE - T)(y - x), and

59

Modules and primitive zdeals

so x E ((Ie - T)(F) because y - z E F and z -=I- (. Hence ((IE - T)(F) = F. By the maximality of Er(W), we have Fe Er(W), and so x E Er(W). (iii) For (E C, write Ee; = Er(C \ {O), and set G = n{E( : (E C \ W}. Clearly Er(W) c G. Now take x E G and z E C \ W. Then there exists y E Ez such that (zIE - T)y = x. Take ( E C \ (W U {z}). By (ii), Y E E(, and so y E G. Hence (zIe - T)(G) = G (z E C \ W). By the maximality of Er(W), we have G C Er(W). Thus G = Er(W). (iv) Set W

= nW v .

nEr(Wv ) v

By (iii), we have

= n {n{Er(C \ {z}): z

E

C \ W v }}

v

= n{Er(C \ {z}): z E C \ W}

= Er(W) , o

as required.

Let E be a linear space, let T E C(E), and take W C
T(ER(W»)

c

Es(W)

(W

c q.

Proof Set G = ER(W), For each z E C \ W, we have

(zIF

-

S)(T(G))

= T(zIe - R)(G) = T(G) ,

and so T(G) C Es(W) by the maximality of Es(W).

o

We now introduce the multiplier algebra of an algebra A. Definition 1.4.25 Let A be an algebra. A left [right] multiplier on A zs an element L [R] zn C(A) such that L(ab) = L(a)b [R(ab) = aR(b)] (a, bE A). A multiplier is a pair (L, R), where Land R are left and right multiplzeTs on A, 'f'espectzvely, and aL(b) = R(a)b (a, bE A). (1.4.17)

The sets of left multipliers, right multipliers, and multipliers on A are denoted by Mi(A), Mr(A), and M(A), respectively. Clearly, they are unital subalgebras of C(A), C(A)OP, and C(A) x C(A)OP, respectively; they are called

60

Algebraic foundations

the left multzplier algebra, the right multiplier algebra, and the mult1,plier algebra of A, respectively. In fact, Mf(A) = £A(A) and Mr(A) = A£(A) when A is regarded as an A-bimodule. Suppose that A is left and right faithful, and let (L, R) E £(A) x £(A) satisfy (1.4.17). Take a, b, c E A. Then cL(ab) = R(c)ab = cL(a)b, and so L(ab) = L(a)b. Thus L is a left multiplier. Similarly R is a right multiplier, and so (L, R) E M(A). Suppose that A is an ideal in an algebra B, and let bE B. Essentially as in (1.2.1), we define Lb : a f---7 ba and Rb : a f---7 ab on A; clearly, (Lb' Rb) E M(A). The map e : b f---7 (Lb, R b), B ----> M(A), is a homomorphism, called the regular homomorphzsm; e is a monomorphism if and only if B is faithful over A, and in this case we regard B as a subalgebra of M(A). Let a E A and (L, R) E M(A). Then (1.4.18) so that, in the case where A is faithful, A is an ideal in M(A), and equations (1.4.18) become (L, R) . a = La and a . (L, R) = Ra. Similarly, in the case where A is left faithful, A is a left ideal in Me(A) and L . a = La (a E A, L E Me(A)). Proposition 1.4.26 Let A be an ideal in an algebra B. Suppose that the map b f---7 L b, B ----> MI(A), zs an isomorphism. Then the regular homomorphism e: b f---7 (Lb' Rb), B ----> M(A), is an isomorphism. Proof The map e is an embedding. Take (L, R) E M(A). Then there exists bE B with L = Lb. For a, c E A, we have Rb(a)c = abc = aL(c) = R(a)c, and so Rb(a) = R(a) because A-L = O. Thus Rb = R, and e is an epimorphism. 0 Proposition 1.4.21 Let A be a fmthful algebra. Then, for each rp E exzsts a unique '15 E (j)M(A) such that '15 I A = 'P.

(j) A,

there

Proof Let (L, R) E M(A). For a, bE A, we have

rp(a)rp(Lb) = rp(aL(b)) = rp(R(a)b) = rp(Ra)rp(b).

(1.4.19)

Now take a E A with rp(a) = 1, and set '15«L, R)) = rp(La). By (1.4.19), rp(Ra) = 'P(La). Suppose also that b E A with rp(b) = 1. Then, again by (1.4.19), rp(Lb) = rp(La), and so '15«L, R)) is well-defined. Certainly '15 E M(AV and '15 I A = rp. Let (LI, Rd, (L2' R 2) E M(A), and take a E A with 'P(a) = 1. Then

'15( (L1' R 1)(L2, R 2)) = rp(Ll (L2a)) = '15( (Ll> Rd )'15 ((L2' R 2)) , and so '15 E (j) M(A)' Clearly '15 is uniquely specified.

o

Now suppose that A is a commutative, faithful algebra. Then each left [right] multiplier is a right [left] multiplier. Let (L1' Rd, (L2' R 2) E M(A). For a, bE A,

b· (L1L2)(a) = L1(bL2(a)) = Ll(R2(b)a) = R2(b)L1(a) = b . (L2Ld(a) , and so L1L2 = L 2L 1; similarly RIR2 = R 2R 1 . Thus M(A) is commutative; also Me(A) e:! M(A), and so we can regard M(A) as a commutative, unital subalgebra of .c(A) and A as an ideal in M(A).

61

Modules and primitive ideals

Definition 1.4.28 Let A be an algebra, and let E be a left A-module. Then E zs non-trivial zf A . ~rJ:-a:naE'is' simple -if it {,9non-trzvw,[ cma Tj 0 -rind E are the only submodules of E. The 'module E zs decomposable if E = F 8 G for some non-zero left A-modules F and G, and E zs indecomposable zf zt is not decomposable. A representation is simple [decomposable], [indecomposable] if the corresponding module is simple [decomposable], [indecomposable].

Similar definitions apply to right A-modules and A-bimodules. A simple left A-module is indecomposable, but the converse is not necessarily true. For each maximal left ideal M in A, A/AI is a simple left A-module. Proposition 1.4.29 Let A be an algebra, let E be a simple left A-module, and let Xo E E-. Then:

(i) A . Xo = E; (ii) the map a + x~

I-->

a . Xo, A/x~

-+

E" is a left A-mod~lle isomorphism;

(iii) x~ is a manmal mod~Llar left ideal in A. Proof (i) Let x E E. Then A . x is a submodule of E, and so A . x = 0 or A . x = E. Let F = {x E E: A . x = o}. Then F is a submodule of E; F i= E because A . E i= 0, and hence F = O. Thus Xo rJ. F and A . Xo = E.

(ii) Define T : a + x~ I--> a . xc, A/x~ -+ E. Then T E AL(A/x~, E) and T is an injection. By (i), T is a surjection, and so A/x~ 3:! E. (iii) Since E is a simple module, x~ is a maximal left ideal. By (i), there exists u E A with u . Xo = xc, and so A(eA - u) . Xo = 0 and A(eA - u) C x~. Thus u is a right modular identity for x~. D 9orollary 1.4.30 Let A be an algebra, let M be a maximal left zdeal in A, let bE A, and set J = {ao E A : ab E AI}. Then ezther J = A or J is a maximal modular left ideal in A. Proof Set E = A/M. Then either Ab eM. and hence J = A, or E is a simple left A-module and b + M E E-. In the latter case, J = (b + M)l. is a maximal modular left ideal in A by 1.4.29(iii). D

There are two standard theorems about simple modules, each to be used several times, which we shall now establish. Theorem 1.4.31 Let A be an algebra, and let E be a simple left A-module. Then AL(E) is a dzvision algebra. Proof We have remarked that AL(E) is a unital sub algebra of L(E). Take T E AL(E)-. Then kerT and T(E) are submodules of E. Since T f 0, we have ker T i= E and T(E) f 0, and so, since E is simple, ker T = 0 and T(E) = E. Thus T is a bijection, with inverse S, say. Certainly S E AL(E), and so T is invertible in AL(E). 0

Algebraic foundations

62

Theorem 1.4.32 (Jacobson's density theorem) Let A be an algebra over lK, and let E be a simple left A-module such that AC(E) = lKIe. Let {Xl, ... ,xn } be a linearly independent set zn E, and let YI,' .. , Yn E E. Then there exzsts a E A with a . Xj = Yj (j E N n ). Proof We first claim that, for each finite-dimensional subspace F of E and each X E E \ F, we have Fl. . X = E. The proof of this claim is by induction on the dimension of F. The case dim F = 0 is 1.4.29(i): if x E E-, then A . x = E. Now suppose that dim F 2': 1, and assume that the claim holds for each subspace G of E such that dim G < dim F. Let Yo E E be such that Fl. . Yo = O. Take Zo E F-, and take a linear subspace G of E with F = G 8lKzo. By the inductive hypothesis, Gl. . Zo = E, and so, for each x E E, there exists a E Gl. with a . Zo = x; set Tx = a . Yo. If also b E Gl. with b . Zo = x, then a - b E FJ.. and (a - b) . Yo E Fl. . Yo = 0, so that a . Yo = b . Yo. Hence T is well-defined. It b easily checked that T E AC(E). By hypothesis, AC(E) = lKls, and so there exists 0:' E lK such that T = O:'lE. For each c E GJ.., we have T(c . zo) = c . Yo, and so c . (Yo - O:'zo) = O. Thus GJ.. . (Yo - O:'zo) = 0, and hence, by the inductive hypothesis, Yo - O:'Zo E G and Yo E G + lKzo = F. Let x E E \ F. It has been shown that Fl. . x =1= 0, and so FJ.. . X = E because E is simple. Hence the claim holds for this F, and the induction continues. Now let Xl, .. " Xn, YI, ... , Yn be as specified in the statement. It follows from the claim that, for each j E N n , there exists aj E A with aj . Xj = Yj and aj . Xi = 0 (i =1= j). Set a = al + ... + an. Then a . Xj = Yj (j E N n ). 0 Definition 1.4.33 Let I be a left zdeal in an algebra A. The quotient of I is I :A

=

{a E A : aA C l} .

The quotient of a maximal modular left ideal is a primitive ideal. The algebra A is primitive if 0 is a primitwe ideal. Each quotient I : A of a left ideal I of A is an ideal in A. Clearly the quotient of a modular left ideal I is maximal in the set of ideals in A which are contained in I. The algebra A itself is not a primitive ideal.

Proposition 1.4.34 Let A be an algebra. (i) An ideal in A is a primitive ideal if and only if it is the kernel of a simple representation of A. (ii) Each primitive ideal in A is the intersection of the maximal modular left ideals which contain it. (iii) Each primitive ideal in A is a prime ideal. (iv) Each maximal modular ideal in A is a primitive ideal, and each modular ideal is contained in a primitive ideal.

Proof (i) Let 1= M : A, where M is a maximal modular left ideal in A. Then AIM is a simple module, and I is the kernel of the left regular representation of A on AIM.

Modules and primitzve ideals

63

Conversely, let p be a simple representation of A on E, take Xo E E·, and set M = X6-. By 1.4.29, A . Xo = E and M is a maximal modular left ideal in A; we have a E ker p if and only if a . (A . xo) = 0, and this holds if and only if aA eM. Hence kerp = lvI: A. (ii) and (iii) Let I be a primitive ideal, say I = E1-, where E is a simple module. Then I = n{x1- : x E E e }, giving (ii). Take ideals J, K in A with JK c I and K ¢. I, say. Then K . E = E, and so J . E = JK . E = 0, whence J C I, giving (iii). (iv) Let lvI be a maximal modular ideal. There is a maximal modular left ideal L with MeL. Clearly lvI C L : A, and so lvI = L : A is primitive. Each modular ideal is contained in a maximal modular ideal. 0 It follows from (i), above, that, in the case where I is a proper ideal in A, the following are equivalent: (a) I is a primitive ideal; (b) AI I is a primitive algebra; (c) there is a simple left A-module E such that I = E1-.

Example 1.4.35 Let E be a non-zero linear space. Then E is a simple left C(E)-module for the map (T, x) f--? Tx, C(E) x E -+ E, and C(E) i::; a primitive algebra. In the case where E is infinite-dimensional, FC(E) is a non-zero, proper ideal in C(E), and so the primitive ideal is not maximal. 0

°

Proposition 1.4.36 Let I be an ideal in a commutative algebra A. Then the following conditions on I are equivalent: (a) I is a primitive ideal; (b) I is a maximal modular ideal; (c) AI I is a field. Proof Suppose that (a) holds, say I = lvI : A, where M is a maximal modular ideal. Then lvI : A = lvI, and so (b) holds. The remainder is immediate. 0 Proposition 1.4.31 Let A be a complex algebra, and let E be a finite-dimensional, simple left A-module wzth representation p. Then p : A -+ £(E) is an epimorphism. Proof By 1.4.31, AC(E) is a division algebra. Since the linear space E is finitedimensional, AC(E) is finite-dimensional, and so, by 1.3.56, AC(E) = Cle. Let {Xl, .. " Xn} be a basis for E, and let T E C(E). By Jacobson's density theorem 1.4.32, there exists a E A with a . x J = TX J (j E N n ). Clearly pea) = T, and so p : A -+ C(E) is a surjection. 0 Corollary 1.4.38 Let I be an ideal of finite codzmenswn in a complex algebra A. Then the following conditions on I are equivalent: (a) I is a prime ideal; (b) I is a maximal modular ideal; (c) I is a primitive ideal; Cd) All ~ Mn for some n E N.

64

Algebrazc foundations

Proof Suppose that I is primitive, and let P be a simple representation of A on a finite-dimensional space E with ker P = I, say dimE = n. By 1.4.37, AI I ~ C(E), and so AI I ~ Mn. By 1.3.51, Mn is a simple algebra, and so I is a maximal modular ideal. Thus (c)=*(d)=*(b)=*(c). By 1.4.34(iii), (c)=*(a), and so it remains to prove that (a)=*(c); it suffices to do this in the case where I = 0 (and A is finite-dimensional). Let J be a non-zero left ideal in A of minimum dimension. Since A is a prime algebra, aAb:j:. 0 (a, bE Ae), and so Ab :j:. 0 (b E Ae). For each b E J, we have Ab C J, and so Ab = J (b E r). Thus J is a simple left A-module, and aJ:j:. 0 (a E Ae). Hence 0 = J 1- is a primitive ideal. 0 Corollary 1.4.39 Let A be a complex algebra, and let PI,"" Pn be szmple, finite-dimensional representations of A on spaces E 1, ... , En, respectzvely, such that ker Pi :j:. ker Pj (i:j:. j). Then the map n

a

f--->

(PI (a), .. . , p,,(a)), A ----

0

C(E;) ,

i=l

zs an epimorphism.

Proof Set M j = ker Pj (j EN,,), so that, by 1.4.38, each M j is a maximal modular ideal in A. Take kEN", and let 1= M1

n··· n M k- 1 n Mk+! n··· n Mn.

Assume that I C Nh. Then Ah ... M k - 1M k+1'" Mn C M k ; by 1.4.34(iii), M j cAlk for some j :j:. k, whence Alj = Mk, a contradiction. Thus I ct- Mk and I . Ek :j:. O. By 1.4.37, for each T E C(Ek), there exists ak E I with Pk(ak) = T. The result follows. 0 Corollary 1.4.40 Let A be a complex, umtal algebra, let E be a szmple, finitedimensional A-bimodule, and let F be a subspace of E such that F is a simple left A-module. Then there exist a1>"" am E A with a1 = eA and such that F ~ F . OJ (j E N m ), E = O~l F . aj, and EJ. = FJ. is a maximal ideal in A. Proof Let a E A. By 1.4.31, either F . a = 0 or the map x f---> x . a, F ---- F . a. is an isomorphism of left A-modules. Since E is finite-dimensional, there is a maximal finite set {F . a1,"" F . am} with each F . aj :j:. 0 and with a1 = eA. Let i E Nm and a E A. Then either F . aia = 0 or F . aia = F . aj for some j E Nm · Thus 2:;"=1 F . aj is an A-bimodule, and so 2:.7=1 F . aJ = E. Since F . ai n F . aj = 0 for i :j:. j, the sum is direct. Clearly EJ. = F1- is an ideal of finite codimension in A; by 1.4.38, this ideal is maximal. D Let A, E, F, and a1, .. ·,am be as in the above corollary, with dimF = k. Then dimE = km. Set M = EJ., so that AIM ~ M k . Let {rll,'" ,rkd be a basis of F, and set riJ = ril . aj (i E N k , j E Nm ). Then we obtain a basis of E that can be written as a (k x m)-array (rij); a generic element x of E is x = 2:.:=12:7=1 Xijrij' Let a E A. Then a + M corresponds to a k x k matrix

Modules and primitive zdeals

65

(aij) E M k , and the element a . x of E is represented (with respect to the basis (rij) of E) by the matrix product

X1m)

X~rn

.

We can also regard the left action of a E A on x E E as being specified by a km x km matrix

(1.4.20) where Em is the identity m x m matrix, and this matrix acts on the element x = (X11' X12,'" ,X1rn, X21,'" ,X2m,··· ,Xk1,.· ., Xkm) of E. Now consider the product x . b for x E E and b E A. This product can be represented by a km x km matrix acting on the left on the km x 1 matrix x. But this matrix commutes with all matrices of the form (1.4.20), and so it has the form

0

rr

B (

0)

1,

where B is an m x m matrix and there are k copies of B. Write the transpose Bt of B as (f3ij), so that the product x . b corresponds to the product given by

In particular, AI ET ~ M m , and the simple right A-modules in E are spanned by the rows of the array (riJ)' We obtain the following result.

Proposition 1.4.41 Let A be a complex, unital algebra, let E be a szmple, finite-dzmensional A-bimod71,le, and let F be a subspace of E such that F is a simple left A -module. Suppose that dim F = k and dim E = km, and let {riJ : i E Nk, j E Nm } be a basis for E, as above. Then there exist families {Pst: s, t E N k } and {quv : U, v E Nm } in A such that: (i) (Pst + EJ.. : s, t E N k ) and (quv + ET : u, v E Nrn) are sets of unital matrix units in AI EJ.. and AI E T, respectively, with L:=l Pss = 2:::'=1 quu = eA;

(ii) for a = Ls t astPst + EJ.., b = Lu v f3uvquv + E T , and x = where (ast) E M k , (f3uv) E M m , and (XiJ) Eo Mk,m, we have

a .x =

L: (2..: aikXkj) rij, 2,J

k

2::i j

Xijrij,

'

o

66

Algebraic foundations

Definition 1.4.42 Let A be an algebra. Then ITA is the set of primitive ideals of A. Let I be a subset of A, and let S be a subset of ITA· Then

~A(I)

= ~(I) = {P E ITA

: p:J I}

and

tA(S)

= t(S) = n{p: PES}

are the hull of I and the kernel of S, respectively (taking t(0) = A).

Let I and J be subsets of A with Ie J. Then ~(J) :J ~(I). :For each S C ITA, t(S) is an ideal in A, and t(S) :J t(T) whenever SeT c ITA. We write ~t(S) for ry(t(S»; clearly S c ~t(S) and t(~t(S)) = t(S) whenever S c ITA, so that ryt(~t(S» = ~t(S) and ryt(0) = 0. For each subset I of A, we have I C try(I) and ~t(ry(I) = ry(I). Let Sand T be subsets of ITA. Then t(SUT) = t(S)nt(T), and so we see that ~t(S U T) :J ryt(S) U ~t(T). For each P E ITA with t(S U T) C P, we have t(S)t(T) C t(S)

n t(T) = t(S U T)

C

P,

and so either t(S) c P or t(T) C P because, by 1.4.34(iii), P is a prime ideal. It follows that ~t(S U T) C ryt(S) U ~t(S), and so ~t(S U

T) = ryt(S) U ~t(T)

(S, T

c

ITA) .

We have shown that the map S 1-+ ~t(S) is a closure operation on P(ITA)' Definition 1.4.43 Let A be an algebra. The hull-kernel topology on ITA is the topology defined by the closure operation S 1-+ ~t(S) on P(ITA); the space ITA with the hull-kernel topology is the structure space of A.

It is certainly possible to have distinct primitive ideals P and Q in A with Pc Q, and so points are not necessarily clol:led in ITA; in general, the hull-kernel topology is only a To-topology. Proposition 1.4.44 Let I be an ideal in an algebra A. Then: (i) the map R : P (ii) the map Q: P

1-+ 1-+

P

n I,

ITA \

PII, ry(I)

~(I) ---->

---->

IT I

,

is a homeomorphism;

ITA/I, is a homeomorphism.

Proof (i) Let p: I ----> L(E) be a simple representation of Ion a space E, and set J = ker p E ITI. Fix Xo E E-. By 1.4.29(i), for each x E E, there exists bEl with b . Xo = x; we define p(a)(x)

=

p(ab)(xo)

(a E A).

For a E A and b, c E I, we have p(c)p(ab) = p(cab) = p(ca)p(b). Thus, if b . Xo = 0, then p(c)p(ab)(xo) = 0 (c E I), and so, by 1.4.29(i), p(ab)(xo) = O. It follows that p(a)(x) is well-defined in E. It is now easily checked that p is a simple representation of A on E and that p I I = p. Set P = kerp. Then PEllA \ 1)(1) and R(P) = J, and so R is a surjection. Suppose that PI, P2 E IIA \fJ(I) with PIn1 = P2 n1. Then PI1 C Pln1 c P2 . But Pz is a prime ideal by 1.4.34(iii) and I ct P2 , and so PI C P2 • Similarly P2 C PI, and so n is an injection.

Modules and primitwe ideals

67

Take Po E ITA \ ~(I) and S C ITA \ ~(I), and set K = £I( {Q n I : Q E S}). Clearly, if £(S) c Po, then K c Po n I, and, if K C Po n I. then £(S)I C Po and £(S) C Po. Thus £(S) c Po if and only if K c Po n I. This proves that n is a homeomorphism. (ii) Clearly Q is an injection. Let p be a simple representation of AI I on a space E. Then p 0 7r is a simple representation of A on E, where 7r : A --+ AI I is the quotient map. Hence Q is a surjection. For each S C ~(I), necessarily tA/I(Q(S)) = Q(t(S)), and so, for Po E ITA, we have £(S) C Po if and only if tA/I(Q(S)) c Q(Po). This proves that Q is a 0 homeomorphism. It follows from 1.4.44(ii) that co dimension in an algebra.

~(I)

is finite whenever I is an ideal of finite

Proposition 1.4.45 Let A be an algebra. and let p E J(A). n: P t-> pPp, ITA \ ~(pAp) --+ ITpAp , is a homeomorphism.

Then the map

Proof Let P E ITA \ b(pAp), say P = EJ.. for a simple left A-module E. Then p ~ P, and so p . E =f O. Suppose that x E (p. E)·. Then A . x = E by 1.4.29(i), and so pAp· x = p . (A . x) = p . E. Thus p . E is a simple left pAp-module; the kernel of the corresponding representation of pAp is pAp n P = pPp. Thus n has range in ITpAp. Suppose that PI, P 2 E ITA \ ~(pAp) with pP1p = pP2P. Then (Ap)PI(Ap) C ApPIP = ApP2P C P 2

.

rt

rt

By 1.4.34(iii), P 2 is a prime ideal in A. Since pAp P2 , certainly Ap P 2 . By 1.3.42. necessarily PI C P 2 . Similarly P 2 C PI, and so n is injective. Let Q E IT pAp . Then there is a maximal modular left ideal 1'v! in pAp such that Q = M : pAp. Consider the left ideal I = M + AM + A( e A - p) in A. Then p is a right modular identity for I, and so there is a maximal modular left ideal J in A with I c J and p ~ J. Define P = J : A; we have P C J and p ~ P, and so P E ITA \ b(pAp). Clearly QAp = QpAp c M c I and QA(eA - p) C I, and so QA C J. Thus Q c pPp. On the other hand, M + pPp c I + Pc J and p E pAp \.1, and so M + pPp =f pAp. Since M is a maximal modular left ideal, this shows that pPp C M. Thus pPp c Q. We have shown that Q = pPp, and so is surjective. Suppose that F is a dosed subset ofITpAp, and set S = n-1(F). To show that S is closed in ITA \ lJ(pAp), we must show that Q E S whenever Q E ITA \ ~(pAp) and Q :::l £(S). But this is immediate. Thus is continuous. Conversely. suppose that S is closed in ITA \ ~(pAp), and set F = n(s). To show that F is closed in ITpAp, we must show that Q E F whenever Q E ITpAp and Q :::l t(F). Set P = n-I(Q). Then pt(S)p = t(S) n pAp = n{pIp : IE S} = t(F) c Q c P, and so (Ap)t(S)(Ap) C P. Again this implies that £(S) C P. Since S is closed, this implies that PES and hence that Q E F. Thus n- l is continuous. 0

n

n

Definition 1.4.46 Let A be an algebra, and take S Jo(S) = {a E A : ua = a for some and J(S) = A# Jo(S)A#.

C

ITA. Then

U E

P(S)} ,

Algebraic foundatwns

68

Thus J(S) is the ideal in A generated by Jo(S). We have J(S) so ~(J(S)) -:) ~e(s).

c reS), and

Proposition 1.4.41 Let A be an algebra, and take S C ITA. Then J(S) for each idPaI I of A wzth ~(I) c S.

c I

Proof Let I be an ideal with ~(I) c S, and take a E .1o(S), i:iay ua = a, where E r(S). Define K = {b E A : ba E I}. Then K is a left ideal of A with K -:) I. Assume that a ~ I. Then u ~ K and u is a right modular identity for K. and so there is a maximal modular left ideal 111 of A with M -:) K and u ~ M: set P = 11.1 : A E IT.4, so that P -:) I and PES. Then 11 E £(S) c P c M, a contradiction. Hence a E I. o It follows that Jo(S) c I, and so J(S) c I. U

Notes 1.4.48 Again, almost all the results in this section arf' contained in standard texts such as (P. M. Cohn 1989), (Hungerford 1980), (Jacobson 1974, 1980), (McCoy 1973), and (Pierce 19i52). Definition 1.4.12 introduces a new notation in an attempt to clarify concepts which are often confused. Our concept of 'split' in 1.4.15 is sometimes referred to as 'splitexact'. There is an extensive theory of injective and divisible modules (and the related 'projective' modules); see (Jacobson 1980, Chapter 3), for example. Every left Amodule can be embedded in an injective left A-module (ibid., 3.18). The left and right multipliers of 1.4.25 are sometimes called centralizers; see (Palmer 1994), where M(A) is called the double centralizer algebra. Our 'primitive ideals' are more accurately termed left primitive ideals; it is not true that each right primitive ideal is left primitive (Bergman 1964). The terminology and early results on primitive ideals in an algebra A are due to Jacobson (1956): the hull-kernel topology on the structure space I1A is also called the Jacobson topology. Other structure spaces are considered in (Palmer 1994, Chapter 7).

1.5

RADICALS AND SPECTRA

In the structure theory of algebras, several different radicals are considered. In the theory of Banach algebras, it is the Jacohson radical which is by far the most important, and so we shall just refer to this radical as 'the radical' for general algebras. There are several different characterizations of this radical; we shall take the notion of primitive ideal a::; the ba::;ic one. We shall also de::;cribe the strong and prime radica]::; of an algebra, and we shall include some cla::;sical structure theorems for finite-dimensional algebras. The section will conclude with the definition of the ::;pectrum of an element in a complex algebra and of the joint spectrum of an n-tuple in a commutative algebra. To simplify the statement of some definitions and results, we adopt throughout the convention that, if the class of subsets of an algebra A satisfying a certain property be empty, then the intersection of all the sets in the class is A itself. Definition 1.5.1 Let A and B be algebras. The (Jacobson) radical, radA, of A is the intersection of the primitive ideals of A. The algebra A is semisimple zf rad A = 0 and radical if rad A = A. A homomorphism 8 : A -+ B is radical if 8(A) c rad B.

69

Radicals and spectra

Thus rad A = t(TIA)' and A is radical if and only if TIA = 0. Let P be a primitive ideal in A. By 1.4.44(ii). AlP is a primitive algebra, and so AlP is semisimple. For example, let E be a non-zero linear space. Then the algebra L(E) is primitive, and hence semisimple. Let R be a commutative, radical algebra. Then R# is a local algebra with maximal ideal R; the unique character on R# with kernel R is denoted by 'PR, so that Inv R# = {a E R# : 'PR(a) of- O}. Theorem 1.5.2 Let A be an algebra. Then:

(i) rad A '/,s the intcrsectzon of the kernels of the simple representatzons of A: (ii) rad A zs the mtersectwn of the maxzmal modular- left zdeals of A; (iii) rad A zs a quasz-inver-hble zdeal, and mel A contams each left quasiinver-tzble left zdeal of A; (iv) radA

=

{a E A : A#a C q-Inv A}.

Proof (i) and (ii) These arc immediate from 1.4.34, (i), (ii), and (iv).

(iii) Let a E rad A. and set I = A(a - e A)' Then I is a modular left ideal with right modular identity a. Assume that a ~ I. Then there is a maximal modular left ideal AI with modular identity a; since a ~ M, we have a ~ rad A by (ii), and this is a contradiction. Thus a E I, and so there exists b E A with b <> a = O. Also, b = ba - a E rad A, and so there exists c E A with c <> b = O. Necessarily c = a, and so a E q- Inv A. Thus rad A is a quasi-invertible ideal. Let I be a left quasi-invertible left ideal of A, let E be a simple left A-module, and take x E E e . AHsume that I . E of- O. Then, by 1.4.29(i), there exists a E I with a . x = x. Take b E A with boa = O. Then x = (b + a - ba) . :E = 0, a contradiction. Thus I . E = 0, and so, by (i), I c rad A. (iv) Suppose that a E radA. Then A#a C radA, and so A#a C q-InvA by (iii). Conversely, suppose that a E A and that A#a C q-InvA. Assume that a ~ rad A. Then there is a simple left A-module E and x E E with a . x of- O. By 1.4.29(i), there exists b E A with ba ..'T = x. Since ba E q-InvA, necessarily x = 0, a contradiction. Thus a E rad A, as required. 0

= radA. rad A = {a E

It follows that radA#

and, in particular, ert

+ radA C

In the case where A iH unital, A : eA

+ Aa C

(1.5.1)

Inv A} ,

Inv A.

Corollary 1.5.3 Let A and B be algebras.

(i) Suppose that a

E

A, b E rad A, and a

(ii) Suppose that B : A

---+

= abo

Then a = O.

B is an epimorphzsm. Then B(rad A)

C

rad B.

Proof (i) By 1.5.2(iii), there exists c E A with b 0 c = 0, and now we have a = a - a(b 0 c) = a - ab - (a - ab)c = O. (ii) Since O(a 0 b) = O(a) oO(b) (a, bE A) and B(A) = B, the result follows 0 from 1.5.2(iv).

Algebraic foundations

70

We have characterized primitive ideals, and hence rad A, in terms of left Amodules. Let J be the ideal defined analogously in terms of right A-modules. Then J contains each right quasi-invertible right ideal of A, and so, by 1.5.2(iii), rad A c J. Similarly J C rad A, and so J = fad A. Thus fad A is independent of the choice of left or right in its definition. In particular, radA

=

{a

E

A: aA#

C

q-Inv A}.

(1.5.2)

The equivalence between the characterizations of rad A in 1.5.2(iv) and in (1.5.2) can also be seen directly: if b E A and if c E A is such that c <) ba = ba 0 c = 0, then ab 0 (acb - ab) = (acb - ab) <) ab = O. Theorem 1.5.4 Let A be an algebra.

(i) Let I be an zdeal in A. Then rad I = In rad A. In particular, if A is semzsimple, then I is semiszmple. (ii) Let I be an zdeal in A wzth I C radA. Then rad(A/1) = (radA)/I. In partic~Llar, A/fad A is semisimple. (iii) Let I and J be zdeals in A with I c J c rad A, and let 7r : A/I -+ A/ J be the natural surjection. Then 7r(rad (A/1)) = rad (A/ J). Proof (i) and (ii) These follow from 1.4.44, (i) and (ii), respectively.

(iii) By (ii), rad(A/ J) ~ (rad(A/ 1))/(J/ 1) because A/ J ~ (A/ 1)/(J/1).

0

Definition 1.5.5 Let A be an algebra with non-zero radical. A minimal radical ideal of A is a minimal ideal of A that is contained in rad A. Proposition 1.5.6 Let A be an algebra.

(i) meA) c q-Inv A. (ii) rad A contains each nil left zdeal and each ml right ideal of A.

(iii) Suppose that A Z8 commutative. Then meA) c rad A. (iv) Suppose that rad A is finite-dzmensional. Then: fad A is a nilpotent ideal; A contains a mmimal radical ideal; for each minimal radzcal ideal K of A, (radA)K = K(radA) = O. (v) Suppose that A zs commutative and rad A is finite-dzmenswnal. Then meA) = radA. (vi) Suppose that rad A has countable dimension over TIC Then rad A is a nil ideal. (vii) Suppose that A/Am is finite-dimenswnal for some m 2': 2. Then A/An is finite-dimensional for each n 2': 2. (viii) Suppose that A is infinite-dimensional and nilpotent. Then AjA2 is infinite-dimensional. Proof Set R = rad A. (i) Take a E m(A), say a n + 1 = 0, and set b boa = a 0 b = 0, and so a E q-Inv A. (ii) This follows from (i) and 1.5.2(iii).

-(a

+ ... + an).

Then

Radzcals and spectra

71

(iii) Since A is commutative, 91(A) is an ideal in A. (iv) Since R is finite-dimensional, the nest (Rn) stabilizes; there exists no EN such that Rn = Rno (n::::: no). Set 1 = R rt ", flO that 12 = 1. Assume that 1 i- O. The family F of left ideals J of A with J c 1 and 1 J i- 0 is non-empty because 1 E F. Let K be a member of F of minimum dimension, and take a E K with 1a i- O. Then 1a E F and 1a c K, so that 1a = K. Take b E 1 with ba = a. Since b E R, there exists c E A with cob = O. But then 0= (cob)a = a, a contradiction. Hence 1 = 0 and R is nilpotent. An ideal that is of minimal dimension in the family of non-zero ideals of A contained in R is a minimal radical ideal. Let K be such an ideal. Then RK E {O,K}. Assume that RK = K. Then R"K = K (n EN). and so K = 0, a contradiction. Thus RK = O. Similarly K R = O. (v) This follows from (iii) and (iv). (vi) Let a E R. Since R# has countable dimension over lK and lK is uncountable, there exist distinct points Zl,"" Zn E lK e a.nd al, .. " an E lK e with L:j=l aj(zJeR - a)~l = O. Set P

=

L" ai II (X i=l

Zj),

#i

so that p is a polynomial over K Then P(Zl) = al TI#l(Zl - zJ) i- 0, and so p i- o. But p( a) = a in A, and so B = alg {a} is finite-dimensional. Since ak+l Beak B (k EN), there exists mEN with a m + 1 B = am B. But then a m+ 1 = a m+1b for some b E B, and so a m + 1 = 0 by 1.5.3(i), thus proving that R is a nil ideal. (vii) Since dim(A/Am) < 00, certainly dim(A/A2) < 00. Assume inductively that dim(A/Ak) < 00, say A = lin{al,"" aT' Ak}. Then

A = lin{ai,aiaj,Ak+l : i,j E NT}' and so dim(A/Ak+1) <

By induction, dim(A/N') <

(n::::: 2). (viii) Take mEN with Am = O. Assume that dim(AjA 2 ) < dim A = dim(AjAm) < 00 by (vii), a contradiction. 00.

00

00.

Then 0

Proposition 1.5.7 Let A be an algebm with radical R.

(i) J(A) n R = {o}. (ii) Suppose that p, q E J(A) with pq = qp and p + R = q + R. Then p = q. (iii) For p E J(A), rad (pAp) = pRp = (pAp) n R. (iv) Suppose that A is commutative. Then Q:(A) n R = {O}. Proof (i) By 1.5.3(i), p = 0 for p E J(A) n R. (ii) Clearly p - pq, q - qp E J(A) n R, and so p - pq = q - qp = a by (i). Since pq = qp, necessarily p = q. (iii) This follows from 1.4.45. (iv) Assume towards a contradiction that there exists a E \E(A) n R with a i- O. By 1.3.23(ii), there is an orthogonal set {PI,··· ,Pn} C ::teA) \ {O}

Algebrazc foundations

72

and 0::1, ... ,O::n E C- with a = L:7=IO::iPi' Since a E R, there exists b E A with b <> 0:: 11 a = 0, and then 0 = (b <> 0:: 11 a)pl = PI, a contradiction. Thus ~(A) n R = {o}. 0 Let A = C 2 with the product (Zb WI)(Z2' W2) = (ZIZ2, WIZ2). Then A is a non-commutative algebra with radA = {(O,w) : W E C}. Set P = (1,1) and q = (1,2). Then p, q E J(A) and P + R = q + R, but P =I- q. Thus the hypothesis that pq = qp is needed in 1.5.7(ii). Let A be a local algebra, as in 1.3.39. Then radA = MA, and it follows from 1.5.7(i) that J(A) = {eA}. We shall require a form of the classical Wedderburn structure theorem for finite-dimensional algebras; it shows that each finite-dimensional, semisimplo, complex algebra is a direct sum of full matrix algebras. As a preliminary to this result, we describe the situation for finite-dimensional, commutative, complex algebras. Proposition 1.5.8 Let A be a non-mlpotent, fimte-dimensional, commutative, complex algebra.

(i) There exzst kEN and an orthogonal set {PI,." ,pd of non-ze7'O idempo tents in A such that ~(A) ~ C k and such that >J1(A) = rad A and A=

~(A) <::)

>J1(A) = lin {PI , ... ,Pk} <::) >J1(A).

In the case where A is umtal, eA = PI

+ ... + Pk.

(ii) Each cP E ell A has the form CPj : L::=l O::iPi

+ >J1(A)

f---t

O::j for a umquf

j E Nk .

(iii) For each cP E J1(A) = radA, and, by 1.5.7(iv), ~(A) n radA = {O}. The result now follows from 1.3.23(iii); in the unital case, we use 1.5.7(ii) to see that eA = PI + ... + Pk·

(ii) Let cP E J1(A) = 0 and cp(pi) E {O, I} (i E N k ). There exists j E Nk such that cp(Pj) =I- 0; also, CP(Pi)CP(pj) = 0 for z =I- j, and so there is a unique j E Nk such that cp(pi) = 6i,J (i E Nk), and then cP = CPJ' (iii) Suppose that cP = CPj in the above notation. If pj>J1(A) = 0, take ao = P)' If pj>J1(A) =I- 0, there exists b E (p/l1(A))- with bpj>J1(A) = 0, and then we take ao = bpj. In each case, ao has the required property. 0 Theorem 1.5.9 (Wedderburn structure theorem) Let A be a non-ze7'O, finitedimensional, semisimple, complex algebra. Then A has an identzty e A and there exist kEN, an orthogonal set {PI, ... ,pd of central idempotents in A, and nl,"" nk E N such that eA = PI + '" + Pk, such that {pjApj : j E Nd zs the family of minimal ideals in A, such that pjApj ~ Mnj (j E Nk), and such that

A

k

k

j=1

j=1

= (0pjApj ~ OMnj

.

73

Radicals and spectra

Proof Since A is semisimple, 0 is an intersection of primitive ideals. Indeed Pj , for there exist finitely many primitive ideals PI, ... , Pk such that = otherwise there would exist a sequence (Qn) of primitive ideals with

° n1=1

QI

n ... n Qn+1

~

QI

n ... n Qn

(n E N) ,

and this is not possible in a finite-dimensional algebra. We may suppose that n'#i Pj i- 0 (i E Nk)' J Let j E N k . By 1.4.38, Pj is a maximal modular ideal and there exists nj E N such that AI Pj ~ M nj ; let Pj : A -+ Mnj be the natural epimorphism. By 1.4.39, the map k

p: a

f--t

(PI (a), ... , Pk(a)),

A -+

0

Mnj ,

j=1

n1=1

is an epimorphism, and ker P = Pj = 0, so that P is an isomorphism. For j E Nk , take Uj to be the identity of M nj , and set Pj = p-I(UJ ). Then {PI, ... ,Pk} is the required set of central idempotents. 0

Corollary 1.5.10 Let A be a non-zero, finite-dimenswnal, semisimple complex algebra. (i) For each non-zero ideal I in A, there is a central idempotent p in I such that p is the identity of I and I = pAp. (ii) Let {PI, .. ' ,pe} be an orthogonal set of minimal idempotents such that eA = PI + ... + Pf. Then dim A :s; £2. Proof (if By 1.5.4(i), I is semisimple, and so, by 1.5.9, 1 is unital; set p = e[. Certainly pAp = 1. For a E A, we have pa E I and ap E I, and so pa = pap = ap, whence p E 3(A).

07=1

(ii) Suppose that A ~ M nJ , a!:l in 1.5.9. Then £ = 1.3.19, and dim A = ni + ... +n~:S; £2.

n1

+ ... + nk by 0

There is a structure theorem related to 1.5.9, which we shall now establish by elementary arguments. The result will be placed in a more general !:letting in §1.9. In the next few pages we write a for a + R in AIR.

Definition 1.5.11 Let A be an algebra with radical R. Then: (i) A is an SBI algebra if, for each x E R, there exzsts Y E R with yo Y = x and {x}C = {yV; (ii) a set U of orthogonal zdempotents in AI R can be lifted to A zf there exists an orthogonal set P in J(A) such that {p : pEP} = U; (iii) orthogonal idempotents can be lifted zf each finite set of orthogonal zdempotents in AIR can be lifted to A. For each x E R, the element Y arising in 1.5.11(i) is unique. For suppose that Y1, Y2 E R, that YI 0 YI = Y2 0 Y2, and that {yd C = {Y2}c. Then we have (YI - Y2)(YI + Y2 - 2eA) = 0 in A#; by (1.5.1), YI + Y2 - 2eA E Inv A#, and so Yl = Y2· We also see that, if ax = 0, then ay = 0, and that, if x belongs to a left or right ideal I, then also y E I.

74

Algebrazc foundatwns

Lemma 1.5.12 Let A be an algebra whose radical zs a nzl ideal. Then A is an

SBI algebra. Proof Let 1-2:j:1 cxJZ j be the Taylor expansion of(1-Z)1/2. Take a E radA. and set b = 2:;'=1 cxja j , where n is the index of a, so that b E rad A. Then bob = (1 and {a}C = {by, and so A is an sm algebra. 0 Lemma 1.5.13 Let A be an SBI algebra 'Unth radical R.

(i) Suppose that 'U E J(A/ R) and (1 E A is such that P E J(A) with p = u and {ay c {p}c. (ii) Orthogonal zdempotents can be lifted.

a=

u. Then there exists

Proof Write e for the identity of A # .

(i) Set x = (12 - a, so that x E R. Since A is an SBI algebra, there exists y E R with (e + y) 2 = e + 4x and {x y = {y y. Set

p=(a-~e)(e+y)-l+~e. Then pEA,

P=

u, and {a}C

C

{py. We calculate:

4p2 = (2a - e)2(e + 4X)-1 + 2(2a - e)(e + y)-l + e = (4a 2 - 4(1 + e)(e + 4X)-1 + 4p - e = (e + 4.c)(e + 4X)-1

= p2

(1.5.3)

+ 4p -

e

= 4p,

J(A). (ii) Take an orthogonal set {U1' ... ,un} in J(A/ R). First, by (i), there exists P1 E J(A) with P1 = U1· Now assume that there exist P1,'" ,Pk E J(A) with Pi = Ui (i E Nk) and Pi 1. Pj (i =I- j), and set P = P1 + ... + Pk E J(A). Choose a E A with a = Uk+1. and set b = (e - p)a(e - p). Then b = Uk+1 because PUk+1 = Uk+1P = 0, and pb = bp = O. By (i), there exists q E J(A) with q = Uk+1 and {by C {qy. Set Pk+1 = (e - p)q(e - p). Then p E {q}C, and so Pk+1 E J(A). Also Pk+1 = Uk+1 and PiPk+1 = Pk+1Pi = 0 (i E Nk)' The result follows by induction. 0 and so p

E

Suppose that R2 = 0 in the above lemma. Then the formula (1.5.3) for p becomes P = 3a 2 - 2a 3 . Lemma 1.5.14 Let A be an algebra. Suppose that (Uij : i, j E Nn ) is a system of matrix units in A/rad A such that the set {Uii : i E N,,} of orthogonal idempotents

can be lifted. Then there is a system (eij : i,j E N n ) of matrzx units in A such that ei] = Uij (z,j E Nn ). Proof By the hypothesis, there exists an orthogonal set {ell, ...• enn } in J(A) such that eii = Uii (i E Nn ). Fix i E {2, ... , n}, and choose a, b E A with a = U1i and b = Uil. Since U11U1iUii = U1i, we may suppose that a E ellAeii, and similarly that b E eiiAell' Since ab = ell, we have ell - ab E rad A, and so there exists r E rad A with r

+ ell

- ab - (ell - ab)r = 0 .

Radicals and spectra

75

Since ella = a, this implies that ab(eA - 1') = ell. Define eli = a and eil = b(eA - r)eu, so that eli = 'Uli, eil = 'Un, eli E ellAeii, and eil E eiiAel1' We see that elieil = ab(eA -1')e11 = ell' Next, set x = eii - eileli. Then x 2 = x and x = Uii - 'lLilUli = 0, and so, by 1.5.7(i), x = O. Thus eileli = eii. Now take i,j E {2, ... ,n}, and define eij = eilelj' Then eij = UilUIJ = Uij' Suppm;e that i, j, k, £ E Nn . If j #- k, then eijeke E AeJjekkA = O. Also, eijej£ = eileljejlel£ = eile11elf = eilelf = eif· Thus we have verified that eijekl = {)j,keil. Hence (eiJ : i,j E Nn ) is the required system of matrix units. 0 Definition 1.5.15 Let A be an algebra with radical R. Then A has a Wedderburn decomposition, or A is decomposable, if there is a subalgebra B of A such that A = B 8 R; in thts case B 8 R lS a (Wedderburn) decomposition of A.

Of course, if A = B 8 R, then B ~ A/ R. The algebra A is decomposable A/ R -; A with 1[' 0 = fAIR' if and only if there is a homomorphism where 1[' : A -; A/ R is the quotient map; the homomorphism e is a splitting homomorphism.

e:

e

Proposition 1.5.16 Let A be an algebra such that A2 n rad A = O. Then A is decomposable. Proof Let E be a linear subspace of A such that A = (A2 8 radA) 8 E, and set B = A2 8 E. Then B is a subalgebra of A, and A = B 8 rad A. 0 Proposition 1.5.17 Let A be a commutative algebra. Suppose that A = B 8 I, where B is a subalgebra of A and I is an ideal with f C rad A. Then ~(A) c B. Proof Take P E J(A), and set P = b + r, where b E Band rEf. Then b + r = b2 + 2br

+ r2 ,

and so b = b2 and r3 = (p - b)3 = p3 - 3p2b + 3pb2 - b3 r2 E rad A, there exists s E R with r2 0 s = O. But now

+ s - r 2 8) = (r - r 3 ) - (r - r3)8 = 0, b E B. Thus J(A) c B, and so ~(A) C B. r

and so p =

= P- b=

=r

r. Since

- r(r 2

0

Theorem 1.5.18 (Wedderburn's principal theorem) Let A be a complex algebra such that A/rad A is the algebra 07=1 M n " where nl,.'" nk EN. Suppose that the standard set of minimal idempotents can be ltfted. Then A is decomposable. Proof Set R = rad A, and let (EW : r, s E Nn j ) be the standard system of matrix units in Mnj for j E N k • By hypothesis, there is an orthogonal set Ro.T· Ro.T}· (j) + R = E(j) J: "'T d' "'T { e(j) r r : r E 1'lnj') E l'lk In A sueh t h a t err rr LOr r E l'lnj an ) E l'lk'

eW,

For j E N k, set Pj = 2:;~ 1 so that {PI, ... ,Pk} is an orthogonal set of idempotents in A. Take j E Nk' By 1.5.7(iii), rad (pJApj) = pJRpj = (pjApj) n R and so pjApj/rad (pjApJ) ~ M nj . By 1.5.14, there is a system (eW : r, S E NnJ of

76

Algebrazc foundations

matrix units in A such that eW + R = EW (r, S E Nnj)j we may suppose that (J) : r, s E 1M} (i)...i. (j) wh enever Z. ,. ...i. • Th l' { e rs !'ilnj C Pj APj, so t h at ers ,. e uv J. e mear map () : A/ R -+ A such that ()(EW) = eW is a splitting homomorphism, and so A is decomposable. 0 Corollary 1.5.19 Let A be a jinite-dimenswnal, non-nilpotent complex algebra. Then A is decomposable, A contains a non-zero idempotent, and there exist kEN and nl, ... ,nk E N such that A ~ 0~=1 Mnj 0 rad A. Proof By 1.5.6(iv), rad A is nilpotent, and so, by 1.5.12. A is an SBI algebra. Thus, by 1.5.13(ii), orthogonal idempotents can be lifted. By 1.5.9, A/ R has the form specified in the theorem, and so the result follows. 0

Let B be a complex, unital algebra with a maximal ideal R such that R2 = O. Then R = rad B. Suppose, further, that R has finite codimension in B. By 1.4.38, there exists n E N such that B/R ~ M n , and we have B ~ Mn 0 R. Let (Eij : i,j E N n ) be the standard system of matrix units for Mn. For i,j E N n , set Bij = EiiBEjJ = {b E B : EiibEjj = b} and Rij = EiiREjj, so that B = L~j=l B ij , R = L~j=l ~j, and Bij = CEiJ + Rij . For each i,j E N n , the map x 1--+ EilxElj , B11 -+ B ij , is a linear isomorphism. Let V be a linear subspace of R 11 , and define W = L~j=l eil Velj' Then W is a subspace of R which is an ideal in Bj if V has codimension k in Rll, then W has co dimension kn 2 in R. Now let A be a complex, unital algebra with a maximal ideal M of finite codimension in A, and set B = A/M2. Then the unique maximal ideal of B is M/M2, and M/M 2 = rad B = R, say. Let V C Rll and W be as above. Then W + M2 is an ideal of A with M2 C W + M2 C Mj W + M2 has finite codimension in A in the case where V has finite co dimension in Ru. We briefly mention a second radical of an algebra. Definition 1.5.20 Let A be an algebra. The strong radical 9l(A) of A is the intersection of the maximal modular ideals of A. The algebra A is strongly semisimple if 9l(A) = O.

It follows from 1.4.34(iv) that rad A C 9l(A). A simple, unital algebra is strongly semisimple. Let E be an infinite-dimensional linear space. Then £( E) is semisimple, but it is not strongly semisimple because, by 1.4.35, 9l(£(E» contains :F£(E). The following result is similar to 1.5.4, and the proof will be omitted. Theorem 1.5.21 Let A be an algebra. (i) Let I be an ideal in A. Then 9l(I) = In 9l(A). In particular, if A zs strongly semiszmple, then I is strongly semisimple. (ii) Let I be an ideal in A with I c 9l(A). Then 9l(A/I) = 9l(A)/I. In particular, A/9l(A) is strongly semisimple. (iii) Let I and J be ideals in A with I c J c 9l(A), and let 7r : A/I -+ AIJ be the natural surjection. Then 7r(9l( AI I» = 9l( AI J). 0

Radicals and spectra

77

The third radical of an algebra that we shall consider in this section is the prime radicaL Definition 1.5.22 Let A be an algebra. The prime radical intersection of the prirne ideals of A.

~(A)

of A is the

Clearly ~(A/~(A)) = O. Since each prime ideal contains a minimal prime ideal, ~(A) is equal to the intersection of the minimal prime ideals of A. Proposition 1.5.23 Let I be an ideal in an algebra A. Then

~(I)

=

In~(A).

Proof Let P be a prime ideal in A, and take a, bEl with alb c In P. Then aAlb c P, and so, by 1.3.42, either a E P or Ib c P. In the latter case, lAb C P, and so either I c P or b E P. In each case, either a E P or b E P, and so In P is a prime ideal in I. Thus ~(I) c I n ~(A). Let a E I \ ~(I), and take a prime ideal Q in I with a f/: Q. Then 1\ Q is an m-system in A, and so, by 1.3.44(i), there is a prime ideal P in A with pn(I\Q) = 0. Since a f/: P, we have a f/: ~(A), and so In~(A) c ~(I). 0 Theorem 1.5.24 Let A be an algebra. Then A is a sernzprzrne algebra if and only if ~(A) = O. Proof Suppose that ~(A) = O. Take a E A with aAa = O. Then a E P for each prime ideal P of A, and so a = O. Thus A is a ::;emiprime algebra. Conversely, suppose that A is a semi prime algebra, so that, by 1.3.43, we have aAa # 0 (a E Ae). First take al E A-, and then successively choose a2, a3,'" in Ae so that a n +1 E (anAa n )- (n EN). For m ::::; n, we have an+l E amAa n , and so {an: n E N} is an m-system. By 1.3.44(i), there is a prime ideal P with al f/: P, and so ~(A) = O. 0 It follows that an ideal is semiprime if and only if it i::; an intersection of prime ideals.

Proposition 1.5.25 The following conditions on an algebra A are eq'uivalent:

(a) A is serniprirne; (b) A contains no non-zero, nilpotent left or right ideal; (c) A contains no non-zero, nilpotent ideal. Proof (a)::::}(b) Assume that I is a non-zero, nilpotent left or right ideal in A, say of index n + 1. Take a E l[n] \ {O}. Then aAa c l[n+l] = 0, and so a = 0, a contradiction. (b)::::}(c) This is trivial. (c)::::}(a) Set] = A.L, an ideal in A with ]2 = O. By (c). ] = O. Thus, if bE A and bA = 0, then b = 0; similarly, if Ab = 0, then b = O. Now take a E A with aAa = O. Then, successively, (AaA)2 = 0, AaA = 0, Aa = 0, and a = O. Hence A is semiprime. 0

Algebrazc foundations

78

Proposition 1.5.26 Let A be an algebra. (i) The prime radical of A is a nil ideal contained in rad A. (ii) SfJ(A) contains each nilpotent left or right ideal of A. (iii) Suppose that A is commutative. Then SfJ(A) = SJt(A). (iv) Suppose that A is commutative and complex, and that IJ1(A) has finite codimension in A. Then SfJ(A) = rad A. (v) Suppose that rad A is finite-dimensional. Then SfJ(A) = rad A and SfJ(A) is the maximum nilpotent ideal in A.

Proof (i) Take a E A \ SJt(A). By 1.3.44(ii), there is a prime ideal P with a tj. P, and so a tj. SfJ(A). By 1.4.34(iii), SfJ(A) c rad A. (ii) Let I be a nilpotent left or right ideal in A. Then I + SfJ(A) is a nilpotent left or right ideal in AjSfJ(A). By 1.5.24, AjSfJ(A) is a semiprime algebra, and so, by 1.5.25, I C SfJ(A). (iii) Let a E IJ1(A). Then A#a C SJt(A) because A is commutative, and so a E A#a C SfJ(A) by (ii). (iv) For each prime ideal P in A, we have P :::) SJt(A) and dim(Aj P) < oc, and so, by 1.3.57, P is a maximal modular ideal. Thus radA C P, and so rad A C SfJ(A). Always SfJ(A) C rad A. (v) By 1.5.6(iv), radA is a nilpotent ideal, and so, by (ii), radA C SfJ(A). Thus SfJ(A) = rad A is the maximum nilpotent ideal. 0

Definition 1.5.27 Let A be a unital, complex algebra, and let a E A. resolvent set of a in A is

PA(a) = {z

E

C : ze A - a

E

The

Inv A} ,

and the resolvent function is the map Ra : z

f-+

The spectrum of a is 0" A(a)

(zeA - a)-I, =

vA(a)

PA(a)

-+

Inv A.

C \ PA (a), and the spectral radius of a is =

sup{lzl : z E O"A(a)}.

We take vA(a) = 0 if O"A(a) = 0, and vA(a) = 00 if O"A(a) is an unbounded set in C. We usually write pea) and O"(a) for PA(a) and O"A(a), respectively. Now suppose that A is a complex, non-unital, algebra, and that a E A. Then we define PA, R a , 0" A (a), and v A (a) by regarding a as an element of A #; in this case, necessarily 0 E 0"( a). If A is either unital or non-unital, then

O"A(a) U {O} = {z E e

: ajz

is not quasi-invertible in A} U {O}

(a E A).

Note that, in the case where A is commutative and unital, z E O"(a) if and only if (zeA - a)A i= A. For example, let E be a finite-dimensional space, and let T E £(E). Then O"(T) consists of the eigenvalues of T. We shall use the following easily checked identity. For each a E A, Ra(w) - Ra(z) = (z - w)Ra\z)Ra(w)

(z,w

E

p(a)).

(1.5.4)

79

Radicals and spectra

~ B be a O"A(a)U{O}, vB(O(a» :::; vA(a),

proposition 1.5.28 Let A and B be complex algebras, let () : A

homomorph2sm, and let a E A. Then O"B(O(a» anda(~A) C O"A(a) U {a}.

C

Proof Take z E PA(a) \ {O}. Then a/z E q-Inv A, and so O(a)/z E q-Inv B, whence z E PB(B(a». For 'P E ~A, 'P(a) =1= z, and so a(A) C O"A(a) U{O}. 0 Proposition 1.5.29 Let A be a complex algebra with radical R, and let a, b E A.

(i) 0" A/R(a + R) U {a} = O"A(a) U {a} and VA/Rea + R) = VA(a). (ii) O"A(ab) U {O} = O"A(ba) U {O} and vA(ab) = vA(ba). (iii) (TB(a) = o"A(a) for each maximal commutative subalgebra B of A wzth aEB. (iv) Suppose that A is unital. Then (J,C(A) (La) = (JC(A) (Ra) = (JA(a). (v) Suppose that E is a left A-module and 0: E C e 2S such that a . x = o:x for some x E Ee. Then 0: E (JA(a). (vi) Suppose that a = 2::7=1 O:jPj, where {PI, ... ,Pn} is an or·thogonal set m J(A) \ {O} and 0:1>"', O:n E e. Then {0:1' ... , O:n} C

(vii) O"A(a)

= U{(J A/p(a + P)

0" A

:P

(a)

E

C

{O, 0:1, ... , O:n} .

ITA}.

Proof (i) By 1.5.28, (JA/R(a + R) C (JA(a) U {a}. Now take z E PA/R(a + R) with z =1= O. Then there exists c E A with co (a/z), (a/z) 0 c E R. By 1.5.2(iii), co (a/z) E q-Inv A, and so a/z E q-Inv A. Thus PA/R(a + R) \ {a} C PA(a). (ii) We have noted that ab E q-Inv A if and only if ba E q-Inv A. (iii) Let B be a maximal commutative sub algebra of A with a E B. Certainly o"A(a) C O"B(a). Let z E PA(a). Then there exists c E A# with

(zeA - a)c = c(zeA - a) = eA . For b E B, (zeA - a)b

= b(zeA - a),

and so bc

= cb.

Thus c E B# and z E PB(a).

(iv) By 1.5.28, O"c(A)(L a ) C O"A(a). Now suppose that La E Inv A, and take T E C(A) with LaT = TLa = lA. Set c = T(eA)' Then ac = (LaT)(eA) = eA. and so LaLc = IA. Thus Lc = T, and hence ca = LaLc(eA) = eA, showing that a E Inv A. It follows that (Jc(A)(L a ) = (JA(a). Similarly O"c(A)(Ra) = O"A(a).

a+

(v) Assume that 0: EPA (a). Since 0: =1= 0, there exists c E A such that o:c - ca = 0, and then o:x = (a + o:c - ca) . x = 0, a contradiction. (vi) By (v), {0:1,'" ,O:n} C (JA(a). For z E c=

t

)=1

e \ {0:1,""

(~)P)' z 0:)

Then zc + a - ca = zc + a - ac = 0, and so z E PA(a). (vii) We may suppose that A is unital. By 1.5.28, U{O"A/p(a+P): P E ITA} C (JA(a).

O:n}, set

80

Algebraic foundatwns

We claim that, if bE A and b+ P has a left inverse in AlP for each P E ITA, then b has a left inverse. For assume that Ab -I- A. Then there is a maximal left ideal M in A with Ab c M. Set P = M : A E ITA. There exists c E A with cb + P = eA + P, and so eA E A1, a contradiction. The claim follows. Now assume that z E aA(a), but that z tf. aA/p(a + P) (P E ITA)' By the claim, zeA - a has a left inverse, say b. Since z tf. aA/p(a + P) (P E ITA), the element b + P has a left inverse in AlP for each P E ITA. By the claim again, b has a left inverse, and so zeA - a = b- l E Inv A, a contradiction. It follows that aA(a) C U{aA/p(a + P) : P E H A }, giving the result. 0 The following result will be crucial for the proof of the Gel'fand-Mazur theorem for topological algebras. Theorem 1.5.30 Let A be a complex dimsion algebra such that a(a) each a E A. Then A = CeA.

-I-

0 for

Proof Let a E A. Since a(a) -I- 0, there exists z E C with zeA - a tf. Inv A. Since A is a division algebra, it follows that zeA -a = 0, and so a E CeA. 0 Definition 1.5.31 Let A be a complex algebra. Then a E A is quasi-nilpotent if vA(a) = O. We write D(A) for the set of quasi-nilpotent elements of A. Clearly a if and only if Ca C q-Inv A.

E

D(A)

Proposition 1.5.32 Let A be a complex algebra. Then:

(i) meA)

c

(ii) radA

=

(iii) I

D(A); {a E A : A#a C D(A)}

c

D(A);

c rad A for each left zdeal I with I c D( A);

(iv) A is radical if and only zf D(A) = A. Proof (i)-(iii) These are immediate from 1.5.6(i), 1.5.2(iv), and 1.5.2(iii), respectively. (iv) This follows from (ii) and (iii).

0

Even if A is commutative it is not necessarily the case that radA = D(A). For let A be a complex field with A -I- CeA. Then D(A) = A \ C·eA, but radA = O. We shall require a generalization for commutative algebras of the notion of the spectrum of an element. Definition 1.5.33 Let A be a commutative, unital, complex algebra, and let a = (aI, ... ,an) E A (n). The joint spectrum of a is aA(a)

= aA(al,'"

,an) = {z = (Zb.'·' zn) E en: (zleA - adA + ... + (ZneA - a)A =1= A}.

polynomwls and formal power senes

81

Let A be a unital algebra, and let a, b E A (k). Then a . b = L7=1 ajb j • The element a is unimodular if there exists bE A(k) with a . b = eA. Clearly, if A is commutative and a E A(k), then

a(a) = {z We define a

=

E

C k : zeA - a is not unimodular in A}.

(a1' ... ,an) E A(n) for a E A (n). Clearly we have

a([>A)

C

aA(a)

(a

(1.5.5)

E A(n).

It is clear that the new definition of a(a) coincides with the old definition in the case where n = 1, and that we always have a(a1, ... , an) C 0;=1 a(aj). Notes 1.5.34 Let A be an algebra. A left A-module is said to be semisimple if it is a direct sum of simple modules. An algebra A is semisimple (in the sense of 1.5.1) if and only if A is semisimple as a left A-module. For the radical of an algebra, see (Hungerford 1980, §IX.2), (Jacobson 1980, Chapter 4), and, in particular, (Jacobson 1956); the calculations on SBI algebras are from (Jacobson 1956, IlL8). (The letters 'SBI' stand for 'suitable for building idempotents'.) A form of Wedderburn's principal theorem is given in the early work (Albert 1939). The strong radical of an algebra is sometimes called the Brown-McCoy radical; it is characterized in (Palmer 1994, §4.5). The prime radical of an algebra is called the Baer radical in (Divinsky 1965) and (Palmer 1994, §4.4); it is a semiprime nil ideal which is included in every semi prime ideal. See also (McCoy 1973). An example of a simple, radical algebra is given in (Sasaida and Cohn 1967), and an example to show that .;p(A) need not contain every nil ideal of A is in (Baer 1943). Several other 'radicals' are considered in the literature; see (Divinsky 1965), for example. A 'history of radicals' is given in (Palmer 1994,4.8.1).

1.6

POLYNOMIALS AND FORMAL POWER SERIES

We shall continue in this section our accumulation of algebraic results that will be required in the later chapters of this book. We shall consider the polynomial algebra A[X] and the formal power series algebra A[[X]] over an algebra A, and their n-variable analogues; this leads to a study of algebraic and transcendental elements of an algebra with respect to a subalgebra, and, in Theorem 1.6.31, to a condition for the extension of a homomorphism defined on a subalgebra of a given algebra. Recall that the ground field OC of an algebra A is always IR or C. Definition 1.6.1 Let A be a unital algebra over K Then:

A[[X]] = {a = (an: n A[X] = {a = (an: n

A}; A[[X]] : an = 0 eventually}.

E Z+) : an E E

Z+) E

For a = (an) and b = (b n ) in A[[X]] and a E lK, set a + b = (an aa = (aa n : n E Z+), and ab = (tarbn-r : n r=O

E

Z+) .

+ bn : n

E Z+),

Algebrazc foundations

82

With respect to these operations, A[[X]] is an algebra over lK, and A[X] is a subalgebra of A[[X]]. The algebra A[[X]] is the algebra of formal power series over A, and A[X] is the polynomial algebra over A; its elements are polynomzals. Clearly IA[X]I = IAI. We identify A with {(a,O,O, ... ): a E A}, so that eA is the identity of A[[X]J, and we write X for (0, eA, 0, 0, ... ); X is the zndeterminate of the algebras A[[X]] and A[X]. Each formal power series a = (an) E A[[X]] can be formally expressed as 00

where XO

= eA,

and each polynomial p

= (an)

E

A[X] can be expressed as

n

p = p(X)

=

2:arXr. r=O

-I- 0, and then the above expression for p is unique; in this case, the degree of p, written op, is n, and the leadzng coefficient of p is an. We set 00 = -00, with the conventions that -00 + (-00) = -00 and that -00 < nand -00 + n = -00 for n E Z+. A non-zero polynomial is monic if its leading coefficient is eA. In particular, we have now formally defined the algebras qX] and q[X]]; the identity of qX] is denoted by 1. We define Ao[X] = {(an: n E Z+) E A[X] : ao = O},

If Pi- 0, then we may suppose that an

and, for n EN,

A(n) [X]

=

{p

E

A[X] : op:S; n}.

(1.6.1)

Let p, q E A[X]. Then o(p + q) :s; max{ op, oq}, and so A(n) [X] is a linear subspace of A[X]. Also, o(pq):S; op+oq, and o(pq) = op+oq in the case where A is a domain. Suppose that A is commutative. Then A[[X]] and A[X] are also commutative; if A is an integral domain, then A[[X]] and A[X] are integral domains. In the latter case, the quotient fields of A[[X]] and A[X] are denoted by A«X)) and A(X), respectively, so that

A(X) = {pjq: p

E

A[XJ, q E A[X]-};

for example, IC(X) is the field of rational functions. Let p = E~=o arXr E A[X], and let bE A. Then p(b) = E~=o arb r (where bO = eA). In the case where A is commutative, the map p 1-+ p(b), A[X] ~ A, is a unital homomorphism. Let A be a commutative, unital algebra, let p be a monic polynomial in A[X] with op = n + 1, where n ~ 1, and let Ap = A[X]jpA[X]. Then each element of Ap corresponds uniquely to an element E~=o aiX i , where ao, al, ... ,an E A. We regard Ap as an algebra containing A as a subalgebra. Now suppose that B is a commutative, unital subalgebra of A and that a E Be. Then:

B[a]

=

{p(a) : p E B[X]};

B(a) = {p(a)q(a)-l : p, q E B[X], q(a) E Inv A}.

Polynomial8 and formal power series Thus B[a] = algA(B U {a}) and B(a) is the smallest inverse-clo~ed subalgebra of A containing B U {a}. Let a E A. Then OCo [a] = {pC a) : p E OCo [Xl}, so that a E OCo[a] and OCo[a] is a commutative subalgebra of A. Let A be a commutative, unital algebra, let p E A[X], and let a E A. Then a is a root of p if pea) = o. Let q E A[X]-. Then the dwi8ion algorithm asserts that there exist b E A- and r, s E A[X] with bp = qr + s and as < aq; if tht' leading coefficient of q belongs to IllV A (and in particular if A is afield), then we may take b = eA, and the representation p = qr + s is then unique. Thus, if a is a root of p, there exist mEN and q E A[X] with p = (X - a)m q and q(a) "10: a is then a root of mult1phcity m. If A is an integral domain and p E A[X]-, then p has at most ap roots (including mUltiplicities). It is immediate from the division algorithm that OC[X] is a principal ideal domain. Let p = L~=o arxr E A[X]. Then the formal denvative of p is the polynomial p', where p' = a1 + 2a2X + ... + na n X n - 1 . The kth formal derivative p(k) is defined by induction: p(j+l) Clearly we have

p

= (p(J)' (j EN).

(n)(o)

= p(O) + p'(O)X + ... + P_ _ xn;

(1.6.2)

n!

this is Taylor'8 formula for polynomials.

Definition 1.6.2 Let A be a commutative, unital complex algebra. A polynomial p E A[X] of the form

(1.6.3) where n ;::- 2 and ao, a2, ... ,an E rad A 1S a Henselian polynomial. The algebra A is Henselian if each Hen8elian polynomial has a root in A. Each root of a Henselian polynomial is necessarily in rad A. Let p be a Henselian polynomial. Then, for each a E A, p' (a) E e A + rad A c Inv A and p(k)(a) E radA (k = 2, ... , n), and also

pCb) - pea) E (b - a) . (eA

+ rad A) c

(b - a) . lnv A

(a, bE A),

(1.6.4)

so that p has at most one root in A.

Proposition 1.6.3 Let A be a Henselian algebra, and let

p = aD

+ alX + a2X2 + ... + anXn ,

where n ;::- 2, ao E rad A, al E Inv A, and a2, ... ,an E A. Then p has a root in radA. Proof We may suppose that a1 = eA. Set q(X) = p(aoX - aD) E A[X]. Then q(O) = p( -ao) E a6A, and q(k) (0) = a~p(k) (-ao) E a6A for k = 2, ... ,n. On the other hand, q'(O) = aop'( -aD) E ao . lnv A, and so q = hr, where bE radA and r is a Henselian polynomial in A[X]. By hypothesis, r has a root, say a, in A, and then aoa - ao is a root of p in rad A. 0

84

Algebraic foundatwns

Proposition 1.6.4 Let A be a local, Henselian algebra, and take a E A and n ~ 2. Suppose that there exists b E A with bn Ea' Inv A. Then there exists c E b . Inv A with cn = a. Proof Take 0: E C· and Cl E MA with bn = a(aeA + cd. Choose (3 E C· with (3n = a, and set p = ((3eA + x)n - (aeA + cd. Then p(O) = -Cl E MA and p'(O) = nBn-leA E Inv A. By 1.6.3, there exists C2 E MA with P(C2) = O. Set c = b((3cA + C2)-1. Then c has the required properties. 0 For P E qX], define Z(p)

= {( E C: p(() = O}.

Proposition 1.6.5 (i) Each character on qX] has the form some (E C.

C( :

P

f-+

p(() for

(ii) Each non-zero ideal m qX] has fimte codimension m qX]. (iii) Each maxzmal ideal m qX] zs the kernel of a character. (iv) Each non-zero, prime ideal in qX] is a maximal ideal. (v) The algebra qX] is semisimple.

(vi) Let PI, ... ,Pn E qX] be such that n?=l Z(Pj) q1, ... , qn E qX] such that 2:.]=1 Pjqj = l.

= 0.

Then there exist

Proof (i) For cp E qX], set (= cp(X). Then cp(p) = p(() = C«(p) (p E qX]). (ii) Let I be a non-zero ideal in qX]. Since qX] is a principal ideal domain, 1= pqX] for some P E qX]·, say P = TI~=l(X - (j)"i, where (1, ... ,(k are distinct points in C and nl, ... , rLk EN. Clearly

I = {q

E

qX] : q((j) = ... = q(nr 1)((j) = 0 (j E Nk)},

and so I has finite codimensioIl in qX]. (iii), (iv) The ideal I in (ii) is maximal (equivalently, prime) if and only if P has the form X - ( for some ( E C, and then I = ker C(. (v) This is now immediate. (vi) Let I = 2:..7=1 pjqX] be the ideal generated by {PI, .. . ,Pn}. Assume that I ii:l a proper ideal. Then I is contained in a maximal ideal of q X]. and so, by (iii), I c ker C( for some ( E C. But this is a contradiction of the fact that n;=l Z(Pj) = 0. Thus I = qX], and the result follows. 0 Definition 1.6.6 Let A be an algebra. An element a of A is algebraic if OCo[a] is finite-dimensional, and A zs algebraic if each element of A is algebraic. Clearly each finite-dimensional algebra is algebraic. Proposition 1.6.7 Let A be a complex algebra, and let a E A. Then the following conditions on a are equivalent: (a) the epimorphzsm p f-+ p(a), Co[X] - 7 Co [a], is an isomorphism; (b) a is not algebraic; (c) the set {an: n E N} is linearly independent.

Polynomials and Jormal power series

85

Proof By lo6.5(ii), the kernel of the epimorphism in (a) is either 0 or has finite codimension Co[X]. The result follows. 0 Corollary 1.6.8 Let A be a complex algebra, and suppose that a E A is not algebraic. Then Co [a] is semism~ple. Proof By lo6.5(v), Co[X] is semisimple. By the implication (b)=>(a), above, Co[X] ~ Cora], and so Coral is semisimple. 0 Proposition 1.6.9 Let A be a commutatwe, complex algebra, and let a E A. Then the Jollowing condztions on a are equzvalent:

(a) (b) (c) (d)

a is algebrazc; a E (E(A) 891(A); a E CeA

+ Q Jor

each przme ideal Q in A#.-

Jor each prime zdeal P in A, there exzsts p E Co[X]- with pea) E P.

Proof (a)=>(b) By lo5.8(i), Coral

=

(E(Ca[a]) 8 91(C o [a]) C (E(A) 891(A).

(b)=>(c) By 1.5.26(iii), we have 91(A) c Q, and so a E (E(A#) 1.3.46(ii), dim ((E(A#)/(~(A#) n Q)) :s; 1, and so a E CeA + Q.

+ Q.

By

(c)=>(d) By 1.3.54(i), there is a prime ideal Q of A# with Q n A = P. Take o E C with a - oeA E Q, and set p = X(X -0). Then p E Ca[X]- and pea) E P. (d)=>(a) Define U = {pea) : p E Ca[X]-}. Then U[2j c U. Assume that U. Then, by 1.3.44(i). there is a prime ideal P in A with P n U = 0, a contradiction of (d). So 0 E U, and (a) follows. 0

o 1.

Proposition 1.6.10 Let A be a commutatwe, complex algebra. Then the Jollowing condztzons on A are equivalent:

(a) A is algebraic; (b) A

=

~(A)

891(A);

(c) each prime zdeal in A zs the kernel oj a character on A; (d) each prime ideal in A has jinite codimenszon. IJ the condl,tzons are satzsjied, then rad A = 91( A).

Proof (a)q(b) This followi:i from the equivalence of (a) and (b) in 1.6.9.

(a)=>(c) Let P be a prime ideal in A. By 1.3.54(i), there is a prime ideal Q in A# with Q n A = P. By 1.6.9, A c CeA + Q, and so, by 1.3.37(ii), there exists

A# with M<.p = Q. Set 'l/J =

A· (c)=>(d) This is trivial. (d)=>(a) Let a EA. For each prime ideal P in A, the set {an + P : n E N} is linearly dependent, and i:iO there exii:iti:i p E Ca[X]- such that pea) E P. By the implication (d)=>(a) in 1.6.9, a is algebraic. Thus A is algebraic. Now suppose that A = (E(A) 891(A). By 1.5.6(iii), 'Jt(A) C rad A, and, by 0 1.5.7(iv), \C(A) nradA = {O}. Thus radA = 91(A).

Algebrazc foundatwns

Proposition 1.6.11 Let A be a unital, complex algebra, and let a E A. (i) For each p E qX], a(p(a»

= p(a(a».

(li) Suppose that a is algebraic. Then a(a) is fimte. Proof (i) We may suppose that op ~ 1. Take z E C. Then there exist 0:0 E C· and (Xl ....• (1:" E C such that z - p = ao(X - ad", (X - an), and HO it is dear that ZPJ\ - pea) E Inv A if and only if a - {tjeA E Inv A (J E N n ). Thus ::; E a(p(a») if and only if a 7 E a(a) for some j. The result follows.

o

(ii) This is immediate from (i).

Definition 1.6.12 An algebm A is spectrally finite ~f the spectrum a(a) is finite for each a E A. It follows from 1.6.11(ii) that each algebraic algebra is spectrally finite.

Proposition 1.6.13 Let A be an algebraic, commutative, unital, complex algebra. SUppOi'Ie that \)"leA) is nilpotent, that 1> A is fimte, and that A is mfinitedimensional. Then there zs a maximal ideal M in A such that M / M2 is mfimtedimensional. Proof Set N = \)"leA). By 1.5.26(iii), N = !.p(A), and so, since A is algebraic, it follows from 1.6.10 that A = (E(A) 8 N and that N = n{M

A}. The space 1>A is finite, say 1>A = {'Pl"'.,'Pn}; N has finite codimension in A, and so dim (E(A) < 00 and dim N = 00. By 1.3.23(iii), there is an orthogonal set {JlI,'" ,Pn} of non-zero idempotents in A such that (E(A) = lin {Pl .... ,Pn}. By 1.5.7(ii), eA = PI + ... + Pn· Since N is nilpotent and dim N = 00, dim (N/N 2 ) = 00 by 1.5.6(viii). We have N = O.?=l pjN and N2 = 0]=1 pj N 2 , and so N/N 2 = 0;'=1 pj N/p j N 2. Thus there exiHts j E Nn such that dim (p j N/p j N 2) = 00; Het M = M'PJ' so that pjM =N. Take rn E N and al,"" am E pjN such that {al + p J N 2 , ... , am + p j N 2 } is a linearly independent set in pj N/pj N 2, and then take al, ... ,a m E C with O'lal + ... + ama m E M2, say a = alaI + ... + ama m . Then a =pja EPj A1 2 =pJ N

2 ,

and so al = ... = am = O. Hence {al +.iVJ 2, ... , am +J1,f2} is linearly independent in M/M2. It follows that dim (M/M2) = 00. 0 We shall be concerned later with modules over the principal ideal domain qX]. The example we have in mind is the following. Example 1.6.14 Let E be a linear space, and let T E C(E). Then E is a unital qXl-module with respect to the map (p, x)

r--t

p(T)x,

qXl x E

-7

E.

Clearly E is a divisible module if and only if (zIe - T)(E) = E for each z E C, where IE is the identity operator on E, and E is a torsion module if and only if,

87

Polynomials and formal power ser'ies

for each x E E, there exists p E crXl- with p(T)x = 0. The algebraic spectral space ET(0) is the maximum divisible submodule of E. Let E and F be linear spaces, let R E C(E), and let S E C(F). Clearly T intertwines (R, S) if and only if T is a crXl-module homomorphism for the module operations on E and F given by Rand S, respectively. 0 Theorem 1.6.15 Let E be a umtal torswn module over C[XJ, and define DC

U{x E E: (X -

Ee, =

on .

X

= O}

«( E q.

n=l

Then each Ec, is a crXl-submodule of E, and E

= O{Ec, : (

E

C}.

Proof Take x E E. Since E is a torsion module, there exists p E crXl- with P . x = O. Since E is unital, necessarily op ~ 1, and so we may suppose that p = rr;=l (X _(J)n j , where (1,"" (k are distinct points in C and nl .... , nk EN. For each j E Nk. define Pj = rri;ij(X - c)n. E crXl. By 1.6.5(vi), there exist ql, .... qk E crXl such that Plql + ... + Pkqk = 1. For j E N k , define x J = PjQj . x; we have (X - (J)n j • x J = qj . (p . x) = 0, and so Xj E Ee,j' Also x = Xl + ... + Xk, and ~o E = L:{Ee, : (E C}. Suppose that x E Ec, and that x = Xl + ... + Xk, where x J E Ec,j and (1. {(l, ... ,(k}. Then there exist nl, ... ,nk E N such that r· X = 0, where r = rr~=l (X - (j)n J • Also, there exists n E N with (X - ()n . X = O. Take p,q E crXl with (X -Onp+rq = 1. Then x = p. ((X -on. x)+q· (r· x) = 0, and so the sum is direct. 0 Let A be a commutative, unital algebra. The process of forming the algebra of formal power series and the polynomial algebra over A can be iterated. Definition 1.6.16 Let A be a commutative, umtal algebra. Then we recursively define:

A[[X1 .... , Xn+lll = (A[[X I , ... Xn]])[[Xn+dJ,} A[X I , ... , Xn+ll = (A[X I1 ···, XTl])[Xn+ll ' I

(n E N).

The algebra~ A[[X1, ... , X nll and A[X I, ... ,XTl ] are the algebra of formal power series and the polynomial algebra in n indeterminates over A, respectively, and A[Xl"" Xnl is a subalgebra of A[[X I , ... , Xnll. Certainly we have A[X l , ... , Xnl ~ (A[X l , ... , X m ])[Xm +1 , ••• Xnl for m < n, and we shall identify these algebras. A generic element of A [[Xl , ... , Xnll will be denoted by I

I

a --

'~ar " xr -- '~arl' " .. ,rn x 1r ,

...

Xrn n

,

(1.6.5)

where the sums are taken over r = (rl,"" 'r,,) E z+n; if a E A[Xl , ... , Xn], then all save finitely many of the coefficients ar are 0. The algebra
88

Algebrazc foundations

Let p = L arX r E A[X I , ... , Xn], and let bl , ... , bn E A. Then we define p( b) = p( bi , ... , bn ) = L arb r , where br = b~' ... b;,n; again the map

p f-t p(b),

A[XI"'" Xn]

---+

A,

is a unital homomorphism. In conformity with the earlier usages lK[a] and lK(a). we denote by lK[S] and lK(S) the smallest unital subalgebra and the smallest inverse-closed subalgebra of A containing a subset S, so that lK[S]

= {p(al,"" an) : p E

lK[XI"'" Xn], 0,1."" an E S. n E N},

lK(S) = {ab- 1 : a E lK[S], bE lK[S] n Inv A}. Also we set lKo[S] = {p(a1'" .,an); p(O) = O} = algS. and write lKo[a1,'" ,an] for lKo[{a1, ... ,an}], etc. Each polynomial p in n indeterminates Xl,"" Xn can be regarded as a polynomial in Xj over A[X1, .... Xj-1,Xj+l, ... 'Xn]; the formal derivative of this polynomial is the formal partzol der"ivatzve of p with respect to X j , and it is denoted by ap/aXj .

qx

Proposition 1.6.17 Let n E N. Then each character- on 1 , ..• , Xn] has the form C( : p 1-+ p«() for some ( E and 1 , ... , Xn] is sernisimple.

en,

qx

Proof For cp E
=

«(1, .. " (71)' D

The algebras lK[[Xll and lK[[X l • ... ,Xnll will have particular importance in this work; throughout they are denoted by ~(lK)

and

~n(lK).

respectively.

(1.6.6)

and we write ~ and ~n for ~(q and ~n(q. respectively. Thus ~ = ~(c, z+) anel ~n = ~(c.z+n) in the notation of §1.2. The identity of ~n(lK) is denoted by 1. For k E z+n, set (1.6.7) Then {71'k ; k E z+n} is the set of coordmate projections from ~n. We give a second prescntation of ~n(lK). Let T = (rl, ... ,Tn) E z+n, and sct 1,,.1 = 7'1 + ... +r n · A monomial is the characteristic function of an element, say r, of z+n, and the degree of the monomial is Irl. For example, Xj is the monomial of the element (OJ.1, ... ,OJ,n) of z+n. Let r E Z+. Then a homogeneous polynomial of degree k is a linear combination (necessarily finite) of monomials of degree k. An element of ~n(.lK) may be identified with a sequence (ak : k E Z+), where each ak is a homogeneous polynomial of degree k. The product of two homogeneous polynomials of degree k and e, respectively, is a homogeneous polynomial of degree k + e. Let A be a commutative, unital algebra, and take n, kEN. Then clearly the set {~Irl>k arXr} is an ideal in A[Xl' ... , Xn], and the quotient algebra is identified with {~lrI9 arX r }. In this space, the product is specified by the rule that xr XS = 0 whenever Ir + 81 > k.

Polynomwls and Jormal power senes Definition 1.6.18 Let a

= (a r : r

E

89 z+n) E In (OC)·. Then the order of a is

o(a) = min{lrl E Z+ : ar =I O}; also, 0(0) =

00.

Further, JOT r E Z+, Mr = {a E In(OC) : o(a) ::::: r}.

It is clear that o(a a, bE In(OC).

+ b) 2

min{o(a),o(b)} and that o(ab) = o(a)

+ o(b)

for

Proposition 1.6.19 Let n E N.

(i) Let a

E

In(OC). Then a is mvert'ible zJ and only zJ o(a)

(ii) The algebra

J n (OC)

= 0.

is a local algebra with rnaximal zdeal M 1 .

(iii) For kEN, Mf = Mk, Jvh has finite codzmension m In(OC), Mk is generated by {xr E In(lK) : Irl = k}, and each zdeal contaming Mk 1.S finitely generated.

°

Proof (i) Certainly o( a) = for a E Inv J n (lK). Now take a = (a r : r E Z+) with o(a) = 0, say ao = 1. Set bo = 1, and inductively define br, for r E N by setting br, = - (a1 br - 1 + ... + arb o). Then each br is a homogeneous polynomial of degree r, and it is easily checked that ab = 1, where b = (b r : r E Z+).

o

Oi) and (iii) These are clear.

Proposition 1.6.20 Let I be a non-zero ideal of J(OC). Then there exzsts r E Z+

such that I

=

Mr = xrJ(OC).

Proof Set r = min{o(a) : a E I}, and take a E I with o(a) = r. Then a = Xrb, where o(b) = 0. By 1.6.19(i), bE InvJ(OC), and so x r E I and I = Mr. 0

Let n E N with n 2 2. For each subset S of z+n such that 7' E S whenever r E z+n and r ~ s for some s E S, the family {a E In : as = (s E S)} is an ideal in In. However, there are many more ideals not of this form. Let a, b E J with 1l'o(b) = O. Then formal composition a 0 b can be defined: for each n E Z+, 1l'n(b T ) = for r > n, and so 1l'o(a 0 b) = 1l'o(a) and 1l'n(a 0 b) = L;'=o 1l'r(a)1l'nW) (n EN).

°

°

Proposition 1.6.21 Let b E J w1.th 1l'o(b)

= 0, and set Ab(a) = a 0 b (a

E

J).

(i) Ab is an endomorphism of J with Ab(Mn) C Mn (n E Z+).

(ii) Suppose that 1l'l(b) =I 0. Then there exists c E J with 1l'o(c) = 0, 1l'l(C) =I 0, and Ac = Ab 1 . In particular, Ab is an automorphism of J. Proof (i) This is clear.

(ii) Define 1'1 = 1l'l(b)-l, and successively define I'n for n 2 2 so that 2:~=1 I'r1l'n(b r ) = O. Then c = 2:::1 I'rxr has the required properties. Take a E J. Then Ab(Ac(a)) automorphism.

=

a, and so Ab is a surjection, and hence an 0

90

Algebraic foundations

Definition 1.6.22 Let £(OC) be the set of sequences a = (an: n E Z) such that an E OC (n E Z) and such that ther-e exists no E Z with an = 0 for n < no. For a = (an) and b = (b n ) in £(OC) and for a E lK, set a + b = (an + bn : n E Z), aa = (WIn: n E Z), and

For each nEZ, a, bn - r -I- 0 for only finitely many values of r, and so ab is well-defined. It is now ellliily checked that £(OC) is an algebra over OC: it is the algebra of Laurent series (in one indeterminate). We write £ for £(q. Each non-zero element of £(OC) can be uniquely expressed in the form L~=no anXn, where no E Z and ano -I- O. The following result is clear. Proposition 1.6.23 The algebra £(OC) is the quotient field of ~(OC).

0

Thus £(OC) = OC«X)) in our previous notation. Note that OC(X) is a subfield of £(OC), and that £(OC) = ~(OC, Z) in the notation of 1.2.24 and 1.3.63; the product in £(OC) coincides with *. The order 0 extends to £(OC). We define the following linear subspace of £: X

=

{t

U

x_jx-j : Xo, X-l,""

X-k E

C, k E Z+} .

(1.6.8)

J=O

Then £ = X + ~ and X n ~ = Cl; the projection L:=no anXn f----t L~=no anXn of £ onto X is denoted by P x . The space X is a unital, torsion ~-module with respect to the map (a,x)

f----t

a . x

=

Px(ax),

~ x

X ---+ X.

(1.6.9)

There is one further subfield of £ that we shall refer to later. introduce C{ X}, the algebra of absolutely convergent power senes:

C{X} =

{~anxn : ~ lanl En <

00

for some E>

First we

o} .

Then qX) is defined to be the quotient field of C{ X}; qX) is the field of meromorphic functions at O. We have qX) c qX) c q(X)) = £. Definition 1.6.24 A subalgebra A of ~ is ordinary if, faT each a E A-, there exists kEN with Xk E aA.

By 1.6.20, J' itself is ordinary; also, the subalgebra C{ X} is ordinary. Proposition 1.6.25 Let A be an ordinary subalgebra of J', and let E be an A-module. Then E is divisible if and only if E is X -divisible. Proof Suppose that E is X-divisible, and take a E A-. Then there exists kEN and bE A with X k = abo For each x E E, there exists y E E with X k . y = x, and then x = ab . yEa· E, and so E is a-divisible. Thus E is divisible. 0

Polynomials and formal power series

91

It follows that X is a divisible ~-module.

Let A and B be commutative, unital algebras, and let () : A --t B be a unital homomorphism. Take p E A[X}, ... , Xn], say p = 2: arxr. Then the element ()p E B[X 1, ... ,XnJ is defined by

(l.6.10) clearly, the map () : p morphism, and

f--->

()p, A[X 1, ... , XnJ ....... B[X 1, ... , XnJ, is a unital homo-

(()p)((}(ad , ... ,(}(an )) = ()(p(al"" ,an))

(a1,'" ,an E A).

Definition 1.6.26 Let B be a unital subalgebra of a commutative, unital algebra A, and let a E A. Then: (i) a is algebraic over B tf there exists p E B[XJe with p(a) = 0; (ii) a is integral over B if there exists a momc polynomwl p E B[XJ with p(a) = 0; (iii) a is transcendental over B if a is not algebratc over B. For example, let B = qXJ and A =~. Then the element exp X E ~ is transcendental over B; here, exp X is defined as 1 L,xn. n. 00

expX=

n=O

A subset S of A is algebraic over B if each element is algebraic over B. The set of elements of A which are algebraic over B is the algebraic closure of B in A, denoted by (l.6.11) B is algebraically closed in A if Alg B = B, and A is an algebratc extension of B if Alg B = A. Clearly IAlgABI = IBI in the case where A is an integral domain. The set of elements of A which are integral over B is the integral closure of B in A, denoted by IntAB or lnt B, and B is integrally closed in A if lnt B = B.

Lemma 1.6.27 Let a E A. Then a tS integral over B ~f and only if there is a finitely generated B-submodule E of A with eA E E and aE c E. Proof Suppose that a is integral over B. Then there exists n E N such that an E E, where E = BeA + Ba + ... + Ba n - 1. Then E is a finitely generated B-submodule, eA E E, and aE C E. Conversely, suppose that E = Bb 1 + ... + Bb n is a B-submodule with eA E E and aE c E. For i E N n , we have abi E E, say abi = 2:;=1 bijbj , and (b 1, ... , bn ) is a solution of the set of simultaneous linear equations n

i)bi] - oi,]a)tj

=

0

(i

E

N n ).

j=1

Let d = det(bij - Oi,ja). Then db i = 0 (i E Nn ) by 1.3.9(iii), and so d . E = O. Since eA E E, necessarily d = deA = O. This shows that a is a root of the monic polynomial (_l)n det(bij - Oi,jX) in B[X], and so a is integral over B. 0

92

Algebrazc foundations

Theorem 1.6.28 Let B be a unital subalgebra of a commutative, umtal algebra A. Then lnt B is an integrally closed s1Lbalgebra of A containing B. Proof Set C = lnt B, and take a, bE C. By 1.6.27, there are finitely generated B-submodules E and F of A such that eA E En F, aE C E, and bF c F. Let G = EF. Then G is a finitely generated B-submodule of A, eA E G, and (a + b)G c G and abG c G. Thus a + band ab are integral over B, and so C is a subalgebra of A. Certainly C ::) B. Let a E A be integral over C. Then there is a finitely generated C-submodule M with eA EM and aM c M, say M = Cal + .. ·+Can . We may suppose that al = eA. For i E N n , there exist Cij E C such that aai = 2:,7=1 CiJaj. Let Eij be a finitely generated B-submodule such that eA E Eij and cijEij C E ij , and let E = fI E ij . Then E is a finitely generated B-submodule, eA E E, and cijE C E. Set F = Ea1 + ... + Ea n , so that F is a finitely generated·B-submodule, eA E F, and aP C 2:,i,J CiJajE C 2:,j Eaj = F. By 1.6.27, a is integral over B, and so a E C. Thus C is integrally closed in A. 0 Corollary 1.6.29 Let B be a unital subalgebra of a umtal integral domain A. Then Alg B is an algebraically closed subalgebra of A. Proof First note that, if a E Alg B, then there exists b E B- such that abc is integral over B for each C E B. For suppose that p( a) = 0, where p = 2:,~0 biXi. Take b = bn , and, for c E B, set q = 2:,;:01 bibn-1-icn-ixi + xn. Then q is a monic polynomial, and q(abc) = O. Now suppose that a, bE Alg B. Then there exists c E B- with ac, bc E lnt B. By the theorem, (a + b)c and abc 2 belong to lnt B, and so a + b, ab E Alg B. Thus AIg B is a subalgebra of A. Finally take a E Alg (AIg B). Then there exist co, Cll ... , Ck E AIg B with Ck -I- 0 and Co + C1a+· .. + Ckak = O. We may suppose that Co, Cl,'" ,Ck E lnt B. Then, as above, Cka E lnt (lnt B), and so, by the theorem, Cka E lntB. Hence a E AIg B, proving that Alg B is algebraically closed. 0 Definition 1.6.30 Let B be a unital subalgebra of a commutative, umtal algebra A, and let a E (AIg B) \ B. Set aa = min{ap: p E B[Xt, pea) = O}. Then a is algebraic of degree aa over B; a polynomial p with p( a) ap = aa is a minimal polynomial for a.

o and

Theorem 1.6.31 Let A and C be commutative, umtal algebras, let B be a unital subalgebra of A, and let () : B -+ C be a unital homomorphism. Suppose that a E Alg B, that q is a minimal polynomial for a, and that Co E C is a root of (jq. Suppose further that either

(i) ()(b)c -I- 0 (b E B-, c E C-); or (ii) A is a integral domain and q(O) E Inv B. Then (j : pea) ~ (()p)(co), B[a] -+ C, is a homomorphism which extends ().

93

Polynomials and formal power senes

Proof We first show that (j is well-defined. Suppose that p E B[Xl e with pea) = o. By the division algorithm, there exist b E Be and r,8 E B[Xl with bp = rq + 8 and < oq. Since sea) = 0, necessarily 8 = O. Thus

as

B(b)(Bp)(co)

=

(Br)(co)(Bq)(co)

= O.

Under hypothesis (i), it follows immediately that (Bp)(co) = O. Now suppose that hypothesis (ii) holds, and set r = L~=o rjX) E B[Xl. Since q(O) E Inv B, it follows by induction on j that there exist 80, ... ,8k E B with rj = b8j (j E zt). Set 8 = L~=o SjXj. Then bp = bq8. Since A is an integral domain, p = q8, and again (Op)(co) = O. We have shown in both cases that (j is well-defined. It is now easily seen that (j is a homomorphism and that (j extends O. 0 The following result is immediate. Theorem 1.6.32 Let A and C be commutative, umtal algebras, let B be a unital subalgebra of A, and let 8 : B ~ C be a unital homomorphism. Suppose that a E A zs transcendental over B, and that c E C. Then the map (j : pea)

f-+

(8p)(c), B[a]

---7

C,

o

is a homomorphism which extends B.

Definition 1.6.33 Let B be a unital subalgebra of a commutatzve, umtal algebra A. A subset S of A zs algebraically independent over B if P(SI, ... , sn) =I 0 for each polynomzal p E B[X1 , ••. , Xn]e and each set {81, ... , sn} of distinct element8 of S. A transcendence basis for A over B is a maximal element in the family of subsets of A which are algebraically independent over B. It follows from Zorn's lemma that each algebraically independent set over B is contained in a transcendence basis for A over B. The cardinality of transcendence bases is an invariant of the algebras A and B. Let A be a commutative, unital, complex algebra. A transcendence basis for A over CeA is a transcendence basis for A; the dimension of such a basis is the transcendence degree of A. Since each algebraically independent set is linearly independent, this degree is at most dim A. Proposition 1.6.34 Let A be a commutative, umtal algebra, let B be a unital subalgebra of A, and let r be a transcendence basis for A over B. Then each a E B [r]- can be written m a unique way in the form a = p( 71, ... , 7 n), where n E N, p E B[X1 , ... , Xnl-, and 71, .. " 7 n are dzstmct elements of f. Further, A is an algebraic extension of B[f]. Proof Set C = B [fl. Certainly each element a E C- has an expression of the given form. Suppose that a = p( 71, ... ,7m ) = q( 7{, ... ,7:,), where {71,' .. ,7m } and {7i, ... , 7~} are sets of distinct elements of r. By relabelling, we may suppose that 71 = T{, . .. ,Tk = T~ and that {Tl, ..• , Tm, T~+I" .. ,<} is a set of distinct elements. Set r(XI, . .. ,Xm+n-k) = p(X1 ,

.• •

,Xm ) - q(XI, . .. ,Xk, X 7n+ L

•.. ,

Xm+n-k) .

94

Algebraic foundatwns

Then r(T1, ... , Tm , T~+l"'" T~) = 0, and so r = 0 because r is algebraically independent. It follows that the representation of a is unique. Take a E A \ C. Then r U {a} is not algebraically independent over E, and so there exist p E B[X I , ... , X n+l ]· and a set {al,"" an} of distinct elements of r with p(al, ... ,an,a) = O. Set q(X) = p(al, ... ,an,X), so that q E C[X]. k . Suppose that p = Li=OPiX~+l' where Po,··· ,Pk E B[X1, ... , Xn] and Pk =I- O. Then Pk(a1, ... , an) =I- 0, and so q E C[X]·. Since q(a) = 0, we have a E Alg C, and hence A is an algebraic extension of C. 0 Definition 1.6.35 A field K zs algebraically closed if each non-constant polynomial m K[X] has a root in K. An integral domam zs algebraically closed zf its quotzent field is an algebraically closed field. A n ordered field K is real-closed if K has no proper algebrmc extension to an ordered field. Let K be an algebraically closed field, and let P E K[X] with ap = n. Then there exist ao E K· and a1, ... , an E K such that p = ao(X - a1)'" (X - an). Of course, the fundamental theorem of algebra is that C is an algebraically closed field. Let U be a free ultrafilter on N. Then the ultrapower ([:f'l jU is algebraically closed. Let A be a unital integral domain. It is easily checked that, if each nonconstant monic polynomial over A has a root in A, then A is algebraically closed. An algebraic closure of a field K is an algebraically closed field L containing K as a subfield and such that L is algebraic over K. We shall use the standard fact that each complex field K has an algebraic closure which is a complex field, and that any two algebraic closures of K are isomorphic as algebras over C. Note that, if E is a unital subalgebra of a field K, then AlgKB is a subfield of K. It is known that an ordered field K is real-dosed if and only if each element of K+ is a square and every polynomial in K[X] of odd degree has a root in K. An ordered field K has a complexification which we denote by Kc: the ordered field K is real-closed if and only if Kc is algebraically closed. For example, the ultrapower (JRN jU, +, . , :S;u) is real-closed for each free ultrafilter U on N. The Artin-Schrezer theorem asserts that every ordered field K has an algebraic extension to a real-closed field RK whose order is an extension of the order of K, and that RK is unique up to an isomorphism that leaves the elements of K fixed. Notes 1.6.36 Most of the theory of polynomial and formal power series algebras can be found in (Hungerford 1980) and (Zariski and Samuel 1958) and other standard texts. The relation between our definition of a Henselian algebra and other definitions is discussed in (Dales 1987). Let A = qx j , ••• , Xn]. By the famous Hilbert basis theorem, the algebra A is Noetherian. Also, the analogues of (iii) and (vi) of 1.6.5 hold for A. Indeed, let J be an ideal in A, and define the variety of J to be

V (J) = {( E en : p( () = 0 (p E l)} . Let S be a closed subset of en, and define J (S) = {p E A : p( () = 0 E S)}. Then Hilbert's Nullstellensatz asserts that J(V(l) = U{ {p E A : pm E I} : mEN} for each ideal J in A. In particular, J = A when V(J) = 0, and so each maximal ideal is the kernel of a character. See (Hungerford 1980, VIII. 7.4). For a study of the existence and uniqueness of the algebraic closure of a field, see (Hungerford 1980, III.3) and (Jacobson 1980, Chapter 8); for the Artin-Schreier theory of ordered fields, see (Jacobson 1980, Chapter 11) and (Dales and Woodin 1996).

«(

Valuation algebras 1.7

95

VALUATION ALGEBRAS

The present section is an introduction to the theory of valuation algebras; this is material that is perhaps not as familiar as some other material in this preliminary chapter. We shall first define valuation algebras, and give their basic properties. The algebra ~ of the previous section is an example of such an algebra, and most of our examples will just be more general versions of J. It will be important for us to know that various complex integral domains are contained in valuation algebras: this is the fundamental theorem for the extenszon of places, to be stated in 1.7.8. We shall then discuss Henselian algebras and algebraically closed valuation algebras, and give an extensive account of the construction of embeddings of certain valuat.ion algebras (and hence of many integral domains) into local algebras; this discussion will culminate in a general algebraic extension theorem 1. 7.42 and a main extension theorem 1. 7.44. These extension theorems will be used mainly in §5.7. Definition 1.1.1 Let V be a local mtegral domain over K. Then V is a valuation algebra zJ, for each x, y E V, ezther x E yV or y E xV, and zf the 1'Eszdue field V/Mv is K It is clear that the set of ideals in a valuation algebra is totally ordered by inclusion. Let V be a valuation algebra. Then there is a unique character on V, denoted by
Definition 1.1.2 Let V be a valuation algebra with quotient field K. Then the group fv = (K-, . )/InvV zs the value group of V. and the quotzent map v: K- --+ rv is the valuation on K. Set v(x):S v(y) for x,y E K- ify E xv. The group operation on rv is denoted by + ; certainly v (ev) = 0, the identity of rv. It is easily checked that the binary relation :S is well-defined on rv. We extend v to be a map from K into rv U {oo} by setting v(O) = 00; we have .5 < 00 (8 E rv). For x, y EMir, we have vex) < v(y) if and only if x -< y in the divisibility order on J.\Iir. Proposition 1.1.3 Let V be a valuation algebra. Then (r v, + :S) is a totally ordered group. Proof We may suppose that Mir i- 0. Set P = v(Mt); p+p c P and 0 f{. P, and so P is a cone. The relation :S is the P-order on rv (see 1.2.14), and so (fv, +,:S) is an ordered group. Let x, y E V-. Then either x E yV or y E xV, and so either v(y) :S vex) or vex) :S v(y). Thus rv is totally ordered. 0

rt-

The positive cone = v(Mir) of the value group of a valuation algebra V for which Mir i- 0 will be denoted by Pv. For example, the algebra ~ = ~(Z+) is a valuation algebra, with value group z; this follows from 1.6.20. Let K be an ordered field. Then the algebra K O # is a (real) valuation algebra, and the value group r K defined in §1.3 coincides with the value group of K O # defined in 1.7.2. Thus, for each totally ordered abelian group G, both ~(G)O# = ~(G+) and ~(1)(G+) are complex valuation algebras, each with value group G.

Algebraic foundatwns

96

Proposition 1. 7.4 Let V be a valuation algebra over lK with quotient field K. Then the valuation v : K ----+ rv U {oo} satisfies the followmg conditions:

(i) v(xy) = vex) + v(y) (x, Y E K·); (ii) v(ax) = vex) (x E K·, a ElK·); (iii) V = {x E K : vex) ~ a}, Mv = {x

E K : vex)

> a},

and

= {x E K : vex) = a}; with vex) = v(y), then there exists a Inv V

(iv) if x, y E vex - ay) > vex);

v·

E lK· such that

(v) for Xl, ... ,Xn E K, V(Xl + ... + x n ) ~ min{v(xt}, ... ,v(xn )}, and equality holds ifv(xi) i- vex)) for each i,j E N n wzth i i- j. Proof (i)-(iv) These are immediate. (v) We may suppose that Xl, ... , Xn E K·. Since r K is totally exists k E N n with Xi E Xk V (i E N n ). Then Xl + ... + Xn E inequality follows. If V(Xi) > V(Xk) (i i- k), then Xi E xkMv Xl +···+X n E Xk· (ev+Mv) C Xk· InvV, so V(Xl +···+x n ) =

ordered, there Xk V, and the (i i- k), and V(Xk). 0

Definition 1.7.5 Let V and W be valuatwn algebras such that V 28 a subalgebra of W. Then W is an extension of V if Mv = Mw n V; an extension W is proper zf V i- W, and immediate if, for each x E W·, there exists y E V· with X - Y E M w. A valuatwn algebra V zs maximal if zt has no proper, immedzate extension. Let V and W have valuations v and w, respectively, and suppose that W is an extension of V. For x, y E V, we have vex) = v(y) if and only if w(x) = w(y), and vex) > v(y) if and only if w(x) > w(y). Thus the map

x . Inv V

f---+

X •

Inv W,

rv

----+

rw ,

is an isotonic morphism. We identify fv with a subgroup of fw, and use the letter v for each valuation. Let V be a subalgebra of a valuation algebra W, and suppose that V is a local algebra. Then Mv = ker (IPW I V) = Mw n V, and so V is a valuation algebra and W is an extension of V. Lemma 1.7.6 Let V and W be valuation algebras, with W an extension of V.

(i) Let x E W \ V. Then vex maximum element.

+ V)

C

v(V) if and only if vex

(ii) The extension W is immediate if and only if fv

=

+ V)

has no

rw.

Proof (i) Suppose that vex + V) c v(V), and take y E V. Then there exists z E V with vex + y) = v(z). By 1.7.4(iv), there exists a E lK· such that vex + y - az) > vex + y), and so vex + V) has no maximum. Conversely, suppose that vex + V) has no maximum, and take y E V. There exists z E V with vex + z) > vex + y), and then vex + y) = v(z - y) E v(V).

97

Valuatzon algebras

(ii) Suppose that W is an immediate extension of V, and let x E W·. Take y E V· with y Ex· (ew + Mw). Then vex) = v(y), and so fv = fw. Conversely, suppose that fv = fw, and take x E W·. There exist yEW· and z E Inv W with x = yz, and then x - CPw(z)y

= y(z - cpw(z)ew)

E yMw

= xMw·

o

Thus W is an immediate extension of V. ~

For example, the algebra C{X} is a valuation subalgebra of ~ = q[X]J, and is an immediate extension of C{X}.

Proposition 1. 7.7 Let G be a totally ordered abelian group. (i) The valuation algebra ~(IK, G+) is maxzmal.

(ii) Both

~(IK, G+) and ~(1) (1K, G+) are Mittag-Leffler algebras.

Proof (i) Set V = ~(IK, G+), and let W be a valuation algebra which is an immediate extension of V. To obtain a contradiction, assume that there exists x E W \ V, and set S = vex + V); by 1.7.6, S ha..'l no maximum element. We define an element f E V. If s E G+ with S « s, set f(s) = O. If s E S+ and it is not true that S « s, then there exists y E V with vex - y) > s; set f(s) = yes). If also z E V with v(x-z) > s, then v(y-z) > s, and so yes) = z(s). Thus f (s) is well-defined. Let (sn) be a decreasing sequence in supp f, and take y E V with 81 < vex - y). Then (sn) C suppy, and so 8 n is eventually constant. Thus supp f is well-ordered, and f E V. Set 8 = vex - f) E S. There exists y E V with vex - y) > s, and then f(s) = yes), whereas v(f - y) = s. This is the required contradiction. Thus W = V, and V is maximal. (ii) Let A be either ~(IK, G+) or ~(1) elK, G+). Take (an) in M A, and set bn = al ... an (n EN). If s E G+ is such that v(b n ) < 8 (n EN), define f(s) = O. If s E G+ and 8 :::; v(b n ), define f(s) = Z=~=1 bk(s). If also s :::; v(b m ), with m > n, then

v (

f

bk)

k=n+l

~ v(b +1) > s, n

and so Z=;;=1 bk(s) = Z=~=l bk(s). Thus f(8) is well-defined. As above, suppf is well-ordered. Since supp f C UbI supp bk, supp f is countable in the case where A = ~(1)(G+). Thus f E A in both cases. Take n E N. Then v (f - Z=~=1 bk ) > v(bn ), and so v (f - Z=~=l bk) E bnMA· Hence MA is a Mittag-Leffler set in A, and so A is a Mittag-Leffler algebra. 0 For example, the algebra Leffler algebra.

~

is a maximal valuation algebra and a Mittag-

Theorem 1. 7.8 (Extension of places) Let A be a complex, unital zntegral domain with a character cp, and let K be a complex field containing A as a sub algebra. Then there is a complex valuation algebra W with quotient field K such that A C Wand
Algebraic foundations

98

Corollary 1.7.9 Let A be a complex, non-unital integral domain, with quotient field K. Then there is a valuation subalgebra W of K such that A c Mw. Proof Let 'Poo be the character on A# with kernel A. By 1.3.55, A# is an integral domain. The result now follows from the theorem. 0 Corollary 1.7.10 Let A be a complex, unital integral domain with a character.

Then there zs an algebraically closed valuation algebra W containing A as a unital subalgebra and such that W is algebraic over A. Proof Let K be the algebraic closure of the quotient field of A. By 1.7.8, there is a valuation algebra W with quotient field K such that A C W. Clearly W is algebraic over A. 0 Proposition 1.7.11 Let V be an algebraically closed valuation algebra. Then

V is Henselian. Proof Denote the quotient field of V by K. Let p = ao + X + a2X2 + ... + anXn be a Henselian polynomial in V[X]. Since K is algebraically closed, there exist Xl, ..• , Xn E K such that

p

=

an(X - xI) ... (X - Xn) .

Assume towards a contradiction that V(Xi

v(p( -aD))

+ ao)

::::; V(Xi) (i E Nn ). Then

= v(a n ) + V(XI + ao) + ... + V(Xn + ao) s v(an ) + v(xd + ... + v(xn) = v(anxI··· Xn)

= V(p(O» = v(ao).

But p( -aD) E a6V, and so v(p( -aD»~ > v(ao), a contradiction. Thus there exists i E N n with V(Xi + ao) > vex;). Set x = Xi. Then x + ao E xMv, and so x E ao . (ev + Mv) c Mv. Thus x is a root of p in V. 0 Definition 1.7.12 Let W be a valuation algebra, and let U be a subalgebra of W. The valuatwn subalgebra of W generated by U is V(U), where

V(U) = {x/y: x E U, YEW, vex) :::: v(y)}. Proposition 1.7.13 The set V(U) is the minim'um valuation subalgebra of W

containing U. Proof Set V = V(U). Take XI,X2 E U and YI,YZ E U· with v(xd:::: V(YI) and V(X2) :::: v(yz), say V(XIY2) :::: v(x2yd. Then V(XIY2 + X2YI) :::: V(YIY2), and so xdYl + X2/YZ E V. Since V(XIX2) :::: V(YIYZ), we have (xdYl)(x2/Y2) E V, and so V is a subalgebra of W. If Xl, Xz, YI, yz are as above, then Xl Y2 E X2YI V and so xdYl E (XZ/Y2)V. Hence V is a valuation algebra. 0 Clearly each valuation subalgebra of W containing U also contains V.

The algebra ~n is a complex, unital integral domain, but it is not a valuation algebra when n:::: 2: neither of the principal ideals XI~n and X2~n contains the other. However, by the above theorem, ~n is contained in a valuation algebra.

99

Valuatzon algebras

Here is a specific example of such a valuation algebra. Let G be the totally ordered group (zn, ::S), where ::S is the lexicographic order. Then ~(G+) is a valuation algebra. \Ve have (rl, ... , rn) t (0, ... ,0) for rl, ... , rn E Z+, and so we may identify ~n as a sub algebra of ~(G+).

Definition 1.7.14 Let n E N. The valuation subalgebra of ~«zn, ::s)+) generated by ~n zs denoted by W n . Note that Xd X 2 E W2, for example, because (1, -1) >-- (0,0). The valuation algebra Wn depends on the choice of the ordering ::S on zn. The following important properties of a valuation algebra will become relevant in §5. 7.

Definition 1.7.15 Let V be a valuatzon algebra. Then V is:

rv of V is an aI-group; < wd, where {Va: < wd

(i) an al -valuation algebra if the value group (ii) a PI-valuation algebra if V = U{Va : (J chain of O!l -valuation subalgebras of V.

(J

zs a

(iii) an TJl -valuation algebra if rv is an 'TIl -group and V is a Mittag-Leffler algebra. For example, the value group of Wn is (zn, ::S), and so Wn is an aI-valuation algebra. Let K = ~(~, G), where G is a totally ordered abelian group. Clearly the following conditions are equivalent: (a) K O # is an aI-valuation algebra; (b) G is an aI-group; (c) K is an arfield. Also the following conditions are equivalent: (a) K O # is an 'TIl-valuation algebra; (b) G is an 'TIl-group; (c) K is an 'TIl-field. Now let L = ~(l) (~, G). Then the following conditions are equivalent: (a) L O # is a PI-valuation algebra; (b) G is a PI-groUP; (c) L is a PI-field.

Proposition 1.7.16 Let V and W be valuation algebras such that W is a proper e.Ttension of V, and let :r E W \ V.

(i) Suppose that x zs algebrazc of degree n over V. Then there exists k E N n such that kv(x) E rt. (ii) Suppose that V is an aJ -valuatzon algebra. Then Alg w V[x] is also an al -valuation algebra.

Proof (i) Take p E V[X]· such that p(x) = 0, say p = 'L7=o yjX), where Yo, .. ·, Yn E V and Yn i= O. By 1.7.4(v), there exist i,J E z;t with i < j, with Yi i= 0, and with V(Yixi) = v(YjxJ). Set k = J - i. Then V(Yi)

= v(Yj) + kv(x)

~ V(Yj) ,

and so there exists Z E V· with Yi = YJz. Clearly kv(x) = v(z) E rt. (ii) By 1.7.10, we may suppose that W is algebraically closed, and hence that rw is divisible. Let y E AlgwV. By (i), v(y) E U{v(V)/k : kEN}, an aI-group, and so we may suppose that V is algebraically closed in W. Let K be the quotient field of V.

100

Algebrazc foundations

rt

First, suppose that vex - K) rK· Take y E K with vex - y)
z E K with v(x-z)
Theorem 1.7.17 Let V be a valuation algebra, and suppose that V has a transcendence basis r with If! :s; ~1' Then V is a /31 -valuatzon algebra. Proof Suppose that r = {x q : a < W1}' First set Vo = Cev, and then recursively define a family {Vq : a < wd of valuation subalgebras of V by setting V q + J = Algv(Vu[x u ]) and Vq = UT
IVI = c.

Proof Now V certainly has a transcendence basis f with

Ifl :s; ~1.

Then V is

0

Theorem 1.7.19 (Mac Lane) Let G be a totally ordered abelzan group. Then the following are equivalent: (a) the group G is diviszble; (b) the complex fields ~(G) and ~(1)(G) are algebraically closed; (c) the real fields

~(lR,G)

and ~(1)(RG) are real-closed.

Proof We prove that (a)=}(b); the remainder is immediate. By 1.7.10, there is an algebraically closed valuation algebra W which is an algebraic extension of ~(G+). Since the group G is divisible, it follows from 1.7.16(i) that v(W-) = G+, and so, by 1.7.6(ii), the algebra V is an immediate extension of ~(G+). By 1.7.7(i), W = ~(G+), and so the field ~(G) is algebraically closed. Let p = "L.;=o f;Xj E ~(1) (G)·, and let H be the smallest divisible subgroup of G containing Uj=l supp Ij· Then H is countable. Since ~(H) is algebraically closed, p has a root in ~(H), and so ~(l)(G) is algebraically closed. 0

101

Valuation algebras

Corollary 1.7.20 (i) The field R %8 real-closed. (ii) The algebra M# zs an algebra%cally closed fil -'TIl -valuation algebra. Proof (i) Since G is a totally ordered, divisible group, R is a real-closed field by 1.7.19. (ii) For each a < WI, the group GO" is an n1-group by 1.2.26(ii), and so the algebra Mt is an nl-valuation algebra. Hence M# = U{Mt : a < WI} is a fil-valuation algebra. By 1.7.19, M# is algebraically closed; by 1.7.7(ii), M# is a Mittag-LefHer algebra; by 1.2.28(ii), G is an 'TIl-group. 0 We now embark on an extreme case of our policy of presenting in this first chapter algebraic results that may not be used until much later: the remaining results in this section will be utilized only in the final section, §5.7. We are concerned with the construction of homomorphisms from certain valuation algebras into local algebras. This construction will proceed step by step; we steadily extend a given homomorphism defined on a subalgebra U of our valuation algebra W, to have as its domain a larger sub algebra of W. Thus, given x E W\ U, we shall seek to extend e from U to U[x]. There are clearly two fundamentally different cases according as x is algebraic or transcendental with respect to U: for these cases see 1.6.31 and 1.6.32, respectively. It is also clear that our final result will require an application of Zorn's lemma to see that there is a maximal subalgebra 'V of W which is the domain of a homomorphism with certain additional properties; our preliminary results will enable us to conclude that in fact the maximal such subalgebra is W itself. 'Ve have explained that the domain of our homomorphisms will be a valuation algebra. such as the algebra M#. The fundamental theorem for the extension of places, stated in 1.7.8. essentially shows that every complex integral domain if> contained in a valuation algebra. and so our results will produce homomorphisms from general integral domains. The codomains of our homomorphisms will eventually he required to he local Henselian algebras; we shall see that many examples satisfy this condition. Let lV be a complex valuation algebra with AIw =I- 0, and let A be a local algebra. We shall work under the hypothesis that we are given a 'framework map' '1/' from the positive cone Pw of the value group r w of W, and we shall construct homomorphisms that are 'compatible with '1//. Suppose that a homomorphism () : U --+ A has been constructed, where U is a subalgebra of W, and that x E W \ U. We shall obtain in 1. 7.42 a general algebraic extenswn theorem that extends () to have domain U[x] in the case where x is algebraic with respect to U. We can always extend () to U[x] in the case where x is transcendental with respect to U, as in 1.6.32, but we must do this in such a way that the extension is still compatible with the framework map, and this is only possible in special circumstances; we shall achieve such a result in our main extension theorem 1.7.44. Our first step is to show in 1.7.25 that each complex valuation algebra has an immediate algebraic extension which is Henselian; in the preliminary lemmas, V and W are complex valuation algebras, and W is a proper extension of V.

e.

Algebmic Joundatzons

102

Lemma 1.7.21 (Esterle) Suppose that x E W \ V and n EN' are such that: (i) vex + V) c v(v); (ii) there exists p E V[X]- with ap

= nand

v(p(x)) ~ v«x - a)p'(x»

(a

E

Then n ~ 2, and there is a Henselian polynomial Po that Po has no root in V.

V).

E

(1.7.1)

V[X] wzth apo ::; n such

Proof We first show that, if p E V[X]- satisfies inequality (1.7.1), then necessarily ap ~ 2. For assume that p = ao +alX, where ao, al E V. If al = 0, then (1.7.1) fails because p'(x) = 0, but p(x) = ao =I O. If al =I 0 and v(ao) ~ v(alx), then ao = alb for some b E V. By (i) and 1.7.6(i), there exists a E V with vex - a) > vex + b). We have

v(p(x»

= v(al(b + x» < v(ad + vex -

a)

= v«x -

a)p'(x»,

and so (1.7.1) fails. If v(ao) < v(alx), then v(p(x» < v(alx) = v(xp'(x», and so (1.7.1) again fails. Thus ap ~ 2. Now let q E V [X]- be a polynomial of minimal degree such that (1. 7.1) holds (with q for p), say aq = m. Then m E {2, ... , n}. For k E N'm, we have q(k) E V[X]- and aq(k) < m, and so there exists ak E V such that v(q(k)(x» < V«X-ak)q(k+l)(x». Take a E {a}, ... ,am } to be such that vex - a) = max{v(x - ad, ... , vex - am)}. Then

v(q(k)(x» < v«x - a)q(k+1) (x»

In particular, q'(x)

=I

(k E N'm).

0, and so, since q satisfies (1.7.1) and vex

(1.7.2)

+ V)

has no

maximum element,

v«x - c)q'(x» < v(q(x»

(1. 7.3)

(c E V).

Take bE V with vex - b) > vex - a), so that vex - a) = v(b - a). Since

q(k)(b) = q(k)(x)

(b

)m-k

+ ... + (: ~ k)!

q(m) (x)

(k E z~),

it follows from (1.7.2) that v(q(k)(b)) = v(q(k)(x» (k E N'm). Also, we have v(q(b» = v«b - x)q'(x» because v«b - x)q'(x» < v(q(x» by (1.7.3) and

v«b - x)q'(x)) < v«b - x)2 q"(X» < ... < v«b - x)mq(m)(.c» by (1.7.2). Set reX) = q«b - a)X + b), and set y = (x - b)/(b - a)j y E Mw because vex - b) > v(b - a). Also r(y) = q(x) and r(k)(O) = (b - a)kq(k) (b) (k E Z;t;.), and so

v(r(k)(O» = v«x - a)kq(k) (x» < v«x - a)k+l q(k+1) (x» = v(r(k+l) (0» for each k E N'm. Finally,

v(r(O»

= v«b -

x)q'(x» > v«b - a)q'(b»

= v(r'(O)).

Thus v(r'(O» < v(r(k)(O» (k = 0,2, ... ,m). Since V is a valuation algebra, there exist be, b2, ... , bm E Mv such that r(k)(O) = k!r'(O)bk (k = 0,2, ... , m).

Valuation algebras

103

Set Po = bo+X +b2X 2+ .. ·+bmXm E V[X]. Then Po is a Henselian polynomial in V[X], apo = m S n. and r'(O)po = r. We now show that the polynomial Po has no root in V. Take e E V. Since v(r'(O)) = v«b - a)q'(b)) = v«b - a)q'(x)), we have

v(r'(O))

+ v(y -

e)

=

v«x - b - eb + ae)q'(x)) ,

and so, by (1.7.3), v(r'(O)) +v(y - e) < v(q(x)). Since q(x) = r(y) = r'(O)po(y), we have v(y - e) < v(Po(y)). By (1.6.4), v(Po(y) - po(e)) = v(y - c), so that v(Po(y) - po(e)) < v(Po(y)). Thus po(e) =f 0, as required. We have shown that V is not Henselian. 0 Lemma 1.7.22 Suppose that W is an immedzate extension of V and that V zs Hen.~elzan.

Then, for each p E V[X]· and each x such that v(P(x)) < v«x - y)p'(x)).

E

W \ V, there exists y

E

V

Proof Since W is an immediate extension of V, condition (i) of the above lemma holds. Since V is Henselian, condition (ii) must fail. 0 Lemma 1.7.23 Suppose that x E W \ V, that x zs algebraic oj degree n over V,

and that, for each q E V[X]- with Oq < n, there exists an element a E V such that v(q(x)) < v«x - a)q'(x)). Then V(V[x]) zs an zrnmediate extension oj V. Proof By 1.7.13, V(V[x]) is a valuation subalgebra of W, and so, by 1.7.6(ii), it suffices to show that v(V(V[x])) c v(V). Take q E V[x]· with aq = m, where m < n. Then there exists a E V with

v(q(k)(x)) < v«x - a)q(k+l) (x))

(k E Z~_l)'

and so v(q(x)) = v(q(a)) E v(V). Now fix p E V[X]· with 8p = n such that p(x) = O. For each q E V[X]·, there exist b E V· and r, s E V[X] with bq = pr + s and as < n. Then bq(x) = sex), and so v(q(x)) E v(V). Thus v(V[x]) C v(V). Take y = al/b l E V(V[x]), where al E V[x], b1 E V[x]·, and Veal) ~ v(bI). Then there exist a2 E V and b2 E V· such that v(a2) = v(ad and v(b2) = v(bI). Also, there exists z E V such that a2 = b2z, and then v(y) = v(z) E v(V). Thus v(V(V[x])) C v(V), as required. 0 Lemma 1.7.24 The valuation algebra V has a maximal immediate extension V

in W. Moreover, zJ W is Henselzan, then

V zs also Henselian.

Proof Let F be the family of all valuation subalgebras U of W such that U ::> V and v(U) = v(V). Clearly (F, c) is a partially ordered set and each chain in F has an upper bound. By Zorn's lemma, F has a maximal element, say V. By 1.7.6(ii), V is a maximal immediate extension of V. Now suppose that W is Henselian. Assume that V is not Henselian, and take a Henselian polynomial p E V[X] with no root in Vj we may suppose that p is of minimal degree among such polynomials. Since W is Henselian, there exists

104

Algebraic foundations

x E W \ V such that p(x) = o. We have vex - a) = v(p(a» (a E V), and so vex + V) c v(V). Take q E V[X]- with aq < ap. If it were the case that v(q(x)) ~ v«x - a)q'(x))

(a E V),

then, by 1.7.21, there would exist a Henselian polynomial Po E V[X] with no root in V and with apo < ap. But this contradicts the assumed minimality of ap. Thus there exists a E V with v(q(x» < v«x - a)q'(x». By 1.7.23, V(V[x]) is an immediate extens~on of V. By the maximal~y of V in (F, C), V(V[x]) = V, and in particular x E V, a contradiction. Thus V is Henselian. 0 Theorem 1.7.25 Each complex valuatwn algebra has an tmmedwte, algebraic extension whzch is a Henselian algebra. Proof Let V be a valuation algebra. By 1.7.10, there is an algebraically closed valuation algebra W which is an algebraic extension of V. By 1.7.11, W is Henselian. By 1.7.24, there is a Henselian algebra V C W such that V is an immediate extension of V. Certainly V is an algebraic extension of V. 0 Corollary 1.7.26 Let G be a totally ordered abelian group. Then J(G+) is a Henselian algebra. Proof By 1.7.7, J(G+) is a maximal valuation algebra. By the theorem, J(G+) is always Henselian. 0 In fact, it is easy to see directly that J(G+) is always Henselian. We now define a 'framework map'. Recall that the positive cone rt; of the value group rw of a valuation algebra W is denoted by Pw , and that the unique maximal ideal of a local algebra A is M A . Definition 1.7.27 Let W be a valuatwn algebra with Mw i= 0, and let A be a local algebra. A map 'lj; : P w ........ MA zs a framework map zf:

(i) 'lj;(s + t) = 'lj;(s)'lj;(t) (.'.I, t E P w ); (ii) 'lj;(s).L = 'lj;(t).L (s,t E Pw). The framework map 'lj;

'U~

freely acting zf:

(ii)' 1f;(s).L = 0 (s E Pw). Let W be a valuation algebra with Mw i= 0, and let 'lj; : Pw ........ framework map. We systematically extend 'lj; to a morphism 'lj; : rt setting 'lj;(0) = eA. Throughout we write

W = 'lj; 0 v,

be a by

W ........ A,

where v is the valuation on W, taking w(O) = w(xy) = w(x)w(y) (x, yEW)

MA

. . . . A-

o.

and

Clearly W(Inv W) = {eA} .

Suppose that U is an immediate extension of W. Then may regard W as a map from U into A.

ru

=

fw, and so we

Valuation algebras

105

Let x, yEW. Then it is clear that lI1(x) E lI1(y)A whenever vex) 2: v(y) and that lI1(x) E lI1(y)MA whenever vex) > v(y). Assume that s < t in Pw and that 1/)(s) = tjJ(t). Then t/J(s)(eA + a) = 0 for some a E MA, and so t/J(s) = 0, a contradiction. Thus t/J is an injection. We shall prove our extension theorems 1.7.42 and 1.7.44, first in the case where t/J is freely acting, and then in the general case; note that a framework map is certainly freely acting in the case where A is an integral domain, and many of our later results will apply in this context. Definition 1.7.28 Let S be a subset of W. A map () : S with a framework map t/J if ()(x) E lI1(x) . Inv A (x E S).

-+

A

tS

compatible

Let () be such a map. Then ()(x) = 0 for xES if and only if x = O. If x, yES with vex) 2: v(y), then ()(x) E ()(y)A. Now suppose that 'l/J is freely acting, that a E A·, and that xES·. We claim that ()(x)a =I O. For, if x E M then vex) E Pw and lI1(x)a =I 0 because t/J is freely acting, and, if x E Inv W, then lI1(x)a = a =I O. In either case, it follows that ()(x)a =I 0, as claimed.

w,

We shall now give a sequence of results which will culminate in our main algebraic extension theorem 1.7.42. Until we state otherwise, Wand A are as in 1.7.27, 1/) is a freely actmg framework map, Uo is a fixed subalgebra of W, and ()o : Uo -+ A is a homomorphism which is compatible with t/J. We denote by F the family of all pairs (V, ()), where V is a subalgebra of W with Uo c V and () : V -+ A is a homomorphism extending ()o such that () is compatible with t/J. Set (VI,()t) =;< (V2,()2) in F if VI c V2 and ()2 I VI = ()l. It is clear that (F, =;<) is a partially ordered set in which each chain has an upper bound, and so, by Zorn's lemma, (F, ~) has a maximal element, say (if, if). The first lemma shows that V is a valuation subalgebra of W. Lemma 1.7.29 Let (V,()) E F. (V(V), /-L) >r (V, ()) in :F.

Then there exists /-L : V(V)

-+

A such that

Proof Take x = ydZI E V(V), where YI E V, Zl E V·, and V(YI) 2: V(ZI)' Then ()(yt) E ()(zl)A: take /-L(x) E A with ()(yt) = ()(Zl)/-L(X). Suppose that also x = Y2/Z2, where Y2 E V, Z2 E V·, and V(Y2) 2: V(Z2), and that ()(Y2) = ()(z2)a for some a E A. Then ()(YIZ2) = ()(Y2Zt), and so ()(zlz2)(a -/-L(x)) = 0, whence a = /-L(x) because t/J is freely acting and ()(ZIZ2) =I O. Thus /-L is well-defined. Further, if x E V, then /-L(x) = ()(x). Take Xl = ydZI and X2 = Y2/Z2 in V(V), and take al. a2 E C. Then zlz2(alxl + a2x2) = alYIZ2 + a2Y2Zlt and so ()(Zlz2)/-L(alxl

+ a2x2) =

()(O:IYIZ2

+ a2Y2 zl)

= ()(zlz2)(al/-L(xd

+ a2/-L(x2)).

Thus /-L(alxl + a2x2) = al/-L(xI) + O:2/-L(X2). Similarly /-L(XIX2) = /-L(Xt}/L(X2). Hence JL : V(V) -+ A is a homomorphism. Take x = y/z E V(V). Then 9(y) E lI1(y) . Inv A and 9(z) E lI1(z) . Inv A, and so ()(z)JL(x) = ()(y) E 9(z)lI1(x) . Inv A. This implies that JL(x) E ll1(x) . Inv A, and so JL is compatible with t/J. It follows that (V(V), JL) >r (V, 0) in F. 0

Al.'Jebrmc foundatzons

106

Let (V, (9) E F with Vi- W, and let x E W \ V. Then a E A is 19-compat~ble with x if the map x + y ~ a + 19(y), x + V ~ A, is compatible with 1/1. Recall from (1.6.1) that, for each n E N, V{n)[x] = {p E v[X] : 8p ::; n}; if x E V, then V{n)[x] = {p(x) : p E V{n)[X]},

Lemma 1. 7.30 Let (V, (9) E F, let x E W \ V, and let n EN. Suppose that a is 19-compatible with x and that, for each p E V{n) [X]-, there e~sts y E V such

that v(p(x)) < v«x - y)p'(x)). Then the map p(x) ~ (19p)(a), V{n)[x] a lmear map which extends 19 and which is compatible with 1/1.

~

A, is

Proof Set J-L(p(x» = (19p)(a) (p E V(n)[X]), Fin;t note that v(p(x» < 00, and so p(x) i- 0, for each p E V(n) [X]-. Thus J-L : V{n) [x] ~ A is a well-defined linear map which extends 19. Take p E V{nl [X]-. It follows from the hypothesis that there exists y E V such that 'I.'(p(k)(x)) < v«x - y)p(k+l) (x» (k E p ). Let k E p • From Taylor's formula (1.6.2), v(p(k)(y)) = v(pCkl(x)), and so

zt

zt

v(p(k)(y)) < v«x _ y)pCk+l)(y». Thus w«x_y)pCk+ 1 )(y)) E W(p(k)(y))MA' Since a-Bey) E lI1(x-y) . Inv A, this implies that (a-B(y))B(pCk+l)(y)) E B(pCk)(y»MA. Again from Taylor's formula, we have (Bp)(a) E B(p(y)) . Inv A, and so J-L(P(x)) E w(P(y)) . Inv A. SincE' v(p(x)) = v(p(y», it follows that w(p(x)) 111 (p(y)), and so IL is compatible with 1/1. 0

Corollary 1.7.31 Let (V, B) E F, and let x E W \ V. Suppose that V is Henselian, that W zs an immediate extension of V, and that a ~s an element

of A wh~ch ~s B-compat~ble with x. Then there is a homomorphism J-L : V[x] such that J-L(x) = a and (V[xJ, J-L) >,;= (V, B) in F.

~

A

Proof It follows from 1.7.22 that, for each p E V[X]-, there exists y E V such that v(P(x)) < v«x - y)p'(x)). Set J-L(P(x)) = (Bp)(a) (p E V[X]). By 1.7.30, J-L : V[x] ~ A is a linear map which is compatible with 1/1. Clearly J-L is a homomorphism satisfying the required conditions. 0 Lemma 1.7.32 Let (V, B) E F, and let x E W \ V be algebm~c of degree n over V. Let q be a minimal polynomial for x, and suppose that a E A is a mot of Bq and that the map ()' : p(x) ~ (Bp)(a), V{n-l)[X] ~ A, is compatible with 1/1. Then there is a homomorphism J-L : V[x] ~ A which extends B' and which is such that (V [x], J-L) >,;= (V, B) in:F. Proof Since 1/1 is freely acting, the homomorphism B : V ~ A satisfies hypothesis (i) of 1.6.31, and so J-L : p(x) ~ (Bp)(a), V[x] ~ A, is a homomorphism which extends B. Certainly J-L extends B'. Take p E V[X]. By the division algorithm, there exist y E V-, r E V[X], and S E V{n-l) [X) with yp = qr+s. We have B(y) (Op) (a) = (Os)(a) E lI1(s(x)) . Inv A and yp(x) = sex), and so (Oy)(Op)(a) E O(y)lI1(p(x)) . Inv A. Thus

J-L(P(x» and so 1-£ is compatible with

E

w(P(x» . Inv A,

1/1. Clearly (V[x),I-£)

>,;=

(V, 0) in F.

o

Valuation algebras

107

We now return to consideration of Henselian polynomials, defined in 1.6.2. Note that, if V is a valuation subalgebra of W, if (V,O) E F, and if p is a Henselian polynomial in V[X], then Op is Henselian in A[X] because 8(ev) = eA and 8(Mv) C MA. Lemma 1. 7.33 Let (V, 0) E F, where V is a valuatzon subalgebra of W, and let E W\ V. (i) Suppose that p is a Henselzan polynomzal m V[X] with p(x) = 0, and that

x

a E A is a root of 8p. Then a is O-compatible with x. (ii) Let n E N. Suppose that a E w(x) . Inv A, and that kv(x) fj. rt (k E N n ). Then the map p(x) t-+ (8p)(a), V(n)[x] ---+ A, is a Imear map which extends 0 and which is compatible with 'Ij;. Proof (i) Let y E V. It follows from (1.6.4) that x - y E -p(y) . (ev and a - 8(y) E -8(P(y)) . (eA + MA), and so

a - 8(y)

E

+ Mv)

w(p(y)) . Inv A = w(x - y) . Inv A.

x + V ---+ A, is compatible with 'Ij;. (ii) Set f.L(p(x)) = (8p)(a) (pE V(n)[X]). By 1.7.16(i), p(x) =1=0 (p E V(n)[Xj-), and so f.L : V(n)[x] ---+ A is a well-defined linear map which extends 8. Take y, z E V- and k E N n . Since kv(x) fj. rt, necessarily v(yxk) =1= v(z). If v(yxk) < v(z), then w(z) E w(Y)W(X)k MA, and so 8(z) E O(y)a k MA. Similarly, if v(yxk) > v(z), then O(y)a k E 8(z)MA. Now take p E V(n) [X]-, say Thus the map x

+ y t-+ a + 8(y),

p(X) =

xm1Xml

+ ... + Xm,.Xmk,

where 0 ::; m! < m2 < ... < mk ::; nand X m1 , ... , x mk E V-. There exists j E Nk such that v(xm,x mj ) < v(xmix m.) (z =1= j). Then v(p(x)) = v(xmjx mj ), and so w(p(x)) = w(xmJx mj ). If i < j, then v(xmjXmj-m;) < v(xmJ, and so

O(amJ E O(xm,)amj-mi M A . Hence O(xmJa mi E 8(xmJa m'MA. This also holds if i », and so J.L(p(x)) E O(xmj)amj . (eA + MA) C w(xmjx m;) . Inv A. Thus f.L is compatible with W. 0 Lemma 1.7.34 Suppose that A and W are both Henselian. Then:

(i)

V is a valuation algebra;

(ii) W is an immediate extension of V; (iii) V is algebraically closed in W; (iv) V is Henselzan. Proof We write V and 0 for V and 8, respectively. (i) By 1.7.29, V = V(V), and so V is a valuation subalgebra of W. (ii) Assume that there exists x E W \ V such that kv(x) fj. rt (k EN), and set a = 1I1(x). By 1.7.33(ii), the map p(x) t-+ (8p)(a), V{n)[x] ---+ A, is a linear map which extends 8 and which is compatible with 'Ij; for each n E N. Define f.L(P(x)) = (8p)(a) (p E V[X]). As in 1.7.31, (V[x], f.L) ~ (V,8) in F. But this

Algebraic foundations

108

contradicts the assumed maximality of (V, 0). Thus, for each x E W \ V, there exists n E N with nv(x) E rt. Next, assume that there exists x E W \ V and n ~ 2 such that nv(x) E rt, but kv(x) fj. rt (k E Nn-d. Then xn E y . Inv V for some y E V. Since W is Henselian, it follows from 1.6.4 that there exists z E W with v(z) = vex) and zn = y. We have w(z)n E w(zn) . Inv A = O(y) . Inv A, and so, since A is also Henselian, there exists a E w(z) . Inv A such that an = O(y). By 1.7.16(i), z is not algebraic of degree less than n over V, and so z is algebraic of degree n. By 1.7.33(ii), the map p(z) ...... (Op)(a), V{n-l)[Z) -+ A, is a linear map which is compatible with t/J. Thus, by 1.7.32, there exists I-' such that (V[z),I-') >r (V,O) in F. By the maximality of (V,O), necessarily z E V, and so vex) E rt, a contradiction. Thus vex) E rt (x E W), and hence W is an immediate extension of V. (iii) Assume towards a contradiction that V is not algebraically closed in W. Then there exist x E W \ V and p E V(X)· with p(x) = O. Certainly

v(p(x» ~ v«x - y)p'(x»

(y

E

V).

(1.7.4)

We may suppose that p and x are chosen so that op is minimal among polynomials satisfying (1.7.4) for any x E W \ V, say op = n. By 1.7.21 there is a Henselian polynomial Po E V(X) with oPo ~ n such that Po has no root in V. Since W is Henselian, Po has a root, say Xo. in W \ v. By the choice of p and x, Xo is algebraic of degree n over V. Since A is Henselian, Opo has a unique root, say a, in A, and, by 1.7.33(i), a is O-compatible with Xo. By 1.7.30, the map q(xo) ...... (Oq)(a), V{n-l)(XO) -+ A, is compatible with t/J. By 1.7.32, there exists I-' such that (V(xoJ, 1-') ~ (V, 0) in:T. But this contradicts the maximality of (V, 0). Thus V is algebraically closed in W. (iv) Since W is Henselian and V is algebraically closed in W, the subalgebra V is Henselian. 0 The above lemma allows us to deduce an embedding theorem for a valuation algebra W into the algebra A = ~(rt). For s E Pw, define 1jJ(s) = as. Then t/J : Pw -+ MA is a freely acting framework map. Let U be a subset of W. Then 0 : U -+ A is compatible with t/J if and only if v(O(x» = vex) (x E U); such a map is said to be valuatwn-preservzng. Let V be a subalgebra of W, and let 0 : V -+ A be a homomorphism which is compatible with t/J. Suppose that V =f. W, that x E W \ V, and that f E A. Then f is O-compatible with x if and only if v(l - O(y» = vex - y) (y E V). Proposition 1.7.35 Let Uo be a subalgebm of a valuatwn algebra W, and let 00 : Uo -+ ~(rt) be a valuation-preserving embeddzng. Then there is a valuationpreserving embedding 0 : W -+ ~(rt) which extends 00 , Proof Set A = ~(rt), and let :F be as above. By 1.7.26, A is Henselian. By 1.7.25, we may suppose that W is Henselian. Let (V, 0) be a maximal element of :F. By 1.7.34, V is Henselian and W is an immediate extension of V. Assume that there exists x E W\ V. Then, essentially as in 1.7.7(i), there exists f E A such that v(l - O(y» = vex - y) (y E V); f is O-compatible with x. By 1.7.31, there exists I-' such that (V[x],I-') ~ (V, 0) in:F,

Valuation algebras

109

a contradiction of the maximality of (V, (J). Thus V = W, and (J is the required m~.

0

Corollary 1.7.36 (Kaplansky's isomorphism theorem) Let W be a maximal, complex valuation algebra. Then there is a valuation-preservzng isomorphism from W onto ~(rtv)· Proof By the above proposition, there is a valuation-preserving embedding (J : W ---+ ~(rtv). Since W is maximal and ~(rtv) is an immediate extension of (J(W), (J is a surjection. 0 We can now show that the algebra M# is unzversal in the class of /31-valuation algebras.

Theorem 1.7.37 (Esterle) Let V be a /31 -valuation algebra. Then there is an embedding of V into M# . Proof Set V = U{Vu : a < wd, where {Vu : a < wd is a chain of aI-valuation subalgebras of V; we regard each rv" as an aI-subgroup of rv. By a transfinite recursion, there is a family {(JT : 7 < WI} such that each (JT : VT ---+ ~(rtJ is a valuation-preserving embedding and (JT I Vu = (Ju whenever a < 7 < Wl. Indeed, given OTl where 7 < WI, it follows from 1.7.35 that there exists (JT+l such that (JT+I I VT = (JT. Now suppose that 7 < Wl is a limit ordinal and that (Ju has been defined for each a < 7; define (J~ on U{Vu : a < 7} by requiring that (J~ I Vu = (Ju for each a < 7, and then extend (J~ to VT by using 1.7.35 again. Finally, define (J : V ---+ ~(I)(rt) by requiring that (J I VT = (JT (7 < wt). Then (J is a well-defined, valuation-preserving embedding. By 1.2.29(i), rv can be identified as an ordered subgroup of G, and so we may regard ~(1)(rt) as a valuation subalgebra of M#. 0 Theorem 1.7.38 Let B be an zntegral domain with a transcendence degree at most ~I. (i) Suppose that B is non-unital. Then there zs an embedding of B in M. (ii) Suppose that B zs unital and has a character. Then there is a unital embedding of B in M# . Proof (i) By 1.7.9, there is a valuation algebra V in the quotient field of B with B c Mv. Clearly the transcendence basis of B is a transcendence basis for V, and so, by 1.7.17, V is a /31-valuation algebra. By 1.7.37, there is an embedding (J : V ---+ M#, and (J(Mv) C M. o (ii) This follows in a similar way; we use 1.7.8 instead of 1.7.9. Let B be an integral domain of cardinality c. Then, with CH, B certainly has transcendence degree at most ~l. The following result will allow us to insert a 'twist' in certain homomorphisms; recall that a pair {a, b} of elements of a commutative algebra is algebraically independent if p{a, b) # 0 whenever p E qx, Yj-.

Algebraic foundatwns

110

Proposition 1.7.39 Let {I, g} be an algebraically independent set m M#. Then there zs an automorphism J.t on M# with J.t(f) = f and J.t(g) =I- g. Proof Choose ao < WI such that f,g E M~o' and then choose hEM- with v(h) » Gto ' say h E M~. For each p E qx, Yj-, we have p(f,g) =I- 0 and p(f,g + h) - p(f,g) E hM#, and so v(p(f,g + h» = v(p(f, g». Thus the map

p(f, g)

()o :

t-+

p(f, 9 + h),

qf, gj

--->

M1a '

is a well-defined, valuation-preserving embedding. By 1. 7.35, there is a valuationpreserving embedding J.tTo : M~ ---> M~ which extends ()o. By a transfinite recursion using 1.7.35, there is a family {J.tT : T < WI} such that each J.tT : M~ ---> M~ is a valuation-preserving embedding and such that J.tq I Mf' = J.tT whenever TO S T < a < Wl· Since each M~ is a maximal valuation algebra, each J.tT is an epimorphism. Define J.t : M# ---> M# by the formulae /1 I M~ = J.tT (TO S T < wd. The map J.t is well-defined, and is the required automorphism. 0 We have given the preliminary results necessary for the proof of our algebraic extension theorem in the case where the framework map 't/J is freely acting. We now explain the modifications that will give the general case. Thus we suppose that 't/J : W ---> A is a framework map; we may suppose that Pw =I- 0. Define I = 't/J(Pw)J... Then I is a proper ideal in A. Write 7r : A ---> AI I for the quotient map, and set B = CeA

+ w(Mw)

. A,

so that W(W) c B. Since A is local, we have 1I'(Inv A) = Inv (AI I) and rad (AI I) = (rad A)I I. It follows that, if A is Henselian, then so is AI I. Note that the only point at which the fact that a framework map satisfies 1.7.27(ii) is used in our theory is in clause (ii) of the following result. Proposition 1.7.40 (i) B is a subalgebra of A. (ii)

11'

IB

:B

--->

AI I is an injection.

(iii) 1I'(B) = CeA/I

(iv)

11'

0

't/J : Pw

--->

+ (11' 0 W)(Mw)

. (AI I). All zs a freely actmg framework map.

Proof (i) Take a,b E A and x,y E M w , say with vex) ~ v(y), so that w(x) E w(y)A. Then w(x)a + w(y)b E w(y)A c w(Mw) A, and hence w(Mw)A = w(Mw) . A. It follows easily that B is a subalgebra of A. (ii) Take b = aeA + W(x)a E B, and suppose that 1I'(b) = 0, so that bEl. Since eA ¢ I and w(x) E M A , necessarily a = O. If x = 0, then b = O. If x =I- 0, then w(x) E w(M and so 0 = bW(x) = W(x)2a = w(x 2)a. By 1.7.27(ii), W(x)a = 0, and so b = O. This shows that 11' I B is an injection. (iii) This is immediate.

w),

(iv) Certainly 11' 0 't/J is a framework map. If a E A and (7r 0 W)(x)1I'(a) = 0, then w(x) = 0 by (ii), and so a E I. Thus 7r 0 'I/J is freely acting. 0

111

Valuatwn algebras

We write 11"-1 : 1I"(B) ---> B for the inverse of the injection 11" I B. Let V be a subset of W. Suppose that a map B : V ---> A is compatible with 'l/J. Then clearly 11" 0 B : V ---> AI I is compatible with 11" 0 'l/J, and B(V) C B, so that 11"-1 011"0 B = B. Proposition 1.7.41 Let () : W ---> AI I be a homomorphzsm which zs compatible with 11" 0 'l/J. Then (}(W) C 1I"(B) and 11"-1 0 () : W ---> B is a homomorphism which is compatible with 'l/J. Proof Let x E W, say x = aew + y, where a E C and y E Mw. Since () is compatible with 11" 0 'l/J. (}(y) = (11" 0 w)(y)1I"(a) for some a E Inv A, and so B(W) C 1I"(B). Since aew + w(y) E B, we have (11"-1 0 (})(x) = aew + W(y)a. If a -1= 0, then we;]:) = eA, and, if 0: = 0, then w(x) = w(y). Thus, in either case, (11"-1 0 B) E w(x) . Inv A, and so 11"-1 0 () is compatible with W. 0 We can now give our algebraic extension theorem. Theorem 1.7.42 Let Uo be a subalgebra of a complex valuation algebra W with Mw -1= 0, let A be a local Henselian algebra, let'l/J : Pw ---> MA be a framework map, and let (}o : Uo ---> A be a homomorphism whzch is compatzble with '1/" Suppose that W is algebraic over Uo. Then there is an embedding of W into A which extends Bo and whzch zs compatzble with 'l/J. Proof By 1. 7.25, W has an immediate, algebraic extension which is a Henselian algebra, and so we can suppose that W is Henselian. Suppose first that the framework map 'l/J is freely acting, and let F be as above. Let (\7,0) be a maximal element of F. By 1.7..:34(iii), V is algebraically closed in W. Since W is algebraic over Uo, we have V = W, and so the result follows in this case. Now consider the general case, and take I, 11", and B as above. Then AI I is a Henselian algebra, 11" 0 'l/J : W ---> AI I is a freely acting framework map, and 11" 0 0 : U ---> AI I is a homomorphism which is compatible with 11" 0 'l/J. Thus there is a homomorphism J.L : W ---> AI I which extends 11" 0 0 and which is compatible with 11" 0 'l/J. By 1.7.41, 11"-1 0 Jt : W ---> B is a homomorphism which extends () and which is compatible with 'I/J; this implies that B is a monomorphism. 0 The above results are also sufficient to cover certain non-algebraic extensions; for this, we require that W be an aI-valuation algebra (see 1.7.15(i» and that a certain subset of A be a Mittag-Leffler set (see 1.3.40). Let W, A, 'l/J, and F be as above, with '1/' a freely acting framework map. The key lemma that we require is the following. Lemma 1.7.43 Let (V, 0) E F, and let x E W\ V. Suppose that v(x+ V) has no maximum element, that W is an aI-valuation algebra, and that 'l/J(Pw) . Inv A is a Mittag-Leffler set. Then there exists a E A which is f)-compatible with x. Proof First suppose that x

E

Mw, so that vex) E Pw.

Algebraic foundations

112

Since v(x + V) has no maximum element and W is an aI-valuation algebra, there exists a sequence (Yn) in V such that (v (x - Yn» is a strictly increasing, cofinal sequence in v(x + V). We have

= v(x -

V(Yn - Yn+l)

Yn) < v(x - YnH)

= V(YnH

- Yn+2)

and we may suppose that v(x - yr) > v(x). Then V(Yl) V(Y2 - yr) > v(yr). Set Zl = Yl, Z2 = (Y2 - Yl)/Yl, and Zn

= (Yn - Yn-d/(Yn-l - Yn-2) (n

~

(n

E

N),

= v(x) > 0 and also 3).

w,

Then (Zn) C M Zl'" Zn+l = Yn+l - Yn, and 'Lj=l Zl'" Zj = Yn (n EN). Set an = (J(zn) (n EN). Then (an) C 1jJ(Pw) . Inv A, a Mittag-LefHer set in MA, and so there exists a E MA such that (1.3.22) holds. We have a - (J(Yn) E (J(YnH - Yn) . Inv A = W(Yn+l - Yn) . Inv A

(n E N).

(1.7.5)

We now show that a is (J-compatible with x. Take Y E V, and choose n E N such that v(x - Yn) > v(x - y). Then V(YnH - Yn) > v(x - Y), and so W(Yn+l - Yn) E w(x - y)MA

.

By (1.7.5), a - (J(Yn) E w(x - y)MA . Also, V(Yn - y) = v(x - Y), and hence (}(Yn) - (J(y) E w(x - y) . Inv A. Thus a - (}(y) E w(x - y) . Inv A, as required. For the general case, let aew + x E W, where a E C and x E Mw. Take a E MA to be (}-compatible with x. Then oeA + a is clearly (J-compatible with oew +x. 0

a

a-

We remark that, if E A is such that a E n{W(Yn+l - Yn)A : n E N} for a sequence (Yn) such that (v(x - Yn» is cofinal in v(x + V), then is clearly also (J-compatible with x.

a

We can now establish the main extension theorem of this section. Theorem 1.7.44 Let Uo be a subalgebra of an 01 -valuation algebra W with Mw =f 0, let A be a local Henselian algebra, let 1jJ : Pw -+ MA be a framework map, and let (}o : Uo -+ A be a homomorphism which is compatible with W. Suppose that 1jJ(Pw) . Inv A ~s a M~ttag-LejJler set in A. Then there ~s an embedding of W into A which extends (Jo and which ~s compatible with 1jJ. Proof As in 1.7.42, we may suppose that W is Henselian. Suppose first that 1jJ is freely acting, and define F as above. Let (V, 8) be a maximal element of:F. By 1.7.34, V is Henselian and W is an immediate extension of V. Assume that there exists x E W \ V; by 1.7.6, v(x + V) has no maximum element, and so, by 1.7.43, there exists a E A which is (}-compatible with x. By 1.7.31, there exists v such that (V[x],v»);= (V,8) in F. But this contradicts the maximality of (V, 8), and hence V = W. The result follows in this case. Now consider the general case, and take I, 71", and B as before. Clearly (71" o'I/J)(Pw) . Inv(A/I) is a Mittag-LefHer set in A/I, and so there is a homomorphism J.L : W -+ A/ I which extends 71" 0 (J and which is compatible with 71" 0 'I/J. The result follows from this, as in 1.7.42. 0

Valuation algebras

113

The above theorem does not cover the case where W = M# because M# is not an aI-valuation algebra. However, M# is a ,a1-valuation algebra, and so we can easily extend 1.7.44 to obtain such an embedding. Theorem 1.7.45 Let A be a local Henselian algebra. Suppose that there is a morphzsm 1j; : (P, +) ---+ (M.A, .) such that al. = bl. (a, bE 1j;(P)) and such that 1j;(P) . Inv A zs a Mittag-Leffier set m A. Let fo E Me and ao E 1j;(P) . Inv A. Then there zs an embedding () of M# into A such that ()(Me) c 1j;(P) . Inv A and ()(fo) = ao· Proof The map 1j; : P

---+

()o :

M.A is a framework map, and p(fo)

1-+

p(ao),

Co[fo]

---+

A,

is a well-defined homomorphism which is compatible with 1j; . Take 7"0 < WI such that fo E Mfa. By a routine transfinite recursion using the main extension theorem 1.7.44, we see that there is a family {()" : 7"0 ~ a < WI} such that each ()" : M~ ---+ A is a homomorphism which is compatible with 1j;, such that (}r I M~ = ()" whenever 7"0 ~ a < 7" < WI. and such that ()" I Co[fo] = (}o for 7"0 ~ a < WI· Define (} : M# ---+ A by requiring that () I M~ = (}" (7"0 ~ a < wt). Then (} is a well-defined homomorphism with the required properties (where we recall that M# = U{M~: a < wd). 0 Notes 1.7.46 For the general theory of valuations, see (Jacobson 1980, Chapter 9). Some texts do not require a valuation algebra to be an integral domain. The connection between valuation algebras and abstractly defined valuations is explained in (Dales and Woodin 1996, Chapter 2); the fundamental theorem for the extension of places 1.7.8 is exactly (ibid., 2.13), where other references are given. Theorem 1.7.17 is essentially the same as (ibid., 2.28). Further properties of the fields Rand C can be found in (tbid.). It is a theorem of J. R. Esterle that each real-closed .81 -1/1-field is isomorphic to R; see (ibid.,2.33). Thus, in the theory ZFC + CH, each ultrapower ~N /U, where U is a free ultrafilter on N, is isomorphic to R. Theorem 1.7.19 was first proved in (Mac Lane 1939); see also (Priess-Crampe 1983, 11.5.6) and (Dales and Woodin 1996, 2.15). Most of the results on the extensions of homomorphisms compatible with a given framework map are abstract versions of calculations given in (Esterle 1977, 1978b, 1979a) and (Dales 1979a); 1.7.21 and 1.7.37 were explicitly given in (Esterle 1984b). Kaplansky's isomorphism theorem 1.7.36 was first proved in (Kaplansky 1942); for a different approach to 1.7.36 and 1.7.37, see (Dales and Woodin 1996, Chapter 2). In fact, each algebraically closed .81-1/1-Valuation algebra V is isomorphic to M#; this follows by a development of 1.7.37, and also follows from the fact that each realclosed .81-1/1-field is isomorphic to R. It follows from 1.7.38 that, with CH, each complex, unital integral domain of cardinality c is isomorphic to a sub algebra of (~N /U)O#. The following related, stronger result, which holds in the theory ZFC itself, is proved in (Dales and Woodin 1996, 5.25). Theorem Each complex, non-unital integral domain A of cardinality c is isomorphic to a subalgebra of co/U for some free ultrafilter U (depending on A) on N. 0 Thus the class of integral domains of the form non-unital integral domains of cardinality c.

eo/U

is universal in the class of all

Algebmic foundations

114 1.8 DERIVATIONS

In this section, we shall develop the algebraic theory of derivations from an algebra A into an A-bimodule E and, in particular, into A itself. The role of point derivations, defined in 1.8.7, will be significant later. We shall see that higher point derivations of infinite order on A correspond to homomorphisms from A into q[X]). We shall establish in Theorem 1.8.14 an important connection between homomorphisms and derivations, and this will lead in Theorem 1.8.15 from 1.6.31 to a condition for the extension of a derivation defined on a subalgebra of a given algebraj we shall also construct derivations from the algebra 3' into its algebraic dual space X. These results will be used in §5.6 to construct discontinuous derivations from certain Banach algebras.

Definition 1.8.1 Let A be an algebm, and let E be an A-bimodule. A linear map D : A ----+ E is a derivation zf

D(ab) = a . Db + Da . b (a, bE A) .

(1.8.1)

Equation (1.8.1) is the derivation identity. The set of derivations from A into E is denoted by Zl(A, E)j it is linear subspace of .c(A, E). For example, take x E E, and set 8x (a) = a . x - x . a (a E A) . (1.8.2) Then, for a, b E A, we have 8x (ab)

=a

. (b . x - x . b)

+ (a

. x - x . a) . b = a . 8x (b)

+ 8x (a)

. b,

and so 8x is a derivation. Derivations of this form are termed inner denvations, and an inner derivation 8x is implemented by Xj derivations which are not inner derivations are outer denvations. The set of inner derivations from A to E is a linear subspace Nl (A, E) of Zl (A, E). Clearly the kernel of a derivation from A to E is a subalgebra of A. Let E and F be A-bimodules, and let D E Zl(A, E) and R E A.cA(E, F). Then RoD E Zl(A,F). Proposition 1.8.2 Let A be an algebm, let E be an A-bzmodule, and let D : A ----+ E be a derivation. (i) Suppose that p E J(A). Then p . Dp . p = o. (ii) Suppose that p E J(A) and p . Dp = Dp . p. Then Dp =

o.

(iii) Suppose that a E A and a . Da = Da . a. Then

D(an ) = nan-I. Da

(n E N) .

(iv) Suppose that A and E are unital and that a E Inv A with a· Da = Da· a. Then D(an ) = nan-I. Da (n E Z) . (v) Suppose that (a" : a E Q+.) is a mtional semigroup in A and that a· Da=Da· a. Then

D(a Ct ) = aaCt -

1 •

Da

(a E Q n (1,00)).

Der'lVations

115

Proof (i) Since p = p2, we have Dp = p . Dp + Dp . p, and so immediately p . Dp = p . Dp + P . Dp . p and p . Dp . p = O. (ii) Now Dp

= 2p

. Dp

= 4p

. Dp, and so p . Dp

= 0 and hence

Dp

= O.

(iii) This is an immediate induction. recalling that aD . x = x (x E E).

(iv) By (iii). the result holds for n E N. By (ii), it holds for n = 0, and it now follows from the derivation identity that it also holds for -n E N. (v) Suppose that a = p/q, where p,q E Nand p/q > 1. By (iii), D(aP/ q ) = pa(p-l)/q . D(a 1 / q ) and Da = qa(q-l)/q . D(a l / q ), and so D(aP/ q)

= pa(p/q)-l

.

al-(l/q) .

D(a l / q )

=

(p/q)a(P/q)-l . Da,

o

as required.

The module operations on £(1, E) in the next result were defined in (1.4.14). Proposition 1.8.3 Let A be an algebm, let E be an A-bzmodule, let I be a left ideal in A, and let D : I --+ E be a der'lVatzon. For each a E A, the map Da : x

1--+

D(ax) - a . Dx,

I

--+

is a right I-module homomorphism, and the map a derivation.

1--+

E, Da. A

--+

£(I,E), is a

Proof Note that Dab(X) = (a . Db + Da x b)(x) (a, bE A, x E I). The results are now obtained by immediate calculations. 0 Theorem 1.8.4 (Gr!/lnbrek) Let I be an ideal in a commutative algebm A, let E be an A-module, and let D : I --+ E be a derivation. Then the map

jj : (a, b)

1--+

D(ab) - b . Da,

I x A

--+

E,

is a bilinear map such that: (i) D(a, b)

=a

. Db (a, bE I);

(ii) for each a E 12, the map b 1--+ D(a, b), A

--+

E, zs a derivation.

Proof Certainly D : I x A --+ E is bilinear, and (i) holds. Take aI, a2 E I and bl , b2 E A. Then D(ala2, bl b2)

= D(ala2blb2) - bl b2 . D(ala2) = alb l . D(a2 b2) + a2b2 . D(alb l ) = bl . (al . D(a 2b2) - b2al . Da2)

bl b2 . D(ala2)

+ b2 . = bl

(blal . Da2 + a2 . D(alb l ) - bl . D(ala2)) (D(ala2 b2) - b2 . D(ala2))

.

+ b2 . b1 and this implies (ii).

.

(D(ala2bl) - b1

.

D(ala2))

D(ala2, b2 ) + b2 . D(ala2, bd

, o

Algebraic foundations

116

A particular case of a derivation arises when E is equal to A; we refer to a derivation on A. For example, the map 8b : a

J---+

ab - ba, A

~

(1.8.3)

A ,

is an inner derivation on A for each b E A. Note that, by induction, 8i:(a)

= ~(_l)k(~)bkabn-k

(a,b E A, n E N).

(1.8.4)

The formula in part (i), below, is Leibnzz's identity. (Our convention is that DO is the identity map.) Theorem 1.8.5 Let D be a derivation on an algebra A, and let n E N. (i) For each a, b E A, we have Dn(ab) =

t (~)Dka

. Dn-kb.

k=O

(ii) Suppose that I is an ideal in A. Then D(I) I I is an ideal in AII and {a E I : Dna E I (n E N)} is an ideal in A. (iii) Suppose that a E A with D 2 a = o. Then Dn(a n ) = n! (Da)n. (iv) Suppose that I is an ideal m A, and that al, ... , an E I. Then Dn(al ... an) - n! (Dal) ... (Dan) E I.

(v) Suppose that r.p E q,A and a E M
Proof (i) This is a straightforward induction. (ii) Take a E A and bEl. Then aD(b) = D(ab) - (Da)b E D(I) + I, and so D(I)I I is a left ideal in AI I. Similarly it is a right ideal. The second part is immediate from (i). (iii) We first claim that D«Da)k) = 0 (k EN). This is true by hypothesis in the case where k = 1. The claim follows by induction because D«Da)k+1)

= D 2a

. (Da)k

+ Da

. D«Da)k)

The formula Dn(a n ) = n! (Da)n is true for n n. Then, using (i) and the claim, we have Dn+l(a n+l ) = D

= Da

. D«Da)k).

= 1. Assume that it holds for

(~(~)Dk(an) . Dn-ka)

= D (n(Dn-1a n ) . Da + Dn(an ) . a) = n (n! (Da)n . Da)

+ n! (Da)n . Da = (n + I)! (Da)n+1,

and so the formula holds for n + 1. Thus the formula holds for each n EN. (iv) We first claim that, for each k ~ 2, we have Di(b l ··· bk) E I whenever j E Nk-l and bI, ... , bk E I. The claim is true for k = 2, for certainly

Derivations

117

D(bl~} = bl . Db2 + Dbl . ~ E I whenever bI. b2 E I. Assume that the claim holds for k and that bll ... , bk+l E I. Then, by (i),

Di(b l · ··bk+l ) =

t G)Di(b

l ..

(j E N),

·bk) . DJ-ibk+l

,=0

and so Di(b l ·· ·bk+l ) E I (j E Nk). By induction on k, the claim holds. We now prove the result by induction on n. The result is certainly true if n = 1. Assume that the result is true for n, and take al, ... , an+l E I. By (i),

D n+l (al· ··an+d

~ (n + k

= ~

1)

D k (al·· ·an) . D n+l-k an+l·

k=O

Each term on the right-hand side of this equality belongs to I, save perhaps for the term (n + l)Dn(al ... an)Dan+t. and so

Dn+l(al ... an+l) - (n + l)Dn(al ... an)Dan+l E I. By the inductive hypothesis, Dn(al ... an) - n! (Dal)··· (Dan) E I, and so Dn+l(al '.' an+l) - (n + I)! (Dat}··· (Dan+d = Dn+l(al ... an+l) - (n + l)Dn(al ... an)(Dan+t)

+ (n + 1)(Dn(al ... an) belongs to I. The induction continues. (v) Apply (iv) with 1= Mtp and al = (vi) This is an immediate calculation.

n! (Dat}··· (Dan))(Dan+d

... = an = a. o

Proposition 1.8.6 Let D be a derivation on an algebra A. Then

D(SfJ(A}) C SfJ(A) . Proof Let P be a prime ideal in A, and set

Q = {a E P : Dna E P (n E N)} . By 1.8.5(iii}, Q is an ideal in A, and D(Q) C Q C P. We claim that Q is prime. For suppose that a, b E A with aAb C Q and that a ¢. Q; set n = minim E Z+ : Dma ¢. Pl. We shall prove by induction on k that Dkb E P (k E Z+). Thus suppose that k = 0 or that kEN and that b, Db, ... , Dk-lb E P. For each x E A, we have

Dn+k(axb) =

'f t J=O

,=0

(n

~ k) (:)Dn+k-ja . Di- i X

•

Dib.

(1.8.5)

J

The left-hand side of (1.8.5) belongs to P, and all the terms on the right-hand side belong to P, save perhaps for the term Dna· X· Dkb, and so this latter term belongs to P for each x E A. Since Dna ¢. P, necessarily Dkb E P, continuing the induction. The claim follows. It follows that D(P) C P whenever P is a minimal prime. Since !.J.l(A) is the intersection of the minimal primes of A, the proposition is proved. 0

Algebraic foundations

118

Definition 1.8.7 Let A be an algebra, and let cp E AU {o}. A linear functwnal d on A tS a point derivation at cp if

d(ab) = cp(a)d(b)

+ cp(b)d(a) (a, bE

A) .

(1.8.6)

As in (1.4.2), lK

A.

(i) A linear functional d on A is a pomt derivation at cp if and only if d I (JKeA + M~) = O. (ii) The space Zi(A,lK
o

(ii) This follows from (i). Let A be a non-unital algebra, and let cp E A. Define

E
= (u -

eA)M
+ M
eA)

+ M;,

(1.8.7)

where u is any modular identity for M
of A tf a.nd only if ), has one of the following forms: (i) ), = (cp, where (E C· and cp E A; (ii) ), = (cp - 'I/J), where (E C· and cp,'I/J are distinct elements ofA; (iii) ), is a non-zero point denvation at cp for some cp E A . Proof Clearly ker), is a subalgebra of A when), has one of the specified forms. Now suppose that ker), is a subalgebra of A. First, consider the case where ),(eA) i=- 0, and set), = (cp, where ( = ),(eA)' For a, bE A, we have «(a-),(a)eA)«b-),(b)eA) E ker)" and so cp(ab) = cp(a)cp(b), :showing that), has the form given in (i). Second, consider the case where ),(eA) = 0, and take v E A with ),(v) = 1, and set u = v-),(v 2 )eA/2. Then ),(u) = 1 and ),(u2 ) = 0. Set J1(a) = ),(ua) (a E A). For a, bE A, we have (a - ),(a)u)(b - )'(b)u) E ker)" and so

),(ab) = ),(a)J1(b)

+ )'(b)J1(a)

(a, bE A) .

(1.8.8)

Substituting bu for b in (1.8.8) gives J1(ab) = ),(a)J1(bu) + J1(a)J1(b), whilst substituting u 2 for a in (1.8.8) gives J1(bu) = )'(b)J1(u2 ), and so

J1(ab) = J1(a)J1(b)

+ ),(a»,(b)J1(u2 ).

(1.8.9)

Derivatwns

119

Take (3 E C with /12 = /1( u 2 ). If (3 =I- 0, set tp = J1 + (3A and '1/) = /1 - j3A. It follows from (1.8.8) and (1.8.9) that tp, 'l/J E if?A, and so A = (tp - 'l/J)/2(3 has the form given in (ii). If (3 = 0, then, by (1.8.9), jL E if?A, and, by (1.8.8), A is a point derivation at /1, so that A has the form given in (iii). 0 Proposition 1.8.10 Let A be a commutative, umtal, complex algebra, and let E if? A. Suppose that A is a non:zer'o lmear functwnal on A such that

tp, 'l/J

A(ab) = tp(a)A(b)

+ ).,(a)'l/J(b)

(a, bE A).

Then either: (i) 'l/J = tp and A is a pomt derivatwn at tp; or (ii) 'l/J =I- tp and)" = (('1/) - 'P) for some (E
c

basts has the form Bl 0

(

.. . 0

where each block B j

IS

an

rlj

0 B2

.. .

0

...

0 0

...

...

(1.8.10)

. ..

... Bk

x nj matrix wzth the upper-tTiangular' fOTm

'Pj(a) rnW(a) ... rn\~~J(a)l C)n (a) tpj(a)··· rn/ ( o .. '.J

. .. . o 0

..

(aE2t)

(1.8.11 )

'Pj(a)

faT some nj E N, each rnW being a linear functional on 2t with rnYs) l.. M~~T+l, and wheTe nl + ... + rtk = n. Proof It is standard that there is a basis of en such that each a = (rnij (a» E 2t is upper-triangular with respect to this basis. Clearly each rnij belongs to III x and each rnii belongs to cI>21; we may suppose that (rnll,"" rn nn ) is listed as

120

Algebraic foundations

(rpl .... , rpl, rp2,···, rp2, ... , rpk,···, rpk), where rpl,··., rpk are distinct dements of cI>2(. By 1.8.1O(ii), the basis of C n can be changed in successive steps so that the matrix of each a E 2l has the block diagonal form (1.8.10), with each block B j upper-triangular and with leading diagonal constantly equal to 'Pj(a). It is now an easy induction to check that 1- M;~r+l in each case. 0

rnW

Let A be a commutative, complex algebra, and let E be an A-bimodule of dimension n, with representations Pi and Pr on the left and right, respectively. Then, as in (1.4.7), the subalgebra S = alg {IE, pe(A), Pr(A)} of £(E) is commutative and unital. It follows that there is a decomposition E = El 0· .. 0 Ek of E into a direct sum of A-submodules corresponding to the specified choice of basis in C n ; the projections Pj : E ---+ E j , which are the projections arising in the decomposition described in 1.5.8, are A-bimodule homomorphisms. Let F be one of the submodules E l , ..• , E k • say dim F = m. Then we may choose a basis {el, ... ,em} of F such that the module products are given by

a . X = pt(a)(x)

=

rp(a) Q12(a) Q13(a) ... Qlm(a) 0 'P(a) Q23(a) ... Q2m(a) rp(a) '" Q3m(a) 0 0 0

0

0

... rp(a)

Xl

X2 X3

(1.8.12)

Xm

and

X . a = Pr(a)(x)

=

'I/!(a) f112(a) 1113(a) ... I1lm(a) 0 '¢(a) 1123(0,)··· 112m(a) ,¢(a) .. , •B:\• 17l. (ll) 0 0 0

0

0

... 7/)( a)

Xl

X2 X3

(1.8.13)

Xm

respectively, for a E A and X = L~l xjej E F, where rp, '¢ E A U{O} and where ars and ,3rs are linear fUIlctionals s11ch that Q,s 1- l\,f;-r+l and I1rs 1- M;,-r+l, respectively. In the case where rp = '¢, we say that F is associated with rp. It is clear that el is a common eigenvector for all the maps pe(a) and Pr(a) and that Cel is a one-dimensional A-submodule of F, identified with Ccp."" Now suppose that T E £(cn), and set 2l = C[T]. Then there exist linear sllbspaces E l , ... , Ek of C n such that C n = El 0· .. 0 Ek and such that, for each j E N k , a basis of Ej can be chosen so that T I E j has the form (j 1 0 '" 0 0 o (j 1 '" 0 0 OO(j'''OO

000"'(j1 o 0 0 '" 0 (j

such a matrix is a Jordan block, and the combination of the Jordan blocks gives the Jordan canonical form of T. We have (J'(T) = {(l, ... , (k}. Thus there eXlst S, N E £(C n ) such that T = S + N, where S has a diagonal matrix and N has

Derivatwns

121

a strictly upper-triangular matrix with respect to the basis; further, SN = NS and N is nilpotent (cf. 1.6.9), and the decomposition is unique.

Definition 1.8.12 Let A and B be algebras. A sequence (Dk .c(A, B) zs a higher derivation of order n if, for each k E z;t,

k

E

z;t) in

k

Dk(ab)

=L

Dja . Dk_jb

(a, bE A).

(1.8.14)

j=O

A sequence (Dk : k E Z+) is a higher derivation of infinite order if (Dk : k E is a hzgher- derivation of order n for each n E N.

z;t)

For example, let A be an algebra, and let D be a derivation on A. Then, by 1.8.5(i), (Dk jk! : k E Z+) is a higher derivation of infinite order. Consider the special case where B = e and

A U {a}. Then a sequence (dk : k E z;t) [( dk : k E Z+)] in A x is a higher- pomt derivation of order n [of infinite order] at

dk(ab)

=L

dj(a)dk_j(b)

(1.8.15)

(a, bE A)

j=O

for k E Nn [k EN]. The higher point derivation (dk) is non-degenerate if d 1 1= o. The point derivations defined in 1.8.7 are higher point derivations of order one in this terminology. Note that it follows from (1.8.15) that, if a E M

A U {a}. Suppose that (d n ) is a higher point derivation of infinite order at
A~~.

(): af---> Ldn(a)X n •

(1.8.17)

n=O

Then () is a homomorphism with 11"0 0 () = <po Conversely, if () : A ~ ~ is a homomorphism, then (1I"n 0 () : n E Z+) is a higher point derivation of infinite order at 11"0 0 () on A. Thus higher point derivations of infinite order on A correspond to homomorphisms from A into~. Similarly, higher point derivations of order n correspond to homomorphisms from A into ~(MnH . Let A be a commutative, complex algebra, and let (do, ... , d n ) be a higher point derivation of order n on A. Let i,j E Nn+ll and define Uij(a) = dj_i(a) if j 2: i and Uij(a) = 0 if j < i, so that (uij(a» E MInH is an upper-triangular matrix which is constant on diagonals parallel to the leading diagonal. The map p: a

f--->

(uij(a)),

A

~

MInH,

is a homomorphism, and so en+! is an A-module. Now suppose that there exists d nH E A x such that (do, ... , dnH ) is a higher point derivation of order n + 1 on A, and define D :a

f--->

«n + 1)d..+1(a), ndn(a), ... , 2d2 (a),d 1 (a»,

A ~

e nH .

(1.8.18)

122

Algebraic foundatwns

Then D is a linear map such that D(ab) = p(a)D(b) so D is a derivation.

+ p(b)D(a)

(a, bE A), and

Proposition 1.8.13 Suppose that there zs a non-degenerate higher point derivation of mfinite order at a character cp on a commutative algebra A. Then there are ao E M

t~

be (d n ). There exist

al E A with dl (al) = 1 and u E A with cp(u) = 1. Set ao = al - cp(al)(2u - u 2). Then ao E M

A;(ao)

Let n E N. Then expao - Z=~=O a~/k! E a~+l A, and so n

J.L(expao) -

L

Xk

T!

E

xn+l~

(1£

E

N).

k=O

Thus J.L(expao) = expX.

0

There is an important connection between homomorphisms and derivations. A generalization of the following construction will be given immediately before 1.9.5. Theorem 1.8.14 Let A be an algebra, and let E be an A-bimodule. Define 21 to be the Imear space A 8 E, with the product (a, x)(b, y) = (ab, a . y + x . b)

(a. bE A, x, Y E E).

(1.8.19)

(i) The Imear space 21 is an algebra with respect to the specified product; 21 is commutative zf and only if A is commutative and E zs an A -module. The map (a, x) J-+ a, 21 ~ A, zs an epimorphism. (ii) The subspace E is a nilpotent zdeal in 21, and rad21 = (radA) 8 E.

(iii) Let D : A ~ E be a map, and define () : a J-+ (a, Da), A ~ 21. Then () is a homomorphism zf and only if D is a derivation. (iv) Suppose that D : A zs a derivatwn.

~

E is a derwatwn. Then (a, x)

J-+

(0, Da), 21

~

21,

Proof (i), (iii), and (iv) These are immediate verifications. E

(ii) Certainly E[2] = 0, and so E is a nilpotent ideal in lXj by 1.5.6(ii), E. 0

c rad lX. By 1.5.4(ii), rad A = (rad 21)/E, and so rad 21 = rad A 8

Let A be a commutative, unital algebra, let E be an A-module, and let D : A ~ E be a derivation. Suppose that p = Z=~=o akXk belongs to A[X). Then, for a E A, we define the element a . Dp E E by the formula n

a . Dp = Dao

+ L a k . Dak. k=l

(1.8.20)

Denvations

123

Theorem 1.8.15 Let A be a unital integral domain, let E be a unital A-module, let B be a unital subalgebra of A, and let D : B - E be a derivatwn. Suppose that a E Alg B, that q is a mimmal polynomial for a, and that a . Dq + q' (a) . x

=0

(1.8.21)

for some x E E. Suppose further that either E is torsion-free or q(O) E Inv B. Then the map D : pea) f--+ a . Dp + p'(a) . x, B[a] - E, is a derivation which extends D, and Da = x. Proof Take S.2[ = A 0 E to be the algebra constructed in 1.8.14. The map () : b f--+ (b, Db), B - Qt, is a homomorphism, and (a, x) is a root of (}q because ((}q) (a, x) = (q(a), a . Dq + q'(a) . x) = (0,0). Suppose that E is torsion-free. For b E B- and (c,z) E Qt-, we have (}(b)(c, z) = (be, b . z + Db . c); if c ::/:- 0, then be ::/:- 0 because A is an integral domain, and, if c = 0 and z ::/:- 0, then b . z + Db . c = b . z ::/:- 0 in this case, and so (}(b)(c,z) ::/:- 0. It follows from 1.6.31 that the map (j : pea) f--+ ((}p) (a, x), B[a] - Qt, is a homomorphism which extends (). Thus the specified map has the required properties. 0 Theorem 1.8.16 Let A be a commutative, unital algebra, let E be a unital Amodule, let B be a unital subalgebra of A, and let D : B - E be a derivation. (i) Let C be the inverse-closure of B in A. Then there zs a derivation D : C - E such that D I B = D. (ii) Suppose that a E A is transcendental over B. Then, for each x E E, there is a derivation D: B[a] - E such that D I B = D and D(a) = x. Proof (i) For b E B and c E B n Inv A, set D(bc- 1 ) = c- 1 Then D is well-defined, and it is the required derivation. (ii) This is immediate from 1.6.32.

.

Db - bc 2

•

Dc. 0

Theorem 1.8.17 Let A be a unital integral domain, let E be a umtal, torsionfree, divisible A-module, let B be a unital subalgebra of A, let D : B - E be a denvation, let a be transcendental over B, and let x E E. Then there is a derivatwn D: A - E such that D I B = D and Da = x. Proof By 1.8.16(ii), there is a derivation D : B[a] - E such that D I B = D and Da = x. Let C be a subalgebra of A with B[a] c C, let D' : C - E be a derivation with D' I B[a] = D, and take b E A \ C. In the case where b E AlgC, let q be a minimal polynomial for b. Since q'(b) ::/:- 0 and E is divisible, there exists y E E with b . Dq + q'(b) . y = O. In the case where b ¢. AlgC, choose y arbitrarily in E. It follows from 1.8.15 and 1.8.16 in the respective cases that there is a derivation D" : C[b] - E with D" I C = D' and D"(b) = y. The existence of the required derivation D : A - E follows by using Zorn's lemma. 0

124

Algebrazc foundations

The unital 3'-module X of the next theorem was defined in (1.6.8). Theorem 1.8.18 (Bade and Dales) Let Xo E X. Then there is a derwation D : 3' ---+ X such that D I qX] = 0 and D(expX) = Xo.

The proof of this theorem will proceed through several lemmas, in each of which we maintain the same notation. A subalgebra A of 3' is X-closed if A :J qX] and if a E A whenever a E 3' and Xa E A. Lemma 1.8.19 Let A be an X -closed subalgebra of 3', and let D : A ---+ X be a derivatzon with D I qX] = O. Suppose that there is a sequence (an: n E Z+) in A such that (1.8.22) Suppose, further, that there are a strictly increasing sequence (nj) m N and a sequence (D J ) such that DJ : A[an;]---+ X is a derwation extending D and

(1.8.23) Then the subalgebra B = U{ A[an ] : n E N} of 3' is X -closed, and there is a derivatzon D : B ---+ X such that D I A[an;] = D j (j EN).

Proof It is immediate from (1.8.22) that B = U{A[a n ;] : j EN}; B is a subalgebra of 3'. Suppose that a E 3' and that Xa E B, say Xa E A[an ]. Then there exist bo , ... , bm E A with Xa = bo +b1an + ... + bma~. By (1.8.22), an = Xa n+1 +al for some 0: E C, and so m

Xa

= bo + b1(Xan+l + 0:1) + ... + bm (Xa n+1 + o:l)m = Lo:Jbj +

Xb

J=O

for some b E A[an+d. We have X(a - b) E A, and so a - b E A because A is X-closed. Thus a E A[an+l] C B, and so B is X-closed. Let j E N. By (1.8.22), an; - xnj+l-nj a nHl E qX], and so Dj+1a nj = Dj+l(xnj+l-nJanj+d = xnj+l-n; . Dj+1anj+l = Dja n;

by (1.8.23), whence Dj+l I A[an ;] = D j . It follows that the map D : B ---+ X which is defined by the requirements that D I A[a n ;] = D j (j E N) is welldefined. Clearly D is a derivation. 0 Lemma 1.8.20 Let A be an X -closed, inverse-closed subalgebra of 3', let D : A ---+ X be a derivation such that D I qX] = 0, and let a E (Alg A) \ A. Then there is an X -closed subalgebra B of 3' with B :J A[a] and a derwatzon D : B ---+ X extending D. Proof Let p E A[Yl be a minimal polynomial for a, say oa = op = m. Then p'ea) i- OJ we set ko = o(p'(a)). Set a = L~=o a( n )xn. Since a fj. qXl, there is a strictly increasing sequence (n J ) in N such that a(nj) i- 0 (j EN); we may suppose that nl ;::: ko + 1. Define a sequence (an: n E Z+) in A by setting an(k) = a(n + k) (k, n E Z+). Then an - Xa n+1 = a(n)1 E Cl, and so the sequence (an) satisfies (1.8.22). We

Derivations

125

shall apply 1.8.19; for this, we must extend D to each A[an;l in such a way that (1.8.23) is satisfied. Take j E N, and temporarily write b for anj and n for nj. Then we have b(O) = a(n) -I 0, and so b E Inv~. Set c = a - Xnb, so that c E qX]. By Taylor's formula for p,

f

p(xny + c) =

i=O

;,-xnip(i) (C)yi = ~.

f

biyi,

say.

i=O

Since Xnp'(e) E Xn(p'(a) + xn A), necessarily o(xnp'(c» = n + ko. For i ;::: 2, o(bi ) ;::: 2n > n + ko. Also p(Xnb + c) = p(a) = 0, and so we conclude that o(p(c» = o(bo) = n+ko. Thus there exist co, ... , Cm E ~ with o(co) = o(cd = 0, O(Ci) > 0 (i = 2, ... ,m), and bi = Xn+ko Ci (i E Zt.). Since A is X-closed, we see that Co, ... , Cm E A. Since p(m) (c) -I 0, necessarily bm -I O. Since bo E A n Inv ~ and A is inverse-closed, bo E Inv A. Now define q(Y) = L:: o cXi , so that xn+koq(Y) = p(xny + c). Then ()q = m, q(b) = 0, and q(O) = bo E lnv A. It cannot be that ()b < m, for this would imply that

q'(b) =

oa

< m. Hence q is a minimal polynomial for b. We have

Cl + L:::2 iCibi-l,

and so o(q'(b» and so there is a unique x E X such that

=

O. This implies that q'(b) E Inv A,

b . Dq + q' (b) . x = 0 .

(1.8.24)

We have verified that all the conditions in one alternative in 1.8.15 are satisfied. By that theorem, the map r(b) ~ b . Dr + r'(b) . x, A[b] ~ X, is a derivation extending D; this is the map D j . It remains to verify that the compatibility condition (1.8.23) is satisfied. Again take j E N, continue to write b for anj , and now write b for anHl' C for the corresponding c, etc. Set = nj+ 1 - nj. Then we have

e

a = Xnb+

C

= XnHb+c;

there exists d E qX] with c = c - Xnd and b = XR.b + d. Note that we have XkOq'(b) = p'(a); similarly XkOq'(b) = p'(a), and so q'(b) = q'(b). We have taken Dja nj to be x, where x is the unique solution of (1.8.24); similarly Dj+1anj+l = x. Thus to verify (1.8.23) amounts to showing that

XR.b . Dq = b . Dq. Clearly q'(Y)

= q'(XRy + d), m-l

(1.8.25)

and so, by Taylor's formula, m-l

L (i + l)~t1yi = L ;'-XR.iq(Hl)yi i=O

j

i=O 't.

equating coefficients of yi-l in this equation shows that

~ = ;'-XR.(i-l)q(i)(d) (i E Nm ).

(1.8.26)

't.

Also, Xn+Hkoeo = bo = p(c) Hence Xi . Deo = D(q(d»).

= p(Xnd + c) = xn+koq(d),

and so XR.eo

= q(d).

Algebraic foundatzons

126

We now verify equation (1.8.25). Recall that b = Xi1) + d. Hence

b . Dq

m = Dc-o + ~)Xi 1) + d)i

i=l

=

f

J=O

XJi 1)J

t (~) t=J

m = Dca + L

. DCi

L i

i=l J=O d!-j DCi

=f

J

= D(q(d» + fXji1)j

Xji 1)j . D

J=O . D

(;!

(')

z, X ji 1)J d i - j . DCi J

(t (~)

Cid!-J)

t=J

q(j) (d»)

J=l

We now see that (1.8.25) indeed holds because it follows from (1.8.26) that 1) . Dq = Dco

+ f1)j j=l

= Dco

. D (\XlU-1)q(j)(d») J

+ I:X l (j-1)1)j j=l

The lemma now follows from 1.8.19.

. D

(~q(J)(d») J

o

Lemma 1.8.21 Let A be an algebraically closed subalgebra of'J wzth A :::) qX], let D : A -+ X be a derivation with D I qX] = 0, let a E 'J \ A, and let Xo EX. Then there zs an X -closed subalgebra B of'J with B :::) A[a] and a derivatzon D : B -+ X extending D such that Da = Xo. Proof As before, we define a sequence (an: n E Z+) in A by again setting an(k) = a(n + k) (k, n E Z+), where a = L:=o a(n)Xn. We note that each an is transcendental over A. Define a sequence (Xn : n E Z+) in X inductively by the requirement that (1.8.27) X . Xn = Xn -1 (n E N), where Xo takes its prescribed value in X. (The value of xn( -k) is uniquely specified by (1.8.27) for kEN, but xn(O) can be chosen arbitrarily.) By 1.8.16(ii), for each n E Z+, there is a derivation Dn : A[an ] -+ X extending D such that D.,an = X n . By (1.8.27), the compatibility condition (1.8.23) is satisfied, and so the result follows from 1.8.19, where we note that A is certainly X-closed. 0

Proof of 1.8.18 We first apply 1.8.21 in the case where A = Alg~qX], D = 0 on A, and a = expX; by 1.8.20, there is an X-closed subalgebra Bo of'J with Bo :::) A[a] and a derivation Do : Bo -+ X with Do I qX] = 0 and Da = Xo. Let F be the family of pairs (B, D) such that B is an X-closed subalgebra of.'J with B :::) Bo and D : B -+ X is a derivation with D I Bo = Do; set (B1, D 1) ~ (B2' D 2) if B1 c B2 and D2 I B1 = D 1, so that (F, ~) is a partially ordered set. Since the union of a chain of X-closed subalgebras of'J is X-closed - - , it follows from Zorn's lemma that (F, ~) has a maximal element, say (B, D). It follows from 1.8.16(i) that B is inverse-closed in .'J, then from 1.8.20 that B is algebraically closed in .'J, and finally from 1.8.21 that B =.'J. Hence jj is the required derivation. 0

127

Cohomology

Notes 1.8.22 The theory of derivations is given in many texts. including (Jacobson 1980, §8.15). The results 1.8.4, 1.8.6, 1.8.9, and 1.8.18 are taken from (Gr!lmbrek 1989a), (Dixmier 1977. 3.3.2), (Rennison 1970). and (Bade and Dales 1989a), respectively. The fact, used in 1.8.11, that a commuting family of matrices can have a simultaneous uppertriangular form is given in (Horn and Johnson 1985, 2.3.3), for example; 1.8.11 also follows from 1.5.8. The canonical form of matrices is described in innumerable texts; see (Horn and Johnson 1985, Chapter 3) and (Palmer 1994, §2.7), for example.

1.9

COHOMOLOGY

This section contains a rudimentary account of the theory of the (Hochschild) cohomology and homology groups of an algebra A with coefficients in an A-bimodule E; these are the linear spaces Hn(A, E) and Hn(A, E), respectively. The group HI (A, E) measures how much the space of all derivations from A to E differs from the space of inner derivations, and H2(A, E) is related, by Hochschild's theorem 1.9.5, to extensions of A by E. The 'Banach' version of this theory will be given in §2.8. Recall that, for n E N, Cn(A, E) denotes the space of n-linear maps from A(n) to E; we take COCA, E) to be E.

Definition 1.9.1 Let A be an algebra, and let E be an A-bimodule. For x E E, define ~O(x) = ~x E C(A, E). and, for n E Nand T E cn(A, E), define ~nT: (a!, ... ,an+l)

1--+

al . T(a2,'" ,an+d + (_1)nHT(all'" ,an) .

anH

n

+ 2) -1)jT(al, ... , aj-ll ajaj+l, aj+2,""

an+d·

j=l

Let n E Z+. It is clear that ~n is a linear map from Cn(A, E) into .cnH(A, E); these maps are the connecting maps. A direct but tedious calculation shows that ~nH 0 ~n = 0, and so im ~n C ker 8n +l . Indeed we have a complex

C·(A.E): 0

~ E ~ C(A,E) ~ C 2(A, E)

L ... }

~ cn(A,E) ~ CnH(A,E) ~ ...

(1.9.1 )

of linear spaces and linear maps. For n E N, the elements of ker8 n and im8n - l are the n-cocycles and the n-coboundanes, respectively; we denote these linear spaces by

zn(A,E)

= ker8n

and

Nn(A,E)

= im8n - I ,

respectively.

Definition 1.9.2 Let A be an algebra. and let E be an A-bimodule. For n E N, the nth cohomology group of A with coefficients in E is

Hn(A, E) = zn(A, E)jNn(A, E); also, HO(A,E)=ker8o={xEE:a· x=x· a (aEA)}. In fact, the cohomology groups Hn(A, E) are linear spaces. We note that Hn(A, E) = Hn(A, F) 8Hn(A, G) whenever F and G are submodules of E with E = F 8 G, and that Hn(A, E) ~ Hn(AOP, E).

Algebraic foundations

128

Our specific results will involve Hl(A, E) and H2(A, E), and we examine these spaces in more detail. For T E £( A, E), we have T E im 60 if and only if there exists x E E with T(a) = a . x - x . a (a E A). Also,

(6 1T)(a, b) = a . Tb - T(ab)

+ Ta

(1.9.2)

. b (a, bE A).

Thus N 1 (A, E) and Zl (A, E) coincide with our previous definitions of these notations (after 1.8.1), and HI (A, E) is the quotient of the space of all derivations by the space of inner derivations; H1(A, E) = {O} if and only if every derivation from A into E is inncr. Now let T E £2(A, E). Then T E Z2(A, E) if and only if a . T(b, c) - T(ab, c)

+ T(a, bc) -

T(a, b) . c = 0

(1.9.3)

(a, b, c E A);

this equation is called the cocycle identity. Furthcr, T E N2(A, E) if and only if T = 61 S for some S E £(A, E). Examples 1.9.3 (i) The map T : (a, b) f-+ a ® b, A x A ~ A ® A, belongs to Z2(A, A®A). In the case where A is unital, T E N2(A, A®A) because T = 61S for the linear map

S: a

1

f-+

"2 (a ® eA + eA ® a),

A ~ A ® A.

However, in the case where A = coo, so that A ® A = coo(N x N), it is easy to see, by considering possible values of S(6n ) for n E N, that there is no linear map S: A ~ A ® A such that (6 1S)(a, b) = a ® b (a, bE A). (ii) Let A be Co [Xl , let E be an A-bimodule, and let T E Z2(A, E). Clearly T is determined by the double sequence (T(xm, xn) : m, n EN). By the cocycle identity (1.9.3), T is in fact determined by the sequence (T(xm, X) : mEN); indeed, for m, n E N, we have T(xm, xn) = Xm+n -1 + Xm+n-2 . X + ... + Xm . X n - 1 _xm . (X n -1 + Xn-2 . X + ... + Xl . X n - 2),

(1.9.4)

where Xm = T(xm. X). Each choice of (xm) in E gives an element of Z2(A, E) by this formula. Now suppose that T = 61S for some S E £(A, E). Then s(xn+1) is defined inductively by the formula

and so, for each n EN, we have s(xn+1) = xn . SeX) + x n - 1 . SeX) . X + ... + SeX) . xn -(Xn + Xn-l . X + ... + Xl • X n- 1).

(1.9.5)

For each choice of SeX), we obtain S E £(A, E) with T = 6 18, and so we have shown that H2(A, E) = {O}. Note that, in the case where E is symmetric, equation (1.9.5) becomes s(xn+1) = (n + l)S(X) . xn - (xn

+ Xn -1

. X

+ ... + Xl

. Xn-

l ).

(1.9.6) o

129

Cohomology

To interpret H 2 (A, E), we consider extensions of A. Definition 1.9.4 Let A be an algebra. An extension of A by 1 sequence L = L(!21; 1) : 0 ~ 1 ~ 21 ~ A ~ 0,

'tS

a short exact

where 1 'tS an ideal m the algebra 21 and t and q = qfJI. are homomorphisms. The extension 2: is: singular if 12 = 0; nilpotent if 1 zs nilpotent; radical if 1 C rad 21; finite-dimensional if 1 is finite-dimensional; annihilator if 1 is an anmhzlator zdeal; and commutative if 21 zs commutative. Two extensions 2:(21; 1) and 2:(1:8; 1) are equivalent if there is an isomorphism 'IjJ : 21- ~ such that 'l/J(x) = x (x E 1) and q'l3 o'l/J = q'.lJ.. The extension 2: splits if there is a homomorphism () : A - 21 such that q 0 () = lA, and () zs then a splitting homomorphism.

Let 2: = 2:(21; 1) be an extension of an algebra A. Then 2: splits if and only if there is a subalgebra ~ of 21 with 21 = ~ 01; 2:(21; rad 21) splits if and only if 21 is decomposable in the sense of 1.5.15. In the case where 1 contains a non-zero idempotent p such that 1 = pI + Ip (which includes the case where 1 has an identity), 2:(21; 1) splits because 21 = (efJI. - p)21(efJI. - p) 0 I. Suppose that 2:(21; 1) is a singular extension of A. Then I is an A-bimodule for the operations x . q(a)

= xa,

q(a)· x

= ax (a

E

21, x E 1) .

(1.9.7)

(The operations are well-defined because 12 = 0.) Conversely, let E be an Abimodule, and let 2: = 2:(21; I) be a singular extension of A such that E is isomorphic to 1 as an A-bimodule. Then we say that 2: is a singular extenswn of A by E. We now give a standard construction extending that in 1.8.14. Let A be an algebra, and let E be an A-bimodule. Take T E Z2(A, E), and define 21T to be the linear space A 0 E with the multiplication (a,x)(b,y) = (ab, a· y+x· b+T(a,b)).

(1.9.8)

The condition that this product be associative is exactly condition (1.9.3) on T, and so 21T is an algebra. Further, L(21T ;E): 0 ~E~21T ~A ~O

is a singular extension of A, where q( a, x) {O} 0 E, so that E is an ideal in 21T .

= a and we are identifying E with

Theorem 1.9.5 (Hochschild) Let A be an algebra, and let E be an A-bzmodule. Then the above map T 1--+ I)21T; E) from Z2(A, E) induces a bzjection between H2(A, E) and the family of equivalence classes of singular extensions of A by E. Proof Take Tl, T2 E Z2(A, E) with Tl - T2 E N 2(A, E), say Tl - T2 = 8 1 S, where S E C(A,E). Then the map (a,x) 1--+ (a,x + Sa), 21Tl - 21T2 , is an isomorphism, and E (21Tl ; E) and E (21T2 ; E) are equivalent extensions of A.

130

Algebmic foundatwns

Let E(~; E) be a singular extension of A by E, and let Q : A --. ~ be a linear map such that q 0 Q = lA. Define T E C2(A,~) by

T{a, b) = Q(a)Q(b) - Q(ab)

(a. bE A) .

(1.9.9)

We have q 0 T = 0, and so we may say that T E C 2 (A, E). By direct calculation using (1.9.7), we see that T E Z2(A, E). As above, we construct the algebra ~T and the corresponding extension E(~T; E). Since

(a - Q(q(a»)(b - Q(q(b») = 0

(a, b E ~),

a calculation shows that the map

1/J : a 1-+ (q(a), a - Q(q(a»),

~

--.

~T,

is an isomorphism, and so the extensions E(~; E) and E(~T; E) are equivalent. Now let E(2t 1 ; E) and E(2t 2; E) be equivalent singular extensions of A by E. Define T 1 , T2 E Z2(A, E) by analogues to formula (1.9.9). Then E«~I)Tl; E) and E«~2)r2; E) are equivalent to E(2t 1 ; E) and E(~2; E), respectively, and so they are equivalent to each other: there is an isomorphism from (~dTl onto (~2)r2 of the form (a,x) 1-+ (a,x + Sa), where S E C(A,E). Again a direct calculation shows that Tl - T2 = 61 S, and so the two extensions define an element of H 2 (A, E). 0

Corollary 1.9.6 Let A be an algebm, and let E be an A-bimodule. Then the following conditions are equivalent: (a) H2(A, E) = {OJ; (b) each singular extension of A by E splits. Proof (a)::::}(b) By the theorem, each singular extension E = E(~; E) is equivalent to the extension given by the zero element of Z2{A, E); the latter extension splits, and so E splits. (b)::::}(a) Take T E Z2(A, E). Then E(~T; E) splits: there is a homomorphism () : A --. ~T such that ()(a) = (a, -Sa) (a E A) for some S E C(A, E), and then T = 61 S. Hence H2(A, E) = {OJ. 0 Let A be a commutative algebra, and let E be an A-module. An element T E C2(A, E) is symmetric if

T(a, b) = T(b, a)

(a, bE A) .

The linear space of symmetric 2-cocycles is denoted by Z; (A, E); clearly N 2 (A, E) c Z;(A, E). The algebra ~T defined above is commutative if and only if T is symmetric. We define

H;(A, E) = Z;(A, E)/N2(A, E); it is easily checked that each commutative, singular extension of A by E splits if and only if H;(A, E) = {OJ. Let E(2l; I) be an extension of an algebra A, as in 1.9.4, and suppose that J is an ideal of 2l with J c I. Then

L)2t/J;l/J): 0 ~ I/J..':.4 2t/J ~ A ~ 0

Cohomology

131

is also an extension of A; here we define qJ(a + J) = q(a) (a E 2l).

~J(x

+ J) =

L(X)

+J

(x E I) and

Proposition 1.9.7 Let 2: = 2:(2l; I) be an extension of an algebm A, and let J be an ideal zn 2l with J c I. Suppose that 2:(2l/ J; 1/ J) splits and that each extension of A by J splzts. Then 2: splits. Proof There is a subalgebra ::D of 2l/ J with 2l/ J = ::D 8 (1/ J) and ::D ~ A. Define It = {a E 2l : a + J E ::D}. Then It is a subalgebra of 2l with It n 1 = J and It/ J =::D. The extension

2)1t; J) : 0 ~ J £

It

~A~0

of A by J splits by hypothesis, and so there is a subalgebra 23 of It with It = 238J and 23 ~ A. Clearly 23 is a subalgebra of 2l. We have 2l = It + 1 = (23

+ J) + 1 =

23

+1

and 23 n 1 = 23 n It n 1 = 23 n J = {O}, and so 2l = 23 8 I. Thus 2: splits.

0

Corollary 1.9.8 Let A be an algebm. (i) Suppose that every singular extension of A splits. Then every nzlpotent extenszon of A splits. (ii) Suppose that every finite-dimensional, singular extenszon of A splzts. Then every finite-dimensional extension of A splits. Proof (i) The proof is by induction on n, the index of the ideal 1 in a nilpotent extension 2:(2l; I) of A. The result holds in the case where n = 2 by hypothesis. Suppose that 1 has index n 2: 3, and assume that the result holds for all nilpotent extensions of index at most n - 1. Set J = 12. Then 2:(2l/J;I/J) is singular, and so it splits. Also, I n - 1 = 0, and so each extension of A by J splits by the inductive hypothesis. By 1.9.7, 2:(2l; I) splits, and the induction continues. (ii) Let 2: = 2:(2l; I) be a finite-dimensional extension of A, and set J = rad 1 = 1 n rad 2l. Then 1/ J is finite-dimensional and semisimple, and so, by 1.5.1O(i), either J = lor 1/ J has an identity; it follows that 2:(2l/ J; 1/.1) splits. By 1.5.6(iv), each extension of A by J is nilpotent, and so, essentially as in (i), each such extension splits. By 1.9.7, 2: splits. 0 Corollary 1.9.9 Let A be an algebm. Then the following conditions are equivalent:

(a) H2(A, E) = {O} for each finite-dimensional A-bimodule E; (b) each finite-dimensional extension of A splits. Proof (b)::::}(a) Let E be a finite-dimensional A-bimodule. Each singular extension of A by E splits, and so, by 1.9.6, H2(A, E) = {O}. (a)::::}(b) By 1.9.6, each finite-dimensional, singular extension of A splits. By 1.9.8(ii), each finite-dimensional extension of A splits. 0

132

Algebrazc foundations

The calculation of cohomology groups is assisted by certain 're>duction-ofdimension' formulae, which we now explain. Let A be an algebra, and let E be an A-bimodule. First we regard C(A, E) as an A-bimodule with respect to the products (a * T)(b) = a . Tb, } (a,b E A, T E C(A,E)). * a)(b) = T(ab) - Ta . b, Note that the product T * a is different from both the products T (T

(1.9.10)

. a and T x a introduced in 1.4.12. Also note that it does not follow from the fact that E· A = 0 that C(A, E) . A = 0, and it does not follow from the fact that A and E are unital that C(A, E) is unital. The definitions in (1.9.10) are arranged so that formula (1.9.13), below, holds. For k,p EN, let Ak,p: Ck+P(A,E) -+ Ck(A, CP(A, E» be the standard identification given by (Ak,p(T)(al,"" ak» (ak+b ... , ak+p) = T(al,"" ak, ak+l,···, ak+p) (al, ... , ak+p E A, T E Ck+P(A, E),

(1.9.11) and write An for An,l. Then we successively obtain a representation of each linear space cn(A, E) as an A-bimodule by setting a * T = A;:;-l(a * An(T» and T * a = A;:;-l(An(T) * a) for a E A and T E cn+1(A, E), where the maps * on the right-hand sides are the bimodule operations on cn(A, C(A, E» defined in the previous step: the products are given by the formulae: (a * T)(al,"" an) = a . T(al, ... , an) j (T * a)(al"'" an) = T(aab a2,"" an) + (-1)nT(a, al,···, an-d' an

)

n-l

+ L( -1)jT(a, ab""

ajaj+1,"" an).

j=l

(1.9.12) It is easily checked that each of the above-defined maps A k •p is an A-bimodule isomorphism for these newly defined products. In general the connecting maps 8n : cn(A, E) -+ cn+l(A, E) are not A-bimodule homomorphisms for the module operations *, but this is the case when A is a commutative algebra and E is an A-module. Proposition 1.9.10 Let A be an algebra, and let E be an A-bimodule. Then Hk+P(A,E) ~ Hk(A, (CP(A, E),

*»

(k,p E N).

Proof Denote the connecting maps for the complex ce(A, (C(A, E), Then An+1 o8n + 1 = bonoAn : C n+1(A, E)

-+

C n+1(A, C(A, E»

*» by bon.

(n E Z+). (1.9.13)

Let n E Z+, and take T E cn+1(A, E). Then T E zn+1(A, E) [T E Nn+1(A, E)] if and only if An(T) E zn(A, C(A, E» [An(T) E Nn(A, C(A, E)]. Thus An induces a linear isomorphism between the spaces Hn+1 (A, E) and Hn(A, C(A, E». The result follows. 0

Cohomology

133

Corollary 1.9.11 Suppose that Hk(A, E) = {OJ for each A-bimodule E. Then 0 Hk+P(A, F) = {OJ for each pEN and each A-bimodule F. Proposition 1.9.12 Let A be a unital algebm, and let E be an (i) Suppose that E . A = 0 or A . E = O. Then Hn(A, E) (ii) Hn(A, E) ~ Hn(A, eA . E . CA) (n EN).

A-b~module.

= {OJ

(n EN).

(iii) Suppose that E is unztaZ and that M is a maximal ideal of codzmension one in A. Then Hn(A,E) ~ H 1t (M,E) (n EN).

Proof (i) We suppose first that A . E First take D E Zl(A, E), and set Xu

D(a)

= D(eAa) = D(eA)

= O. = -D(eA)' Then

. a = -;]:0

.

a

= clxo(a)

(a

E

A),

and so D E N1(A,E). Hence H1(A,E) = {OJ. Now take kEN, and assume that Hk(A, F) = {OJ for each A-bimodule F for which A . F = O. Then Hl(A . .ck(A, E)) = {OJ. By 1.9.10, Hk+l(A,E) = {OJ. The result follows by induction. In the case where E . A = 0, the final result follows by regarding E as an AOP-bimodule; in this case, AOP . E = O. (ii) The complex .ce(A, E) is the direct sum of four complexes corresponding to the decomposition (1.4.4) of E, and this leads to a similar decomposition of Hn(A, E). Since the action of A on all the bimodules other than eA . E . eA is trivial on the left or right, it follows from (i) that Hn(A, E) ~ Hn(A, eA . E . eA). (iii) By 1.8.2(i), D(CA) = 0 for each D E Zl(A,E). It follows that the map D I---> DIM, Zl(A, E) ---+ Zl(M, E), is a linear bijection. Clearly DE NI(A. E) if and only if DIM E NI(M, E). Thus RI(A, E) ~ Hl(M, E). Set F = £(A, E) * eA, a submodule of .c(A, E). We clazm that the map

W:T

I--->

TIM,

F

---+

.c(M, E) ,

is an A1-bimodule isomorphism. Certainly W is an M-bimodule homomorphism. Suppose that weT * eA) = 0 in .c(M, E). Since E is unital, (T * cA)(eA) = 0, and so T * eA = 0 and W is an injection. Finally take T E .c(M, E), and extend T to T E .c(A, E) by setting TeA = O. Then T * eA = T, and so T E F. Thus w is a surjection. Now take kEN, and assume that Hk (A, F) ~ Hk (M, F) for each unital A-bimodule F. Then Hk+I(A, E) ~ Hk(A, .c(A. E)) by 1.9.10 ~ Hk(A, .c(A, E)

* eA)

~ Hk(M, .c(A, E)

* c,t)

~

Hk(M,.c(M,E)) ~ H k +1(M,E) The result follows by induction.

by by by by

(ii)

hypothesis the claim 1.9.10.

o

Easy examples show that the fact that HI(A, E) = {OJ does not imply that HI (M, E) = {OJ in the case where E is non-unital.

134

Algebraic foundations

The following is a standard result on long exact sequences of cohomology. Theorem 1.9.13 Let A be an algebra, and let

O----+E~F~G----+O be a short exact sequence of A-b~modules and A-bimodule homomorphisms. Then there are connecting lmear maps Un, Vn , and ~Vn such that the sequence

o ----+

HO(A, E) ~ HO(A, F) ~ HO(A, G)----+

~ Hl(A, E)

____+ . . . ____+

}

Hn(A, E) ~ Hn(A, F)----+

~ Hn(A, G) ~ Hn+l(A, E)

(1.9.14)

____+ • • •

is an exact sequence of linear spaces and lmear maps.

Proof Throughout, we identify E with its image in F. Note that we have Uo8 n T = 8n (UoT) (T E .cn(A,E»), etc., because U, V, and Ware A-bimodule maps. There is a linear map Q : G -+ F such that V 0 Q = Ia. We take Wo(x) = 8Q (x) (x E HO(A, G)), and set Uo = U I HO(A, E) and Va = V I HO(A, F). Trivially, we see that the complex is exact at HO(A, E) and HO(A, F), and exactness at HO(A, G) is easily checked. We now take n EN. Define: Un :Tf---+ UoT, Wn : T

f---+

8n (Q

0

Vn :Tf---+ VoT, T) - Q 0 (8 nT),

.cn(A, E) .cn(A, F) .cn(A, G)

-+ -+ -+

.cn(A, F); .cn(A, G) ; .cn+l(A, F).

Clearly Un, Vn , and Wn are linear maps. For each T E .cn(A, G), we have Vo Wn(T)

= Vo (8 n (Q 0

T» - (V

0

Q)

0

(8 n T)

= 8n (V 0

Q 0 T) - 8n T

= 0,

so that Wn(T)(a) E ker V = U(E) (a E A(n+l), and hence we may regard Wn as a map into .cn+1(A, E). We claim that, taking restrictions, we have maps • • • ____+

zn(A, E) ~ zn(A, F) ~ zn(A, G) ~ zn+l(A, E)

____+ • • • •

The images of each Un and Vn certainly lie in the claimed spaces. Now take T E zn(A,G), so that 8n T = O. Then 8n+1(Wn(T» = (8 n +1 0 8n )(Q 0 T) = 0, and so the image of Wn lies in zn+l(A, E). We similarly claim that, taking restrictions, we have maps . . . ____+

Nn(A,E) ~ Nn(A, F) ~ Nn(A, G) ~ Nn+1(A,E)

Again, the images of each Un and Vn lie in the claimed spaces. T E Nn(A,G), say T = 8n- 1S, where S E .cn-1(A,G). Then Wn(T)

= 8n (Q

0

= 8n (Q

0

____+ . . . .

Now take

8n- 1 S) - Q 0 (8 n 0 8n- 1S) = 8n (Q 0 8n- 1 S) 8.n-1S - 8n- 1 (Q 0 S» E Nn+l(A, E).

Thus we obtain the linear maps (also called Un, Vn , W n ) required in (1.9.14).

Cohomology

135

We show that (1.9.14) is exact at Hn(A, F). For take T E zn(A, F) such that "\;.,(T) = 8n - 1 S for some S E .en - 1 (A, G), and set

S' = Q 0 S E .en- 1(A, F). Since Vn(T') = 0, in fact T' E zn(A,E). Then T = Un (T')+8 n- 1S'. Conversely, suppose that T has this form, where T' E zn(A, E) and S' E .en - 1(A, F). Then Vn(T) = 8n - 1(V 0 S') E Nn-1(A, G). We show that (1.9.14) is exact at Hn(A, G). For take T E zn(A, G) such that 8n(Q 0 T) = 8nS for some S E .en(A, E), and set T' = T - 8n - 1(Q 0 S) E zn(A, F)

and

T' = Q 0 T - Un(S) E zn(A, F) and S' = 0 E .en-l(A, G). Then T = Vn(T') + 8n- 1S'. Conversely, suppose that T has this form, where T' E zn(A,F) and S' E .en-1(A,G), and set S = Q 0 Vn(T') - T' - 8n- 1(Q 0 S') + Q 0 8n- 1S'. Then 8n (Q 0 T) = 8nS and V 0 S = 0, so that S E .en (A, E). We show that (1.9.14) is exact at Hn+1(A,E). For take T E zn+l(A,E) such that Un+l(T) = 8n S for some S E .en(A,F), and set

T' = Vn(S) E Zll(A, G) and S' = S - Q 0 Vn(S) E .en(A, E). Then T = 8n (Q 0 T') + 8n S'. Conversely, suppose that T has this form, where T' E zn(A,G) and S' E .en(A, E). Then Un+1(T) = 8n (Q 0 T' + U 0 S'). Similarly, (1.9.14) is exact at H1(A, E). We have shown that (1.9.14) is an exact complex of linear spaces.

0

Corollary 1.9.14 Suppose that n E Z+ and that Hn(A, E) = Hn(A, G) = {O}. Then Hn(A, F) = {OJ. 0 Corollary 1.9.15 Let n E N, and let A be a commutative algebm such that Hn(A, E) = {OJ for each one-dimensional A-bimodule E. Then Hn(A, E) = {OJ

for each finite-dimensional A-bimodule E. Proof We prove by induction on dim E that Hrt(A, E) = {OJ for each finitedimensional A-bimodule E. The case where dim E = 1 is the hypothesis. Take kEN, assume that HlI(A, E) = {OJ for each A-bimodule E with dim E ~ k, and let Fo be an A-bimodule with dim Fo = k + 1. The maps x 1--+ a . x and x 1--+ x . a for a E A have a common eigenvector, say Xo E Fa. Set Eo = CTo and Go = Fo/ Eo, so that Eo and Go are A-bimodules with dim Eo = 1 and dim Go = k. We have HlI(A, Eo) = Hn(A, Go) = {OJ, and so, by 1.9.14, Hn(A, Fo) = {OJ. The induction continues. 0 Let A be a unital algebra, let cp,t/J E A U {OJ, and let J.L E Z2(A,Ccp,tJ1). Then the cocyle identity for J.L is

cp(a)J.L(b, c) - J.L(ab, e)

+ J.L(a, be) -

J.L(a, b)1jJ(e) = 0 (a, b, e E A) ;

(1.9.15)

it follows that

t/J(eA)J.L(a, eA) = cp(a)J.L(eA' eA),

cp(eA)J.L(eA, a) = t/J(a)J.L(eA, eA)

(a

E

A) .

(1.9.16)

Algebraic foundations

136

In the case where c.p = 'IjJ E cf> A, the cocycle identity for JJ, on M

JJ,(ab, c) = JJ,(a, be)

(a, b, c E M
(1.9.17)

in this case, JJ, E N2(Mcp, q if and only if there exists ,x E M'; with

JJ,(a, b) = -,x(ab)

(a, bE M
(1.9.18)

Proposition 1.9.16 Let A be a commutatzve, unztal algebra. (i) Let c.p,'IjJ E
(1.9.19)

Proof (i) That H1(A, Ccp.",) = {a} follows immediately from 1.8.10. Take u E Mcp with 'IjJ(u) = 1, and set v = eA - u. Let JJ, E Z2(A, C
JJ,(a, b) = JJ,(u, ab)

+ JJ,(ab, v)

(a E Mcp, bE !vI",).

(1.9.20)

Now define ,x(eA) = JJ,(eA' eA) and ,x(a) = -/-leU, a) - /-lea, v) (a E Mcp), so that ,x E A x. It is easily checked, using (1.9.16) and (1.9.20), that «5 1 ,x = /-l. It follows that H2(A, Ccp,,,,) = {o}. (ii) Suppose that (1.9.19) holds. Let E be a one-dimensional A-bimodule. Then E = Ccp,,,, for some c.p,'IjJ E cf>A U {a}. We have H 2(A,Ccp,.p) = {a} in the cases: where c.p = 0 or 'IjJ = 0, by 1.9.12(i): where c.p, 'IjJ E

A®···®A®E, where there are n copies of A; in particular, .c 1 (A, E) = A®E. We take .co (A, E) to be E. Define

(a®x)

* b = a®x . b,

b * (a®x) = ba®x-b®a . x

(a, bE A, x E E). (1.9.21)

It is easily checked that .c 1 (A, E) is an A-bimodule with respect to the above products. There is an obvious identification of .cn+1 (A, E) with .c1(A, .cn(A, E)), and this gives a representation of each .cn(A, E) as an A-bimodule: the products are given by the formulae

a

*

(al ® ... ® an ® x) * a = al ® ... ® an ® x . a, (al ® ... ® an ® x) = aal ® ... ® an ® x n-l

+ L( -l)ja ® al ® ... ® ajaHI ® ... ® an ® x j=l

+ (-l)n a ®al®···®an _l®an · x. Note that, if E . A = 0, then T

*a=

0 (T

E

.cn(A, E), a

E

A).

(1.9.22)

Cohomology

137

There exists do E C(A ® E, E) such that do : a ® x 1---+ X • a - a . x, and, for each n E N, there exists dn : C n+1 (A, E) --+ Cn(A, E) such that dn : al ® ... ® a:+1 ® x

1---+

a2 ® ... ® an+l ® x . al

)

+ 2) -l)kal ® ... ® ak ak+1 ® ... ® an+! ® x

(1.9.23)

k=l

+ (-l)n+1 al ®···®an ®a n+1 . x. In particular, d l (a ® b ® x) = b ® x . a - ab ® x + a ® b . x and d 2(a ® b ® e ® x) = b ® e ® x . a - ab ® e ® x + a ® be ® x - a ® b ® e

. x.

The maps dn are linear, and we have the following complex of linear spaces: C.(A,E): 0 ~ E ~ C 1 (A,E) ~ C2(A,E) ~ ... }

~ Cn(A,E) ~ Cn+1(A,E) ~ ... .

(1.9.24)

Definition 1.9.17 Let A be an algebm, and let E be an A-bimodule. For n E N, the nth homology group of A with coefficients in E zs Hn(A, E)

= kerdn _ l / imdn j

also, Ho(A,E) = E/imdo = E/lin{a . x - x . a: a E A, x E E}.

Proposition 1.9.18 Let A be an algebm, and let E be an A-bzmodule. Then: (i) forn E Z+, Cn(A,E)X ~ cn(A,Ex); (ii) for n E N, Hn+l(A, EX) ~ Hl(A, Cn(A, E)X) as linear spaces; (iii) the complex C·(A,EX) is the algebmic dual complex of C.(A,E); (iv) for n E Z+, Hn(A, E) = {O} if and only if Hn(A, EX) = {O}. Suppose, further, that A is unital. Then: (v) Hn(A, A ® E)

= {O}

(n EN).

Proof (i) We first show that Cl(A,E)X ~ Cl(A,EX). Define f: Cl(A,E)X = (A®E)X --+ C(A,EX) by (x, (fA)(a»

=

(a®x, A)

(a E A, x E E, A E (A®E)X).

Certainly f is a linear isomorphism. To show that f is an A-bimodule homomorphism, take a, bE A, x E E, and A E (A ® E) x. Then (x, (b

* fA)(a» = (x, b . (fA)(a» = (x . b, r(A)(a» = (a®x· b, A) = «a®x) * b, A) = (a®x, b·

A) = (x, f(b· A)(a»

and (x, (fA

* b)(a» = (x,

(fA)(ba) - (fA)(b) x a) = (ba ® x, A) - (a . x, (fA)(b» - b ® a . x, A) = (b * (a ® x), A) = (a ® x, A . b) = (x, f(A . b)(a».

= (ba ® x

Thus b * fA = r(b . A) and fA * b = r(A . b), as required. The result follows by induction on n.

138

Algebmic foundations

(ii) This follows from 1.9.10. (iii) This follows from the routine calculation that d~ = 8n (n E Z+). (iv) This follows from (ii) because a sequence is exact at n if and only if its algebraic dual is exact at n. Now suppose that A is unital. (v) For each k E Z+, there is a linear map Sk : Ck(A, AI8lE) such that

-->

C k+ 1(A. AI8lE)

Sk(al ® ... I8l ak ® a I8l x) = al I8l '" I8l ak I8l a ® eA ® x (al •... , ak. a E A, x E E).

Let n EN. Then we see that

(0: E Cn(A, A I8l E));

(dns n - Sn-ldn-d(a) = (-l)na

if a E kerdn-l. then a = d n ((3), where (3 = (-l)nsn(a) E C n+1(A.A I8l E). It follows that kerd n- 1 = imd n , and so Hn(A, A ® E) = {a}. 0 We shall now characterize the algebra':! A such that Hl(A, E) = {a} for each A-bimodule Ej to do this we first introduce the notion of a diagonal. Recall that AI8lA is an A-bimodule for products determined by the conditions a . (b I8l c) = ab I8l c and (b ® c) . a = b I8l ca for a, b, c E A. and that 71"A denotes the induced product map from A I8l A to A.

Definition 1.9.19 Let A be a unital algebm. A diagonal for A is an element u E A I8l A such that 7l"A (u) = e A and a . U = U . a (a E A). A diagonal for A is an element u = L-J=1 a] I8l bj in A ® A such that n

n

and

Lajbj = eA j=l

n

Laaj I8lb) = La) I8lbj a )=1 j=1

(a E A).

(1.9.25)

Let L-J=l a) ® bj be such a diagonal, and suppose that E is an A-bimodule and T E C(A, E). Then there is a linear map A I8l A --> E such that a I8l b ~ a . Tb. and so n

n

Laa) . Tbj = Laj . T(bja) j=1 j=1

(a E A).

(1.9.26)

Suppose that AI, ... ,An are algebras and that Uj E Aj ® Aj is a diagonal for Aj for each j E N n . Then the element UI $ .. '$u n is a diagonal for Al $ ... $ AnLet A be an algebra, and let 7l"A be the induced product map. Then we have a complex of A-bimodules and A-bimodule homomorphisms:

LA :0

---+

ker7l"A

---+

A I8l A ~ A

---+

0j

(1.9.27)

the complex is exact if and only if A factors weakly. We exemplify the notion of a diagonal with a calculation for the full matrix algebra Mn. A group G is an irreducible n x n matrix group if G is a finite subgroup of G L(n) and lin G = Mn. For example, the set of matrices (O!ij) such

Cohomology

139

that (Xij E {-I,O, I} (i,j E Nn ) and each column and each row contains exactly one non-zero term is such a group. As in §1.3, the standard system of matrix units in Mn is (Eij : i,j E N n ), and the identity of Mn is En.

Proposition 1.9.20 Let G be any irreducible n x n matrix group, and set

d=

1~ll)x®X-l: x E G}.

Then d is the unique element of Mn ® Mn which is a diagonal for both Mn and 1 n d = - L Eij ® Eji . n.

~,J=l

Proof Set A = Mn. For each a E G, we have

Lax®x- 1 = Lax®(ax)-la xEG

xEG

y®y-1 a = Lx®x- 1a.

= L yEa·G

xEG

Since lin G = A, this holds for each a E A, and so a . d = d . a for each a E A. Also 7rA(d) = En, and so d is a diagonal for A. Similarly, d is a diagonal for AOP. Let d and d' both be diagonals for both A and A oP, say d = Lr ar ® br and d' = Ls a~ ® b~. Then

r,s

s

r,s

and so d is uniquely determined. Set do =

Ekl . do

=

(L~j=l Eij ® E ji )

(.t

~,J=l

EklEij ® Eji)

/

=

n. Then 7r A (do) = En and

t

)=1

Ekj®Ejl

=

(.t

',)=1

EiJ ® EjiEkl\

= do . Ekl

)

e

for k, E Nn ; it follows that do is a diagonal for A. Similarly, do is a diagonal for AOP, and so do = d. 0

Theorem 1.9.21 Let A be a complex algebra. Then the following conditwns are equivalent: (a) H1(A, E) = {o} for every A-bimodule E; (b) A is unital and has a diagonal in A ® A; (c) A is unital and the short exact sequence LA of A-bimodules splits; (d) A is semisimple and finite-dimensional. Proof (a)g.(b) Set E = A x A. Then E is an A-bimodule with respect to the products a . (b, c) = (ab, 0) and (b, c) . a = (0, ca) for a, b, C E A. The map

140

Algebmic foundations

D : a ~ (a,a), A ---+ E, is a derivation, and so, by (a), there exists (r,s) E E with Da = a . (r,s) - (r,s) . a (a E A). But then ar = a = -sa (a E A), and so A has an identity, say e. The map D : a ~ a ® e - e ® a, A ---+ ker 'irA, is a derivation, and so there exists v E ker'lrA with Da = a . v - v . a (a E A). Set u = e ® e - v E A ® A. Then u is the required diagonal. (b)~(a) Let u =

let D : A

---+

'L-J=l aJ ® bJ be a diagonal, let E be an A-bimodule, and E be a derivation. Set Xo = 'L-;=l aj . Dbj . Using (1.9.26), we have a . Xo - Xo . a =

n

n

j=l

j=l

L aaJ . DbJ - L aj . Dbj . a n

=

L aJbj . Da = eA . Da

(a E A) .

j=l

Set D = D - eA . D. Then D is a derivation and a . Db Xl = -DeA. Then Da

= D(eAa) = eA

and so Da = a . (xo

. Da + DeA . a

= 0 (a, b E A). Set

= -Xl· a = a . Xl - Xl . a

(a E A) ,

+ Xl) - (xo + xr) . a (a E A). Thus D is inner.

(b)~(c) Let u be a diagonal for A, and define Q : a ~ a . u, A Then Q E ACA(A, A ® A) and Q is a right inverse to 'irA. (c)~(b)

---+

A ® A.

Take Q E ACA(A, A ® A) to be a right inverse to 'irA, and then set

u = Q(eA). We see that u is a diagonal. (d)~(b) By Wedderburn's structure theorem 1.5.9, each finite-dimensional, semisimple algebra is isomorphic to a finite direct sum of matrix algebras, and each matrix algebra is unital and, by 1.9.20, has a diagonal.

(b)~(d) Let u = 'L-J=l aj ® bj be a diagonal for A, and let P E C(A) be a projection of A onto lin {bI> ... ,bn }. Define n

T :a ~

L ajP(bja),

A

---+

(1.9.28)

A.

j=l

J

For each a E A, Ta = 'L-J=l aaJP(bj ) by (1.9.26), and so Ta = a 'L- =l ajbj = a. Thus T is the identity on A, and hence A = T(A) = lin{ aibj : i, j E N n } is finite-dimensional. Now set K = rad A, and let P E £(A) be a projection of A onto K. Define T by equation (1.9.28). For each a E K and j E N n , P(bja) = bJa, and so Ta = 'L-j=l ajbja = a, showing that T is also a projection onto K. Let a E A. Then Ta = a . TeA, whence TeA E ::Y(A). It follows from 1.5.7(i) that T = 0, and so K = o. Thus A is semisimple. 0 Corollary 1.9.22 Let A be a finite-dimensional, semisimple complex algebm. Then Hn(A, E) = {O} for each A-bimodule E and each n E N. Proof This follows from the theorem and 1.9.11.

o

141

Cohomology

The above theorem leads to an alternative proof of Wedderburn's principal theorem for finite-dimensional algebras (cf. 1.5.18). For let B be a finitedimensional algebra with radical R, and set A = B / R, so that A is semisimple and finite-dimensional. By 1.9.21, HI (A, E) = {O} for every A-bimodule E, and so, by 1.9.11, H2(A,E) = {O} for every such E. By 1.9.9, B is decomposable. Example 1.9.23 Theorem 1.9.21 shows, in particular, that every derivation on Mn is inner. This is not necessarily true of every sub algebra 2l of Mn. Indeed a derivation on 2l is not necessarily implemented by an element of Mn. For let 2l = lin{El1,E22,E33,E44, E 31 ,E32 ,E41, E 42 } , where (Eij) is the standard

system of matrix units for M 4 ; the elements of 2l are represented by matrices of the form

(~

o o o

x x x

x

o

It is immediately checked that 2l is a subalgebra of M 4. Define D E .c(2l) by setting D(E31 ) = E31 and D(Eij) = 0 ((i,j) -I- (3,1)). Then D E ZI(2l, 2l). Now assume towards a contradiction that there exists X E M4 such that D(A) = AX - XA (A E 2l). Then clearly EiiX = XEii (i E N4 ), and so X is a diagonal matrix, with diagonal (Q,{3,'"Y,~), say. Since E 32 X = XE32 , necessarily {3 = '"Y. Since E 4I X = XE41 , necessarily 0: =~. Since E 42 X = XE42 , necessarily (3 =~. Thus X = o:E4. But now D(E3d = 0, a contradiction. Thus HI(2l,M4) -I- {a}. D Notes 1.9.24 The Hochschild cohomology theory of algebras was introduced in (Hochschild 1945, 1946); for accounts, see (Cartan and Eilenberg 1956), (P. M. Cohn 1991, §6.7), (Jacobson 1980, Chapter 6), and (Weibel 1994, Chapter 9). In these accounts, the cohomology groups Hn(A, E) and homology groups Hn(A, E) are defined to be Extli(A,E) and Tor:(A,E), where Ext and Tor are certain functors and B is the algebraic enveloping algebra A # ® A #oP. The long exact sequence of cohomology 1.9.13 follows from the homology version in (Jacobson 1980, §6.3); cf. (Weibel 1994, §1.3). Our 'diagonal' of 1.9.19 is sometimes called a separability idempotent for A (Pierce 1982, §1O.2). Proposition 1.9.20 is taken from (Gr(llnbrek et al. 1994). A result related to 1.9.21 is in (Pierce 1982, §§1O.2. 11.5) and (Weibel 1994, 9.2.11). The algebras A such that H2(A, E) = {O} for every A-bimodule E are called quasi-free, and the commutative algebras A such that H~(A, E) = {O} for every A-module E are called smooth; see (Weibel 1994, §9.3). For example, each polynomial algebra qxI , .•• ,Xnl is smooth. For an algebra A, define the maps is,. : A®n+I -+ A@n to be linear maps such that n

c5n (al ® ... ® an+d =

' " ~

. I aI ® ... ® ajaj+I ® '" ® an+l (-1)'-

j=l

Then A is homologically unital or H-unital if the sequence

0.....- A~ A®A~ A®A®A<- .. · is exact. This notion was introduced by Wodzicki (1989), where it is shown that A is homologically unital if and only if A has the 'excision property' in various homology theories. The property has been shown by Wodzicki to be important for Banach algebra theory.

142

Algebraic foundatwns

1.10 INVOLUTIONS

Our final section of algebraic preliminaries deals with involutions on an algebra; we shall also consider positive linear functionals. Definition 1.10.1 Let E be a complex lmear space. A map is a linear involution if"

(i) (ax + fJy)* = ax* + f3y* (a, fJ (ii) (x*)* = x (x E E).

E

* :x

1-+

x*, E

~

E,

C, x, Y E E)i

The element x* is the adjoint of Xi x is self-adjoint if x* = x, and the set of self-adJoint elements of E is denoted by Esa. For a subset F of E,

F* = {a* : a E F}, and F is *-closed if F* c F.

The self-adjoint elements of E are also termed the hermztian elements. It is easily seen that Esa is a real-linear subspace of E and that E = Esa 8 iEsa: for each x E E, we have x = ~x + i<;S x E Esa 8 iEsa, where

~x = ~(x + x*)

and

<;Sx

= 4(x* -

x)

are the real and imaginary parts of x, respectively. Let E and F be linear spaces with linear involutions, both denoted by *. We define a linear involution T 1-+ T<J on C(E, F) by setting T<J(x) = (T(x*))*

(x E E)

(1.10.1)

for T E C(E, F). In particular, for ,X E EX, we have ,X<J(x) = 'x(x*) (x E E), so that ,X 1-+ ,X<J, EX ~ EX, is also a linear involution. Clearly,X is self-adjoint if and only if ,X(Esa) C JR. Definition 1.10.2 Let A be a complex algebra. A l,tnear involution * on A ~s an involution if (ab)* = b*a* (a, bE A). A complex algebra wtth an involution is a *-algebra.

The complex field C is always taken to be a *-algebra with respect to conjugation. Let A be a *-algebra, and let S be a *-closed subset of A. Then se and see are *-closed, and so every *-closed, commutative subset of A is contained in a maximal, *-closed, commutative subalgebra. In the case where A is unital, e:4. = eA and (a- 1)* = (a*)-l (a E Inv A); in general, the formula * : (a, a) 1-+ (a, a*) extends the involution to an involution on Ab, and hence on A #. The multiplier algebra M(A) of A is a *-algebra with respect to the involution (L, R) 1-+ (R
143

Involutions

the conjugate transpose matrix, and the term 'adjoint' is used elsewhere for the transpose of the matrix of cofactors.) When we refer to MIn as a *-algebra, we are specifying the above involution. A collection (eij : i, j E N n ) of elements in an arbitrary *-algebra A which satisfy the conditions n

Eeu = en,

eiJekl = 8j,kei€

(i,j,k,l E Nn),}

(1.10.2)

i=l

and

eij = eji

(i,j E N n ),

is a system of *-matrix units for A. We note that the standard system of matrix units in Mn is such a system. We collect a number of elementary facts about involutions. Throughout, A is a *-algebra. First we record the following identities, which hold for all a, b, c E A: 4ab = (b

4acb = (b

+ a*)*(b + a*) - (b - a*)*(b - a*) + i(b + ia*)*(b + ia*) - i(b - ia*)*(b -

}

(1.10.3)

ia*);

+ a*)*c(b + a") - (b - a*)*c(b - a*) + i(b + ia*)*c(b + ia*) - i(b - ia*)*c(b -

}

ia*) .

(1.10.4)

It follows that A2 = lin {a*a : a E A} and A3 = lin {a*ba : a,b E A}. Definition 1.10.3 Let (A, *) be a *-algebra, and let a E A. Then a is normal if aa" = a*a. In the case where A has an identity, a is unitary if aa* = a*a = eA. Suppose that a = b + ic E Ass. 0 iAss.. Then a is normal if and only if bc = cb. The set of unitary elements of a unital *-algebra A is a subgroup of (Inv A, . ) called the unitary group U(A). In particular, the unitary group U(n) = U(Mn) consists of the unitary matrices. Proposition 1.10.4 Let A and B be *-algebras. Then A ® B is a *-algebra for an involutzon whzch satisfies (a ® b)* = a* ® b* (a E A, bE B).

a;

a

Proof We show that E7=1 ® b; = 0 whenever E;=l J ® bj = 0 in A ® B. Choose linearly independent elements Cl, ... ,em E B with the same linear span as b], . .. , bn , say bj = E:: 1 aijCi (j E N n ). Then E::l E;=l aijaJ ® Co = 0, and so E;=l aiJaj = 0 (i E Nm ). It follows that

f;n

a; ® b;

n

m

m(n)* ® ci = 0,

= ~ ~ a; ® Qijci = ~ ~ aijaj

as required. The remainder is straightforward.

o

An ideal [a subalgebra] of a *-algebra is a *-ideal [*-subalgebra] if it is *closed. A maximal modular *-ideal is a *-ideal which is maximal in the family of modular *-ideals. Let {A')' : 'Y E r} be a family of *-algebras. Then TI')'Er A')' is a *-algebra for the map (a')') 1-+ (a;), and 0')'Er A')' is a *-subalgebra of TI')'Er~'

144

Algebraic foundations

Proposition 1.10.5 Let A be a *-algebra. (i) For each maximal modular *-ideal I m A, there is a maximal modular ~deal M in A such that I = M n M* . (ii) The intersection of the maximal modular *-ideals of A is the strong radical 9t(A) of A.

Prior

(i) Let M be a maximal modular ideal with I C M. Then I C AI* because I is a *-ideal, and so I c M n M*. But M n M* is a proper *-ideal and I is a maximal modular *-ideal, and so I = M n M* . (ii) This follows immediately from (i) and 1.5.20. 0 Let () : A ---+ B be a homomorphism between *-algebras. Then (}
(a E A) ;

in this case, ker () is a *-ideal in A. Let I be a *-ideal in a *-algebra A. Then A/I is a *-algebra for the involution defined by setting (a + 1)* = a* + I (a E A). and the quotient map is then a *-homomorphism. The involution transforms left ideals of A into right ideals and vice versa, and preserves modularity and maximality. Thus rad A is a *-ideal of A, and A/rad A is a *-algebra. Let I be a left ideal in A such that I is *-closed, and take a E I and b E A. Then a* E I and ab = (b*a*)* E I, and so I is also a right ideal, and hence an ideal. Let P be a prime ideal in A. Then P* is also a prime ideal, and so ~(A) is a *-ideal in A. The following result is easy to verify. Lemma 1.10.6 Let A be a *-algebra, and let a E A. Then: (i) O"(a*) = {z ; z E O"(a)} and v(a*) = v(a); (ii) a E q-InvA zf and only if a* <> a, a <> a* E q-InvA.

o

Let A be a *-algebra. Then we define A+ = {ta;aj ;a1, ... ,an EA,

nEW}.

(1.10.5)

3=1

Clearly (A+,+) is a semigroup over JR+; if a,b E A+ and a,(J E JR+, then aa + (Jb E A+. We have A2 = linA+; also A+ - A+ = Asa n A 2, and so Asa = A+ - A+ if and only if A = A2. In the case where A = C, we have A+ = JR+. Let Asa have the A+e-order (defined in 1.2.14), so that, for a, b E Asa, we have a ~ b if and only if b - a E A +. The binary relation ~ is a preorder on Asa. Definition 1.10.7 A *-algebra A is: (i) ordered ~f A+ n (-A+) = {O}; (ii) proper if a = 0 whenever a*a = 0 in A; (iii) very proper if a1 = ... = an = 0 whenever L:j=l ajaj

=0

in A.

145

Involutions

\Ve say that the involution * is ordered or proper or very proper when the *-algebra (A, * ) has the corresponding property. The preorder is a partial order if and only if A is ordered; in this case, Asa is an ordered real-linear space. Clearly a very proper algebra is both proper and ordered, and an algebra which is both proper and ordered is very proper. If a*a in a very proper algebra, then a = o. The involution on the full matrix algebra Mn is very proper. There is an involution on Mn which is not proper: for example, this is the case for the map

:s:

:s:

°

However, * is essentially the unique proper involution on M n , as the next result shows. Lemma 1.10.8 Let morphzc to (Mn, *).

t

be a proper involution on Mn.

Then (Mn'

t)

is *-iso-

Proof The map T >--+ (Tt)* is an automorphism on M n , and so there is an invertible matrix P such that (Tt)* = P- 1TP (T E Mn). Set Q = (p*)-l, so that Tt = Q-1T*Q and T = Ttt = (PQ)-lT(PQ) for all T E Mn. We have PQ E 3(M n ), and so, by 1.3.51, P = aP* for some a E C; necessarily, lal = 1, and so, by replacing P by (3P throughout, where (3 -=1= and a(3 = /3, we may suppose that P = P*, and hence that Q* = Q. The eigenvalues of Q are real. Assume that some are positive and some are negative. Then there is a non-zero vector x E cn with [Qx, xJ = (where [ ., .J is the usual inner product in C n ). Let R be the orthogonal projection of C n onto Then RQR = 0, and so Rt R = Q-l RQR = 0, a contradiction because t is proper and R -=1= O. Thus the eigenvalues of Q are all positive, say, and there is an invertible matrix 8 with Q = 8*8. We have Tt = 8- 1(8*)-lT*8*8 and 8Tt 8- 1 = (8T8- 1 )* for T E M n , giving the result. 0

°

°

ex.

Proposition 1.10.9 Let (A, *) be a non-zero, jinzte-dimenszonal, proper *algebra. Then A zs *-zsomorphzc to Mnl 0 ... 0 Mnk for some n1, ... ,nk E N. Proof We first note that the algebra A is semisimple. For take a E rad A. By 1.5.6(iv), (a*a)2m = for some mEN, and so a = 0 because * is a proper involution. Let J be a minimal ideal in A. Then J* is also a minimal ideal in A. Assume that J* -=1= J. Then J* nJ = {O}, and so a*a = 0 (a E J), whence J = 0 because the involution is proper, a contradiction. Thus J* = J. The result now follows from 1.5.9 and 1.10.8. 0

°

We also define Apos

= {a

E

Asa : a(a)

C

JR.+}.

(1.10.6)

In general, little can be said about Apos (save that aa E Apos when a E JR.+ and a E Apos); its relation to A+ in the case where A is a Banach algebra will be clarified in Chapter 3.

146

Algebraic foundations

Definition 1.10.10 Let A be a *-algebra. A projection in A is a self-adjoint zdempotent; the set of projections in A is denoted by Jsa(A). An element zn Jsa(A) n 3(A) is a central projection.

Thus Jsa(A) = J(A) nAsa = {p E A : p2 = p* = p}j we regret using the same word 'projection' as that used in §1.3 for an element of .c(E), but no confusion should arise. We have already (in 1.3.18(ii» defined a partial order j on J(A). Clearly p :5 q in Jsa(A) if and only if qp = p. Proposition 1.10.11 Let A be a *-algebra. (i) Suppose that a ::; b m Asa and that c E A. Then c*ac ::; c*be in Asa.

(ii) S-uppose that p j q in Jsa(A). Then p ::; q in A~a' Now suppose that A is very proper. (iii) Suppose that b*b ::; amAsa and that a

= ac.

Then b = bc.

(iv) Suppose that p, q E J sa (A). Then p j q if and only if p ::; q. Proof (i) This is immediate.

(ii) Since p j q, we have qp = p, and so q - p = (q - p)*(q - p) E A+. Now suppose that A is very proper. Take e to be the identity of A#. (iii) By (i), (e - e)*b*b(e - e) ::; (e - e)*a(e - c) A is very proper.

= 0,

and so b = be because

(iv) Suppose that p ::; q. Then e - q ::; e - p, and so, by (i), (p - pq)(p - pq)* = pee - q)2p::; pee - p)p = O.

Thus p = pq hecause A is very proper. The result now follows from (ii).

0

Definition 1.10.12 Let A and B be *-algebras. A linear map () : A ~ B is positive if ()(A+) C B+; a Imear functzonal A on A is a positive functional if A(A+) C jR+. The set of positive functzonals on A zs denoted by PA. Let A EPA. Then A is a positive trace if (aa*, A) = (a*a, A)

(a E A).

A positive functional A on a unital *-algebra A is a state if (eA, A) = 1: the state space of A, denoted by SA, zs the set of states on A; a state whic~ is also a trace is a tradal state.

Let A and B be *-algebras, and let () : A ~ B be a *-homomorphism. Then () is positive, and () : (Asa, ::;) ~ (Bsa, ::;) is order-preserving. Clearly (PA,+) is a unital subsemigroup of (AX)sa, and (PA,+) is a scmigroup over jR+.. Let::; be the PA-order on PA, so that A ::; J1- if and only if Jl. - A EPA. Then (PA I A2,::;) is a partially ordered set, and (PA,::;) is a partially ordered set if and only if A = A2. In the case where A is unital, the state space SA is a convex subset of AX . Let A EPA. Then clearly Ab : a ~ (b*ab, A) is also in PA for each b E A.

Involutions

147

It follows from (1.10.3) that (ab, A) = (ba, A) (a, b E A) for each positive trace A on A, and so A is a trace in the sense of 1.3.6. For example, the trace map Tr : (O'.ij) --+ L}=l fl'.jj of (1.3.16) satisfies n

Tr(AA*) = Tr(A* A) =

L

(A = (Cl'ii) E Mn),

/nij/2 2: 0

i,j=l and so Tr is a positive trace on the *-algebra Mn in the above sense. Proposition 1.10.13 Let A be a *-algebra. and let A EPA. Then:

(i) (a*,A) = (a,X) (a E A2); (ii) (a*b, A) = (b*a, A) (a, bE A); (iii) (Cauchy Schwarz inequality) /(a*b,A)/2::; (a*a,A)(b*b,A) (a,b E A). Proof (i) Let a E A2. By (1.10.3), a = L}=l O'.jajaj for some 0'.1, ... 'O'. n E 0 (0'., {3 E q. In particular, take {3 = (a*b, A). Then 0'.2 (a*a,

A)

+ 20'. /(a*b, A)/2 + /(a*b, A)/2 (b*b, A) 2: 0

(O! E JR),

o

and this implies the Cauchy-Schwarz inequality. Corollary 1.10.14 Let A be a umtal *-algebra, and let A E SA. Then

/(a,A)/2::; (a*a, A)

(a

E

A).

Proof This is immediate from the Cauchy-Schwarz inequality 1.10.13(iii).

0

Let A be a *-algebra. For A EPA, set h = {a E A: (a*a,A) = a}. It is clear from the Cauchy-Schwarz inequality that 1)" is a left ideal in A and that

(ba, A)

= (a*b, A) = 0

(a E lA, bE A) .

(1.10.7)

In the case where A is a trace on A, I A is an ideal in A. Definition 1.10.15 Let A be a unital *-algebra. Then *-rad A = nih :A E SA} is the *-radical of A. The algebm is H;emisimple if *-radA = 0 and *-radical if *-rad A = A.

Thus A is *-radical if and only if SA = 0. The following proposition shows that the *-radical has the properties that we would wish it to have. Proposition 1.10.16 Let A be a unital *-algebra. Then:

(i) *-radA = n{ker.x: .x E SA}; (ii) *-rad A is a *-ideal in A; (iii) AI( *-rad A) is *-semisimple.

148

Algebrazc foundations

Proof (i) Suppose that a E *-radA, and take ..\ E SA. Then a E ker..\ by 1.10.14. Suppose that a E n{ker..\ : ..\ E SA}, and take ..\ E SA. For each ( E C, we have ((eA +a*)a((eA +a),..\) = 0, and so ((a 2 ,..\) +((a*a,..\) = O. Taking (= 1 and (= i, we see that (a*a,..\) = 0 and a E l;... Thus a E *-radA.

(ii) Certainly *-rad A is a left ideal in A, and so it suffices to show that *-radA is *-closed. Let a E *-radA, and take ..\ E SA' Then (aa*aa*,..\) = 0, and so (aa*,..\) = 0 by 1.10.14, i.e., a* E l;... Thus a* E *-radA. (iii) This follows easily.

0

Example 1.10.17 The space A = C 2 , with coordinatewise multiplication, is a commutative, unital *-algebra for the involution * : (z, 'W) ---t (ill, z). The only proper *-ideal in A is {O}, and so A is strongly semisimple. The *-algebra A is neither proper nor ordered. We have Asa = Apos = {(z, z) : z E C}, and it follows easily that PA = {O} and SA = 0. Thus A is *-radical. 0 Definition 1.10.18 Let A be a *-algebra. Then: (i) A is symmetric if -a*a E q-InvA (a E A);

(ii) A is hermitian if CT(a)

C

lR (a E Asa).

We say that the involution * is symmetric or hermztzan when the *-algebra (A, *) has the corresponding property. For example, the conjugation map on CS is a symmetric involution, as is the involution on Mn. A unital *-algebra is symmetric if and only if eA + a*a E Inv A (a E A). Clearly A# is hermitian if and only if A is hermitian.

Proposition 1.10.19 (Doran) Let A be a symmetric *-algebra. Then Ab is also a symmetric *-algebra. Proof Take b = ae b + a E Ab, and set [3 = 1 + lal 2, c = d = _[3-1c, so that 1 e b - d = _(e b + b*b). [3

aa + aa* + a*a,

and

(1.10.8)

= 1 + ([3 -1)X 2 - [3X3 E qX]. Since d* = d, we have p(d) = e b + ([3 - 1)d2 - [3d 3 = e b - d2 + d(eb + b*b)d = e b + (bd)*bd. Now assume that e b + b*b £/:. Inv Ab. By (1.10.8), 1 E CT(d), and so, by 1.6.11(i), 0 = p(l) E CT(p(d)). Thus e b + (bd)*bd ¢ Inv Ab. But bd E A, and

Set p

so we have -(bd)*bd E q-Inv A because A is symmetric, a contradiction. Thus -b*b E q-Inv Ab, and so Ab is symmetric. 0 We defined the equivalence relation

rv

on J(A) in 1.3.18(iii).

Definition 1.10.20 Let A be a *-algebra, and let p, q E Jsa(A). Then p and q are Murray-von Neumann equivalent, denoted by p ~ q, if there exists a E A such that p = a*a and q = aa*.

Involutions Clearly, if P

149 ~

q, then p

rv

q. As in 1.3.20(i),

~

is an equivalence relation on

Jsa(A). We shall see later, in 3.2.1O(ii), that, in the class of C*-algebras, p if and only if p

rv

~ q

q.

Proposition 1.10.21 Let A be a unital, symmetric *-algebra. (i) Let p E J(A). Then there exists q E Jsa(A) such that q rv p. (ii) Suppose that A is properly mfinite. Then there eX'tst p, q E Jsa(A) such that p ..1 q and p rv q rv eA.

Proof (i) Set a = eA + (p - p*)*(P - p*) = eA + p*p + pp* - p - p*. Then a E Asa , and a E Inv A because A is symmetric. We have pa = pp*p = ap. Set q = pp*a- l = a-Ipp*. Then q2 = a-Ipp*pp*a- I = q, and so q E Jsa(A). We have pq = q and qp = a-Ipp*p = p, and so q rv p. (ii) There exist PI,P2 E J(A) such that PI ..1 P2 and PI '" P2 rv eA. Set

aj qi ql

= eA + (Pj - pj)*(Pj - pj) (j = 1,2), so that all a2 E InvA. Also set = piPlal l and q2 = P2P2a21. Then, essentially as in (i), QI,q2 E Jsa(A) and rv q2 rv eA. Further, we have qIq2 = a1IpipIP2P2a2I = 0 and q2qI = O. Set

p = qi and q = q2. Then p, q E Jsa(A) with p ..1 q and P rv q rv eA.

0

Proposition 1.10.22 Let A be a *-algebra. (i) Suppose that A is hermitian and cp E q,A. Then cp(a*) = cp(a) (a and cp is a positive trace. (ii) Suppose that A zs unital and hermztian. Then q, A C ex SA. (iii) Suppose that A 1S symmetric. Then A is hermitwn.

E A),

Proof (i) For a E A, set b = 3!a and c = s:fa. Since A is hermitian, we have a(b),a(c) C JR, and so cp(b),cp(c) E JR. Thus

cp(a*) = cp(b - ic) = cp(b) - icp(c) = cp(a) , and so cp(aa*) = cp(a*a)

= Icp(a)12

~ O.

(ii) By (i). q,A c SA. Suppose that cp E q,A and

0= 4cp(a 2) -4cp(a)2 = 2«(a 2, AI) + (a 2,A2») - «(a,AI) + (a,A2»)2 ~ 2( (a, AI)2 + (a, A2)2) - «(a, AI) + (a, A2»)2 = «(a, AI) - (a, A2»)2 , whence (a, AI) = (a. A2). Thus Al = A2 = cp and cp E ex SA. (iii) By 1.10.19, we may suppose that A is unital. Take a E Asa and z = a + i(3 E C with (3 =J 0, and set b = (a - aeA)/(3. Then b* = b, and so eA + b2 E Inv A, say (CA + b2)c = c(eA + b2) = eA. We have

(a - zeA)(a - zeA)c = «a - aeA)2

+ j3 2eA)r = (32(eA + b2)c = (32cA'

and similarly c(a - zeA)(a - zeA) = (32 eA . Thus a - zeA E Inv A and z E pea). This shows that a(a) C JR, and so A is hermitian. 0

A 1gebrmc fo'Undation,~

150

\Ve have one result concerning derivations on a *-algebra: it is for later use.

Proposition 1.10.23 Let A be a 'Umtal *-algebra, let E be a umtal A-bzrnodule, fLnd let D : A ~ E be a derivation. Then v* . D(vav*) . v - Da = a . Dv* . v - Dv* . va for each a E A and

Proof Take a and

l' 11

E

A wzth v*v = eA.

v* . D(uav*) . v

= v*

. (1!a . Dv*

= a . Dv* . v

giving the result.

+ Dv* . v* + Dv

a.<.; specified. By l.R2(i), v* . Dv

+v

+ Da -

. Da

.

/I

= 0, and so

. av*) . v

Dv* . va,

o

Notes 1.10.24 For a substantial account of algebras with an involution, see (Palmer 2000). The notions of ordered, proper. and very proppr *-algebra.."l. and of algebras with some related properties, are discussed in (zbid., §9. 7); an example of a comlIlutative Banach *-algebra which is an integral domain, and hence proper, but which is not a very proper algebra is noted in (ibid., 9.7.38). The Murray-von Neumann equivalence relation ~ on JRa(A) is a foundation stone of the comparison throry of projections in a von Neumann algebra; see (Kadisoll and Ringrose 1986, Chapter 6). The *-radical of a unital *-algebra A is called the reducing ideal of A in (Naimark 1964, IV.18.2) and the pre-reducing ideal in (Palmer 2000, 9.7.14). It seems to be unknown whether or not rad A C *-rad A for every *-algebra A. Some of the discussion of symmetric and hermitian involutions is taken from (Fell and Doran 1988a); for example, 1.10.19 is a result of Doran (ibid., VI.2.4). For many further results, see (Palmer 2001, §9.8). For a hermitian algebra A, each primitive ideal is a *-ideal (zbid .. 9.8.2). An example of a hermitian *-algebra which is not symmetric is given in (ib!d., 9.8.13).

2

Banach and topological algebras

We now turn to the definitions and fundamental properties of (in general, noncommutative) Banach algebras and some more general topological algebras. Thus we are moving from the foundations given in Chapter 1, and shall now examine the implications of the additional assumption that the algebras that we are considering also have a topology; in each case. the topology is connected to the algebraic structure by the simple requirement that the algebraic operations be continuous. In §2.1, we shall introduce normed and Banach algebras, and discuss the normability and automatic continuity questions that are the theme of this book; in §2.2, we shall give the definitions of more general classes of topological algebras. We shall then continue to develop the theory of Banach and topological algebras. giving the classical Gel'fand theory of commutative Banach algebra..') in §2.3 and the (analytic) functional calculus theorems (for one and several variables) in §2.4. In §2.5, we shall discuss some important Banach algebras which are subalgebras of the Banach algebra B(E) of all bounded linear operators on a Banach space E. In §2.6-§2.8, particula.r stress will be placed on Banach Abimodules and the cohomology theory of a Banach algebra A, the cohomological groups having coefficients in a Banach A-bimodulej our theory is a 'Banach' version of the algebraic theory of §1.9. The chapter concludes in §2.9 with the surprisingly important theory of bounded approximate identities in Banach algebras. Throughout this chapter, a few basic examples will be noted, but a more substantial study of specific Banach algebras will be left to Chapters 3 and 4. In Chapters 2-5, all linear spaces will be taken to be over C, and all algebras will be complex algebras, unless we state otherwise. 2.1

INTRODUCTION TO NORMED ALGEBRAS

This first section gives fundamental definitions related to normed algebras, and then discusses the normability, automatic continuity, and uniqueness-of-norm questions for Banach algebras that are at the heart of this book; we shall obtain some preliminary results, including some examples of non-normable algebras. We shall conclude the section by proving the simple fact that characters on Banach algebras are automatically continuous; it will be seen in due course that this result is a foundation stone of our theory. Seminorms on a linear space are defined in Appendix 3.

]52

Banach and topological algebms

Definition 2.1.1 Let A be an algebm. An algebra seminorm [algebra norm] on A is a seminorm [norm] p on A such that p(ab)

~

p(a)p(b)

(a, bE A).

A pair (A. p) is a seminormed [normed] algebra if p is a non-zero algebm seminorm [algebm norm] on A; A is normable if there is an algebm norm on A, and A zs seminormable zf A = 0 or zf there is a non-zero algebm seminorm on A. A complete normed algebm is a Banach algebra. Let II· such that

II

be a norm on all algebra A, and suppose that there exists

Ilabll

~

e Ilallllbll (a, bE A) .

e

> 0

(2.1.1)

Set Illalll = ellall (a E A). Then 111·111 is an algebra norm on A. and 111·111 is equivalent to II . II· Thus A is normable, and (A, III . III) is a Banach algebra in the case where (A, 1\ . 1\) is complete. We shall often say that 'A = (A, II . II) is a Banach algebra' whenever (2.1.1) holds. Let (A. 11·11) be a llormed algebra. Then rnA : (a. b) I---> ab, A x A ~ A, is a continuous bilinear map. and (2.1.2) Each normed algebra A is isometrically isomorphic to a dense subalgebra of a Banach algebra: this is the completion of A. The completion is unique up to isometric isomorphism. Note that the completion has the same cardinality as A. Vo.le can now give some of the major themes with which we shall be concerned ill this book. First, we wish to know which algebras are normable. The main result that we shall eventually obtain is Theorem 5.7.19(ii): with the assumption of the continuum hypothesis (CH), each integral domain A (with a character if A is unital) of cardinality c = 2"0 is normable. However, many algebras are not normable (see 2.1.14,2.1.15.2.3.17, and 4.10.27). or even seminormable (see 2.1.14. the examples after 2.2.43, and 4.10.27), and there are algebras which are not even topological algebras for any Hausdorff topology (see 2.2.50(i) and 4.10.32). Second, we wil:;h to know which algebras are Banach algebras with respect to some algebra norm. We shall give many examples of Banach algebras. and several examples of normed algebras which are not complete. For example, it will be shown in 4.6.2 that the algebra ~ of formal power series in one indeterminate is not a Banach algebra with respect to any norm, although ~ is llormable (see 5.7.2). Third, we wish to know which algebras are the homomorphic images of a Banach algebra: for example, it will be shown in 5.5.19 that there is an epimorphism from certain commutative Banach algebras onto~. Finally. we are interested in the special form of Banach algebras which satisfy certain algebraic conditions. A fundamental result of this type is the Gel'fand- Mazur theorem 2.2.42(ii), which asserts that the only seminormable division algebra is C; as another example, it will be shown in 2.6.39 that a Banach algebra for which every closed left ideal is finitely generated is finite-dimensional. Let p be an algebra seminorm on an algebra A, and let I be an ideal in A. Then the quotient seminorm on AII is an algebra semi norm. Suppose that I is a closed ideal in a normed algebra A. Then AII is a normed algebra with

153

Introduction to normed algebras

respect to the quotient norm; AI I is a Banach algebra in the case where A is a Banach algebra. The question of the norm ability of an algebra A can also be phrased in terms of the existence of homomorphisms from A into Banach algebras, as the first proposition shows.

Proposition 2.1.2 Let A be a non-zero algebra. Then A is seminormable [normable 1 zf and only if there is a non-zero homomorphzsm [an embeddtng1from A tnto a Banach algebra. Proof Let 0 be a non-zero homomorphism from A into a Banach algebra. Then the function a f-+ IIO(a)II, A ---+ IR. is clearly an algebra semi norm on A; it is a norm if 0 is an embedding. Let p be a non-zero algebra semi norm on A, and set I = ker p. Then I is a proper, closed ideal in (A,p), and the formula Iia + III = pea) defines an algebra norm 11·11 on the quotient algebra AI I. Let B be the completion of (AI I, II ·11). Then B is a Banach algebra, and the quotient map a f-+ a + I is a non-zero homomorphism from A into B; it is an embedding if p is a norm. 0 We shall use the following tests for normability in 2.1.14, in 4.1.21, and in 4.1.42(ix).

Proposition 2.1.3 (i) Let A be an algebra. Suppose that there exzst a E A and {b n : n E N} c A- such that ab n = nbn

(n E N) .

Then A zs not nornwble. If, further, A#bnA# = A (n EN), then A is not semtnormable.

(ii) Let (A, II ·11) be a Banach algebra, let I be an ideal in A such that AI I is nornwble, and suppose that (an) is an orthogonal sequence in!. Then a~ E I eventually. If, further, an E anI (n E N), then a; E I eventually.

Proof (i) Assume towards a contradiction that 11·11 is an algebra norm on A. Then n Ilbnll = IIabnll ~ lIallllbnll (n E N); since Ilbnll =I- 0, necessarily Iiall 2:: n for each n E N, a contradiction. Thus A is not normable. Suppose, further, that A#bnA# = A (n EN), and assume that p is a non-zero algebra seminorm on A. Set I = ker p. Then AI I is normable. But {b n + I : n E N} is contained in (AI 1)-, and so this contradicts the result we have just proved. Thus A is not seminormable. (ii) Let the algebra norm on AI I be p. Assume towards a contradiction that the result fails. Then we may suppose that p( a~ + 1) =I- 0 (n E N); we may also suppose that p(a n + 1) = p(a~ + 1) (n EN). Take a sequence (b n ) in I with n 3 11an lilian - bnll < 1 (n EN), and define 00

a=

2: nan (an n=l

bn ) EA.

154

Banach and topologzcal algebras

For each n EN, we have ana = na~(an - bn ). and hence p(a n

+ I)p(a + I) 2:

np(a~

+ I)

because bn E I. Thus p(a + I) 2: n for each n E N. the required contradiction. Now suppose that an E anI (n EN). The result follows by a similar argument: to obtain a contradiction, assume that p(a n +I) = p(a; +I) f:. 0 (n EN). choose (b n ) in I with n 3 IIa n - anbnll < 1, and set a = n(a n -anbn ). 0

r::=l

A second major theme of this book is the automatic ('ontmuity problem for homomorphisms and derivations. Let A and B be Banach algebras, and let e be a homomorphism from A into B. Under what algebraic conditions on A and B is () automatically continuous? For which Banach algebras A is it the case that each homomorphism from A into a Banach algebra is automatically continuous, and for which Banach algebras B is it the case that each homomorphism from a Banach algebra into B is automatically continuous? An important result that we shall obtain in 5.4.13 is that every homomorphism from the C*-algebras B(H) and lC(H). where H is a Hilbert space, is continuous. The dominant theme in §5.4 and §5.7, however, will be that homomorphisms from commutative C*algebras are continuous on a dense subalgebra, but that, with CH. there are discontinuous homomorphisms from each infinite-dimensional, commutative C*algebra. For certain Banach algebras, there may be homomorphisms which are very discontinuous in the following sense. Definition 2.1.4 Let A be a Banach algebra. and let E be a topological linear space. An operator T : A -+ E zs very discontinuous iJ the restriction oj T to every injinzte-dimensional subalgebra oj A zs discontinuous.

We have the following minor result in connection with the automatic continuity problem for homomorphisms. Proposition 2.1.5 Let I be a closed ideal in a Banach algebra A. and let B be a Banach algebra. Suppose that all homomorphisms Jrom A into B are continuous. Then all homomorphisms J7'Om AI I mto B are contznuous. Proof Let () : AI I -+ B be a homomorphism, and let Q : A -+ AI I be the quotient map. Then the homomorphism () 0 Q : A -+ B is continuous, and so () is continuous because Q is open. 0

A special case of the automatic continuity problem for homomorphisms is the uniqueness-oj-norm p7'Oblem, which asks which Banach algebras have a unique complete algebra norm. Recall first that two norms on a linear space E are equivalent if they determine the same topology on E. Thus two norms n·111 and II ·112 are equivalent if and only if there exist constants m,!vI > 0 such that

m IIxlll

~

IIxl12

~ M

Ilxlll

(x

E

E).

In the case where (E, 11·111) and (E, 11.11 2 ) are Banach spaces, it follows from the closed graph theorem A.3.25 that the norms 11·111 and 11.11 2 are equivalent if and only if the identity map £ : (E, 11.11 1) -+ (E, 11.11 2 ) is continuous.

Introductzon to normed algebras

155

We shall not be concerned with the geometric properties of norms on Banach algebras, but only with the topologies which they define. Thus our interpretation of 'uniqueness-of-norm' is the following. Definition 2.1.6 Let (A, 11·11) be a Banach algebra. Then A has a unique complete norm [unique norm] if each norm with respect to which A is a Banach algebra [normed algebra] is equivalent to the given norm 11·11.

Thus, in the case where A is a Banach algebra with a unique complete norm, the topological and algebraic structures of A are intimately linked. We shall prove in Corollary 2.3.4 that each commutative, semisimple Banach algebra has a unique complete norm, and we shall later describe several classes of Banach algebras with this property. In particular, we shall give three proofs (5.1.6, 5.1.9(iii), and 5.2.28(iv)) of the famous theorem of Johnson that each semisimple Banach algebra (not necessarily commutative) has a unique complete norm. Other Banach algebras with a unique complete norm are certain convolution algebras Ll(W) (see 5.2.18(i)) and Banach algebras of power series (see 5.2.20(ii)). On the other hand, one may easily form Banach algebras which do not have a unique complete norm. For example, let A be a linear space which is a Banach space with respect to two inequivalent norms, and set ab = 0 (a, bE A). Then A is a Banach algebra with respect to each of the two norms. Less trivial examples will be given in §5.1, which is devoted to the question of the uniqueness of complete norms. However, we do not have any tractable characterization of exactly which Banach algebras have a unique complete norm. For example, it is not known whether or not each commutative Banach algebra which is an integral domain has a unique complete norm; see Question 5.1.B and 5.3.13. The property of having a unique norm is a stringent condition. Of course, finite-dimensional algebras satisfy this condition; infinite-dimensional examples, such as the Banach algebras B(E) and K(E) for certain Banach spaces E, will be presented in 5.4 12. The following remark will be useful in determining when a Banach algebra has a unique norm. Proposition 2.1.7 Let (A, 11·11) be a Banach algebra such that, for each algebra norm 111·111 on A, there exists C > 0 with lIall ::; C IlIalil (a E A). Then A has a umque norm if and only if each homomorph%sm from A mto a Banach algebra is contmuous. Proof Suppose that () is a discontinuous homomorphism from A into a Banach algebra, and set IIlalil = max{lIall, 1I()(a)lI} (a E A). Then 111·111 is an algebra norm on A, and III· III is not equivalent to II . II· Conversely, suppose that III· III is an algebra norm on A not equivalent to 11·11. By hypothesis, the identity map from A into the completion of (A, III· liD is a discontinuous homomorphism. 0

The automatic continuity problem for derivations is the following. Let A be a Banach algebra. Under what conditions on A is it true that all derivations from A into some or all Banach A-bimodules (see 2.6.1) are automatically continuous? For example, all derivations from a C* -algebra A into a Banach A-bimodule

Banach and topological algebras

156

are continuous; see 5.3.7. In fact, we shall consider more general maps than derivations; these are the intertwining maps of Definition 2.7.1. We shall see in 2.7.5 that, if all homomorphisms from a Banach algebra A are continuous, then all derivations (and, indeed, by 2.7.7, all intertwining maps) from A are also continuous. The converse is not true; for example, it will be shown in 5.3.7 that all intertwining maps from each C(O) are continuous, but, with CR, there are discontinuous homomorphisms from C(O) whenever 0 is infinite. Definition 2.1.8 A normed algebra (A, II· II) is unital if A has an identity eA and IleAIi = 1.

Let (A,p) be a seminormed algebra. Then the formula

p(ael>

+ a)

= lal

+ pea)

(a E C, a E A)

defines an algebra seminorm on AP; it is a [complete] norm if and only if p is a [complete) norm on A. Thus, in the case where A is a Banach algebra, so are AI> and A#. Proposition 2.1.9 Let A be an algebra, let 11·11 be a norm such that (A, 11·11) is a Banach space and multzphcation on A is separately continuous, and let S be a bounded semigroup in (A, . ). Then there zs a norm III . III equivalent to II . II on A such that (A, III· liD is a Banach algebra and IIlslll :s; 1 (s E S). Proof By A.3.39, there exists C ~ 1 such that lIabli :s; C Ilalillbli (a, b E A). Also there exists M ~ 1 such that IIsli :s; M (s E S). Set pea) = sup{lIall, IIsali : s E S} (a E A), so that p is a norm on A. Then lIall :s; pea) :s; CM lIall (a E A), and so p is equivalent to 11·11. Clearly:

p(sa) p(ab) p( s)

:s; pea) :s; Cp(a)p(b) :s; M

(a E A, s E S); (a, bE A); (s E S) .

Define IIlalll

= sup{p(aa. + ab) : Cl'

E

C, bE A, M lal

+ pCb) :s; I}

(a E A).

Then 111·111 is a norm on A and p(a)/M :s; IIlalll :s; Cp(a) , and so 111·111 is equivalent to p, and hence to II . II. For a, b E A, we have IIlablli

:s; sup{p(aab + abe) : a E C, c E A, p(ab + be) :s; IIlblll} :s; sup{p(ad) : d E A, p(d) :s; I} IIlblll :s; Illallllllblil ,

and so (A, 111·111) is a Banach algebra. Finally, take s E S. For a E C and b E A with M lal + pCb) :s; 1, we have peas + sb) S; 1, and so IIlslll s; 1. 0 In particular, we may always suppose that a Banach algebra with an identity is a unital Banach algebra.

Corollary 2.1.10 Let A be a unital Banach algebra, and let G be a bounded subgroup of (Inv A, . ). Then there is a Banach algebra norm III . Ilion A equivalent to the given norm such that IIlalil = 1 (a E G). 0

Introduction to norrned algebras

157

There is a further renorming result that is sometimes useful. Let (A, 11·11) be a commutative, unital normed algebra, and let 8 be a unital subsemigroup of (A, . ) with 8 c A[k], where k > O. Define Illalll

= inf {~lla.11 : a =

~aisi'

Sl, ... ,Sm

E 8, mEN} .

(2.1.3)

Then (A, 111·111) is a unital normed algebra, and Iiall jk ::::; Illalll ::::; Iiall (a Also IIlslll ::::; 1 (s E 8).

E

A).

Definition 2.1.11 Let A be a norrned algebra, let a E A-, and let K be a subsemigroup oj (re, +) with 1 E K. A K -semzgroup (a( : ( E K) zs continuous [bounded] zJ the map ( f---+ a(, K ---+ A, is continuous [bounded]. In the case where K zs an open set in re, the semzgroup zs analytic zJ thzs map is analytic onK. Proposition 2.1.12 Let A be a normed algebra, let K be an open semzgroup m (!C. +) with 1 E K, and let (a,( : ( E K) be an analytic semigroup in A. Then:

(i) a(A = aA (ii) (a( + I: (

«( E K); E

K) is an analytzc semigroup in A/I Jor each closed ideal I

in A. Proof (i) Fix (0 E K, and take A E A' such that A I a(o A = O. For each b E A, the function Fb : ( f---+ (a(b, A), K ---+ re, is analytic. For each ( E (o+K, we have a(b E a(oA, and so Fb«() = O. Hence Fb = 0 on K, and so A I a(A = 0 E K). It follows from the Hahn-Banach theorem A.3.18(i) that a("A C a(oA E K). The result follows. (ii) This follows immediately from A.3.76. 0

«( «(

Many specific examples of Banach algebras will be extensively discussed in this book. At this stage, we wish to list a few well-known examples to establish notation and to allow us to make remarks as we progress. All the examples will be considered more fully later. Examples 2.1.13 (i) Let 8 be a non-empty set, and let £00(8) denote the Het of complex-valued, bounded functions on 8. Then £00(8) is a unital sub algebra of res (with the pointwise operations of (1.3.7». We define

IJls

= sup{IJ(s)1 : s E 8} (f E £'>0(8».

(2.1.4)

Then I· Is is an algebra norm on loo(8), called the unzJorm norm on 8, and (£00(8), I· Is) is a commutative, unital Banach algebra. Let X be a non-empty topological space. Then C(X), Cb(X), and CoCX) are subalgebras of rex. Also Cb(X) = £oo(X) n C(X) and Co(X) are commutative Banach algebras which are closed subalgebras of (£oo(X), I· Ix ); Coo(X) is a dense ideal in Co(X). The algebra Co(X) is unital if and only if X is compact. As a special case of the algebras Co(X), note that (eo, I· IN) is a commutative Banach algebra. The subspace C(X, 1R) of C(X) is a closed, real subalgebra.

158

Banach and topological algebms

For a study of the Banach algebras Cb(X) for a completely regular topological space X and of Co(O) for a non-empty, locally compact space 0, see §4.2. We shall see that algebraic properties of Co(O) correspond to topological properties of O. For example, it is clear that J(Co(O)) consists of the functions XK for K a compact and open subset of O.

(ii) Let A (iD) = {J E C (iD)

: f IIDl is analytic} ,

e : Izl < I} is the open unit disc, as in Appendix 2. Then is a commutative, unital Banach algebra; it is called the disc algebm (see 4.3.11). Each f E A(iD) has a Taylor expansion:

where

IDl

= {z E

(A(iD) , I· Ii»)

f(z) =

~

L.J

n=O

f(n) (0) zn n!

(z

E

iD) .

However, this series need not converge in the uniform norm on iD. Let f E A(iD). For each r < 1, set fr(z) = f(rz) (z E iD). Then fr --> fin A(iD) as r --> 1-, and each such fr has a uniformly convergent Taylor series. Thus the restrictions to iD of the polynomials are dense in A (iD) . We denote by A+ (iD) the set of functions f = 2::=0 QnZ n in A(iD) which have an absolutely convergent (and hence uniformly convergent) Taylor expansion on iD. It is easy to see that A+ (iD) is a commutative, unital Banach algebra with respect to the pointwise operations and the norm II . 111' where

it is called the algebra of absolutely convergent Taylor series. We exhibit some specific functions in A+(iD). Let (3 E jR+. \ N. We claim that 00

(1 - z)13 = 1 +

L Q13,nZn

(z E iD)

,

(2.1.5)

n=l

where

IQ13,nl '" n- 1 -{3 as n --> 00.

Indeed, we have

Q{3,n = (-I)n(3«(3 - 1) ... «(3 - n

+ 1)/n!

= (n - 1 - (3)(n - 2 - (3) ... (1 - (3)( -(3)/n!

= r(n - (3)/f(n + l)f(-(3) = f(n - (3)/nf(n)f( -(3), where f is the gamma function. By (A.2.3), f(n - (3)(n - (3)13 /f(n) --> 1 as n --> 00, and so Q13,nn1+13 --> f( _(3)-1 as n --> 00. Thus (1- Z)13 E A + (iD). Note that, in the case where 0 < (3 < 1, we have Q13,n < 0 (n EN). Clearly «1- Z){3 : (3 E JR+.) is a continuous real semigroup in A+(iD). (iii) Let E be a non-zero Banach space, and let 8(E) be the algebra of all bounded linear operators on E. Then 8(E) is an algebra with identity IE, and it is a Banach algebra with respect to the operator norm II . II. An idempotent P in 8(E) is a projection onto the closed subspace P(E) of E. In the case where E = en, we have 8(E) = C(E) ~ Mn.

Introduction to normed algebms

159

Let A be a finite-dimensional algebra. Then there is a norm II . II on the finitedimensional space A#, and the map a 1-+ La, A -+ B(A#), is an embedding. Set Illalil = IILall (a E A). Then (A, 111·111) is a Banach algebra. (iv) Let 8 be a non-empty set, let w : 8 -+ R+ e be a function, and define £1(8, w) = {I =

L l(s)08 E C

S :

II/lIw =

sES

L I/(s)1 w(s) < oo}.

(2.1.6)

sES

Then (P1(8, w), II·IU is a Banach space; we write £1(8) and 11·111 in the case where w = 1, and we write £1 for £l(N), as in Appendix 3. The space £1(8, w) is separable if and only if 8 is countable. If the function w is bounded below, then ((1(8, w), ., 1I·lI w ) is a commutative Banach algebra with respect to the pointwise product . on 8; this algebra is unital if and only if 8 is finite. (v) Now suppose that 8 is a semi group, not necessarily abelian. A weight on 8 is a map w : 8 -+ R+e such that

west) ::; w(s)w(t)

(s, t E 8) j

we also require that w(e 8 ) = 1 in the case where 8 is unital. It follows from A.1.26(iii) that limn----+oow(sn)l/n exists for each s E 8. A weight on Z+ is a weight sequence. For example, w : s 1-+ exp(-s2) is a weight on (R+,+), and w : p/q 1-+ q, where p and q are coprime in N, is a weight on (tQ+e, +). Here is a recipe for constructing rather strange weights on tQ+.. Let U be a subset of tQ+e. Then a function 1j; : U -+ R is subadditzve if 1j;(u) ::; E;=l1j;(Uj) whenever u, U1, ... ,Un E U and U = Ul + ... + Un. Let 1j; : U -+ R be subadditive, and take v E Q+. \ U. Then there is a subadditive function ;p : U U {v} -+ R such that 7jj I U = 1j;. Now let {Pit P2, ... } be an enumeration of the prime numbers (with Pn < Pn+1 (n EN)), and let Uo = {1/Pn : n EN}. Suppose that u, UIt ... , Un E Uo and U = U1 + ... + Un. Then necessarily n = 1, and so every function 1j; : Uo -+ R is subadditive. Each such 1j; has an extension to a subadditive function :;j; on Q+e j exp:;j; is then a weight on tQ+e. By starting with the function which takes the value -n at I/Pn, we obtain a weight w on Q+e such that liminfs----+o+w(s) = O. Let w be weight on a semigroup 8. We define a product *, called convolution, in £1 (8, w) such that Os * Ot = Ost (s, t E 8). First note that, for I, 9 E £ 1(8, w), we have

~ L~t I/(r)llg(s)l} wet) ::; ~ I/(r)1 w(r) ~ Ig(s)1 w(s) = II/l1w IIgllw ' and so £1(8, w) is a Banach algebra with respect to the product defined by (f

* g)(t) =

L

I(r)g(s)

(t

E

8).

(2.1.7)

rs=t

(If there are no elements r, s E 8 with rs = t, set (f * g)(t) = 0.) The algebra (£1(8, w), *, 1I·ll w ) is the weighted semigroup algebm for the weight w on 8j in the case where 8 has an identity e, Oe is the identity of the algebra. The algebra £1(8, w) is commutative if and only if S is abelian.

Banach and topological algebras

160

The algebraic semigroup algebra coo(S), described in 1.3.7(iii), is a dense subalgebra of each weighted semigroup algebra £l(S, w). For example, by taking S = z+n, we obtain subalgebras of the algebra ~n of formal power series in n indeterminates by identifying the element O(Tl, .,Tn) with the monomial X[l '" X~n for (r1,"" rn) E z+n; these algebras will be studied in §4.6. In particular, the constant function 1 is a weight on S; the corresponding algebra, denoted by £ 1 (S) = (£ 1(S), *, II . 111)' is the (discrete) semzgroup algebra of S. It is easy to see, for example, that (£l(Z+), *, 11'//1) is isometrically isomorphic to (A + (ii)) , . , 1/ . "1)' We refer to the weighted group algebra £ 1(G, w) in the case where G is a group; in this case, the formula for the convolution product of f,g E fl(G,w) is (J

* g)(t) =

L

f(s- l t)g(s)

(t E S).

(2.1.8)

sEG

In particular, £ 1 (G) is the group algebra of the group G.

o

We begin by showing that certain algebras are not normable. Proposition 2.1.14 (i) Let X be a non-empty topological space. Then C(X) is normable if and only if X is pseudocompact. (ii) Let E be a non-zero linear space. Then C(E) is seminormable if and only if E zs finite-dimensional. (iii) Let A be a unital algebra containing elements a, b such that ab - ba = eA. Then A is not seminormable. Proof (i) Suppose that X is pseudocompact. Then C(X) = Cb(X) is a Banach algebra with respect to the uniform norm / . /x on X. Conversely, suppose that X is not pseudocompact, and take f E C(X) with f(X) unbounded; we may suppose that f(X) C ~+. Inductively choose a sequence (xn) in X such that f(xn+l)

> f(xn) + 2 (n E N),

and set In = [J(Xn) - 1, f(xn) + 1] (n EN). Then the pairwise disjoint, closed intervals In are such that there exist h E C(JR) with h(t) = n (t E In) for each n E N, and (gn) C C(JR) with 9n(J(Xn» = 1 and supp gn C In for each n E N. Set a = h 0 f and bn = gn 0 f (n EN). Then a E C(X), {bn : n E N} c C(X)-, and abn = nbn (n EN). By 2.1.3(i), C(X) is not normable. (ii) Set 2l = C(E). Suppose that E is finite-dimensional. Then 2l = B(E) is a Banach algebra with respect to the operator norm. Conversely, suppose that E is infinite-dimensional. Then there is a sequence (En) of subspaces of E such that E = O{En : n E N} and En ~ E (n EN). For n E N, take Pn : E --+ En to be a projection and take Qn E 2l such that Qn / En : En --+ E is a linear isomorphism and Qn / Em = 0 (m =1= n). Then the map QnPn : E --+ E is an epimorphism, and hence there exists Un E 2l with QnPnUn = IE· Thus we have !2lPn !2l =!2l. Define A E !2l by requiring that A / En = nPn (n EN). Then APn = nPn (n EN), and so, by 2.1.3(i), !2l is not seminormable.

Introduction to normed algebras

161

(iii) By an immediate induction, anb - ban = na n -

(n E N) .

1

(2.1.9)

Assume that p is a non-zero algebra seminorm on A. It cannot be that peak) = 0 for any kEN, for, by (2.1.9), this would imply that p(eA) = 0 and hence that p = O. From (2.1.9), np(a n - 1 ) ~ 2p(a)p(b)p(a n - 1 ) (n EN), and so n ~ 2p(a)p(b) for each n E N, a contradiction. 0 It is clear that each algebra A with A2 = 0 is normable: every norm on A is an algebra norm. However, we now !ihow that this result does not extend to the case of algebras A such that A3 = o. Proposition 2.1.15 (Dales) (i) There is a commutative algebra A with A3 = 0 such that A is not normable.

(ii) There zs a commutatwe algebra A with a countable basis such that A is not normable. Proof (i) Take A to be the algebra which, as a linear space, has as a basis the set {e s : S E [0, I)}, and for which the product is defined by the formula: _ { eo/ Is - tl eset 0

(s, t E (0,1), s (otherWlse .) .

=I t) ,

It is clear that the product of any three elements of A is 0, and so A is associative and A3 = O. Assume that there is an algebra norm /I . /I on A. Then

0< IIeoil ~ Is - tl lies II IIetil

(s,t E (0,1), s

=I t).

For n E N, set Un = {s E (0,1) : IIesil > n}. If s E (0,1) and t E (0,1) with Is - tl < IIeo/i /n IIesll, then t E Un, and so each Un is open and den!ie in (0,1). By the category theorem A.1.21, Un =10. Take s E Un. Then IIesil > n for each n E N, a contradiction. Thus A is not normable. (ii) Let S = Ul:+)<w be the free abelian semigroup on countably many generators, as in 1.2.2(ii), and let A = coo(S), the algebraic semigroup algebra, as in 1.3.7(iii). We regard the elements of A as polynomials with zero constant term in the countable family {Xn : n EN}. Let I be the ideal in A generated by the elements X 1 X n - nXn for n 2 2, and set 2{ = A/I and bn = Xn + I (n EN), so that 2{ is a commutative algebra with the countable basis ibn : n EN}. We claim that bn =I 0 for each n 2 2. For assume that Xn E I for some n 2 2. Then there exist kEN and PI, ... , Pk E A with Xn = L~=2(XIXj - jXj)Pj. Evaluate this expression at the point «(J) where (1 = n, (n = 1, and (j = 0 otherwise; we obtain a contradiction, and so the claim holds. Clearly b1bn = nbn (n EN), and so, by 2.1.3(i), 2{ is not normable. 0

n:=1

n:=1

Let A be a unital normed algebra. Then the map a ....... La, A ~ B(A), is an isometric, unital homomorphism which identifies A as a closed subalgebra of 8(A) such that fA EA.

162

Banach and topological algebras

Proposition 2.1.16 Let A be a unital normed algebra. Then eA is an extreme point of A[l]' Proof It suffices to show that IE E ex8(E)[1] for each Banach space E. Set a = a(E',E) and K = Efl]' so that K is convex and compact in (E',a)j

by the KreIn-Mil'man theorem A.3.30(i), (exK) (0-) = K. Take T E 8(E)[1] with IIIE ± Til ~ 1. Then IIIEI ± T'" ~ 1 in 8(E'). Let A E exK, and set Al = A + T'(A) and A2 = A - T'(A). Then AI. A2 E Efl] and --(0-)

2A = Al + A2, and so Al = A2 = A. Thus T'(A) = O. Since (exK) = K and T' is continuous on (E', a), it follows that T' = O. Thus T = O. Now suppose that 2IE = R + S, where R, S E 8(E)[1], and set T = IE - R. Then IIIE ± Til ~ 1, and so T = 0, R = S = IE, and IE E exB(E)[l]' 0 Definition 2.1.17 Let (A, II "D be a normed algebra. Then a Banach algebra extension of A is a Banach algebra (B, ",, III) such that there is an isometric embedding of (A, 11·11) in (B, ",, III).

In this case we regard (A, II· II) as a closed subalgebra of (B, ",, "'). We now give some ways of building new Banach algebras from old ones. Examples 2.1.18 (i) Let A and B be Banach algebras, let I be a closed ideal in B, and let () : A - B be a continuous homomorphism. Set 21 = A Eel I as a Banach space, and define

(aI, bl )(a2, b2) = (ala2, bl b2 + ()(al)b 2 + bl ()(a2)

(aI, a2 E A, bI, b2 E I). (2.1.10) Then 21 is a Banach algebra extension of Aj 21 is commutative if both A and B are commutative. (ii) Let A be a unital Banach algebra, and set 21 = MIn(A), the algebra of n x n matrices over A. Then 21 is easily checked to be a unital Banach algebra with respect to the norm given by

II(aij)'1 = max{IIailil

+ ... + IIainil

: i E N n}

and identity" = (oi,jeA); 21 is a Banach algebra extension of A with respect to the map a 1-+ (oi,ja), A - 21. In particular, the full matrix algebra MIn is a Banach algebra with respect to the norm //(ai)// = max{laill + ... + lainl : i E Nn }. The algebra MIn is also a Banach algebra with respect to the norm "'lI p ' formed by regarding MIn as an algebra of (bounded, linear) operators on (en, "./1 ) for p E [1, oo)j the case p = 2 gives the C*-norm on MIn, to be discussed in §3.~. (iii) Let {A-y : I E r} be a family of Banach algebras, and take p E [1,00). Then the Banach spaces fP(r,A-y), co(r,A-y), and fOO(r,A-y) (c/. Appendix 3) are Banach algebras for the coordinatewise product. Clearly co(r, A-y) is a closed ideal in .e 00 (r, A-y ) j the quotient .e 00 (r, A-y ) / Co (r , A-y) is sometimes useful.

163

Introduction to normed algebras

(iv) Let A be a Banach algebra, and let n be a non-empty, locally compact space. Then the Banach space (eb(n, A), II ·11) is a Banach algebra with respect to the pointwise product, and eo(n, A) is a closed ideal therein; here,

IIFII = sup{IIF(x) II : x E n} (F E eben, A)). (v) Let (A, II· II) be a unital Banach algebra, let n E N, and let W

be a weight

on z+n. Set !!w

=

{a

=

LarX r E A[[X1, ... ,Xn]] : IIall w = L

IIarilwr <

oo} .

Then !!w is a subalgebra of A[[Xl, . .. , Xn]], and (!!w, II ·IL) is a Banach algebra extension of A. Let kEN, and set I = {a E !!w : ar = 0 (Irl > k)}. Then I is a closed ideal in !!w, and !!w/ I is a Banach algebra that we identify with {~lrl:5k arX r }; !!w/ I is also a Banach algebra extension of A with respect to the norm given by (vi) Let (A,

\\~Irl~k arX r \\ = ~Irl~k IIarll·

11·11)

be commutative, unital Banach algebra, and let p = ao

+ a1X + ... + anX n + X n+ 1

be a monic polynomial in A[X]. Set Ap = A[Xl/pA[X], so that Ap is a commutative, unital algebra. Each elemcnt of Ap has a unique expression in the form ~~=o biXi, where bo, b1 , ... , bn E A. Choose t > 0 such that t n +1 ;::: ~;=o Ilajll tj, and then consider

!!t = {a E LarX r E A[[X]] : lIali t = L

liar II t r < oc} ,

so that, as in (v), (!!t, II . lit) is a Banach algebra. It is easy to check that IIrp + slit;::: IIsli t (r, s E !!t), and so p!!t is a closed ideal in !!t. Also, wc have Ap 9:! !!t/p!!t. and so Ap is a commutative, unital Banach algebra which is a Banach algebra extension of A; Ap is called the Arens-Hoffman extension of A by p. In the case where p = Xk+1, we identify Ap with

{~~=o arxr }.

0

Definition 2.1.19 Let A be a normed algebra wzth unzt sphere S, and let a E A. Then a zs a topological divisor of 0 ifinf{lIabll + llball : bE S} = o. Theorem 2.1.20 (Arens) Let A be a unital, commutative Banach algebra, and let a E A. Then there is a Banach algebra extension!! of A such that a E Inv!! if and only if a is not a topological divisor of o. Proof The unit sphere of A is denoted by S. Suppose that a is a topological divisor of O. Then there exists a sequence (b n ) in S such that abn ---+ 0 as n ---+ 00. Assume that!! is a Banach algebra extension of A such that there exists c E !! with ac = e21. Then 1 = IIbnll ::; Ilclillabnll ---+ 0 as n ---+ 00, a contradiction. Now suppose that a is not a topological divisor of 0, say lIabll 2': 1 (b E S). Set B = {b = E'f=oa,Xj E A[[X]] : IIbll = E'f=o lIajll < oo}, as in 2.1.18(v), so that B is a unital Banach algebra which is an extension of A. Set J = (eA - aX)B,

Banach and topological algebras

164

a closed ideal in B, and set 21 = B/J. We claim that the natural embedding c 1---+ c + J, A --+ 21, is an isometry. Indeed, take p = Ej=o ajXj E B. Then (eA - aX)pl/ = I/a - aol/ + Iial - aaoll + ... + Ilan - aan-lll + Ilanll I/all-Ilaol/ + Ilaaoll-llalll + ... + lIaan-III-llanll + Ilaanil ~ lIall because Ilaaj II ~ Ilaj II (j E Z;i) The claim follows from this inequality. Clearly a + J E Inv 21, and so we may regard 21 as the required extension. 0

I/a -

~

Proposition 2.1.21 Let kEN, and let 11·11 be an algebra norm on M k • Then there is an automorphi,sm r : Mk --+ Mk such that (2.1.11) Proof Let the linear subspace

E={(~:~·::~): QI, ... ,QkEC} ak

0 ···0

of Mk have the relative norm 11·11, and let 111·111 denote the corresponding operator norm on £(E). Clearly Mk acts on E by left multiplication, and we have an isomorphism Mk ~ £(E) in this way; certainly IIITIII :::; IITII (T E Mk)' By A.3.5, there is a linear homeomorphism L : (C k ,II'II) --+ (C k ,II'112) with l v'k. Define an isomorphism

IILlillc l1 : :;

r :T

1---+

LoT

0

L-l,

£(E)

--+

£((C k , 11·lb)).

By identifying Mk with £(E) and £((C k , 11·lb)), respectively. we obtain a.n automorphism r : Mk --+ Mk which satisfies (2.1.11). 0 Let AI, ... , An be algebras. The tensor product ®~l Ai was defined in 1.3.11; the projective tensor norm II· 1/11" and the projective tensor product are defined in A.3.65 and A.3.67, respectively. Theorem 2.1.22 Let AI"'" An be normed algebras. The norm

algebra

on the

®~l Ai zs an algebra norm, and (®~=l Ai, 11·1111") is a Banach algebra.

Proof Let a = ®~=l Ai' Then r

Ila bll

1/ ·1I 7r

7r :::;

E;=l al,j I8l ... 18l an,j and b= E~=l bl,k I8l ... I8l bn,k belong to 8

r

8

L L al,Jbl,k I8l ... I8l an,Jbn,k : :; L L j=lk=l

: ; (t

I/al,jll" 'lI an ,jl/)

I/al,jbl,kli" 'lIan,jbn,kll

j=lk=l

(1; I/b1,kll" 'lI bn,kll),

where we are using A.3.66, and so

lIabll 7r :::; Ilall7r IIbll ....

The result follows.

0

Introduction to normed algebms

The Banach algebra

165

(®~=1 Ai, II . lin) is the proJectzve~ensor product of the

normed algebras AI,"" An; it is sometimes denoted by ®nAi' A particularly important example is given in the next definition. Definition 2.1.23 Let A be a normed algebr·a. The enveloping algebra of A is the Banach algebm Examples 2.1.24 Let 8 and T be non-empty sets, and consider the Banach space [1(8) 0£1(T). The map (8 s , 8 t ) f-+ 8(s.t) defines a continuous bilinear map from £1(8) x £l(T) into £1(8 x T), and, by A.3.69, there is a unique continuous linear map W : £1(8) 0£1(T) -+ £1(8 x T) with w(os ® 8t ) = 8(s.t) (s, t E 8). Clearly W is an isometric isomorphism. (i) Let £1(8) have the pointwise product· of 2.1.13(iv). Then the algebra [1(8) 0[1(8) is the Banach space ([1(8 x 8), 11·111) with the pointwise product. (ii) Now suppose that 8 is a unital semigroup: (£1(8), *,11,111) is the semigroup algebra of 8. as in 2.1.13(v). Then the enveloping algebra of £1(8) is the semigroup algebra of 8 x 8, where the semigroup operation on 8 x 8 is given by (s, t)(u. v)

= (su, Vi) (.'I. t. U. v

E

8).

o

Let (A. II· II) be a normed algebra, and let It, ... .In be idealr:; in A. Then It ... In is an ideal in A; we transfer the projective norm from It ® ... ® In to 11 ... In. so that, for a E II ... In, we have

lIall.

~ in!

{t,

Ila,.,II·· ·1Ia"., II ' a

~

t,

a,,'" anJ with a;J E I; } .

(2.1.12) Clearly II· lin is an algebra norm on It· .. In with Iiall :::; Iialin (a EIt ... In); the norm II· lin is again called the projective norm. In particular. we may consider the projective norm II· lin on A2; note that

lIablln :::; lIalln Ilbll

E A 2 , bE A). related to II· lin' First (a

There is a norm which is closely 'Banach' version of the induced product map 7fA of 1.9.19.

(2.1.13)

we introduce a

Definition 2.1.25 Let A be a Banach algebm. Then the contm'ILOllS lmear map 7rA: A0A -+ A surh that 7fA(a®b) = mA(a,b) = ab (a,b E A) is the projective induced product map. and In = ker7fA. The existence of 7fA follows from A.3.69; In is a closed linear subspace of A0A, and I" is a left ideal in the algebra A0Aop, (~alled the projectzve diagonal ideal. The quotient norm on the image 7fA(A0A) e:: (A0A)jln is denoted by 111·lIl n, so that

IIlalil. ~ in!

{t,

lIa, II 11M '

a~ t, a,b,} (a ~A(A®A)) E

.

(2.1.14)

Banach and topologzcal algebras

166 We have

(2.1.15) Suppose that A is unital. Then clearly 7rA is an admissible epimorphism, is a norm on A, and

111" = lin {a ® b - e A

(g

111·11111"

(2.1.16)

ab : a, b E A} .

Definition 2.1.26 A normed algebra A has the S-property [7r-property] if there is a constant C > 0 8uch that Ilal 11" ::; C lIall [lllalll11" ::; C Iiall] (a E A2).

Clearly, if A has the S-property, then A has the 7r-property, and a Banach algebra A has the 7r-property if and only if 7rA(A0A) = A2, so that A has the 7r-property whenever A factors weakly. An example of a commutative, separable Banach algebra which has the 7r-property, but which does not have the S-property, will be given in 4.1.43. Proposition 2.1.27 Let A be a Banach algebra such that A is a finitely generated left ideal in A #. Then A has the S -property. Proof Take all"" an E A[1] with A = L;=l A#aj, and set E = A(n). Define T : (b i , ... , bn ) ~ L~~=l bjaj, E --+ A, and set 1 = T(E) = Lj=l Aaj. Then T E 8(E,A), I has finite codimension in A, and I C A2. By A.3.24, I is closed in A, and so, by the open mapping theorem A.3.23, there exists C > 0 such that, for each eEl, there exist bl, ... ,bn E A with L;=lllbjil ::; Cllell and e = Ej=l bjaj. Thus lIell11" ::; C llell (e E 1). By A.3.42(i), the identity map (A2,1I'11) --+ (A2, 11·1111") is continuous, and so A has the S-property. 0 Proposition 2.1.28 Let (A, 11·11) be a norm,ed algebra. Then there is an algebra norm 111·111 on A such that IIlalil = II a 1111" (a E A2) and lIall ::; IlIalil (a E A). Proof There is a linear subspace E of A such that A = A 2 0 E. For an element a = (b,x) E A2 0 E, define IIlalil = IIbll11" + IIxli. Then it is immediately checked that III . III has the required properties. 0

We conclude this first section with a result that is the foundation stone of automatic continuity theory for Banach algebras. Let A be a Harmed algebra, and let a E A. Since lIam+nll ::; lIa"'lIlIanll for m, n E N, it follows from A.1.26(iii) that lim

n-+oo

lIanll l /n

= inf {lIanlll/n : n E

N} .

(2.1.17)

Theorem 2.1.29 Let A be a Banach algebra. (i) Suppose that a E A with infnEN

lIanll l / n < 1.

Then a E q-Inv A wzth

00

a = - La q

k .

k=l

FUrther, o-(a) C ])(0; IIall)· In the case where A is unital and (1

+ lIall)-1 ::;

lIall < 1,

lI(eA - a)-III ::; (1- 110.11)-1.

we have (2.1.18)

Topologzcal algebms

167

(ii) Let r.p be a chamcter on a Banach algebm A. Then cp is continuous. and In the case wher-e A zs unital, IIcpll = r.p(eA) = 1. (iii) Let a E A. Then Ip(z)1 ::; IIp(a) II (z E O'(a), p E qX]).

IIcpll ::; 1.

Proof (i) Since limn-+oc.lla n Il 1 / n < 1, the series E:=l ak converges, say to -b. Clearly aob = boa = 0, and so b = aq • For Izl > lIall, we have zeA -a E Inv A#, and so v(a) < 14 Thus O'(a) C Jl)l(0; Ilall). The estimate (2.1.18) is immediate. (ii) Assume towards a contradiction that there exists a E A with lIall < 1 and Ir.p(a) I 2: 1, and set b = a/cp(a). Then IIbll < 1, and so, by (i), there exists c E A with b + c = be. But cp(b) = 1, and so 1 + cp(e) = r.p(c), a contradiction. Thus IIcpll ::; 1. If A is unital, then r.p(eA) = 1 = IlcAII, and so Ilcpll = 1. (iii) This follows from 1.6.11(i) and (i). 0 Notes 2.1.30 The elementary theory of Banach algebras can be found in many texts; for example, (Bonsall and Duncan 1973), (Gel'fand et al. 1964), (Helemskii 1993), (Naimark 1964). (Palmer 1994), (Rickart 1960), (Rudin 1973), and (Zelazko 1973). Questions of automatic continuity were stressed at It very early stage in the history of the subject; for example, Rickart focused on whether a semisimple Banach algebra has a unique complete norm in (1960, II, §5). An account of the historical origins of Banach algebra theory is given in (Palmer 1994). The semigroup algebra 1 (8) (for 8 abelian) is discussed in (Hewitt and Zuck('rmann 1956) and (Barnes and Duncan 1975). Propositions 2.1.14. (i) and (ii), are from (Meyer 1995). ThE' Weyl algebra is the algebra of polynomials in two variables X, Y such that XY - YX = 1; by 2.1.14(iii), the Weyl algebra is not normable. Proposition 2.1.15 is from (Dales 1981a). where it is also proved that each commutative, radical algebra with a countable basis is normable. The Arens-Hoffman extensions of 2.1.18(vi) were introduced in (Arens and Hoffman 1956). It is shown in (ibid.) that

e


= {(A

X

!I))(O;t):
= O}.

See also (Lindberg 1964) and (Zame 1984). The methods of Arens and Hoffman were developed by Lindberg (1973) to show that each commutative. unital Banach algebra has a Banach algebra extension which is integrally closed. Theorem 2.1.20 is from (Arens 1958c); for more on topological divisors of 0, see (Bonsall and Duncan 1973, §2), (Palmer 1994, §2.5), and (Rickart 1960, I. §5), For a study of tensor products, see (Palmer 1994, §1.1O and 3.2.18). 2.2

TOPOLOGICAL ALGEBRAS

Although we shall concentrate throughout this book on Banach algebras, we shall also consider more general topological algebras. We shall discuss which algebras are topological algebras with respect to certain types of topology, and we shall study, to some degree, automatic continuity theory for linear maps from topological algebras. A certain amount of the theory of topological algebras will also be required for our main study of Banach algebras. In particular, we shall define {F)-algebras, pliable algebras, (Q)-algebras, topologically simple algebras, and locally multiplicatively convex (LMC) algebras, and we shall discuss elements of finite closed descent and inversion in topological algebras. For the theory of topological linear spaces, see Appendix 3; note that, for us, the topology of a topological linear space is necessarily Hausdorff.

Banach and topolog2cal algebras

168

Definition 2.2.1 Let S be a semigroup, and let T be a topology on S. Then (S, T) is a topological semigroup 2f T 2S Hausdorff and the multiplication map (s, t) ~ st, S x S -+ S, 2S continuous. A topological semigroup (S, T) is an (F)-scmigroup if 1,ts topology 28 given by a complete metric. A topological group is a group G which is also a topological semzgroup and is such that the map s ~ s-l, G -+ G, zs continuous. A topological group (G, T) 1,S an (F)-group 2f its topology zs gwen by a complete metric. Let G be a topological group, and let Nc be the family of open neighbourhoods of the identity ec of G. The topology of a topological group G is translation-invariant, in the sense that, for each s E G, {8 . U : U E N c } is the family of open neighbourhoods of 8. It follows that a group morphism from G into a topological group is continuous if it is continuous at ee.

Lemma 2.2.2 Let G be a topolog2cal group.

(i) For each U ENe, there is a symmetrzc V ENe wzth V[2j cU. (ii) For each non-empty, compact subset K and each open subset U of G with K c U, there exists V ENe with (K . V) U (V . K) cU. (iii) Suppose that K and L are non-empty, compact subsets ofG. Then K . L is compact. 0 The following theorem will be required later; it is a first example of an argument using the Mittag-Leffler theorem, a modest extension of the Baire category theorem which is very powerful in our subject. For the notion of a projective system, see 1.1.5. We first introduce a subset of a topological semi group that will be important later.

Definition 2.2.3 Let S be a topolog2cal semigroup, and let T be a closed subsem2group of S. Then

n8(T) =

{t

c t:""B} .

E T :T

We write n(S) for ns(S) = {s E S : S = s . S}. Clearly n(S) is a semigroup WhE'lleVer it is non-empty; also, if Sl, 82 E S with 8182 E n(S), then 81 E n(s).

Theorem 2.2.4 Let S be an (F)-semigroup.

(i) Suppose that T zs a closed subsemigroup of S. Then ns(T) 2S a Go-subset of S, and ns(T) . Inv S# c ns(T); zfn(S) i- 0, then n(S) 'tS an (F)-semigroup. (ii) Suppose that n(S) i- 0. Then, for each sequence (8 n ) in n(S), we have n{Sl'"

8n .

S: n

E

N} = S.

(iii) Suppose that n(S) = S. Then: n(n(S) = n(s); for each sequence (sn) in n(S), there is a sequence (tn ) in n(S) with tn = sntn+1 (n EN); for each s E n(S), the set n{8 n . n(8) : n E N} is dense in S.

Proof (i) Let d be a complete metric giving the topology of S. To prove that ns(T) is a Go-set, we may suppose that ns(T) =1= 0, say to E ns(T).

Topologzcal algebras

169

For n E Nand s E S, set Un,s = {t E T : d(to, ts) < lin}; for n E N. set Un = USES Un,s; set U = n::='=l Un· Then each Un is open, and so U is a C,,-subset of S. For each t E Os(T), we have to E 08, and so t E Un for each n E N. Thus t E U. Now take t E U. For each n E N, there exists Sn E S with d(to, ts n ) < lin, and so to = lim n -+ oc tS n E 08. Thus T c to . S C t . S, and hence t E Os(T). We have shown that Os(T) = U, and so Os(T) is a Cd-set. Take n E N, t E Un, and u E Inv S#. There exists s E S with d(to. ts) < lin, and now tu E Un because d(to, (tu)(u-ls)) < lin. Since Os(T) = n::='=l Un, it follows that Os(T) . Inv S# c Os(T). That O(S) is an (F)-semigroup when O(S) -::f. 0 is a consequence of A.1.17. (ii) For n E N, set Xn = S, so that Xn is a complete metric space, and define ()n : x f---> SnX, X n+1 ---+ X n , so that ()n is a continuous map such that ()n(X n+1 ) is dense in X n . By the Mittag-Leffler theorem A.1.25, ?fleX) is dense in S, where X = limproj{Xn ; ()n} and ?fl is the coordinate projection. But clearly ?fleX) c n{Sl" 'Sn • S: n EN}. (iii) For s E O(S), we have -s-.=O'"7"":(S=) = S . S = S because n(S) = S, and so S E O(O(S)). Now take (sn) in O(S). As in (ii), there exists (t n ) in limproj{O(S); ()n}, and we have tn = 8 ntn+l (n EN); further, it follows that n{8 n . O(S) : n E N} is dense in O(S), and hence in S, for each S E O(S). 0 Definition 2.2.5 Let A be an algeb'm, and let r be a topology such that (A, r) is a topologicallznear space. Then (A, r) zs a topological algebra zf the semigroup (A, .) zs a topological semzgroup wzth respect to the topology r. The topological algebra (A, r) zs complete [an (F)-algebra], [a locally convex algebra], if, as a topological linear space, (A, r) zs complete [an (F)-space], [a locally convex space]. An algebra zs topologizable if it zs a topologlcal algebra for some topology. Of course, each normed algebra is a topological algebra. Various other examples of topological algebras will be noted in 2.2.46, below, and further (commutative) examples will be given in §4.7. However, we note that there are algebras which are not topologizable; see 2.2.50 at the end of this section for a non-commutative example, and 4.10.32 for a commutative example. We shall always assume that the family of seminorms which defines the topology of a locally convex space is separating and saturated (see Appendix 3). Thus, in the case where A is a locally convex algebra whose topology is defined by a family P of seminorms, multiplication in A is continuous if and only if, for each pEP. there exist q E P and C > 0 such that p(ab) ~ Cq(a)q(b) (a, bE A). Let A be a topological algebra. Then AOP is a topological algebra; if A belongs to a class of these algebras that we have defined, then AOP belongs to the same class. Clearly the closure of a sub algebra of A is a subalgebra, and the closure of an ideal is an ideal. Further, the closure of a commutative subset of A is commutative, so that maximal commutative subalgebras of A are closed. For each subset S of A, the commutant se is closed; in particular, the centre 3(A) of A is closed. The subalgebra pAp is closed in A for each p E J(A). Finally, let I be a closed ideal in A. Then AI I is a topological algebra; it is an (F)-algebra, a locally convex algebra, a normed algebra or a Banach algebra in the case where A has the corresponding property.

Banach and topological algebms

170

The following result is a special case of A.3.39.

Proposition 2.2.6 Let A be an algebm which Z8 a~m an (F)-8pace 8uch that multzplicatwn i8 8epamtely continuou8. Then A is an (F)-algebm. 0 Let A be a topological algebra. Then Ab is a topological algebra for the product. topology, and Ab is complete or locally convex or metrizahle if and only if A has the corresponding property.

Definition 2.2.7 Let A be a topological algebm, and let S be a subset of A. Then: (i) the .subalgebm of A polynomially generated by S is Co[S]: (ii) the subalgebm of A# unit ally polYllomially generated by S is qS]; (iii) the subalgebm of A# rationally generated by S is qS). An element a E A is a polynomial generator of A [rational generator of A#] if Co [a] = A [q a) = A #]. The algebm A is singly generated if it has a polynomial genemtor. Let A be a topological algebra and I be a closed ideal in A. Then, as in 2.2.3. WE' have

OA(I) = {a E I : aA =

I}

and

O(A) = {a E A : aA = A} .

Recall from (1.3.20) that Ia = n{ an A : n E N} for an elE'ment a of an algebra A. Now let A be a topological algebra. and take a E O(A). Then we define

ia=n{an.O(A):nEN}. Clearly ia is a subsemigroup of la, and

(2.2.1)

fa . O(A) cia.

Theorem 2.2.8 Let A be a commutative (F)-algebm, and let a E O(A). (i) Then: Ia is dense in A.; the map Lb I aA is inJectwe for each b E O(A); La I Ia E Inv.e(Ia)' (ii) Suppose that A is non-umtal and that O(A) = A. Then: ia is dense in A; is a cone; La I E Inv.e(L).

(Ia, .)

Ia

Proof (i) That Ia is dense in A is a special case of 2.2.4(ii). Let b E O(A). Suppose that x E aA and that bx = 0, say x = ay, where yEA. Then (abA) y = 0 and so x = 0 because a E abA. Thus Lb I aA is injective. Now suppose that x E Ia. For each n E N, there exists Xn E A with x = anxn . Take n ~ 3. Then a(ax2 - anxn+d = 0, and so aX2 = an Xn+l because La I aA is injective. Thus aX2 E la, and hence x E ala, which shows that La : Ia -+ Ia is a surjection. Since it is an injection, La I Ia E Inv.e(1a). (ii) By 2.2.4(iii), ia is dense in A. Certainly ia is an abelian semigroup, and it is non-unital because A is non-unital. Take b E n(A). By (i), Lb is injective on i a , and so fa is cancellative. As in (i), La is a bijection on i a. 0 Let A be a local algebra which is a topological algebra. Then Inv A is dense in A. For take a E MA· Then a = limn--+oo(eA/n + a), so that a E Inv A.

171

Topological algebms Corollary 2.2.9 Let R be a commutative, mdzcal (F)-algebm.

(i) Let I be a closed ideal in R such that nRC!) =I- 0. Then nRC!) is a dense G8-subset of I. La

(ii) Suppose that a E nCR). E Inv £(1:).

Then: fa zs dense m R; (la, .) is a cone;

I fa

Proof (i) By 2.2.4(i), nRC!) is a G.s-set. Take a E nR(I). Then, by 2.2.4(i), a . Inv R# C nRC!)' and so I = aR = a . Inv R# = nR(I).

(ii) By (i), nCR)

o

= R, and so this follows from 2.2.8(ii).

We now introduce the important notion of an element of finite closed descent. Definition 2.2.10 Let A be a commutative topologzcal algebm, and let a E A. Then a has finite closed descent if there exists n E N such that

(2.2.2) a has finite closed descent k if a has finite closed descent and k is the mmimum element n E N such that (2.2.2) holds. We write 8A(a) = k if a has finite closed descent k and 8A(a) = 00 if a does not have finite closed descent. Note that, if A is unital and a E Inv A, then 8A (a) = 00, and that, if A is radical, then an E a n+1A only if an = O. Proposition 2.2.11 Let A be a commutative (F)-algebm, and let a EA.

(i) Suppose that kEN is such that a k E ak+1A, and set b = a k and I = bA. Then bEl = Ia = anA (n ~ k) and an . Inv A# C n(!) (n ~ k). (ii) Suppose that a has finite closed descent. Then am f/. a m+1A (m EN), and the map p t-> pea) + la, qX] ---+ AIla, is an embedding. (iii) Suppose that 8A (a) = k. Then k = min {n EN: an E a n+1A}. (iv) Suppose that A zs mdical and that a E n(A). Then 8A(a) = 1. Proof (i) For each j E Z+, we have aJ+1 I = ak+H1 A = I. For n ~ k, an E I, and so an . Inv A# C n(I). Also bI = a 2k A = I. By 2.2.8(i), bn I is dense in I, and so Ia = I. (ii) Set k = 8A(a), and assume that am E a m+1A. Then clearly m > k. There exists c E A with am(eA - ac) = OJ we have amA(eA - ac) = 0, and so, by (i), ak(eA - ac) = 0, whence a k E a k+! A, a contradiction. The map p t-> pea) + Ia is a homomorphism. Assume that p E qx]e with pea) E la, say 8p = n. Then pea) E an+! I, a contradiction. Hence the map is an embedding.

n:'=l

(iii) This is immediate from (ii). (iv) We have a E aA = a 2A, and a 8A (a) = 1.

rt a 2A because A is radical. Thus 0

172

Banach and topological algebms

Proposition 2.2.12 Let A be a commutatzve (F)-algebm. (i) Let B be a closed subalgebm of A, and suppose that a E B that ~B(a) < 00. Then ~A(a) ::; 8B (a).

(ii) Let I be a closed ideal in A, and let a

n rad A

and

E I. Then 8r (a) = ~A(a).

Proof (i) Set n = ~B(a). Then an E a n+ 1 Be a n+1 A, and an ¢ a n+1 A because a E radA and an -# o. So 8A(a) ::; n. (ii) Suppose that ~A(a) = kEN. Then a k E ak+l A = ak+2 A c ak+l I and k a ¢ ak+l I, and so ~r(a) ::; k. Suppose that ~I(a) = kEN. Then a k E ak+lI C a k+1A. Assume towards a contradiction that a k E a k+1 A. Then a k E a k+2A C a k+ l I, a contradiction. So ~A(a) ::; k. Thus ~r(a) = ~A(a), as required. 0 Definition 2.2.13 Let A be a topological algebm. Then A is essential if A2 is dense in A. Theorem 2.2.14 Let A be a commutative (F)-algebm.

(i) Suppose that A zs essential. Then n{An : n E N} is dense in A. (ii) Suppose that A = alA + ... + akA for some aI, . . . ,ak E A. Then

n{{L:

a? ... a~k A: r}, ... ,rk E Z+, rl

+ . " + rk =

n} : n EN}

is dense in A. Proof (i) Let ~ be the given metric on A. For each mEN, the Cartesian product space A (m) is also a complete metric space with respect to the product metric, which we denote by ~m' Choose a E A and c > O. We shall make a certain inductive choice involving an increasing sequence (k n ) in N. Having defined (kn ), set Xn = A(k n ) (n EN). Then each (Xn, 8k n ) is a complete metric space. We shall specify for each n E N a complete metric dn on Xn defining the same topology as ~kn' a point Xn E X n , and a map On : X n +l - - Xn such that

(2.2.3) and

(2.2.4) To start the inductive construction, take kl = 1, Xl = a, and d l =~. Now take n E N, and assume that k n E N, Xn E X n , and d n have been specified. Temporarily, we write k for kn' and set Xn = (al,"" ak). Since A2 = A, for each j E Nk there exist mj E Nand aJ,I, ... , aj,mj , bJ,I, ... ,bj,mj E A such that dn(xn,Yn) < c/2n , where

Topologzcal algebms

173

Set kn+l = E;=l mj and Xn+l = (al.l,"" al,ml"'" ak,l,"" ak,mk)' and then define On: X n+1 -+ Xn by the formula

Since addition and multiplication in A are continuous, On is continuous. Set dn+1(x, y) = 8k n + 1 (x, y)

+ dn(On(x), On(Y))

(x, y E Xn+d.

Then d n +l is a complete metric defining the topology of X n +l, and (2.2.3) is satisfied. Since On(xn +1) = Yn, (2.2.4) is also satisfied. Thus the inductive construction continues. It follows from (2.2.3) and (2.2.4) that the hypotheses of the Mittag-Leffler theorem A.1.24 are satisfied, and so there exists (zn) E limproj{Xn ; On} with 8(a, zd :::; E~=l c/2n = c. Clearly Zl E n~=l An. It follows that n~=l An is dense in A, as required. 0 (ii) This is a similar argument. A related result applies just to sepamble (F)-algebras; it follows from the theory of analytic spaces in Appendix 5.

Theorem 2.2.15 Let A be a sepamble (F)-algebm. (i) Suppose that A2 has countable codzmension in A. Then A2 is closed and has finite codimension in A, and there exzsts mEN such that each a E A2 can be written as a = Ej:l ajbj for some al,"" am, bl , ... , bm E A. Further, An is closed and has fimte codzmension in A for each n E No (ii) Suppose that each element of A zs the sum of fimtely many squares of elements of A. Then there exzsts mEN such that each element of A is the sum of m squares. Proof The space A2 is analytic. Thus (i) follows from A.5.19(ii) and A.5.20(i), and (ii) follows by applying A.5.20(i) with 'P : a ~ a 2 , A -+ A. 0 Theorem 2.2.16 (Loy) Let A be a sepamble Banach algebm. (i) Suppose that A2 has countable codimension in A. Then: An is closed and has finite codimenswn in A for n EN; A has the S -property; there exist mEN and M > 0 such that, for each a E A 2 , there exzst al, ... ,am,bl, ... ,bm E A with m m a=

L ajb

J

j=l

and

L Ilaj 1IIIb II :::; 111 Iiall . j

j=l

(ii) Let h, .. . , In be close.d ideals in A. Suppose that h ... In has countable codimenswn in A. Then h··· In is closed and has finite codimension in A, and • 11·11 and 11·11.". are equivalent on h ... In.

Proof (i) By 2.2.15(i), A2 has finite co dimension in A, and so An has finite codimension for each n E N by 1.5.6(vii). By A.5.19(ii), each An is closed in A. The remainder is now a special case of A.5.21. (ii) This also follows from A.5.19(ii) and A.5.21. 0

174

Banach and topological algebms

Proposition 2.2.17 Let A be a sepamble Banach algebm which factors. Then there exists 111 > 0 such that, for each a E A, there exzst b1 , b2, Cl, C2 E A with a = b1c]

+ b2C2

IIb1 11ilclil + IIb21111c211 S M lIall . = {a E A : lIall < 1}; S is an analytic set

and

Proof Set S = U!21, where U in A, and S is absorbing because A factors. Thus S is non-meagre in A, and so, by Pettis's lemma A.5.15, S - S is a neighbourhood of O. The result follows. 0

Recall that Mittag-Leffler sets and Mittag-Leffler algebras were defined in 1.3.40; we give some examples related to these notions. Proposition 2.2.18 Let M be a commutatwe (F)-algebm, and let I be a proper, dense ideal in M such that Mil is a mdical algebm. Then (M11)# is a MittagLeffter algebm. Proof Set A = (MI1)#, so that A is a local algebra and MA = Mil. Let d denote an invariant metric defining the topology of 111. Take (an) in MA. We inductively choose a sequence (xn) in M such that, for each n E N, we have Xn + I = an and max{ d(Xl ... Xn , 0), d(X2 ... Xn , 0), ... , d(Xn-IX n , 0). d(xn, O)} < 1/2n .

For n E Z+, define Yn = Z=~l Xn+l '" Xn+k; each series converges in M, and Yo - Z=~=l Xl' .. Xk = Xl' .. XnYn' Set a = Yo + I. Then a satisfies condition 0 (1.3.21), and so MA is a Mittag-Leffler set. Proposition 2.2.19 Let A be a commutative (F)-algebm such that n(A) =I- 0. Then n(A) is a Mittag-Leffter set. Proof Take (an) in n(A). By A.3.28, applied with Tn : X ~ anx, A -+ A, and Un = an+!, there exists b E A with b - Z=~=l al'" ak+! E al '" a1l +IA (n EN). Set a = b + al. Then a - Z=~:~ al ... ak E al ... an+IA (n EN), and so n(A) is a Mittag-Leffler set. 0 Proposition 2.2.20 Let R be a commutative, mdical (F)-algebm, and let a E R have finite closed descent k. Set A = R# a,nd

60

= {b E la : bak A =

akA} ,

6

= 6 0 u U {an.

Inv A: n E N} .

Then: (i) 6 is a subsemigroup of (A, . ) with a E 6 and 6 . Inv A = 6; (ii) 6 0 =I- 0, and n{al" ·an . 6: n EN} =I- 0 for each sequence (an) in 6: (iii) 6 is a Mittag-Leffler set in A; (iv) if {an}.L = 0 (n EN), then {b}.L = 0 for each b E 6.

Proof Set 1= akA. By 2.2.11(i), we have anA = I (n ~ k), and so

60U

U{an. Inv A: n ~ k} c n(I).

(i) Clearly 6 0 • 6 0 c 6 0 and a E 6. Also a . 6 c 6, and so 6 is a subsemigroup of (A, .), with 6 . Inv A = 6.

Topological algebras

175

(ii) Since fl(1) =I 0, fl(1) = I by 2.2.9(i). Take a sequence (an) in ~, and define bn = a(n-I)k+l ... ank (n EN). Since ~ c aA, we have (b n ) C akA C I, and so (b n ) C fl(1). By 2.2.4(iii), there exists (cn ) C fl(1) with Cn = bnCn+1 (n EN). Clearly (c n ) C la, and so (c n ) C ~o, which is thus not empty. Also E b bn . ~ = n~=1 al ... an . ~, and so n~=1 al ... an . ~ =I 0. (iii) Take (an) in~. By A.3.28, there exists bEl with

CI n::1 l...

n

b - ~)al ... aj ) (aj+l ... aj+k) E al ... an+lI J=l

(n E N) .

Set a = b + L:;=I al ... aj. Then, for each n E N, we have a-

Thus

~

n

k+n

j=l

j=n+l

L al ... aj = L

al'" aJ

+ al ... an+lI C

al ... anA.

is a Mittag-Leffler set.

(iv) Take C E A with bc = O. If b E ~o, then b E fl(1), and so ak+ 1 c C = O. If bEam . Inv A, then amc = 0, and so c = O.

= 0 and 0

We noW' introduce a rather strange condition which will nevertheless be seen later to be important in automatic continuity theory.

Definition 2.2.21 A topological algebra A is pliable if, for each closed

~deal

J

of mfinite codimension in A, there are sequences (an) and (b n ) in A such that (bnal ... an) C J and (bn+1al ... an) C A \ J.

Proposition 2.2.22 Let A be an (F)-algebra. Suppose that, for each closed ideal J of infimte codimension m A, there exu;ts a sequence (bn ) ~n A such that at least one of the followmg holds:

(i) bmbn E J (m

=I n)

and b~

f/- J

(n EN);

(ii) bmbn - bm/\n E J (m, n E N) an1 bn+ 1

-

bn

f/- J

(n EN).

Then A is pliable.

Proof In both cases, choose a sequence (Qk) C 1R+- such that L:~l Qk < 00 and such that (L:~=l Qkbk) is a Cauchy, and hence convergent, sequence in A, say a = L:~l Qkbk· Set f3n = L:~n+l Qk (n EN). In case (i), set an = L:~n+l Qkbk (n EN). Let n E N. For m ?: n + 1, bma n E Qmb~ + J, and, for m ~ n, bma n E J, and so (bnal ... an) C J and bn+1al ... an E Q~+lb~t} + J, so that (bn+lal ... an) C A \ J. In case (ii), set an = a - abn (n EN). For each n E N, abn - bna E J, and so (bnal ... an) C J. Since (bn + 1 - bn)a E f3n(b n+ 1 - b.,) + J, we have b"+lal ... an E f3;:+I(b n +l - bn ) + J, and so (bn+Ial ... an) C A \ J. 0 Let A be a unital (F)-algebra with a continued bisection of the identity {(Pn), (qn)}. Then A is pliable: (qn) satisfies the conditions on (bn ) in 2.2.22(i).

Banach and topologzcal algebras

176

Definition 2.2.23 Let A be a topological algebra. Then A zs the directed union of a family {Aa} of closed subalgebras if: (i) for each Q,{3, there exists'Y wzth Aa U A,6 CAy;

(ii)

Ua

Aa zs dense in A.

Theorem 2.2.24 (Willis) Let A be a unital topological algebra which is a dzrected union of a famzly {Aa} of pliable subalgebras. Suppose that Aa/ J is semisimple for each Q and each closed ideal J with finite codimension in Aa. Then A is pliable. Proof We may suppose that eA E Aa for each Q. Let I be a closed ideal of infinite co dimension in A. Suppose that I n Aa has infinite codimension in Aa for some Q. Then the appropriate sequences exist in A a , and hence in A. Thus we may suppose that In Aa has finite codimension in ACt for each Q. Since In (U Aa) has infinite co dimension in U A a , there is a sequence (O:n) such that dim(Acl!n/(I n Aan )) -+ 00 as n -+ 00 and Ae>n C Aan+l (n EN). Write An for A a ", and set In = I n An; the embeddings are 7rm ,n :

Am/1m

-+

An/In

(m ~ n).

Choose an orthogonal set Sl = {P1,1, ... ,P1,k]} of minimal idempotents in Ad11 with L7~1 P1,j = eA + 11, Then {7r1.2(P1,1), . .. ,7r1.2(P1.k t )} is an orthogonal set of idempotents in Ad12. Each element 7r1,2 (P1,J) in this set can be expressed as a finite sum of an orthogonal set of minimal idempotents in Adh the children of 7r1,2(P1.J)' The family of all children is an orthogonal set S2 = {P2,b ... ,P2.k2} of minimal idempotents in A2/I2 with L7~1 P2,j = eA +h We continue in this way, obtaining an orthogonal set Sn = {Pn,1, ... , Pn,k" } of minimal idempotents in An/In with L~~l Pn,j = eA + In for each n E N. Since An/In is semisimple and dim(An/ In) -+ x, it follows from 1.5.1O(ii) that kn -+ 00 as n -+ 00. The collection of idempotents obtained from a given idempotent in one set Sn by taking its children, their children, etc., are the descendants of the given idempotent; an idempotent P ha..., dn(P) descendants in the nth subsequent generation, and P is said to be infinzte if dn(p) -+ 00 as n -+ 00. Clearly there is an infinite idempotent in Sl, and each infinite idempotent has an infinite child. Thus we may choose (nr) in N such that, for each r E N, we have nr+1 > n r . Snr contains two distinct elements Pr and qr with Pr infinite, and Pr+1 and qr+ 1 are descendants of Pr. Note that two elements which are descendants of distinct idempotents in an earlier generation are orthogonal, and so 7rnr ,n r+l (qr) 1. qr+]' For each r E N, choose br E Anr with br + Ir = qr' Then brb s E I (r #- s) and b~ ¢ I (r EN). By 2.2.22(i), A is pliable. 0 We shall shortly discuss inversion in topological algebras. However, as a preliminary, we note the following result. Let kEN. Then InvMk = {T E Mk : detT

#- O},

Topolog~cal

177

algebras

and so InvMk is an open subset of (M k , 11.11 2 ), and the map T f-t T-l is continuous on (Inv M k , 11·112) because det is a continuous function on (Ck 2 • Also the trace map Tr is continuous on (M k , 11·112). Proposition 2.2.25 Let A be an (F)-algebra, let {7rn : A ---+ (Mk n , 11·11 2 )} be a of contznuous epimorphisms for some sequence (k n ) zn N, and let no EN. Suppose that ker 7rm -=I ker 7rn when m, n E N with m -=I n. Then there exists ao E A such that 7rn (ao) = 0 (n E N no ), 7rn (ao) E InvMkn (n > no), and Tr7rm (ao) -=I Tr7rn (ao) (m,n > no, m -=I n).

fam~ly

Proof Let B = {a E A : 7rn (a) = 0 (n E Nno)}' a closed subalgebra of A. For n > no, set Un = {a E B : 7rn (a) E InvM kn }, and, for m,n > no with m -=I n, set Um,n = {a E B : Tr7rm (a) -=I Tr7rn (a)}. Each Un and each Um,n is an open subset of B, and clearly each Un is dense in B. For m -=I n, the ideal 7r n (B n ker 7rm ) of the simple algebra Mk n is non-zero, and so it follows that 7rn (B n ker 7rm ) = Mk n • Consequently B \ Um •n is a proper linear subspace of B, and hence Um,n is also dense in B. Let U = n{Un : n > no}nn{Um,n: m,n > no, m -=I n}. By the category theorem A.1.21, U is dense in B. Choose any ao E U.

0

Corollary 2.2.26 Let {'Pk : kEN} be a family of d~stinct contznuous characters on an (F)-algebra A. For each no E N, there exists ao E A such that 'Pk(ao) = 0 (k E Nno ), 'Pk(ao) -=I 0 (k > no), and 'Pl(aO) -=I 'Pk(ao)

(j, k ~ no, j -=I k) .

o

For each infinite set S of continuous characters on A, there exists ao E A such that (lo(S) is infinite. In particular, if A is spectrally finite, then q, A is finite. Definition 2.2.27 Let A be a topological algebra. q-InvA zs open in A.

Then A is a Q-algebra zf

For example, suppose that A is a topological algebra which is a division algebra. Then Inv A = A· is open, and so A is a Q-algebra. It is easy to see that A is a Q-algebra if and only if q-InvA has a non-empty interior in A. For suppose that U is a non-empty, open set with U c q-Inv A. For a E A, define 'Pa : b f-t a <> b and 1/Ja : b f-t b <> a, so that 'Pa,1/Ja : A ---+ A are continuous. If bE q-InvA and 'Pa(C) = b, then (b q <> a) <> c = bq <> b = 0, and so c has a left quasi-inverse. Thus 'P;;l (U) is left quasi-invertible. Similarly, 1/J;;1 (U) is right quasi-invertible. Now take a E q-InvA and bE U. Then 'Pboa.(a) = 1/Ja Qob(a) = b,

and so q- Inv A is open because a E 'P~~. (U) n 1/J-;q~b(U) .

Let A be a Banach algebra. Then it follows from 2.1.29(i) that q-Inv A contains the open ball {a E A: lIall < I}, and so q-Inv A has non-empty interior;

178

Banach and

topolog~cal

algebras

by the above remark, this shows that A is a Q-algebra. Thus Banach algebras are the main examples of Q-algebras, but we ::;hall not.e some other examples later (see 4.4.9, for example). Some incomplete normed. algebras are Q-algebras: the algebra (Coo(X), !·!x) is a Q-algebra for each completely regular ::;pace X, but it is only complete for special spaces X. Indeed. it is clear that a normed algebra (A, II·!D is a Q-algebra if and only if :E:'=1 an converges in A for each a E A with lIall < 1. The next result shows that certain ideal::; in a Q-algebra are necessarily closed. As in §1.4, ITA is the structure space of an algebra A; ITA con::;ists of the primitive ideals of A and has the hull-kernel topology; for the definition of J(0), see 1.4.46.

Theorem 2.2.28 Let A be a Q-algebm. (i) Each maximal modular left or right ideal and each maX'tmal modular ideal in A is closed. (ii) Each chamcter on A ~s continuous.

(iii) Each primitive ~deal 2n A is closed. (iv) The (Jacobson) mdical radA and the strong ideals in A. (v) Suppose that I is an ideal zn A. Then ~(I) = ~

md~cal

(7)

9t(A) are closed

in ITA, and

rad (A/1) = {a + I : a E A, a + 1 E rad (A/I)} "J 1/1. (vi) Suppose that I is a dense ideal in A. Then A/ I is a md~cal algebm and J(0) C I.

Proof (i) Let M be a maximal modular left ideal with right modular identity u. Assume towards a contradiction that u E 1M. Since q-Inv A is a neighbourhood of 0, there exists a E M with u - a E q-Inv A, say b + u - a - b(u - a) = O. But then u = a - ba - (b - bu) E M, a contradiction. Hence u rt. M, and so M = M. A ::;imilar argument applies to both right ideals and ideals.

(ii) Let r.p E q, A . Then M'f' = ker r.p is a maximal modular ideal, and hence closed.. Thus r.p is continuous. (iii) By 1.4.34(ii), each primitive ideal is the intersection of the maximal modular left ideals which contain it. Thus each primitive ideal in A is closed. (iv) By 1.5.1, rad A is the intersection of the primitive ideals in A; by 1.5.20, 9t(A) is the intersection of the maximal modular ideals of A. Thus both radA and 9t(A) are closed. in A. (v) By (iii), each primitive ideal in A containing I also contain::; 1. (vi) It is immediate from (v) that A/ I is a radical algebra. Since ~(I) = 0. it follows from 1.4.47 that J(0) C I. 0 It follows from 2.2.28(ii) that the space q, A U {O} for a Q-algebra A can be regarded as a subset of the dual space A'; unless otherwise stated, it is supposed to have the relative weak* topology 0"( A' , A), called the Gel 'fand topology. Let A be a Banach algebra. Then A/radA is a semisimple Banach algebra. For each S C ITA, f(S) is a closed. ideal in A. A closed subset S of IIA is a set

Topological algebras

179

of 8yntheszs if e(s) is the only closed ideal of A with hull equal to S; 8pectral analysts holds for A if each proper closed ideal of A is contain<->d in a primitive ideal. The algebra A is weakly Wzener if spectral analysis holds for A: ill the case where A is not weakly Wiener, there is a proper closed ideal [ in A with ~(I) = 0, and then AI[ is a radical Banach algebra. These notions will be explored for commutativc Banach algebras in §4.1. 'We have explained that we shall be concerned with the question when all homomorphisms from a topological algebra are automatically continuous. \Ve give a condition for this in the cru:;e where the range algebra is finite-dimensional.

Lemma 2.2.29 Let A be a unital Q-algebra, and let AI be a maximal ideal of fimte codzmenswn in A such that M2 has infinite codimenswn. Then there 't~ an zdeal [ in A of finite codimension such that M2 C [ ~ M and 7 = M. Proof Let B = AIM2. as in the remarks after 1.5.19, and let R, E iJ • R'J' V, and IV be as in those remarks. Since 1\,12 has infinite codimension in ,M. the radical R is infinite-dimensional, and so Rl1 is infinite-dimcnsional. Thus V + E11/o.12 Ell is a dense subspace of codimension 1 in the closed subspace Ell JvIE u . The ideal [ = W + M2 has the required properties. 0 Theorem 2.2.30 (Dales and Willis) Let A be a unital Q-algebra. following conditions on A are equivalent:

Then the

(a) each homomorphism from A into a fimte-dzmensional Banach algebra zs continuous;

(b) each zdeal in A of fimte codimenszon is closed; (c) [2 is closed and of fimte codzmension m A for each closed ideal [ of finite codimension in A.

Suppose, further, that A zs either a separable (F) -algebra or 1,S such that AI[ is semi.mnple for each closed ideal [ of finite codimension in A. Then the above conditwns are also equivalent to: (d) M2 is of fimte codimenswn in A for each maximal zdeal M of finite codimension in A.

Proof (b)::::}(a) This follows immediately from A.3.8(i). (a)::::}(b) Assume that (b) fails, and take [to be an ideal of finite codimension such that [ is not closed. The finite-dimen.."ional algebra AI [ is a Banch algebra, and the quotient map from A onto AI[ is discontinuous, a contradiction of (a). (c)::::}(b) Let [ be an ideal of finite codimension, and set R = rad (AI1) and J = 1f- 1 (R), where 1f : A ---> AI [ is the quotient map. Then J is an intersection of maximal left ideal:; in A, and so, by 2.2.28(i), J is closed in A. By (c), J2 n is closed in A for each n E N. Since AI[ is finite-dimensional, R m = 0 eventually, and so J2 n C [ eventually. Thus [ is closed. (c)::::}(d) This holds because each maximal ideal in A is closed. (b)::::}(c) To obtain a contradiction, assume that [ is a closed ideal of finite codimension such that the conclusion of (c) fails. By (b), [2 has infinite codimension in A; we may suppose that I has minimum codimension among the

180

Banach and topolog'tcal algebras

ideals with this property. As above, set J = 7r- 1 (rad (Aj I). Then Jk C [2 eventually; if J2 has finite codimension, then so does J k , and hence so does [2. Thus J = I by the minimality condition on I, and Aj [ is semisimple. By the Wedderburn structure theorem 1.5.9, Aj [ ~ 0~=1 Mnj for some kEN. Assume that k ? 2, and set L1 = 7r- 1 (M nl ) and L2 = 7r- 1 (0;=2 M n , ). Then L1 and L2 are ideals of finite codimension in A. each properly containing [, and L1 n L2 = L1L2 = L2L1 = I. Choose pEA to be such that p + [ is the identity of MIn" and set q = eA - p. Next define K = L~ n L~. Then K = (p + q)3K c L~L~ c K, and so K = L~L~ and 12 = L1L2L1L2 = LiL~ = Li n L~.

If both of L~ and L~ were of finite codimension, then the same would be true of [2, a contradiction. Thus k = 1 and [ is a maximal ideal. By 2.2.29, this is a contradiction of (b). Now suppose that A satisfies either of the additional conditions; we shall prove that (d)===>(c) . Assume that there is a closed ideal [ in A of finite co dimension such that [2 is either not closed or not of finite codimension; again, suppose that [ has minimum co dimension among the ideals with this property, and set J = 7r- 1 (rad(Ajl). If J i=- [, then J2 is of finite codimension, and so, in the case where A is a separable (F)-algebra, Jk is closed and has finite codimension for each kEN by 2.2.15(i), a contradiction because Jk C 12 eventually. Thus under either of the additional hypotheses. J = [ and AjI is semisimple. As before, it follows that I is a maximal ideal in A, and so we have a contradiction of (d). 0 Proposition 2.2.31 Let A be a Banach algebra, let kEN, and let {In : n E N} be a family of distinct ideals such that Ajln ~ Mk (n EN). Then there are a famtly {7r n : n E N} of eptmorphisms from A onto Mk and ao E A such that: (i) In = ker 7rn (n EN);

(ii) l17rn (a)1I2 ~ v'k lIall (n E N, a E ~); (iii) 7rn (ao) E Mk has upper-trzangular form for each n E N; (iv) there exists j E Nk such that {(7r n (ao»j,j : n E N} c C is mfinite. Proof By 2.2.28(i), each ideal In is closed. For n E N, choose an epimorphism ?Tn : A -> Mk with ker ?Tn = In, and let II· lin denote the quotient norm on Mk induced by 7rn . Then, by 2.1.21, there is an automorphism f n : Mk -> Mk with

(2.2.5) By 2.2.25, there exists ao E A such that Tr Vm i=- Tr Vn (m, n E N, m i=- n), where Vn = (fn 0 7Tn )(ao) (n EN). For each n E N, there is a unitary matrix Un in Mk such that the matrix Un VnU~ E Mk has upper-triangular form; define 7rn : a t-+ Un(f n 0 7Tn)(a)U~, A -> M k • Then it is immediate that the family {7r n : n E N} satisfies (i) and (iii), and (2.2.5) implies that (ii) holds as well. Furthermore, since the trace of a matrix remains unchanged under conjugation with unitaries, it follows that Tr7rm (ao) i=- Tr7rn (ao) (m,n E N, m i=- n), and so the set {Tr 7rn (ao) : n E N} is infinite; from this (iv) follows. 0

Topologzcal algebras

181

We now turn to a very mysterious class of topological algebras.

Definition 2.2.32 Let A be a topologzcal algebra. Then A zs topologically simple zf A2 i= 0 and if 0 and A are the only closed idmls in A. For example,
Proposition 2.2.33 (i) Let A be a topologically simple topological algebra. Then A Z8 a przme algebra. (ii) Let A be a topologzcally simple Q-algebra. Then A is either przmitive or radical. Proof (i) Set 1 = {a E A : AaA = a}. Then 1 is a closed ideal in A, and I i= A because A is essential. As in 1.3.52(i), it follows that A is a prime algebra. (ii) Suppose that A has a primitive ideal P, so that P i= A. By 2.2.28(iii), P is a closed ideal in A, and so P = 0 and A is a primitive algebra. If A has no primitive ideals, then A is a radical algebra. 0 Proposition 2.2.34 Let A be a topological algebra, and suppose that I is a non-nilpotent zdeal whzch zs mimmal in the family of non-zero, closed ideals of A. Then I is topologically simple. Suppose, further, that 1 zs semtprzme. Then IT is a closed ideal whzch is a minimal prime ideal in A; also, I ct IT and InI T = o. Proof Set J = {a E I : IaI = a}. Then J is a closed ideal of A with J c I. Suppose that I is not nilpotent. Then 1 3 i= 0 and so J i= I and J = O. Let K be a non-zero, closed ideal in I, and take a E K-. Then a ~ J, and so IaI i= O. Thus IKI i= O. Since IKI is a closed ideal of A with IKI c I, we have IKI = I. But IKI C K. and so K = I. This proves that I is topologically simple. Now suppose that I is scmiprime, so that I is not nilpotent. Set P = IT, a closed ideal in A. Take a, b E A such that aAb C P, and suppose that a ~ P. Then Ia i= 0, and so, by 1.5.25, (Ia)2 i= O. Thus IaA i= 0, and so IaA = I and Ib = (IaA) b = 0, whence b E P. By 1.3.42, P is a prime ideal. Let Q be a prime ideal of A with Q c P. Then I P = 0 c Q. Assume that 1 C Q. Then 12 C 1Q c IP = 0, a contradiction. Thus P C Q, and so P is a minimal prime. Finally, 1 ct P, for otherwise 12 = 0, and now 1 n P is a closed ideal of A with 1 n P S;; I, and so In P = o. 0 Examples of topologically simple (and even simple) primitive Banach algebras will be given in §2.5. We cannot give an example of a topologically simple, radical Banach algebra because no such example is known. The existence of such algebras, particularly in the commutative case, is, in my opinion, the most significant open problem in general Banach algebra theory; we raise it as a formal question.

Banach aTld topological algebras

182

Question 2.2.A Is there a topologically simple, radical Banach algebra? Is ther-e a topologically 8imple, commutative Banach al,qebra which is not isomorphic to C? Is there a singly-generated. topologically simple. commutative Banach algebra 'u'!nch ~s not zsomorphzc to C '?

The above questions go back to the beginnings of Banach algebra theory. Implications of a negative answer to the questions will be given in §5.3. My guess is that there is an infinite-dimensional, singly-generated, topologically simple, comulIltatiw Banach algebra. However, its com,truction will surely be a formidable task. To indicate this, assume that A = Coral is such an algebra. Then the operator La on A is quasi-nilpotent and has no invariant subspace. (For each infinitedimensional, topologically simple, commutative Banach algebra A. the operator La on A is quasi-nilpotent and has no hyper-invariant subspace for each a E A.) Although non-zero, quasi-nilpotent operators with no invariant subspaces are now known, their construction is a major undertaking; the construction of a topologically simple, singly-generated, commutative Banach algebra will surely be even more complicated. It is also an open question whether there is a simple, radical Banach algebra; by 1.3.52(ii), such an algebra is necessarily non-commutative. Finally, we note that it is not known whether every commutative Banach algebra A other than C contains a closed subalgebra different from both 0 and A. Definition 2.2.35 Let S be a unital topological semigroup. Then inversion is continuous for S if the map s t-t s-l, Inv S - t Inv S, is contmuous. Let A be a unital topologzcal algebra. Then inversion is continuous for A if mversion is continuous for the topological semigroup (A, . ). Proposition 2.2.36 Let A be a unital algebra wzth a non-zero algebra semmorm p, andleta.bElnvAwithp(b-a):S;1/2p(a- 1 ). Then p(b- 1

_

a-I) :S; 2p(a- 1 )2p(b - a) .

Proof Since p is non-zero, p(eA) inequality

f. 0, and so p(a- 1 ) f. O. From the elementary (2.2.6)

I

we obtain Ip(b- 1 ) - p(a- I ) :S; p(b- I )/2 by using the hypothesis. and hence we see that p(b- I ) :S; 2p(a- I ). We obtain the result by again using (2.2.6). 0 It follows that inversion is continuous for each normed algebra.

Lemma 2.2.37 Let G be a group which is also a complete metric space. Suppose that, for each a, bEG, the map x t-t axb, G - t G, is continuous. Then the map x t-t x-I, G - t G, is continuous, and G is a topological group. Proof Let a complete metric on G be d, and let e be the identity of G. Suppose that Xn - t e in G. We claim that there exist a E G and a strictly increasing sequence (nk) in N such that ax~; - t a. To see this, set Uj = {x E G: inf{d(xx;;l,x): n ~ j} < Iii}

(j EN).

Topologzcal algebras

183

Then each Uj is open in G. Fix.i E N, anrllet V be a ball in G with centre b and radius e > 0, say. Then there exists n ;:::: j such that d( b, bx n ) < min {I Ii, e}, and the element bX n of G belongs to Uj n V. This shows that Uj is dense in G. Uj f= 0, say a E U j • Inductively choose (nk) eN so that By A.I.21, nk+1 > nk and d(ax;;;, a) < 11k for each kEN. Then ax;;; ~ a, establishing the claim. Let Yn ~ bin G. Then, by hypothesis, b-IYn ~ e, and so, by the claim, there exist a E G and (nk) in N such that aY;;kIb ~ a. But then, again by hypothesis, y;;l ~ b- I . This implies that the map x 1---+ x-I is continuous at b. 0

n;:1

n;:1

Theorem 2.2.38 (Banach) Let S be a unital (F)-semzgroup. Then inversion zs continuous for S if and only if Inv S is a Go-set in S. Proof Let

f : Inv S

~

S be the inversion map, and, as in (A.I.l), write

Of

= {a E Inv S

: oscf (a)

= o} .

so that Of is a Go-set in S. We first claim that Of C Inv S. For, if a E Of, then a = limn--->oo an for some sequence (an) in InvS. Since oscf(a) = 0, the sequence (a;;l) is Cauchy in S, and so a;;1 ~ b, say. It follows that ana;;1 ~ ab, and so a = b- I E Inv Sand Of C Inv S, as required. Suppose that inversion is continuous for S. Then Inv S C Of, and so necessarily Inv S = Of. Thus Inv S is a Go-set in S. Conversely, suppose that Inv S is a Go-set in S. By A.I.I7, there is a complete metric defining the relative topology on Inv S. and so Inv S satisfies the hypotheses on the group G in 2.2.37. Thus inversion is continuous for S. 0 Corollary 2.2.39 Let A be a unital (F)-algebra which is a Q-algebra or, in particular, a ditnszon algebra. Then inversion is contmuous for A. 0

We shall see in 4.10.29 that there is a commutative, unital (F)-algebra for which inversion is not continuous, and in 4.10.30 that there is a commutative, unital (F)-algebra for which Inv A is a Go-set which is not open. The spectrum a(a), the resolvent set pea), the spectral radius v(a), and the resolvent function Ra of an element a of an algebra A were defined in 1.5.27. For example, let 0 be a non-empty, compact space, and let A = C(O). Then a(l) = f(O) and v(l) = If In for f E C(O). Lemma 2.2.40 Let A be a topological algebra with identity, and let a E A. (i) Suppose that A is a Q-algebra. Then pea) zs open and a(a) zs compact. (ii) Suppose that inver'sion is continuous for A and that pea) is open and unbounded. Then Ra is an analytic A-valued function on pea), and Ra(z) ~ 0 as z ~ 00 in pea).

Proof (i) Define j : z 1---+ zeA - a, C ~ A. Then j is continuous, and so the set pea) = j-I(InvA) is open and a(a) is closed in C. Since eA - ajz E InvA for Izi sufficiently large, u(a) is bounded, and hence compact.

Banach and topological algebras

184

(ii) Take Z E p(a). It follows from (1.5.4) and the fact that inversion is continuous for A that

(Ra(w) - Ra(z))/(w - z)

-->

-Ra(z)2

as w --;

Z

in p(a) \ {z},

and so Ra is analytic at z. Since Ra(z) = z-l(eA - a/z)-1 (z E C e ). we have Ra(z) -~ 0 as Z --> 00 in p(a). 0

Theorem 2.2.41 Let A be a locally convex algebra with identzty Jor which inversion Z8 continuous. Then 0"(0.) zs non-empty Jor each a E A. Proof Assume towards a contradiction that there exists a E A with 0-( a) = 0. Since A is a locally convex space, there exists ..\ E A' with (a -1, ..\) =I- O. Set

J(z) = (Ra(z),..\)

(z E

q.

By 2.2.40(ii), J is an entire function with J(z) --> 0 as z --> 00, and so J = 0 by Liouville's theorem. In particular, J(O) = _(a- 1 ,..\) = 0, a contradiction. Thus O"(a) =I- 0 (0. E A). 0 A further theme of this book is that topological algebras whose topologies satisfy certain algebraic conditions must have a special form. We now give our first example of a result of this type: it is a general version of the celebrated GelJand-Mazur theorem.

Theorem 2.2.42 Let A be a dwzs'tOn algebra. Suppose that either: (i) A zs a locally convex (F)-algebra; or (ii) A is serninorrnable. Then A = CeA. Proof Suppose that A satif:,fies (i). By 2.2.39, inversion is continuous for A. Suppose that A satisfies (ii), and let p be a non-zero algebra seminorm on A. ThC'n kerp = {O}, so that (A,p) is a normed algebra, and inversion is continuous for (A,p) by 2.2.36. In either case, 2.2.41 shows that O"(a) =I- 0 (a E A), and so the result follows from 1.5.30. 0 Corollary 2.2.43 Let A be a topologically simple, commutative, locally convex (F)-algebra with A '1- C. Then A zs radical and an integml domain. Proof By 1.4.36. a primitive commutative algebra is a field, and the only field which is a locally convex (F)-algebra is C. Thus A is radical; by 2.2.33(i), A is an integral domain. 0 Consider the following three fields, which were all introduced in §1.6: the field .c = C((X)) of Laurent series; the field C(X) ofrational functions; the field C (X) of meromorphic functions at O. These fields are not isomorphic to C, and so the theorem shows that there is no topology with respect to which they are locally convex (F)-algebras, and that they are not seminormable. In §4.1O. we shall give various examples of commutative topological algebras which show that the conditions imposed in the above theorem cannot be removed completely.

Topological algebras

185

We now introduce a special class of topological algebras. Definition 2.2.44 Let A be a topological algebra. Then A 2S a locally multiplicatively convex algebra or an LMC algebra if there is a base of nezghbourhoods of the orzgin cons2sting of sets which are absolutely convex and multiplicative. A complete, metrzzable LMC algebra zs a Frechet algebra. Note that our terminology is such that a Frechet space which is a topological algebra is not necessarily a Frechet algebra. Clearly a subalgebra of an LMC algebra is LMC, and the quotient of an LMC algebra by a closed ideal is also LMC; closed subalgebras of Frechet algebras and quotients of Frechet algebras by closed ideals are Frechet algebras. Let A be an algebra. The Minkowski functional of an absolutely convex, multiplicative neighbourhood of 0 in A is clearly an algebra semi norm on A, and so the topology of an LMC algebra can be specified by a separating, saturated family of algebra seminorms. In particular, the Gel'fand-Mazur theorem 2.2.42(ii) shows that each LMC algebra which is a division algebra is isomorphic to and A# have the corresponding propl'rty. An LMC algebra A is umtal if A has an identity CA and p(eA) = 1 for each p in some family of algebra seminorms defining the topology of A. It follows easily from 2.1.9 that each LMC algebra with identity may be supposed to be unital.

Examples 2.2.46 (i) Let In = C[[X1 , ... , Xnll be the algebra of formal power series in n indeterminates, as in §1.6. For k E Z+, define Pk(a)

= L{/ar /: r

E z+n, /r/::; k}

(a

=

LarX r E

In) .

Then each Pk is an algebra semi norm on In, and In is a Frechet algebra with respect to the sequence (Pk : k E Z+) of seminorms. The specified topology on In is the topology of coordinatew2se convergence; it will be denoted by Te. The unique maximal ideal of the local algebra J .. is clearly closed in (In, Tc), and so (In, Tc) is a Q-algebra. The algebra (J711 Tc) is unit ally polynomially generated by Xl, ... ,Xn. This algebra will be studied further in §4.6. (ii) Let E be a Banach space. For each pEN, we write projective tensor product

-p

®

-p

®

E or the injective tensor product

-0

E for either the

-P

®

E of p copies

-p

of E; also, we set ® E =
E)

-p+q

®

E)

E; we have

lIu®vllp+q=lIullpllvllq

(UE®PE, vE®qE).

(2.2.7)

186

Banach and topological algebras

We define a unital algebra

Q9E=

II {®PE:PEZ+}={U=(Up):UpE®PE

(PEZ+)},

where the product is formally defined by (1.3.13), so that

L

(up) ® (v q ) = (

up ® Vq : r E Z+) .

(2.2.8)

p+q=r

= L:Z=o lIupllp (u = (up) E ®E). Then each Pk is an algebra seminorm on ®E, and ®E is a unital Frechet algebra with respect to For k E Z+, define Pk(U)

the sequence (Pk : k E Z+) of seminorms. The two separate algebras are called the pro)ectwe tensor algebra and the m)ective tensor algebra of E, respectively;

®E,

they are denoted by @E and respectively. They both contain ®E as a dense subalgebra, and the sequence (I, 0, 0, ... ) is the identity of each algebra. Let pEN. For each cr E 6 p , the map (j of (1.3.5) is an isometry on the normed space (®PE, An element

U

E

II· lip)' and so it extends to an isometry,

® pE

is symmetric if (j(u) = -p

U

(cr E 6

p );

-p

also (j, on

®

E.

we write \ / E for

the closed linear subspace of ® E consisting of the symmetric elements. The symmetrizing map of (1.3.6) extends to a bounded linear projection of norm 1 on ® pE with range pE; it is also denoted by Sp. Essentially as above, we define

V

-

-

so that VE is a closed linear subspace of ®E. As in §1.3, VE is a commutative, unital Frechet algebra with respect to the product defined by (1.3.15). The two separate algebras are called the projective and injective symmetric algebras of E, respectively; they are denoted by E and E, respectively. They both contain VE as a dense subalgebra. There are some important Banach Rubalgebras of the abov~ examples. Let W be a ~eight on Z+, and let A be one of the algebras ®E or VE. Define ®wE and VwE, respectively, as

V

{ U

= (up) E A : Ilull =

V

~ lIupllpw(p) <

00 }

;

we obtain two unital Banach algebras, with VwE being commutative. These algebras are called wezghted tensor algebras and weighted symmetric algebras, respectively. (iii) Let X be a completely regular topological space. For each non-empty, compact subset K of X, define PK(f) = IflK (f E C(X». Then PK is an algebra seminorm on C(X). The family {PK : K E Kx} of seminorms defines the compact-open topology on C(X), and, with respect to this topology, C(X) is an LMC algebra. In general, C(X) is neither complete nor metrizable.

187

Topologzcal algebras

For each non-empty, open subset U of en, the topology of C(U) can be given explicitly as follows. Let (Km) be a compact exhaustion of U (see A.1.9(v)), and set Pm(f) = IflK (f E C(U)). Then (Pm) is a ~equence of algebra seminorms which defines th;~ompact-open topology, and C(U) is a Frechet algebra. The algebra C(JR) b a unital Frechet algebra (with respect to the compactopen topology), and Coo(JR) is a proper, dense ideal, and so C(JR) is not a Qalgebra. (iv) Let U be a non-empty, open subset of en. As in Appendix 2, we denote by O(U) the algebra of analytic functions on U. We note in Appendix 2 that O(U) is a clo~ed subalgebra of C(U), and so O(U) is a Frechet algebra. In particular, O( en) is the algebra of all entire functions in n variables; this algebra is unit ally polynomially generated by the coordinate functionals Zl,' .. , Zn. Further properties of the Frechet algebras O(U) will be given in §4.1O. Let K be a non-empty, compact subset of en. We denote by OK the algebra of germs of analytic functions on K, with the inductive compact-open topology. Then OK is a complete LMC algebra; OK is ea~ily seen to be a Q-algebra. D

'Ve now present a structure theorem for LMC algebras. This result allows many questions about these algebras to be reduced to analogous questions about Banach algebras. Theorem 2.2.47 (Michael) Let A be a complete LMC algebra. Then there ex~sts a projective system {Ap; 7r pq; P} of Banach algebras Ap and norm-decreasing homomorph~sms 1ipq such that: (i) A is topolog~cally isomorphic to.li = limproj{Ap;7rpq; P}; (ii) 7rp (.A:) is dense in Ap for each p, where 7rp is the coordinate projection.

Proof Let P be a (separating, saturated) family of algebra seminorms defining the topology of A. Then P is a directed set for the relation: p ::; q if and only if pea) ::; q(a) (a E A). For each pEP, let Ip = ker p, let Qp : A ....... Allp be the quotient map, let IIQp(a)llp = pea) (a E A), and let Ap be the completion of the normed algebra (Allp, II· lip)' so that (Ap, II· lip) is a Banach algebra. Suppose that p ::; q in P. Then Iq C Ip. and so

7rpq : Qq(a) ....... Qp(a) ,

Allq

---+

Allp,

is a norm-decreasing homomorphism. The map 1ipq has an extension to a normdecreasing homomorphi~m, also denoted by 7rpq , from Aq into A p, and then {Ap; 1ipq ; P} is a projective system of Banach algebras and norm-decreasing homomorphisms because 7rpq 0 7rqr = 7rpr (p::; q ::; r). Let

.Ii = limproj{Ap; 7rpq ; P} so that

=

{(a p) E I1

.Ii is a closed subalgebra of I1 Ap. 7r : a

Then 7r(A)

C

..4,

f--->

and 7r : A

PE 'P

Ap : 7rpq (aq) = ap (p::; q)} ,

Define

(Qp(a) : pEP), ---+

E

A

---+

I1PE 'P Ap.

7r(A) is a topological isomorphism.

Banach and topological algebms

188

Let (7r(a ll )) be a net in 7r(A) with 7r(a ll ) --+ x in nAp. Then (all) is a Cauchy net in A. Since A is complete, (all) is convergent in A, say all --+ a. Clearly x = 7r(a), and so 7r(A) is closed in A. Take (ap : pEP) E A, and let U be a basic neighbourhood of (ap : pEP) in nAp, say U = Up, where Up = Ap for p E P\{Pl,'" ,Pn}. Set q = PlV", vpn and V = 7rp;~q(Upj)' Then V is a neighbourhood of aq in A q. Choose a E A with Qq(a) E V. Then 7r(a) E unA, and so 7r(A) is dense in A. Thus 7r(A) = A, and 7r : A --+ A is a topological isomorphism. 0 Clearly 7rp(A) is dense in Ap for each P because 7rp(A) = Allp.

n;=l

n

Conversely, each projective limit of a family of Banach algebras and continuous homomorphisms is a complete LMC algebra. We shall systematically identify a complete LMC algebra A with an algebra limprojAp = limproj{Ap; 7rpq ; P}, where {Ap; 7rpq ; P} is a projective system of Banach algebras and norm-decreasing homomorphisms satisfying condition 2.2.47(ii). For pEP, set pea) = II7rp(a)1I (a E A). Then {p: pEP} is a family of seminorms defining the topology of A. Note that (2.2.9) Suppose that A is a Fnkhet algebra. Then the topology of A is determined by a countable family of algebra seminorms. Thus we have the following corollary. Corollary 2.2.48 Let A be a Prechet algebm. Then A is topologically isomorphzc to the projective limit of a projective sequence of Banach algebms and norm-decreasing homomorphisms. 0

Thus, in the case where A is a Frechet algebra, we shall identify A with limprojAn = limproj{An; 7rmn }, where {An; 7rmn } is a projective sequence of Banach algebras and norm-decreasing homomorphisms. Let n E N. We write 7rn : A --+ An for the coordinate projection, so that 7rn (A) = An, and we set Pn(a) = I!7rn (a)I! (a E A). Hence an abstract Frechet algebra A arrives equipped with a sequence (Pn : n E N) of algebra seminorms such that (2.2.10)

Proposition 2.2.49 Let A = limproj Ap be a complete LMC algebm. Then:

(i) the algebm A has an identity if and only if each Ap has an identity; (ii) an element a E A zs invertible if and only if 7rp(a) zs invertible in Ap for each p. Proof (i) Suppose that A has an identity e. Then 7rp(e) is the identity of Ap. Conversely, let ep be the identity of Ap for each p. For q ;::: p, we have

7rpq(x)7rpq(eq) = 7rpq(eq)7rpq(x) = 7rpq (x)

(x

E

Aq),

and so, by (2.2.9), Y7rpq (eq) = 7rpq (e q)y = y (y E Ap). Thus 7rpq (e q) (e p) E limproj Ap is the required identity of A. (ii) Suppose that a E InvA. Then 7rp(a) E InvAp for each p.

= ep, and

Topological algebras

189

Conversely, suppose that 11"p (a) E Inv Ap for each p, say bp = 11"P (a) -1. Since the inverse is unique, 1I"pq(bq) = bp (q 2: p), and so b = (bp) E limproj Ap. Clearly b = a-I, and so a E Inv A. 0 We conclude this section by exhibiting an algebra which is not topologizable. Indeed, the following theorem shows that C(E) is not topologizable for any infillite-dimensionallinear space E. Theorem 2.2.50 (Zelazko) Let E be an injinite-dzmenswnallinear space. Then the algebra E ® EX = FC(E) zs not topologzzable. Proof Assume towards a contradiction that E 0 EX is a topological algebra, and let V be a base of neighbourhoods of 0 in E 0 EX consisting of balanced, absorbing sets. Take Xo E E and AO E EX with (xo, AO) = 1. Then there exists V E V with Xu 0 AO ~ V, and there exists W E V with W . We V. Define K = {x E E: x0Ao E W}, so that K is a balanced, absorbing subset of E. For each A E EX, there exists rnA > 0 with Xo 0 A E rnA W, and then

(x, A) Xo ® AO = (xo 0 A) (x 0 AO)

E rnA W

.W

C rnA V

(x

E

K) ,

so that I(X.A)I :::; rnA (x E (K), A E EX). Let PK be the Minkowski functional of the absolutely convex, absorbing set (K). Then (2.2.11) For each x E E·, there exists A E EX with (x, A) #- 0, and so it follows from (2.2.11) that PK is a norm on E and that all linear functionals on (E,PK) are continuous. But this contradicts the fact that E is infinite-dimensional. 0 Notes 2.2.51 For an introduction to the theory oftopological groups, see (Hewitt and Ross 1979, Chapter Two). An early source on topological algebras is (Zelazko 1965); more recent texts include (Mallios 1986) and (Helemskii 1993). Some authors use the term Prechet algebra for our '(F)-algebra', and Polish authors refer to a locally convex (F)-algebra as aBo-algebra. Proposition 2.2.6 is from (Arens 1947). An algebra which is a topological linear space such that multiplication is separately continuous is a sem~-topological algebra. A semi-topological algebra which is a complete locally convex space is not necessarily a topological algebra. For let A = £ 00 with the topology given by the scminorms PA : (Qk) 1--+ E JAkQkJ for A = (Ak) E £1. Then A is a semi-topological algebra and a complete locally convex space. Assume that multiplication is continuous. Then, for each A E £ 1, there exists 11- E £ 1 and C > 0 with (2.2.12) and n = {3 = (Ilk." : kEN). Then it follows from (2.2.12) that a contradiction of the fact that 11- E t 1. The important notion of an element of finite closed descent is due to Allan (1972, 1973); we have modified the definition slightly. The significance of this property in the construction of discontinuous homomorphisms will become apparent in §5.7. Theorem 2.2.16 is from (Loy 1976). It is striking that the hypothesis of separability occurs in 2.2.15 and 2.2.16. Clearly the whole approach to these results through the theory of analytic spaces requires this hypothesis, but it is not obvious that it is required for the final results. In fact, an ingenious construction of Dixon (1977b) shows that But take>.

1/n2 S C

= (1/k 2 )

JI1-nJ 2 (n EN),

190

Banach and topological algebras

these results may fail if the condition that the algebra be separable is dropped. The main theorem that Dixon proves is the following.

Theorem Let X be an infinite-dimensional Banach space, and let Y be any linear subspace of X. Then there exists a Banach algebm A containing a copy of X such that A2 n X = Y and such that A2 + X = A. 0 By using modifications of the construction that gives the Banach algebra A, Dixon obtained the following results. (i) Let kEN. Then there is a Banach algebra A such that A2 ha..'! codimension k in A, but such that A2 is not closed in A. (ii) Let kEN. Then there is a Banach algebra A such that A2, ... , Ak are closed, but Ak+l = Ak+2 = .. , is of finite codimension and not closed in A. (iii) There is a Banach algebra A with A2 = A, so that A factors weakly and A has the 7r-property, but such that A does not have the S-property. Further, there is no finite upper bound to the number of summands that are required to express an element of A2 as a sum of products; in particular, A does not factor. These examples are neither commutative nor semisimple, and it is not known what happens if either of these conditions be imposed. The examples also show that, for a non-separable Banach algebra A, A2 need not be a Borel set. Whether or not A2 is necessarily a Borel set for a separable Banach algebra A is an open question. The term 'pliable' in 2.2.21 is new; 2.2.24 was shown to me by G. A. Willis. The terminology for Q-algebras in 2.2.27 is due to Kaplansky (1947); it seems useful to introduce this generalization of Banach algebras to stress just which property is important. Let p be an algebra seminorm on an algebra A. Then (A,p) is a Qalgebra if and only if p is a spectral seminorm in the sense of (Palmer 1994, 2.2.4). The term 'Q-algebra' is also used in a different sense in (Bonsall and Duncan 1973, §50) to denote a commutative Banach algebra which is a quotient of a uniform algebra by a closed ideal; see (Diestel et al. 1995, Chapter 18) for an account of these algebras. Let A be a normed algebra. Then an example to be given in §4.7 will show that radA is not necessarily closed in A. This example is taken from (Dixon 1997), where a theory of radicals in various topological algebras is developed. Theorem 2.2.30 is from (Dales and Willis 1983). Question 2.2.A on the existence of topologically simple, commutative Banach algebras goes back to the beginning of the subject. Although this question is open for Banach algebras, it has been proved by Atzmon (1984) that there are topologically simple, commutative, locally convex (F)-algebras. An example of an operator on a Banach space which is quasi-nilpotent and has no invariant subspace is given in (Read 1997); perhaps related examples could lead to the construction of a singly-generated, topologically simple, commutative Banach algebra. Theorem 2.2.38 is due to Banach (see (Zelazko 1965. Theorem 7.4»; the proof of 2.2.37 was shown to me by J. R. Esterle. The argument in 2.2.41 goes back at least as far as (A. E. Taylor 1938) and (Gel'fand 1941a). The Gel'fand-Mazur theorem 2.2.42 for normed division algebras was announced in (Mazur 1938). The first published proof is in (Gel'fand 1941a); Mazur's original proof (for real normed division algebras) is given in (Zelazko 1973, 5.5). See also (Bonsall and Duncan 1973, §14). The algebras of 2.2.46(ii) are discussed in (Leptin 1969) and (Dales and McClure 1977b). The seminal memoir on LMC algebras is that of Michael (1952), and our initial account still follows this source. Michael calls our 'Frechet algebras' oF-algebras. The structure theorem 2.2.47 is from (ibid.); the application in 2.2.49(ii) is Arens's invertibility criterion. The final result 2.2.50 is taken from (Zelazko 1996).

Spectra and Gel 'fand theory 2.3

191

SPECTRA AND GEL'FAND THEORY

In this section, we shall study the character space CPA of a Banach algebra A, the spectrum a(a) of an element a in A, and Gel'fand theory for A. In particular, we shall prove the fundamental theorem of Banach algebras, that the spectrum of each element in A is compact and non-empty, and prove the spectral radius formula. We shall then give a medley of applications of this formula. We shall conclude the section with some weak automatic continuity results. Recall that throughout CPA U {O} is taken to have the Gel'fand topology a(A. A') unless we state otherwise. Let A be a unital Banach algebra. We define

KA

=

{A

E

A' :

11>'11

=

(eA, >.) = I};

(2.3.1)

we have shown in 2.1.29(ii) that CPA c KA. The set KA is non-empty and convex, and it is closed, and hence compact, in (Ahl' a(A', A». By the KreIn-Mil'man theorem A.3.30(i), KA = (exKA). Let a E A. Then V(a) = {(a, >.) : >. E KA}

is the numerical range of a; clearly, ~T(a) is a non-empty, compact, convex subset ofC, and V(a) C lIJ)(O; Ilall). Let z E a(a). Then (z+1] E a((a+17eA) ((,1] E q, and so, by 2.1.29(i), I(z + 1]1 ~ II(a + 1]eAIi. Thus

>'0 : (a + 1]eA 1-+ (z + 1], lin{ a, eA} -+ C, is a well-defined, continuous linear functional with (a, >'0) = z, (eA' >'0) = 1, and 11>'011 = 1. Take>. E A'to be a norm-preserving extension of >'0. Then>. E KA and (a, >.) = z. This shows that a(a) C V(a). Theorem 2.3.1-f.~ _~ be a <;orrl:mutatwe Banach alg~bra. Then the maxzmal modular ideals of A are the kernels of the characters of A. Suppose that A is unital. Then CPA 1:- 0. Proof By 1.3.37(i), it suffices to show that each maximal modular ideal of A has codimension one. Let M be such an ideal. By 2.2.28(i), M is closed in A, and so AIM is a Banach algebra. Since AIM is a field, the Gel'fand-Mazur theorem 2.2.42(ii) shows that AIM ~ C, and so M indeed has codimension one in A. A unital algebra contains a maximal ideal, and so CPA 1:- 0. 0

Let A be a commutative Banach algebra. By the above theorem the maximal modular ideals of A have the form Mcp for some cp E CPA; for this reason, the character space CPA of A is sometimes called the maximal ideal space of A. It now follows from 1.5.2(ii) that radA

= n{Mcp : cp E CPA}.

We explained earlier that one theme of this book is to decide which algebras are seminormable. We now have a further condition for seminormability; an example of a commutative integral domain failing the test will be given in 4.10.27(iii) .

Banach and topological algebras

192

Theorem 2.3.2 Let A be a commutative, unital algebra. normable zf and only if CPA =1= 0.

Then A is serm-

Proof Suppose that cp E CPA. Then a I-t Icp(a)l, A ~ jR+, is a non-zero algebra seminorm on A. Conversely, suppose that A is seminormable. By 2.1.2, there is a non-zero homomorphism 8: A ~ B for a Banach algebra B, and we may suppose that B is commutative and that Band 8 are unital. Since cP B =1= 0, there exists 'Ij; E cP B, and then 'Ij; 0 () E CPA. 0

We now give two basic automatic continuity results for commutative Banach algebras; they follow easily from the characterization of the radical that has just been given. Theorem 2.3.3 (Silov) Let A be a Banach algebra, let B be a commutative, semiszmple Banach algebra, and let 8 : A ~ B be a homomorphism. Then B zs automatzcally continuous. Proof Suppose that an ~ 0 in A and 8(a n ) ~ bin B, and take cp E CPR. Then cp 0 () E CPA U {O}, and so cp 0 () is continuous. Thus

cp(b) = lim (cp n--+oo

0

B)(a n ) = O.

This shows that b E n{ Mcp : cp E cP B} = rad B. Since B is semisimple, b = 0, and so B is continuous by the closed graph theorem A.3.25. 0 Corollary 2.3.4 Let A be a commutatwe, semzszmple Banach algebra. Then A has a unique complete norm. Proof Suppose that II· II and III . III are each norms with respect to which A is a Banach algebra. By the theorem, the identity map (A, 11·11) ~ (A, 111·111) is continuous, and so II . II and III . III are equivalent. 0

The generalization of 2.3.4 to all semisimple Banach algebras will be given in 5.1.6. Proposition 2.3.5 Let A be a local algebra which is a normed algebra. Then MA is the kernel of a continuous character on A, and A is a Q-algebra. Proof Let B be the completion of A. Then B is a unital, commutative Banach algebra, and so there exists cp E CPB; cp I A is a continuous character on A. Clearly M