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Chapter IX Quotients of Products of the Integers
CHAPTER IX QUOTIENTS OF PRODUCTS OF THE INTEGERS We shall say that an abelian group N is a product if N is isomorphic to Z>. for some cardinal A. In this chapter we concern ourselves primarily with homomorphic images of products. In the first section we deal with the question of when such a homomorphic image is itself a product. It turns out that a sufficient, and in some cases necessary, condition for Z>. I A to be a product is that A be a direct summand of Z>.. In section 2 we consider arbitrary quotients of ZW; this leads us to Nunke's characterization of slender groups. In the third section we consider quotients of uncountable products of Z, where current knowledge is more fragmentary. In the last section, we deal with a related subject, that of radicals of groups and how their properties are affected by large cardinal hypotheses. Throughout this chapter M* will denote the dual group Homz(M, Z), and "reflexive" will mean Z-reflexive.
§l. Perps and products We begin by investigating necessary and sufficient conditions for a subgroup A of a group M to be a direct summand of M conditions which arise from the fact that if A is a summand of M then A * may be regarded as a summand of M*. In this connection we introduce the convenient "perp" notation which we shall use repeatedly later. If A is a subgroup of M and t: A --7 M is the inclusion map, denote by [* the induced map: M* --7 A* which takes y E M* to its restriction, yo c, to A. Let A.L - read "A perp" - denote the kernel of r. Thus A.L
= {y
E M*:
(y, a) = 0 for all a E A}.
The canonical projection of M onto MIA induces an embedding of (MIA)* into M* whose image is A.L j thus A.L is canonically
277
IX.l Perps and products
isomorphic to (MIA)*. Hence if M = A EB B we can identify A1. with B* via the map which takes y E A1. to its restriction to B. In this case it is easy to check that M* is the internal direct sum of B1. and A1.; indeed, for all y E M*, y = y IA + y IB, where in an abuse of notation - Y IB denotes the element of A 1. which agrees with y on B, and similarly for Y I A. In the same spirit, if M = A EB B we shall write M* = A* EB B* and regard the elements of B* (= A1.) as either functions on B or as functions on M which are zero on A. Now A1. is a subgroup of M* so we can define (A1.)1., a subgroup of M**. In an abuse of notation we shall denote by A 1.1. the preimage of (A1.)1. under aM: M -+ M**. Thus A1.1.
= {x
E M: Vy E A1.( (y,
x)
= On.
Clearly A ~ A 1.1.. It is not hard to see that MIA 1.1. is always torsionless. 1.1 Lemma. (i) If MIA is torsionless, then A = A1.1.. In particular, if M is torsionless and A is a summand of M, then A = A1.1.. (ii) If A1.1. is a direct summand of M J then (A1.1. I A)* = O.
(i) Suppose x E M \ A. Since MIA is torsionless there exists y E M* such that yEA 1. and (y, x) =J. 0, so x ¢:. A 1.1.. If M is torsionless and A is a summand of M, then MIA is torsionless since it is isomorphic to a summand of M. (ii) Say M = A1.1. EBD. If f E (A1.1.IA)* and 7r is the canonical projection: A 1.1. -+ A 1.1. I A, then f 0 7r E (A 1.1.)* = D1. ~ M*. But f 0 7r E A1., so by the definition of A1.1., f 0 7r _ 0; since 7r is surjective, this implies that f is zero. 0 PROOF.
In view of this lemma, we seek conditions that imply that A1.1. is a direct summand of M. 1.2 Lemma. Suppose M is reflexive and A is a subgroup of M such that M* = A1. EB N. Then M = A1.1. EB C where C ~ (A1.)*J and A1.1. ~ N*.
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IX. QUOTIENTS OF PRODUCTS OF THE INTEGERS
By the remarks at the beginning of this section, M** = (A 1.)1. ED N 1.. But then, since M is reflexive, this decomposition of M** pulls back to a decomposition of M: M = Al.l. ED C, where C is isomorphic to N 1.. The final assertions then follow from the identification of Nl. with (Al.)* and of (Al.)l. with N*. 0 PROOF.
Let us denote the group ZW by P. Say that A is a countable product if it is isomorphic to P or to zn for some finite n. By III.2.5, P is reflexive. 1.3 Proposition. For any subgroup A of P, Al.l. is a direct summand of P. Moreover, Al.l. and PjAl.l. are both countable products.
By Lemma 1.2 it's enough to show that Al. is a direct summand of P*. By defini tion of A 1. we have a short exact sequence PROOF.
where B is the image of P* under the map t": P*-t A* induced by inclusion. Now P* is a countable free group (d. before III.2.5); so B is a countable subgroup of A* and hence is free, since A* is ~l-free by IV.2.10. Therefore the short exact sequence splits, and A 1. is a direct summand of P* : P* = A 1. ED N where N ~ B. By 1.2 P = Al.l. ED C where C ~ (Al.)* and Al.l. ~ B*; hence Al.l. and C are countable products since B and A 1. are countable free groups. 0 Proposition 1.3 fails to hold for uncountable products: see Exercise 1. 1.4 Theorem. For any cardinal K which is not w-measurable and any subgroup A of Z", the following are equivalent: (1) A is a product and Al.l. = A; (2) A is a direct summand of Z". If these conditions hold, then Z" j A is a product. PROOF.
Assume (1) and consider the short exact sequence
IX.l Perps and products
279
where B is the range of the homomorphism z": (ZIt)* -+ A* induced by the inclusion of A in Zit. Since A* is free by 111.3.7, B is free and thus the sequence splits. Since A.L is a summand of (ZIt)*, A.L.L is a summand of Zit by Lemma 1.2. Thus (2) is proved since A.L.L = A. Now assume (2): say Zit = A EEl D. Then (ZIt)* = A.L EEl D.L, so by 1.2 Zit = A.L.L EEl C where C ~ (A.L)* and A.L.L ~ (D.L)*. Now (ZIt)* is free by 111.3.7, so A.L and D.L are free, and hence C and A.L.L are products. By Lemma 1.1(i) A = A.L.L so (1) is proved. Finally, notice that Zit/ A is a product since it is isomorphic to C.
o We want to consider when the converse of the last assertion of the theorem is true. Recall that a Whitehead group is a group G such that Ext(G, Z) = O. 1.5 Theorem. Let K; and A be as in 1.4 and suppose that Z"/ A is a product. Then A = A.L.L; moreover, if either (a)K;=w, (b) every Whitehead group of cardinality S K; is free, or (c) Zit / A is a countable product, then A is a direct summand of Zit. PROOF. That A.L.L = A follows, by Lemma 1.1(i), from the fact that Zit / A is torsionless since it is a product (d. 1.2.1). So it remains to show that A.L.L is a direct summand of Zit. For assumption (a) this is Proposition 1.3. Assume (b); let M denote Zit. The short exact sequence
o -+ A~M~M/A -+ 0 induces a short exact sequence
where B is the image of t: M* -+ A *. The latter sequence induces the Cartan-Eilenberg sequence 0-+ B*
-+
M**~(M/A)** -+ Ext(B,Z)
-+
Ext(M*,Z) = 0
280
IX. QUOTIENTS OF PRODUCTS OF THE INTEGERS
where the last term, Ext(M**, Z), is zero because M** ~ Z(tt) is free. Now MIA is reflexive because it is a product and /\, is not w- measurable (d. Exercise III. 9). Hence 71"** is surjeetive because 71" is surjective and 71"** 0 at« = aM/A 071". Therefore Ext(B, Z) = 0, so by hypothesis, B is free. But then A.l is a direct summand of M* since B ~ M* I A.l. Lemma 1.2 then finishes the proof in this case. Now assume (c), and retain the notation of case (b). By assumption, (MIA)* is countable (d. before III.2.5). Thus since B rv M* I(MIA)* and M* is free, B is isomorphic to C EB F where C is countable and F is free. Hence Ext(B, Z) = Ext(C, Z), so if Ext (B, Z) = 0, then C is free because every countable Whitehead group is free (d. XII.1.2). We then finish the proof as in case (b).
o 1.6 Corollary. (V = L) For all cardinals r: and all subgroups A of L", A is a direct summand of L" if and only if L" I A is a product.
Recall that V = L implies that there are no w-measurable cardinals (d. VI.3.15). Then the corollary follows from Theorem 1.5(b) and VII.4.10. 0 PROOF.
1. 7 Example. Suppose there is a Whitehead group B of non-w-
measurable cardinality, «, which is not free. There is a short exact sequence
o --t J( --t F
--t
B
--t
0
for some free group F ~ Z(tt). This induces the Cartan-Eilenberg sequence
o --t B* ~F* --t J(* --t Ext(B, Z) =
O.
Without loss of generality we may suppose that v is an inclusion map, so B* is a subgroup of F* ~ Ztt. Thus F* I B* is a product because J( is free. We claim that B* is not a direct summand of F*. If it were, then (*) would be a splitting exact sequence and hence so would be
281
IX.1 Perps and products
o -+ K"' -+ F** -+ B** But then since F and J{ are reflexive, F and B would be free.
J{
-+
O.
would be a summand of
It follows from Example 1.7 and XII.1.11 that Corollary 1.6 is not provable in ZFC. Thus, for example, it is undecidable in ZFC whether for every subgroup A of ZWl, A is a direct summand of ZWI whenever ZWI / A is a product.
For future reference we include the following result, which can be proved by a straightforward application of the definitions and remarks at the beginning of this section. 1.8 Lemma. (i) If M = BtBCffiD, then C1- = (BffiC)1-ffi( CffiD) 1- . (ii) If M = EBkEW C k, then M* is canonically isomorphic to
I1nEw( ffik#nCk)1- .
0
We conclude this section by considering some applications to reflexivity. By 1.2.2, there is a homomorphism p: M*** -+ M* such that p 0 au- = 1M*. Thus M*** = im( aM*) ffi ker(p) and M* is reflexive if and only if ker(p) = O. From the definition of p it is easy to see that ker(p) = aM[M]1-. 1.9 Theorem. For any group A, A* is reflexive if and only if A** is reflexive. PROOF. Let a1: A* -+ A*** and a2: A** -+ A**** be the canonical homomorphisms defined in section 1.1. Let P3: A*** -+ A* (respectively P4: A**** -+ A**) be the splitting of a1 (respectively, (2) defined in 1.2.2. Then A** is reflexive if and only if a1[A*]1- = 0 and A* is reflexive if and only if ker(P3) = o. But A*** = a1[A*]ffiker(p3) so ker(P3)* ~ adA *] 1-. Since ker(P3) is torsionless, ker(P3) = 0 if and only if ker(P3)* = 0, so the theorem is proved. 0
1.10 Theorem. A direct summand of a reflexive group is reflexive. PROOF. Suppose M is reflexive and M = A ffi B. Then M** = A ** ffi B**, with the identifications as described at the beginning of
282
IX. QUOTIENTS OF PRODUCTS OF THE INTEGERS
this section. Moreover, aM = aA + ae, i.e., for all a E A, b E B, aM((a, b)) = (aA(a), aB(b)) E A** ED B**. It follows easily that if aM is an isomorphism, then so is aA. 0 The preceding generalizes in an obvious way to H-reflexivity. We will see in X.I.1 that direct sums and products (over nonmeasurable index sets) of reflexive groups are reflexive. We make use of this and III.2.12 to derive the following. 1.11 Corollary. If H is a free R-module of infinite non-w-measurable rank and M is a projective R-module with a generating set which is not co-measurable, then M is H -reflexive.
By hypothesis there is an infinite non-ce-measurable /'\, and a module P such that M ED P ~ R(K) ~ H(K). Thus by III.2.12 (with N = H), M ED P is H-reflexive. So by the remark above, M is H-reflexive. 0 PROOF.
§2. Countable products of the integers In this section we will look at quotients of ZW and use the information gained to characterize slender Z-modules by their subgroups. As before we will denote ZW by P and say that a group is a countable product if it is isomorphic to P or to for some finite n. Let us give P the product topology where Z is given the discrete topology. Thus {Un: nEw} is a basic system of neighborhoods of 0, when Un = {x E P: x(m) = 0 for all m < n}. For any subset A of P, let A denote the closure of A in P; i.e., A = the set of all x in P such that for every n, x agrees with some element of A in the first n coordinates.
zn
2.1 Lemma. For any subgroup A of P of infinite rank, morphic to P and AIA is a coiorsion group.
A
is iso-
For each n let 7r n : P ---+ Z be projection on the nth coordinate. Then 7rn[A n Un] is a cyclic subgroup of Z, say dnZ; let an be an element of A n Un chosen so that 7r n(an) = dn and an = 0 if dn = O. For all a E A we can choose kn E Z by induction so that PROOF.
283
IX.2 Countable products of the integers
a = EnEw knan (cf. the argument below that 0 is onto A). Hence since A has infinite rank, infinitely many of the an,s must be nonzero. Let {m(n): nEw} enumerate in increasing order those n so that an i- o. Define O:P ~ P by O((kn)nEw) = En knam(n) , which is a well-defined element of P because for all i E w, En knam(n) (i) is a finite sum. Since {am(n): nEw} is linearly independent, 0 is oneone. Clearly O[Z(w)] ~ A and O[P] ~ A. We claim that O[P] = A. Given x E A we shall define kn inductively so that for all i x(i) =
L: knam(n)(i). n:5i
Suppose that we have defined kn for all n < £ so that (*) holds for all i < m(£). Since x E A there exists yEA such that y(i) = x(i) for i ::; m(£). Then by (*), y-En
Let t be the inclusion of A into Al.l.. Then we have a short exact sequence PROOF.
o ~ (Al.l.)*~A* -+ N -+ 0 because by 1.3 and 1.1(ii) kertz") = (ALLI A)* = O. (Here N is the cokernel of t*.) This in turn induces the Cartan-Eilenberg sequence
o ~ N*
~ A**~(Al.l.)*" ~ Ext(N, Z) ~ Ext (A*, Z)
=0
284
IX. QUOTIENTS OF PRODUCTS OF THE INTEGERS
where the last term is zero because by 2.1 A ~ P, so A* is free. Now A and Al.l. are both reflexive since by 2.1 and 1.3, respectively, they are both countable products (d. III.2.5). Thus A l.l. / A is isomorphic to Ext(N, Z) and hence, by V.2.3, is cotorsion. 0 2.3 Theorem. For any subgroup A of P, PIA is the direct sum of a cotorsion group and a countable product. PROOF. By 1.3 P = Al.l. EB C where Al.l. and C are countable products; so PIA ~ (Al.l. /A) EB C, and it remains to prove that A l.l. / A is cotorsion. We first dispose of the case when A has finite rank. In this case the pure closure, A*, of A is a summand of P since P is separable by IV.2.8. Since A* is clearly contained in Al..1., we have A* = A;l. = Al.l. by 1.1(i). Then Al.l./A is cotorsion (even pure-injective) since it equals A*/A and hence is finite. Now assume A has infinite rank. By Lemmas 2.1 and 2.2, Al.l./A and A/A are cotorsion; but then by V.2.2 Al.l./A is cotorsion. 0
It is not the case that every quotient of ZWI is of the form given in the previous theorem: see Exercise 2. We shall deal further with quotients of Z" when K is uncountable in the next section. Now we derive, as a consequence of 2.3, Nunke's characterization of slender groups. Recall that Z(p) is the cyclic group of order p and Jp is the group of p-adic integers. 2.4 Corollary. An abelian group H is slender if and only if it does not contain a copy of Q, P, Z(p) or i, for any prime p. PROOF. If H is slender, then by V.2.1 it does not contain any pureinjective group and hence does not contain a copy of Q, Z(p) or J p ; clearly also it does not contain a copy of P. Conversely, suppose H does not contain a copy of any of these groups, and consider a homomorphism r.p: P --+ H. By 2.3, r.p[Pj is of the form G EB C where C is a countable product and G is cotorsion. By V.2.9, H is cotorsion-free, so G = O. Since H does not contain a copy of P, C must be isomorphic to T" for some finite m. Thus r.p[Pj is a finitely-generated free group, T", By IIL1.10, zm is slender, so
IX.2 Countable products of the integers cp( en) = 0 for all but finitely many n. Since this holds for all is slender. 0
285 ip,
H
In other words, 2.4 says that H is slender if and only if it is cotorsion-free and does not contain a copy of P. 2.5 Corollary. (i) Every Nl-free group which does not contain a copy of P is slender. (ii) Every strongly Nl-free group is slender. Part (i) is immediate from the theorem. For part (ii) we need to prove that a strongly Nl-free group H cannot contain a copy of P. But, by definition, any countable subgroup S of H is contained in a countable Nl-pure subgroup, which must contain the Z-adic closure of S (ef. §I.3). However, P has a countable subset, namely Z(w), whose Z-adic closure in P is uncountable. So H cannot contain a copy of P. 0 PROOF.
Later (in Chapter XIV), we shall construct non-free strongly Nl-free (even Nl-separable) groups which are dual groups. Hence we shall have examples of dual groups which are slender but are not free. We conclude with a characterization, in the same spirit as 2.4, of almost slender groups (see before 111.1.11) . 2.6 Corollary. An abelian group H is almost slender if and only if it does not contain a copy of P or of any unbounded cotorsion group.
If H is almost slender, then it clearly does not contain a copy of P. Suppose that H contained an unbounded cotorsion group G; by V.2.6, G is the homomorphic image of a pure-injective group A; say cp: A -+ G is an epimorphism. It is then easy to construct 'ljJ: Z(w) -+ A such that for all m =I 0, {mcp('ljJ{e n ) ) : nEw} is infinite. Since A is pure-injective, we can extend 'ljJ to ;p: P -+ A; then cp 0 ;p shows that H is not almost slender. Conversely, if H does not contain a copy of P or of any unbounded cotorsion group, consider a homomorphism cp: P -+ H. By 2.3 cp[P] is of the form G EEl C, where G is cotorsion and C is a countable product. By PROOF.
286
IX. QUOTIENTS OF PRODUCTS OF THE INTEGERS
hypothesis, G is bounded and C is a finite product. For some r E Z \ {a}, rr.p[P) ~ C, we can then finish the proof as in 2.4. 0
§3. Uncountable products of the integers We are interested in a generalization of Theorem 2.3 to the case where P is replaced by an uncountable product of Z. By Exercise 2, the full analog of 2.3 fails, but we do have the following result.
3.1 Theorem. Let I\, and ,\ be cardinals which are not w-measurable. Suppose A is a subgroup OfZK which is a homomorphic image ofZ>'. Then ZK / A is the direct sum of a cotorsion group and a product.
Let us denote Z>' by M and ZK by N. Note that M and N are reflexive by Exercise III.9. Let f: M - t N be a homomorphism such that A = f[M). Let i: A - t N be the inclusion map. Then we can factor fast 0 g where g: M - t A is a surjection. We have induced maps ,,*: N* - t A* and q": A* - t M"; the latter is a monomorphism, so A * is free since M* C>! Z(>.) is free. The kernel of t * is denoted A.L (d. section 1), and is a (free) direct summand of N*('::::. Z(K)) because N* jA.L is isomorphic to a subgroup of A*. Therefore by 1.2, N = A.L.L EEl C, where C ~ (A.L)*; hence NjA ~ A.L.L j A EEl C. Now C is a product since A.L is free; so it remains to prove that A.L.L / A is cotorsion. Let B denote the image of the map [": N* - t M*, i.e., B = ]*[N*). Now 1* factors as j 0 h where h: N* - t B is a surjection and j is the inclusion of B into M*. Let D = M* j B, so we have a short exact sequence PROOF.
a - t B .i,M* - t D
-t
a
which induces the top row of the following commutative diagram: M*
1£1:\/ M
L
B*
.i;
N
-t
Ext(D, Z) - t (M*, Z) =
a
1
0
where 0 = ai/ 0 h*. Notice that aM and aN are isomorphisms and that the square commutes because = h* 0 j* and aN 0 f =
r:
287
IX.2 Uncountable products of the integers
1** 0 (J'M.
We claim that the image of () is A1.1.. In fact, this follows from the remarks in the first few paragraphs of section 1 (where we let N* play the role of M and let A1. play the role of A). The map h may be identified with the canonical projection of N* onto N* / A 1. since ker(h) = ker(J*) = A1.. Then h* in turn may be identified with the induced embedding of (N* / A1.)* into N**, whose image is (A1.)1.. Hence the image of h* equals (J'N[A1.1.]. Now () is a monomorphism since it is the composition of two monomorphisms, so Ext(D, Z) ~ B** / im(j*) ~ ()[B*]/()[j*[M**]]
rv
()[B*]/f[M]
and the last term, as we have just seen, is A 1.1./ A. Thus A 1.1./ A is isomorphic to Ext(D, Z), which is cotorsion by V.2.3. 0 In fact, the proof of the theorem gives us a little more information, which we state as a corollary. 3.2 Corollary. For any homomorphism f: Z>" ~ ZK where /'I, and ,\ are non-co-measurable cardinals, there are groups C and D such that C is a product, ker(J) ~ D*, and coker(J) ~ C EB Ext(D, Z).
The only new thing that requires checking is the assertion about the kernel of f. But it is easily verified from the commutativity of the diagram above that (J'M induces an isomorphism between ker(J) and D*. 0 PROOF.
3.3 Corollary. If /'I, is not w-measurable and A is a subgroup of ZK which is a product, then ZK/A ~ C EB Ext(D, Z) where C is a product and D* = O. 0
We will see in Chapter XII that we have, if we assume V = L, a complete structural description of groups of the form Ext(D, Z) and hence of groups of the form ZK / A, where A is as in the results above. (See also Exercise 8.) We-conclude with some notions that will be of use in the next section. 3.4 Definition. For any /'I, ~ ~o, let Z
288
IX. QUOTIENTS OF PRODUCTS OF THE INTEGERS
Otherwise said, Z,.. is the reduced power Z'" IC,.. where C,.. is the CO-I\, filter on I\, (d. II.2.5 and II.3.1). Say that an abelian group G is strongly cotorsion-free if Hom(Z,.., G) = 0 for all regular I\, which are not w-measurable. The structure of Zw is discussed in Exercise VA and ZWI is discussed in Exercise IV.11. 3.5 Lemma. For every PROOF.
K,
of uncountable cofinality, Z,.. is
~l-free.
If A is a countable subgroup of Z,.. = L" IC,.. then by 11.304, ~l-complete. But
A is isomorphic to a subgroup of Z,.. since C,.. is Z,.. is ~l-free (d. IV.2.8), so A is free. 0
3.6 Lemma. IfHom(Zwl' G) = 0, then G is cotorsion-free. Hence, a strongly cotorsion-free group is cotorsion-free. PROOF. Suppose to the contrary that there is a non-zero cotorsion subgroup C of G. Then by V.2.6, C is a homomorphic image of a pure-injective group H; say 'IjJ: H ---+ C is an epimorphism. By 3.5, ZWI is ~l-free, so it contains a countable pure free subgroup F. Let a E C \ {OJ. There is a homomorphism <.p: F ---+ H such that a E ('IjJ 0 <.p) [F]. This homomorphism extends to ZWI since H is pure-injective. But then it is clear that there is a non-zero homomorphism from ZWI to C ~ G, a contradiction. 0
In fact, G is cotorsion-free if and only if Hom(Zw, G) = 0: see Exercise 10. Wald 1983b has proved that, assuming V = L, every cotorsion-free group of cardinality ~1 is strongly cotorsion-free. A group H is called residually slender if for all a E H \ {OJ, there is a subgroup N of H such that a t/:. Nand HI N is slender. For example, any separable group is residually slender, but may not be slender (e.g. ZW). 3.7 Lemma. Every residually slender group is strongly cotorsionfree.
Suppose, to the contrary, that there is a residually slender group H and a non-zero homomorphism <.p: Z,.. ---+ H for some regular, non-w-measurable K,. Say a E im(<.p) \ {OJ. By hypothesis PROOF.
289
IX.4 Radicals and large cardinals
there is a subgroup N of H such that a ~ Nand H/N is slender. If 1r is the canonical surjection: H --+ H / N, then 1r 0 c.p is a non-zero homomorphism: Z" --+ H/N. Hence there is a non-zero homomorphism from Z" to H/N which is zero on Z<"; but this is impossible by 111.3.4, since H/N is slender and /'C is not w-measurable. 0 §4. Radicals and large cardinals We take up a theme which was first introduced in Chapter II: the effect of large cardinal hypotheses on group-theoretical properties. (The results of this section will not be used later, so the reader not interested in them for their own sake may safely skip this section.) Let A denote the class (or category) of abelian groups. A function T: A --+ A is called a preradical (or subfunctor of the identity) if for all groups A, and all group homomorphisms h: A --+ B, T A ~ A and h[TA] ~ T B. (In this context we usually write TA instead of
T(A).) If T is a preradical, it is easy to see that ToT ~ T, i.e., for all A, T 2 A ~ T A. We say that T is a socle if ToT = T. We define a "co-composition", S: T, of preradicals Sand T as follows: for any preradicals Sand T, if p: A --+ A/TA is the canonical surjection, (S: T)A = p-l[S(A/TA)]. It is easy to see that T ~ T: T. We say that T is a radicalifT = T: T, i.e., T(A/TA) = 0 for all A. Using the canonical projections and injections, one can show that for any preradical T, and any groups Ai, T(ffiiEI Ai) = ffiiEI T Ai and T(I1iEI Ai) ~ I1iEI T Ai (see Exercise 9). 4.1 Examples. For any class of groups X and any group A define
SxA = L:{im(c.p):c.p E Hom(X, A), X E X} and RxA
= n{ker(c.p):c.p E Hom(A,
X), X EX}.
We will write SG instead of S{G} and R G instead of R{G}' It is left to the reader to check that Sx is always a socle and R x is always a radical.
290
IX. QUOTIENTS OF PRODUCTS OF THE INTEGERS
For example, if G = Z/pZ, BaA = A[p]def{a E A:pa = O}. If G = EBp,n Z/pnz, BaA = At, the torsion part of A. Also, RQA = At. For any A, AIRzA is torsionless, and, in fact, A is torsionless if and only if RzA = O. The following is easy to verify.
4.2 Proposition. Suppose X is a class of groups and Y -is the class of direct sums of elements of X. Then Rx = Ry. In particular, if F is any non-trivial free group then R F = R z. 0 Let X be the class of N1-free groups. The radical Rx is called the Chase radical and denoted by u, In the rest of this section we shall concentrate mainly on the Chase radical and on Rz.
4.3 Proposition. (i) For any countable group C, vC = RzC. (ii) If C is countable, t/C = C if and only if C* = o. (iii) v is a socle. PROOF. (i) Let F be the free group of countable rank. If ep is any map from C to an N1-free group, then the image of ep is a free group. So vC = RFC, which, by Proposition 4.2, equals RzC. (ii) For any group, C* = 0 if and only if RzC = C, by definition. (iii) We must prove that v ~ v 0 t/, For this it suffices to prove that for all groups A, AlvvA is N1-free, for then a E vA implies that the image of a under the canonical surjection of A onto AIvvA is 0, which means that a E vvA. Now we have a short exact sequence
o ~ vAlvvA ~ AlvvA ~ AlvA ~ 0 and the middle term will be ~l-free if vAlvvA and AlvA are ~l free (d. Exercise IV.5). Thus it suffices to show that for all G, GlvG is N1-free. Let S = {l<: K is a subgroup of G s.t. GIl< is ~l-free}. Then vG = nKes K, Define (): G ~
IT GIl<
KeS
by ()(g)(l<) = g + K, The kernel of () equals vG, so () induces an embedding of GI vG into IlKes GI J(. Since a product of N1-free
IX.4 Radicals and large cardinals
291
groups is Nt-free (d. Exercise IV.5), it follows that GlvG is Nrfree.
o 4.4 Definition. Let T be a preradical and /\, an infinite cardinal. Define T[K)A = 'E{TB: B is a subgroup of A of cardinality < /\'}. It is easy to check that T[K) is a preradical. We say that T satisfies the cardinal condition if there is a /\, such that T[K) = T.
First we will prove that v satisfies the cardinal condition. Then we shall show that whether or not R z satisfies the cardinal condition depends on whether a certain large cardinal exists.
4.5 Theorem. The Chase radical satisfies the cardinal condition; in fact, v = V[Wl). Suppose, to the contrary, that there is a group A and an element a of vA such that a 1:. E{vC: C is a countable subgroup of A}. Let I be the set of all a: C, ---t Z such that C; is a countable subgroup of A, a E Cu, and a(a) #- O. For each x E A, let Sx = {a E I: x E dom a}. We claim that every intersection of countably many of the Sx's is non-empty. Indeed, given X n E A(n E w), let C be a countable subgroup of A containing {x n : nEw}; since a 1:. R z C, there exists a: C ---t Z such that a( a) #- 0; but then a E I and a E nnEw SX n • Thus there is an wt-complete filter F such that Sx E F for all x E A. Define '1/;: A ---t ZI as follows: for all x E A and a E I, PROOF.
'I/;(x)(a)
= {a(x) o
if a E ~x otherwise
Composing 'I/; with the canonical surjection onto ZI I F we obtain a map if;: A ---t ZI/ F which is a homomorphism because Sx+ynSxnSy belongs to F for all x, yEA. Now ZI IF is Nt-free since F is WIcomplete (d. Exercise 11.10), and if;(a) #- 0 by construction, so a 1:. v A, a contradiction. 0
292
IX. QUOTIENTS OF PRODUCTS OF THE INTEGERS
4.6 Corollary. v = Rr d. 0 Now we turn our attention to radicals of the form Rc, and consider the question of whether they satisfy the cardinal condition.
4.7 Theorem. Assume there are no measurable cardinals. Then for every non-zero strongly coiorsion-free G, R c does not satisfy the cardinal condition. Suppose to the contrary that Rc = Rgl for some A. Choose a regular cardinal K 2: A and let A = Z" (= Z" /Z<": see 3.4). Then by hypothesis on G, Hom(A, G) = 0, so RcA = A. However, if B ~ A and IBI < A ~ K, B is isomorphic to a subgroup of Z" by II.3.4. Since RcZ" ~ (RcZ)" = 0 (since G is torsionfree), we conclude that ReB = O. Hence Rg1A = 0 -=I RcA, a contradiction. 0 PROOF.
This result shows that the absence of certain large cardinals implies the failure of the cardinal condition. Another result of Dugas and Gobel 1985a says that if 0# does not exist (see VI.3.15) and GCH holds, then for any cotorsion-free G -=I 0, R c does not satisfy the cardinal condition. On the other hand Eda 1989 has proved that Vopenka's Principle - a very large cardinal hypothesis - implies that every preradical satisfies the cardinal condition. In the next results we shall pin down exactly which large cardinal is needed in order for Rz to satisfy the cardinal condition.
4.8 Theorem. If K is an LAw-compact cardinal and G is a group of cardinality < A, then R c = R~l.
A and let P,,(A), Uy and F,,(A) be defined as in 11.3.9. Let I = P,,(A). Then F,,(A) is a x-complete filter on I, so by hypothesis, there is a A-complete ultrafilter D on I which contains F,,(A). Define 'l/J: A --t IlYEl(Y) as follows: for all a E A and Y E I, PROOF. Fix a group
a Y = {a if a E Y 'l/J( )() 0 otherwise
Then the composition of'l/J with the canonical surjection onto the ultraproduct yields a homomorphism 1[J: A --t IlYEI(Y) / D.
IXA Radicals and large cardinals
293
Now suppose that a ¢:. R~] A; then for all Y E I such that a E (Y), a ¢:. R a (Y) so there exists [v : (Y) - t G such that fy(a) =J O. Note that fy is defined for all Y in a member Uz of D: choose Z so that a E Z. Then the fy's induce a homomorphism f: I1YEl(Y) I D - t GIlD. Now 8a : G - t GIlD is an isomorphism by II.3.2. The composition 8(/0 f 0 if; is a homomorphism from A to G which is non-zero on a. Therefore, a does not belong to RcA. D
4.9 Theorem. Let G be a residually slender group of cardinality ,\ ~ ~o where ,\ is not w-measurable. Let", be an infinite cardinal. Suppose Ra = R~]. Then", is L>..+w-compact. It suffices to show that every x-complete filter F on a set I is contained in an wI-complete ultrafilter D on 1. (For, by 11.2.11, D is then '\+-complete.) Let A = G1IF. We claim that R~] A = O. Indeed, if B is a subgroup of A of cardinality < «, then by 11.3.4, B is embeddable in G1 , so RaB = O. Therefore RaA = O. Hence there is a non-zero homomorphism
f: GI_~G
-t
GIN
such that f(x) =J O. Now we apply Theorem II1.3.2: there are WIcomplete ultrafilters Db ... , D'; on I such that for all a E G 1 , if for all k, supp( a) ¢:. D k then f( a) = O. If we can prove that F is contained in one of the Dk's, then we will be done. So suppose, to the contrary, that this is not the case, i.e., for all k = 1, ... , n, there exists s, E F \ o; Then Sk belongs to F but not to any D k • Define
nk
x'(i)
= {X(i) o
nk
if i E s, otherwise
Then p(x') = p(x), so f(x') = f(x) =J O. But supp(x') ¢:. D k for any k, which contradicts Theorem III.3.2. D
IX. QUOTIENTS OF PRODUCTS OF THE INTEGERS
294
4.10 Corollary. Rz satisfies the cardinal condition if and only if there is an LW1W-compact cardinal. 0
See Exercises 14 and 17 for more conditions equivalent to the existence of an LW1W-compact cardinal. We now turn our attention to the question of when a radical commutes with direct products. Recall that always T TIl Ai ~ TIl T Ai (d. Exercise 9). 4.11 Proposition. If G is cotorsion-free, then RG commutes with countable products.
We must show that if a E TInEw RoAn, then we have a E RoTInEwAn. Given ip: TIn An ---7 G, let () = ep~TIn(a(n)). Since each a(n) E RoAn, the kernel of () contains EBn(a(n)), and hence () induces 8:TIn(a(n))/EBn(a(n)) ---7 G. Now the domain of 8 is pure-injective by V.1.16, so the range of 8 is cotorsion. But then by hypothesis on G, the range of 8 is 0, so ep(a) = O. 0 PROOF.
The converse of 4.11 holds for torsion-free G: see Exercise 13. There are cotorsion-free groups which don't commute with uncountable products: see Exercise 15. 4.12 Theorem. IfG is strongly cotorsion-free, then RG commutes with TIll: whenever /'i, is not w-measurable. PROOF. The proof is by induction on «, The case /'i, = ~o is 4.11. The case of a singular /'i, is straightforward, so we assume that /'i, is regular (not w-measurable) and that for A < /'i, and all groups A tL , TItL<,\ RGAtL ~ Ro TItL<,\ A w Suppose, to obtain a contradiction, that there is an element a of TItL<1I: RoA tL which is not in R G TItL<1I: A w Thus there exists f: TItL<1I: A tL ---7 G such that f(a) =/:. O. Let 'lj;: ZII: ---7 TItL<1I: A tL be defined by 'lj;((ntLL) = (ntL a(J1. ))w The
composition () ~f f 0 'lj; is a homomorphism: 111: ---7 G such that ()( (etL) tL) = f( a) =/:. O. By the inductive hypothesis, () ~Z
---7
G, which is a con-
Measurability is a real barrier, as we see from the next results.
IX.4 Radicals and large cardinals
295
is the first measurable cardinal and G is a strongly cotorsion-free group of cardinality < K, then Ra does not commute with Ill'> .
4.13 Theorem. If
K
Let D be a normal x-complete ultrafilter on K; then S~f{a < K: a is a regular cardinal} belongs to D (d. II.2.15). Let 1= UoEsa x {a}. For X ~ I, let PROOF.
Wx ~{a E S: la \ {I: (J, a) E X}I < a} and let F = {X ~ I: Wx ED}. It is routine to check that F is a x-complete filter on 1. We claim that
(*)
there is a x-complete ultrafilter U on I which contains F.
Assuming this for a moment, let us finish the proof. (Note that the claim follows immediately if K is strongly compact.) Now RaZo = Zo for every a E S because Hom(Zo, G) = 0 by hypothesis on G. We shall show that IloEs ssr; =I- R a IloEs t; by defining a non-zero homomorphism from IloEs t; to G. Since G is torsion-free, we can identify Z with a subgroup of G. Given z E IloEs ZO , let z((3, a)( for (3 < a) denote the (3thcoordinate of z(a) E ZO; so z((3, a) E Z. Then since U is Kcomplete, there is a unique x E Z ~ G such that {((3, a) E I: z((3, a) = x} belongs to U (d. proof of II.3.2). Define ep(z) = x. Since F ~ U, ip: IloEs ZO ---t G is identically 0 on IloEs Z
C x = {, E
K:
{a E S: (J, a) EX} ED}
and let U = {X ~ I: Cx ED}. We leave it to the reader to verify that U is a x-complete ultrafilter on 1. To show that F ~ U we use the normality of D. Let X E F. Define {}: K ---t K by:
{}(a)
= { sup(a \ o
{, E a: (J,a) E X} if a E ~ otherwise
296
IX. QUOTIENTS OF PRODUCTS OF THE INTEGERS
Since X E F, () is a regressive function, so there exists (3 E '" and T ~ S such that TED and ()(a) = (3 for all a E T (d. 11.2.16). Now one can check easily that I E Cx if I > (3; hence x \ (3 ~ Cx , so Cx ED, and therefore X E U. 0 4.14 Corollary. If « is co-measurable, then R z does not commute with TIK' 0
In contrast, we have: 4.15 Theorem. For ansj «, if A is a group whose cardinality is not w-measurable, then Rz(AK) = (RzA)K.
We must show that (RzA)K ~ Rz(AK), i.e., if f: AK - t Z and x E (RzA)K, then f(x) = O. Now f = ((gD)DEV) for some gD: A K / D - t Z where D is an wI-complete ultrafilter on '" (d. III.3.6). But 0A: A - t AK/ D is an isomorphism by 11.3.2, so for all D, gD(XD) = (gD 0 OA)(X(J.l)) for some J.l E «, and therefore gD(XD) = 0 since x(J.l) E RzA. 0 PROOF.
The Chase radical, i/, commutes with countable products: see Exercise 16. Eda 1987 has shown that v does not commute with uncountable products. EXERCISES 1. If M = ZWl, then there is a subgroup A of M such that ALL is not a direct summand of M. [Hint: start with K ~ M* = Z(Wl) such that M* [K is torsionless but not free.] 2. (i) ZWI is not the direct sum of a cotorsion group and a product. [Hint: Assume false. By Exercise IV.Il, (ZWl)* = 0, hence ZWI is cotorsion. Then use the fact that ZWI is ~I- free to obtain from this a contradiction using the results of Chapter V.] (ii) Prove that ZWI has no free summand. 3. (i) If '" is non-co-measurable and A is a subgroup of ZK such that A = AJ..1., then A is a dual group, i.e., A ~ B* for some B. [Hint: let B = the image of t": (ZK)* - t A*.] (ii) For any group B, B* is isomorphic to a subgroup A of ZK for some", such that A = A 1.1..
IX. Exercises
297
(iii) Hence, A is a dual group of non-co-measurable cardinality if and only if it can be embedded in a product ZK so that K is not w-measurable and ZK I A is torsionless.
4. (i) For any group H such that H* = 0, there is a subgroup A of ZK for some K such that if L = ZK lA, then H '" ker(O"L)' [Hint: by 111.3.8 there are arbitrarily large free groups which are also dual groups. So by 3(ii), for some K there is a subgroup F of ZK such that F = Fl..l.. and F has a subgroup A such that FIA C;:,! H.] (ii) There is a non-co-measurable K and a subgroup A of ZK such that if L = ZK I A, L is not the direct sum of a product and a group H such that H* = 0. [Hint: let H be slender such that H* = 0, and use (i).] 5. Use Corollary 2.4 to give another proof that a direct sum of slender groups is slender.
6. (i) A subgroup A of ZW is a countable product if and only if A is a closed subgroup of ZW (topology as in section 2) if and only if there exist elements {an: nEw} of ZW such that for all m, an E Urn for almost all n and A = U:=n rnan: rn E Z}. (ii) There is a closed subgroup A of zw such that ZW fA ~ Jp • [Hint: consider the function cp:Zw ---? J p defined by cp((xn)n) = LnXnpn.] This provides a counterexample to Lemma 95.1 of Fuchs 1973. 7. Prove the converse of 3.2: for any product C and group D both of non-w-measurable cardinality, there exist non-ce-measurable cardinals K and ,\ and a homomorphism f: Z>' ---? ZK such that ker(J) ~ D* and coker(J) ~ C EB Ext(D, Z). 8. Prove the following generalization of 3.1: let R be any ring, let H be any R-module, let M and N be reflexive R-modules and let f: M ---? N be a homomorphism such that f[M]l.. is a direct summand of N* and Ext(M*, H) = O. (Here M* = HomR(M, H).) Then there is a module D, which is a homomorphic image of M* such that ker(J) ~ D* and coker(J) ~ C EB Ext(D, H), where C is a direct summand of N.
298
IX. QUOTIENTS OF PRODUCTS OF THE INTEGERS
9. 1fT is a preradical, then for any groups Ai, T(EBl Ai) = EBl T(A) and T(Ill Ai) ~ III T(A i). [Hint: for ~ use 7rj: Il Ai --+ Aj; for ;2 use £j: A j --+ EBAd 10. A is cotorsion-free if and only if Hom(Zw, A) = O. [Hint: for ({=) use the structural description of Zw in Exercise VA]
11. Let X be a set of groups and let G = EBXEX X and H = IlXEX X , Show that Ra = Rx = R H and SG = S». 12. (i) An annihilator class of groups is a class of groups which is closed under direct products and subgroups. Show that the map which takes a radical R to {A: RA = O} is a one-one correspondence between annihilator classes and radicals. (ii) Formulate an analogous result for socles. 13. If G is torsion-free and R G commutes with countable products, then G is cotorsion-free. [Hint: to show that Jp is not contained in G, use the fact that Jp is a direct summand of a product of cyclic p-groups; to show that G is reduced, use the fact that !lQ is the torsion subgroup functor.]
14. Prove that the following are equivalent. (See Exercise 11.8 for the definition of 11:- torsionless.) (1) Rz = R~l; (2) for every group A, A x-torsionless =} A torsionless; (3) for every group A =I- 0, A x-torsionless =} A* =I- OJ (4) II: is LW1W-compact. [Hint: for the proof of (3) =} (4), use 111.3.2; cr. 4.9.] 15. There is a cotorsion-free group G such that RG does not commute with IlWl' [Hint: Let Ga(a E WI) be a rigid system of groups, i.e., Hom(Ga, G{3) = 0 if a =I- (3; let G = Il, Ga/Il~Wl Ga. Show that RaGa = Ga.] 16. The Chase radical v commutes with countable products. [Hint: cr. proof of 4.11.] For any radical R, define inductively Ra+l = R RJ.1. = na
0
Ra and
299
IX. Notes (i) JlOO is well-defined and is a radical which is a socle; (ii) Ra(A) = 2:{B ~ A: Hom(B, G) = O}; (iii) Hom(R(; A, G) = 0; (iv) if K, is regular, K, is LW1W-compaet if and only if R [Hint: show that (R'Z)[K] = (R~K])OO.]
z = R~[K].
NOTES 1.1 through 1.5 are based on Nunke 1962a. Parts (b) and (c) of Theorem 1.5 come from Dugas-Gobel 1979b. Theorem 1.9 and Corollary 1.11 are due to Huber 1983a. Section 2 is from Nunke 1962b, except for 2.6 which comes from Rychkov 1980 and Cobel-Richkov-Wald 1981. Theorem 3.1 and Corollaries 3.2 and 3.3 are due to Dugas-Gobel 1979b. For further generalizations see also Huber 1979, Dugas-Gobel 1981, and Huber-Warfield 1981. The notion of strongly cotorsion-free and Lemmas 3.6 and 3.7 are from Dugas-Gobel 1985a. For more on homomorphic images of ZK and Z
§l. Perps and products We begin by investigating necessary and sufficient conditions for a subgroup A of a group M to be a direct summand of M conditions which arise from the fact that if A is a summand of M then A * may be regarded as a summand of M*. In this connection we introduce the convenient "perp" notation which we shall use repeatedly later. If A is a subgroup of M and t: A --7 M is the inclusion map, denote by [* the induced map: M* --7 A* which takes y E M* to its restriction, yo c, to A. Let A.L - read "A perp" - denote the kernel of r. Thus A.L
= {y
E M*:
(y, a) = 0 for all a E A}.
The canonical projection of M onto MIA induces an embedding of (MIA)* into M* whose image is A.L j thus A.L is canonically
277
IX.l Perps and products
isomorphic to (MIA)*. Hence if M = A EB B we can identify A1. with B* via the map which takes y E A1. to its restriction to B. In this case it is easy to check that M* is the internal direct sum of B1. and A1.; indeed, for all y E M*, y = y IA + y IB, where in an abuse of notation - Y IB denotes the element of A 1. which agrees with y on B, and similarly for Y I A. In the same spirit, if M = A EB B we shall write M* = A* EB B* and regard the elements of B* (= A1.) as either functions on B or as functions on M which are zero on A. Now A1. is a subgroup of M* so we can define (A1.)1., a subgroup of M**. In an abuse of notation we shall denote by A 1.1. the preimage of (A1.)1. under aM: M -+ M**. Thus A1.1.
= {x
E M: Vy E A1.( (y,
x)
= On.
Clearly A ~ A 1.1.. It is not hard to see that MIA 1.1. is always torsionless. 1.1 Lemma. (i) If MIA is torsionless, then A = A1.1.. In particular, if M is torsionless and A is a summand of M, then A = A1.1.. (ii) If A1.1. is a direct summand of M J then (A1.1. I A)* = O.
(i) Suppose x E M \ A. Since MIA is torsionless there exists y E M* such that yEA 1. and (y, x) =J. 0, so x ¢:. A 1.1.. If M is torsionless and A is a summand of M, then MIA is torsionless since it is isomorphic to a summand of M. (ii) Say M = A1.1. EBD. If f E (A1.1.IA)* and 7r is the canonical projection: A 1.1. -+ A 1.1. I A, then f 0 7r E (A 1.1.)* = D1. ~ M*. But f 0 7r E A1., so by the definition of A1.1., f 0 7r _ 0; since 7r is surjective, this implies that f is zero. 0 PROOF.
In view of this lemma, we seek conditions that imply that A1.1. is a direct summand of M. 1.2 Lemma. Suppose M is reflexive and A is a subgroup of M such that M* = A1. EB N. Then M = A1.1. EB C where C ~ (A1.)*J and A1.1. ~ N*.
278
IX. QUOTIENTS OF PRODUCTS OF THE INTEGERS
By the remarks at the beginning of this section, M** = (A 1.)1. ED N 1.. But then, since M is reflexive, this decomposition of M** pulls back to a decomposition of M: M = Al.l. ED C, where C is isomorphic to N 1.. The final assertions then follow from the identification of Nl. with (Al.)* and of (Al.)l. with N*. 0 PROOF.
Let us denote the group ZW by P. Say that A is a countable product if it is isomorphic to P or to zn for some finite n. By III.2.5, P is reflexive. 1.3 Proposition. For any subgroup A of P, Al.l. is a direct summand of P. Moreover, Al.l. and PjAl.l. are both countable products.
By Lemma 1.2 it's enough to show that Al. is a direct summand of P*. By defini tion of A 1. we have a short exact sequence PROOF.
where B is the image of P* under the map t": P*-t A* induced by inclusion. Now P* is a countable free group (d. before III.2.5); so B is a countable subgroup of A* and hence is free, since A* is ~l-free by IV.2.10. Therefore the short exact sequence splits, and A 1. is a direct summand of P* : P* = A 1. ED N where N ~ B. By 1.2 P = Al.l. ED C where C ~ (Al.)* and Al.l. ~ B*; hence Al.l. and C are countable products since B and A 1. are countable free groups. 0 Proposition 1.3 fails to hold for uncountable products: see Exercise 1. 1.4 Theorem. For any cardinal K which is not w-measurable and any subgroup A of Z", the following are equivalent: (1) A is a product and Al.l. = A; (2) A is a direct summand of Z". If these conditions hold, then Z" j A is a product. PROOF.
Assume (1) and consider the short exact sequence
IX.l Perps and products
279
where B is the range of the homomorphism z": (ZIt)* -+ A* induced by the inclusion of A in Zit. Since A* is free by 111.3.7, B is free and thus the sequence splits. Since A.L is a summand of (ZIt)*, A.L.L is a summand of Zit by Lemma 1.2. Thus (2) is proved since A.L.L = A. Now assume (2): say Zit = A EEl D. Then (ZIt)* = A.L EEl D.L, so by 1.2 Zit = A.L.L EEl C where C ~ (A.L)* and A.L.L ~ (D.L)*. Now (ZIt)* is free by 111.3.7, so A.L and D.L are free, and hence C and A.L.L are products. By Lemma 1.1(i) A = A.L.L so (1) is proved. Finally, notice that Zit/ A is a product since it is isomorphic to C.
o We want to consider when the converse of the last assertion of the theorem is true. Recall that a Whitehead group is a group G such that Ext(G, Z) = O. 1.5 Theorem. Let K; and A be as in 1.4 and suppose that Z"/ A is a product. Then A = A.L.L; moreover, if either (a)K;=w, (b) every Whitehead group of cardinality S K; is free, or (c) Zit / A is a countable product, then A is a direct summand of Zit. PROOF. That A.L.L = A follows, by Lemma 1.1(i), from the fact that Zit / A is torsionless since it is a product (d. 1.2.1). So it remains to show that A.L.L is a direct summand of Zit. For assumption (a) this is Proposition 1.3. Assume (b); let M denote Zit. The short exact sequence
o -+ A~M~M/A -+ 0 induces a short exact sequence
where B is the image of t: M* -+ A *. The latter sequence induces the Cartan-Eilenberg sequence 0-+ B*
-+
M**~(M/A)** -+ Ext(B,Z)
-+
Ext(M*,Z) = 0
280
IX. QUOTIENTS OF PRODUCTS OF THE INTEGERS
where the last term, Ext(M**, Z), is zero because M** ~ Z(tt) is free. Now MIA is reflexive because it is a product and /\, is not w- measurable (d. Exercise III. 9). Hence 71"** is surjeetive because 71" is surjective and 71"** 0 at« = aM/A 071". Therefore Ext(B, Z) = 0, so by hypothesis, B is free. But then A.l is a direct summand of M* since B ~ M* I A.l. Lemma 1.2 then finishes the proof in this case. Now assume (c), and retain the notation of case (b). By assumption, (MIA)* is countable (d. before III.2.5). Thus since B rv M* I(MIA)* and M* is free, B is isomorphic to C EB F where C is countable and F is free. Hence Ext(B, Z) = Ext(C, Z), so if Ext (B, Z) = 0, then C is free because every countable Whitehead group is free (d. XII.1.2). We then finish the proof as in case (b).
o 1.6 Corollary. (V = L) For all cardinals r: and all subgroups A of L", A is a direct summand of L" if and only if L" I A is a product.
Recall that V = L implies that there are no w-measurable cardinals (d. VI.3.15). Then the corollary follows from Theorem 1.5(b) and VII.4.10. 0 PROOF.
1. 7 Example. Suppose there is a Whitehead group B of non-w-
measurable cardinality, «, which is not free. There is a short exact sequence
o --t J( --t F
--t
B
--t
0
for some free group F ~ Z(tt). This induces the Cartan-Eilenberg sequence
o --t B* ~F* --t J(* --t Ext(B, Z) =
O.
Without loss of generality we may suppose that v is an inclusion map, so B* is a subgroup of F* ~ Ztt. Thus F* I B* is a product because J( is free. We claim that B* is not a direct summand of F*. If it were, then (*) would be a splitting exact sequence and hence so would be
281
IX.1 Perps and products
o -+ K"' -+ F** -+ B** But then since F and J{ are reflexive, F and B would be free.
J{
-+
O.
would be a summand of
It follows from Example 1.7 and XII.1.11 that Corollary 1.6 is not provable in ZFC. Thus, for example, it is undecidable in ZFC whether for every subgroup A of ZWl, A is a direct summand of ZWI whenever ZWI / A is a product.
For future reference we include the following result, which can be proved by a straightforward application of the definitions and remarks at the beginning of this section. 1.8 Lemma. (i) If M = BtBCffiD, then C1- = (BffiC)1-ffi( CffiD) 1- . (ii) If M = EBkEW C k, then M* is canonically isomorphic to
I1nEw( ffik#nCk)1- .
0
We conclude this section by considering some applications to reflexivity. By 1.2.2, there is a homomorphism p: M*** -+ M* such that p 0 au- = 1M*. Thus M*** = im( aM*) ffi ker(p) and M* is reflexive if and only if ker(p) = O. From the definition of p it is easy to see that ker(p) = aM[M]1-. 1.9 Theorem. For any group A, A* is reflexive if and only if A** is reflexive. PROOF. Let a1: A* -+ A*** and a2: A** -+ A**** be the canonical homomorphisms defined in section 1.1. Let P3: A*** -+ A* (respectively P4: A**** -+ A**) be the splitting of a1 (respectively, (2) defined in 1.2.2. Then A** is reflexive if and only if a1[A*]1- = 0 and A* is reflexive if and only if ker(P3) = o. But A*** = a1[A*]ffiker(p3) so ker(P3)* ~ adA *] 1-. Since ker(P3) is torsionless, ker(P3) = 0 if and only if ker(P3)* = 0, so the theorem is proved. 0
1.10 Theorem. A direct summand of a reflexive group is reflexive. PROOF. Suppose M is reflexive and M = A ffi B. Then M** = A ** ffi B**, with the identifications as described at the beginning of
282
IX. QUOTIENTS OF PRODUCTS OF THE INTEGERS
this section. Moreover, aM = aA + ae, i.e., for all a E A, b E B, aM((a, b)) = (aA(a), aB(b)) E A** ED B**. It follows easily that if aM is an isomorphism, then so is aA. 0 The preceding generalizes in an obvious way to H-reflexivity. We will see in X.I.1 that direct sums and products (over nonmeasurable index sets) of reflexive groups are reflexive. We make use of this and III.2.12 to derive the following. 1.11 Corollary. If H is a free R-module of infinite non-w-measurable rank and M is a projective R-module with a generating set which is not co-measurable, then M is H -reflexive.
By hypothesis there is an infinite non-ce-measurable /'\, and a module P such that M ED P ~ R(K) ~ H(K). Thus by III.2.12 (with N = H), M ED P is H-reflexive. So by the remark above, M is H-reflexive. 0 PROOF.
§2. Countable products of the integers In this section we will look at quotients of ZW and use the information gained to characterize slender Z-modules by their subgroups. As before we will denote ZW by P and say that a group is a countable product if it is isomorphic to P or to for some finite n. Let us give P the product topology where Z is given the discrete topology. Thus {Un: nEw} is a basic system of neighborhoods of 0, when Un = {x E P: x(m) = 0 for all m < n}. For any subset A of P, let A denote the closure of A in P; i.e., A = the set of all x in P such that for every n, x agrees with some element of A in the first n coordinates.
zn
2.1 Lemma. For any subgroup A of P of infinite rank, morphic to P and AIA is a coiorsion group.
A
is iso-
For each n let 7r n : P ---+ Z be projection on the nth coordinate. Then 7rn[A n Un] is a cyclic subgroup of Z, say dnZ; let an be an element of A n Un chosen so that 7r n(an) = dn and an = 0 if dn = O. For all a E A we can choose kn E Z by induction so that PROOF.
283
IX.2 Countable products of the integers
a = EnEw knan (cf. the argument below that 0 is onto A). Hence since A has infinite rank, infinitely many of the an,s must be nonzero. Let {m(n): nEw} enumerate in increasing order those n so that an i- o. Define O:P ~ P by O((kn)nEw) = En knam(n) , which is a well-defined element of P because for all i E w, En knam(n) (i) is a finite sum. Since {am(n): nEw} is linearly independent, 0 is oneone. Clearly O[Z(w)] ~ A and O[P] ~ A. We claim that O[P] = A. Given x E A we shall define kn inductively so that for all i x(i) =
L: knam(n)(i). n:5i
Suppose that we have defined kn for all n < £ so that (*) holds for all i < m(£). Since x E A there exists yEA such that y(i) = x(i) for i ::; m(£). Then by (*), y-En
Let t be the inclusion of A into Al.l.. Then we have a short exact sequence PROOF.
o ~ (Al.l.)*~A* -+ N -+ 0 because by 1.3 and 1.1(ii) kertz") = (ALLI A)* = O. (Here N is the cokernel of t*.) This in turn induces the Cartan-Eilenberg sequence
o ~ N*
~ A**~(Al.l.)*" ~ Ext(N, Z) ~ Ext (A*, Z)
=0
284
IX. QUOTIENTS OF PRODUCTS OF THE INTEGERS
where the last term is zero because by 2.1 A ~ P, so A* is free. Now A and Al.l. are both reflexive since by 2.1 and 1.3, respectively, they are both countable products (d. III.2.5). Thus A l.l. / A is isomorphic to Ext(N, Z) and hence, by V.2.3, is cotorsion. 0 2.3 Theorem. For any subgroup A of P, PIA is the direct sum of a cotorsion group and a countable product. PROOF. By 1.3 P = Al.l. EB C where Al.l. and C are countable products; so PIA ~ (Al.l. /A) EB C, and it remains to prove that A l.l. / A is cotorsion. We first dispose of the case when A has finite rank. In this case the pure closure, A*, of A is a summand of P since P is separable by IV.2.8. Since A* is clearly contained in Al..1., we have A* = A;l. = Al.l. by 1.1(i). Then Al.l./A is cotorsion (even pure-injective) since it equals A*/A and hence is finite. Now assume A has infinite rank. By Lemmas 2.1 and 2.2, Al.l./A and A/A are cotorsion; but then by V.2.2 Al.l./A is cotorsion. 0
It is not the case that every quotient of ZWI is of the form given in the previous theorem: see Exercise 2. We shall deal further with quotients of Z" when K is uncountable in the next section. Now we derive, as a consequence of 2.3, Nunke's characterization of slender groups. Recall that Z(p) is the cyclic group of order p and Jp is the group of p-adic integers. 2.4 Corollary. An abelian group H is slender if and only if it does not contain a copy of Q, P, Z(p) or i, for any prime p. PROOF. If H is slender, then by V.2.1 it does not contain any pureinjective group and hence does not contain a copy of Q, Z(p) or J p ; clearly also it does not contain a copy of P. Conversely, suppose H does not contain a copy of any of these groups, and consider a homomorphism r.p: P --+ H. By 2.3, r.p[Pj is of the form G EB C where C is a countable product and G is cotorsion. By V.2.9, H is cotorsion-free, so G = O. Since H does not contain a copy of P, C must be isomorphic to T" for some finite m. Thus r.p[Pj is a finitely-generated free group, T", By IIL1.10, zm is slender, so
IX.2 Countable products of the integers cp( en) = 0 for all but finitely many n. Since this holds for all is slender. 0
285 ip,
H
In other words, 2.4 says that H is slender if and only if it is cotorsion-free and does not contain a copy of P. 2.5 Corollary. (i) Every Nl-free group which does not contain a copy of P is slender. (ii) Every strongly Nl-free group is slender. Part (i) is immediate from the theorem. For part (ii) we need to prove that a strongly Nl-free group H cannot contain a copy of P. But, by definition, any countable subgroup S of H is contained in a countable Nl-pure subgroup, which must contain the Z-adic closure of S (ef. §I.3). However, P has a countable subset, namely Z(w), whose Z-adic closure in P is uncountable. So H cannot contain a copy of P. 0 PROOF.
Later (in Chapter XIV), we shall construct non-free strongly Nl-free (even Nl-separable) groups which are dual groups. Hence we shall have examples of dual groups which are slender but are not free. We conclude with a characterization, in the same spirit as 2.4, of almost slender groups (see before 111.1.11) . 2.6 Corollary. An abelian group H is almost slender if and only if it does not contain a copy of P or of any unbounded cotorsion group.
If H is almost slender, then it clearly does not contain a copy of P. Suppose that H contained an unbounded cotorsion group G; by V.2.6, G is the homomorphic image of a pure-injective group A; say cp: A -+ G is an epimorphism. It is then easy to construct 'ljJ: Z(w) -+ A such that for all m =I 0, {mcp('ljJ{e n ) ) : nEw} is infinite. Since A is pure-injective, we can extend 'ljJ to ;p: P -+ A; then cp 0 ;p shows that H is not almost slender. Conversely, if H does not contain a copy of P or of any unbounded cotorsion group, consider a homomorphism cp: P -+ H. By 2.3 cp[P] is of the form G EEl C, where G is cotorsion and C is a countable product. By PROOF.
286
IX. QUOTIENTS OF PRODUCTS OF THE INTEGERS
hypothesis, G is bounded and C is a finite product. For some r E Z \ {a}, rr.p[P) ~ C, we can then finish the proof as in 2.4. 0
§3. Uncountable products of the integers We are interested in a generalization of Theorem 2.3 to the case where P is replaced by an uncountable product of Z. By Exercise 2, the full analog of 2.3 fails, but we do have the following result.
3.1 Theorem. Let I\, and ,\ be cardinals which are not w-measurable. Suppose A is a subgroup OfZK which is a homomorphic image ofZ>'. Then ZK / A is the direct sum of a cotorsion group and a product.
Let us denote Z>' by M and ZK by N. Note that M and N are reflexive by Exercise III.9. Let f: M - t N be a homomorphism such that A = f[M). Let i: A - t N be the inclusion map. Then we can factor fast 0 g where g: M - t A is a surjection. We have induced maps ,,*: N* - t A* and q": A* - t M"; the latter is a monomorphism, so A * is free since M* C>! Z(>.) is free. The kernel of t * is denoted A.L (d. section 1), and is a (free) direct summand of N*('::::. Z(K)) because N* jA.L is isomorphic to a subgroup of A*. Therefore by 1.2, N = A.L.L EEl C, where C ~ (A.L)*; hence NjA ~ A.L.L j A EEl C. Now C is a product since A.L is free; so it remains to prove that A.L.L / A is cotorsion. Let B denote the image of the map [": N* - t M*, i.e., B = ]*[N*). Now 1* factors as j 0 h where h: N* - t B is a surjection and j is the inclusion of B into M*. Let D = M* j B, so we have a short exact sequence PROOF.
a - t B .i,M* - t D
-t
a
which induces the top row of the following commutative diagram: M*
1£1:\/ M
L
B*
.i;
N
-t
Ext(D, Z) - t (M*, Z) =
a
1
0
where 0 = ai/ 0 h*. Notice that aM and aN are isomorphisms and that the square commutes because = h* 0 j* and aN 0 f =
r:
287
IX.2 Uncountable products of the integers
1** 0 (J'M.
We claim that the image of () is A1.1.. In fact, this follows from the remarks in the first few paragraphs of section 1 (where we let N* play the role of M and let A1. play the role of A). The map h may be identified with the canonical projection of N* onto N* / A 1. since ker(h) = ker(J*) = A1.. Then h* in turn may be identified with the induced embedding of (N* / A1.)* into N**, whose image is (A1.)1.. Hence the image of h* equals (J'N[A1.1.]. Now () is a monomorphism since it is the composition of two monomorphisms, so Ext(D, Z) ~ B** / im(j*) ~ ()[B*]/()[j*[M**]]
rv
()[B*]/f[M]
and the last term, as we have just seen, is A 1.1./ A. Thus A 1.1./ A is isomorphic to Ext(D, Z), which is cotorsion by V.2.3. 0 In fact, the proof of the theorem gives us a little more information, which we state as a corollary. 3.2 Corollary. For any homomorphism f: Z>" ~ ZK where /'I, and ,\ are non-co-measurable cardinals, there are groups C and D such that C is a product, ker(J) ~ D*, and coker(J) ~ C EB Ext(D, Z).
The only new thing that requires checking is the assertion about the kernel of f. But it is easily verified from the commutativity of the diagram above that (J'M induces an isomorphism between ker(J) and D*. 0 PROOF.
3.3 Corollary. If /'I, is not w-measurable and A is a subgroup of ZK which is a product, then ZK/A ~ C EB Ext(D, Z) where C is a product and D* = O. 0
We will see in Chapter XII that we have, if we assume V = L, a complete structural description of groups of the form Ext(D, Z) and hence of groups of the form ZK / A, where A is as in the results above. (See also Exercise 8.) We-conclude with some notions that will be of use in the next section. 3.4 Definition. For any /'I, ~ ~o, let Z
288
IX. QUOTIENTS OF PRODUCTS OF THE INTEGERS
Otherwise said, Z,.. is the reduced power Z'" IC,.. where C,.. is the CO-I\, filter on I\, (d. II.2.5 and II.3.1). Say that an abelian group G is strongly cotorsion-free if Hom(Z,.., G) = 0 for all regular I\, which are not w-measurable. The structure of Zw is discussed in Exercise VA and ZWI is discussed in Exercise IV.11. 3.5 Lemma. For every PROOF.
K,
of uncountable cofinality, Z,.. is
~l-free.
If A is a countable subgroup of Z,.. = L" IC,.. then by 11.304, ~l-complete. But
A is isomorphic to a subgroup of Z,.. since C,.. is Z,.. is ~l-free (d. IV.2.8), so A is free. 0
3.6 Lemma. IfHom(Zwl' G) = 0, then G is cotorsion-free. Hence, a strongly cotorsion-free group is cotorsion-free. PROOF. Suppose to the contrary that there is a non-zero cotorsion subgroup C of G. Then by V.2.6, C is a homomorphic image of a pure-injective group H; say 'IjJ: H ---+ C is an epimorphism. By 3.5, ZWI is ~l-free, so it contains a countable pure free subgroup F. Let a E C \ {OJ. There is a homomorphism <.p: F ---+ H such that a E ('IjJ 0 <.p) [F]. This homomorphism extends to ZWI since H is pure-injective. But then it is clear that there is a non-zero homomorphism from ZWI to C ~ G, a contradiction. 0
In fact, G is cotorsion-free if and only if Hom(Zw, G) = 0: see Exercise 10. Wald 1983b has proved that, assuming V = L, every cotorsion-free group of cardinality ~1 is strongly cotorsion-free. A group H is called residually slender if for all a E H \ {OJ, there is a subgroup N of H such that a t/:. Nand HI N is slender. For example, any separable group is residually slender, but may not be slender (e.g. ZW). 3.7 Lemma. Every residually slender group is strongly cotorsionfree.
Suppose, to the contrary, that there is a residually slender group H and a non-zero homomorphism <.p: Z,.. ---+ H for some regular, non-w-measurable K,. Say a E im(<.p) \ {OJ. By hypothesis PROOF.
289
IX.4 Radicals and large cardinals
there is a subgroup N of H such that a ~ Nand H/N is slender. If 1r is the canonical surjection: H --+ H / N, then 1r 0 c.p is a non-zero homomorphism: Z" --+ H/N. Hence there is a non-zero homomorphism from Z" to H/N which is zero on Z<"; but this is impossible by 111.3.4, since H/N is slender and /'C is not w-measurable. 0 §4. Radicals and large cardinals We take up a theme which was first introduced in Chapter II: the effect of large cardinal hypotheses on group-theoretical properties. (The results of this section will not be used later, so the reader not interested in them for their own sake may safely skip this section.) Let A denote the class (or category) of abelian groups. A function T: A --+ A is called a preradical (or subfunctor of the identity) if for all groups A, and all group homomorphisms h: A --+ B, T A ~ A and h[TA] ~ T B. (In this context we usually write TA instead of
T(A).) If T is a preradical, it is easy to see that ToT ~ T, i.e., for all A, T 2 A ~ T A. We say that T is a socle if ToT = T. We define a "co-composition", S: T, of preradicals Sand T as follows: for any preradicals Sand T, if p: A --+ A/TA is the canonical surjection, (S: T)A = p-l[S(A/TA)]. It is easy to see that T ~ T: T. We say that T is a radicalifT = T: T, i.e., T(A/TA) = 0 for all A. Using the canonical projections and injections, one can show that for any preradical T, and any groups Ai, T(ffiiEI Ai) = ffiiEI T Ai and T(I1iEI Ai) ~ I1iEI T Ai (see Exercise 9). 4.1 Examples. For any class of groups X and any group A define
SxA = L:{im(c.p):c.p E Hom(X, A), X E X} and RxA
= n{ker(c.p):c.p E Hom(A,
X), X EX}.
We will write SG instead of S{G} and R G instead of R{G}' It is left to the reader to check that Sx is always a socle and R x is always a radical.
290
IX. QUOTIENTS OF PRODUCTS OF THE INTEGERS
For example, if G = Z/pZ, BaA = A[p]def{a E A:pa = O}. If G = EBp,n Z/pnz, BaA = At, the torsion part of A. Also, RQA = At. For any A, AIRzA is torsionless, and, in fact, A is torsionless if and only if RzA = O. The following is easy to verify.
4.2 Proposition. Suppose X is a class of groups and Y -is the class of direct sums of elements of X. Then Rx = Ry. In particular, if F is any non-trivial free group then R F = R z. 0 Let X be the class of N1-free groups. The radical Rx is called the Chase radical and denoted by u, In the rest of this section we shall concentrate mainly on the Chase radical and on Rz.
4.3 Proposition. (i) For any countable group C, vC = RzC. (ii) If C is countable, t/C = C if and only if C* = o. (iii) v is a socle. PROOF. (i) Let F be the free group of countable rank. If ep is any map from C to an N1-free group, then the image of ep is a free group. So vC = RFC, which, by Proposition 4.2, equals RzC. (ii) For any group, C* = 0 if and only if RzC = C, by definition. (iii) We must prove that v ~ v 0 t/, For this it suffices to prove that for all groups A, AlvvA is N1-free, for then a E vA implies that the image of a under the canonical surjection of A onto AIvvA is 0, which means that a E vvA. Now we have a short exact sequence
o ~ vAlvvA ~ AlvvA ~ AlvA ~ 0 and the middle term will be ~l-free if vAlvvA and AlvA are ~l free (d. Exercise IV.5). Thus it suffices to show that for all G, GlvG is N1-free. Let S = {l<: K is a subgroup of G s.t. GIl< is ~l-free}. Then vG = nKes K, Define (): G ~
IT GIl<
KeS
by ()(g)(l<) = g + K, The kernel of () equals vG, so () induces an embedding of GI vG into IlKes GI J(. Since a product of N1-free
IX.4 Radicals and large cardinals
291
groups is Nt-free (d. Exercise IV.5), it follows that GlvG is Nrfree.
o 4.4 Definition. Let T be a preradical and /\, an infinite cardinal. Define T[K)A = 'E{TB: B is a subgroup of A of cardinality < /\'}. It is easy to check that T[K) is a preradical. We say that T satisfies the cardinal condition if there is a /\, such that T[K) = T.
First we will prove that v satisfies the cardinal condition. Then we shall show that whether or not R z satisfies the cardinal condition depends on whether a certain large cardinal exists.
4.5 Theorem. The Chase radical satisfies the cardinal condition; in fact, v = V[Wl). Suppose, to the contrary, that there is a group A and an element a of vA such that a 1:. E{vC: C is a countable subgroup of A}. Let I be the set of all a: C, ---t Z such that C; is a countable subgroup of A, a E Cu, and a(a) #- O. For each x E A, let Sx = {a E I: x E dom a}. We claim that every intersection of countably many of the Sx's is non-empty. Indeed, given X n E A(n E w), let C be a countable subgroup of A containing {x n : nEw}; since a 1:. R z C, there exists a: C ---t Z such that a( a) #- 0; but then a E I and a E nnEw SX n • Thus there is an wt-complete filter F such that Sx E F for all x E A. Define '1/;: A ---t ZI as follows: for all x E A and a E I, PROOF.
'I/;(x)(a)
= {a(x) o
if a E ~x otherwise
Composing 'I/; with the canonical surjection onto ZI I F we obtain a map if;: A ---t ZI/ F which is a homomorphism because Sx+ynSxnSy belongs to F for all x, yEA. Now ZI IF is Nt-free since F is WIcomplete (d. Exercise 11.10), and if;(a) #- 0 by construction, so a 1:. v A, a contradiction. 0
292
IX. QUOTIENTS OF PRODUCTS OF THE INTEGERS
4.6 Corollary. v = Rr d. 0 Now we turn our attention to radicals of the form Rc, and consider the question of whether they satisfy the cardinal condition.
4.7 Theorem. Assume there are no measurable cardinals. Then for every non-zero strongly coiorsion-free G, R c does not satisfy the cardinal condition. Suppose to the contrary that Rc = Rgl for some A. Choose a regular cardinal K 2: A and let A = Z" (= Z" /Z<": see 3.4). Then by hypothesis on G, Hom(A, G) = 0, so RcA = A. However, if B ~ A and IBI < A ~ K, B is isomorphic to a subgroup of Z" by II.3.4. Since RcZ" ~ (RcZ)" = 0 (since G is torsionfree), we conclude that ReB = O. Hence Rg1A = 0 -=I RcA, a contradiction. 0 PROOF.
This result shows that the absence of certain large cardinals implies the failure of the cardinal condition. Another result of Dugas and Gobel 1985a says that if 0# does not exist (see VI.3.15) and GCH holds, then for any cotorsion-free G -=I 0, R c does not satisfy the cardinal condition. On the other hand Eda 1989 has proved that Vopenka's Principle - a very large cardinal hypothesis - implies that every preradical satisfies the cardinal condition. In the next results we shall pin down exactly which large cardinal is needed in order for Rz to satisfy the cardinal condition.
4.8 Theorem. If K is an LAw-compact cardinal and G is a group of cardinality < A, then R c = R~l.
A and let P,,(A), Uy and F,,(A) be defined as in 11.3.9. Let I = P,,(A). Then F,,(A) is a x-complete filter on I, so by hypothesis, there is a A-complete ultrafilter D on I which contains F,,(A). Define 'l/J: A --t IlYEl(Y) as follows: for all a E A and Y E I, PROOF. Fix a group
a Y = {a if a E Y 'l/J( )() 0 otherwise
Then the composition of'l/J with the canonical surjection onto the ultraproduct yields a homomorphism 1[J: A --t IlYEI(Y) / D.
IXA Radicals and large cardinals
293
Now suppose that a ¢:. R~] A; then for all Y E I such that a E (Y), a ¢:. R a (Y) so there exists [v : (Y) - t G such that fy(a) =J O. Note that fy is defined for all Y in a member Uz of D: choose Z so that a E Z. Then the fy's induce a homomorphism f: I1YEl(Y) I D - t GIlD. Now 8a : G - t GIlD is an isomorphism by II.3.2. The composition 8(/0 f 0 if; is a homomorphism from A to G which is non-zero on a. Therefore, a does not belong to RcA. D
4.9 Theorem. Let G be a residually slender group of cardinality ,\ ~ ~o where ,\ is not w-measurable. Let", be an infinite cardinal. Suppose Ra = R~]. Then", is L>..+w-compact. It suffices to show that every x-complete filter F on a set I is contained in an wI-complete ultrafilter D on 1. (For, by 11.2.11, D is then '\+-complete.) Let A = G1IF. We claim that R~] A = O. Indeed, if B is a subgroup of A of cardinality < «, then by 11.3.4, B is embeddable in G1 , so RaB = O. Therefore RaA = O. Hence there is a non-zero homomorphism
f: GI_~G
-t
GIN
such that f(x) =J O. Now we apply Theorem II1.3.2: there are WIcomplete ultrafilters Db ... , D'; on I such that for all a E G 1 , if for all k, supp( a) ¢:. D k then f( a) = O. If we can prove that F is contained in one of the Dk's, then we will be done. So suppose, to the contrary, that this is not the case, i.e., for all k = 1, ... , n, there exists s, E F \ o; Then Sk belongs to F but not to any D k • Define
nk
x'(i)
= {X(i) o
nk
if i E s, otherwise
Then p(x') = p(x), so f(x') = f(x) =J O. But supp(x') ¢:. D k for any k, which contradicts Theorem III.3.2. D
IX. QUOTIENTS OF PRODUCTS OF THE INTEGERS
294
4.10 Corollary. Rz satisfies the cardinal condition if and only if there is an LW1W-compact cardinal. 0
See Exercises 14 and 17 for more conditions equivalent to the existence of an LW1W-compact cardinal. We now turn our attention to the question of when a radical commutes with direct products. Recall that always T TIl Ai ~ TIl T Ai (d. Exercise 9). 4.11 Proposition. If G is cotorsion-free, then RG commutes with countable products.
We must show that if a E TInEw RoAn, then we have a E RoTInEwAn. Given ip: TIn An ---7 G, let () = ep~TIn(a(n)). Since each a(n) E RoAn, the kernel of () contains EBn(a(n)), and hence () induces 8:TIn(a(n))/EBn(a(n)) ---7 G. Now the domain of 8 is pure-injective by V.1.16, so the range of 8 is cotorsion. But then by hypothesis on G, the range of 8 is 0, so ep(a) = O. 0 PROOF.
The converse of 4.11 holds for torsion-free G: see Exercise 13. There are cotorsion-free groups which don't commute with uncountable products: see Exercise 15. 4.12 Theorem. IfG is strongly cotorsion-free, then RG commutes with TIll: whenever /'i, is not w-measurable. PROOF. The proof is by induction on «, The case /'i, = ~o is 4.11. The case of a singular /'i, is straightforward, so we assume that /'i, is regular (not w-measurable) and that for A < /'i, and all groups A tL , TItL<,\ RGAtL ~ Ro TItL<,\ A w Suppose, to obtain a contradiction, that there is an element a of TItL<1I: RoA tL which is not in R G TItL<1I: A w Thus there exists f: TItL<1I: A tL ---7 G such that f(a) =/:. O. Let 'lj;: ZII: ---7 TItL<1I: A tL be defined by 'lj;((ntLL) = (ntL a(J1. ))w The
composition () ~f f 0 'lj; is a homomorphism: 111: ---7 G such that ()( (etL) tL) = f( a) =/:. O. By the inductive hypothesis, () ~Z
---7
G, which is a con-
Measurability is a real barrier, as we see from the next results.
IX.4 Radicals and large cardinals
295
is the first measurable cardinal and G is a strongly cotorsion-free group of cardinality < K, then Ra does not commute with Ill'> .
4.13 Theorem. If
K
Let D be a normal x-complete ultrafilter on K; then S~f{a < K: a is a regular cardinal} belongs to D (d. II.2.15). Let 1= UoEsa x {a}. For X ~ I, let PROOF.
Wx ~{a E S: la \ {I: (J, a) E X}I < a} and let F = {X ~ I: Wx ED}. It is routine to check that F is a x-complete filter on 1. We claim that
(*)
there is a x-complete ultrafilter U on I which contains F.
Assuming this for a moment, let us finish the proof. (Note that the claim follows immediately if K is strongly compact.) Now RaZo = Zo for every a E S because Hom(Zo, G) = 0 by hypothesis on G. We shall show that IloEs ssr; =I- R a IloEs t; by defining a non-zero homomorphism from IloEs t; to G. Since G is torsion-free, we can identify Z with a subgroup of G. Given z E IloEs ZO , let z((3, a)( for (3 < a) denote the (3thcoordinate of z(a) E ZO; so z((3, a) E Z. Then since U is Kcomplete, there is a unique x E Z ~ G such that {((3, a) E I: z((3, a) = x} belongs to U (d. proof of II.3.2). Define ep(z) = x. Since F ~ U, ip: IloEs ZO ---t G is identically 0 on IloEs Z
C x = {, E
K:
{a E S: (J, a) EX} ED}
and let U = {X ~ I: Cx ED}. We leave it to the reader to verify that U is a x-complete ultrafilter on 1. To show that F ~ U we use the normality of D. Let X E F. Define {}: K ---t K by:
{}(a)
= { sup(a \ o
{, E a: (J,a) E X} if a E ~ otherwise
296
IX. QUOTIENTS OF PRODUCTS OF THE INTEGERS
Since X E F, () is a regressive function, so there exists (3 E '" and T ~ S such that TED and ()(a) = (3 for all a E T (d. 11.2.16). Now one can check easily that I E Cx if I > (3; hence x \ (3 ~ Cx , so Cx ED, and therefore X E U. 0 4.14 Corollary. If « is co-measurable, then R z does not commute with TIK' 0
In contrast, we have: 4.15 Theorem. For ansj «, if A is a group whose cardinality is not w-measurable, then Rz(AK) = (RzA)K.
We must show that (RzA)K ~ Rz(AK), i.e., if f: AK - t Z and x E (RzA)K, then f(x) = O. Now f = ((gD)DEV) for some gD: A K / D - t Z where D is an wI-complete ultrafilter on '" (d. III.3.6). But 0A: A - t AK/ D is an isomorphism by 11.3.2, so for all D, gD(XD) = (gD 0 OA)(X(J.l)) for some J.l E «, and therefore gD(XD) = 0 since x(J.l) E RzA. 0 PROOF.
The Chase radical, i/, commutes with countable products: see Exercise 16. Eda 1987 has shown that v does not commute with uncountable products. EXERCISES 1. If M = ZWl, then there is a subgroup A of M such that ALL is not a direct summand of M. [Hint: start with K ~ M* = Z(Wl) such that M* [K is torsionless but not free.] 2. (i) ZWI is not the direct sum of a cotorsion group and a product. [Hint: Assume false. By Exercise IV.Il, (ZWl)* = 0, hence ZWI is cotorsion. Then use the fact that ZWI is ~I- free to obtain from this a contradiction using the results of Chapter V.] (ii) Prove that ZWI has no free summand. 3. (i) If '" is non-co-measurable and A is a subgroup of ZK such that A = AJ..1., then A is a dual group, i.e., A ~ B* for some B. [Hint: let B = the image of t": (ZK)* - t A*.] (ii) For any group B, B* is isomorphic to a subgroup A of ZK for some", such that A = A 1.1..
IX. Exercises
297
(iii) Hence, A is a dual group of non-co-measurable cardinality if and only if it can be embedded in a product ZK so that K is not w-measurable and ZK I A is torsionless.
4. (i) For any group H such that H* = 0, there is a subgroup A of ZK for some K such that if L = ZK lA, then H '" ker(O"L)' [Hint: by 111.3.8 there are arbitrarily large free groups which are also dual groups. So by 3(ii), for some K there is a subgroup F of ZK such that F = Fl..l.. and F has a subgroup A such that FIA C;:,! H.] (ii) There is a non-co-measurable K and a subgroup A of ZK such that if L = ZK I A, L is not the direct sum of a product and a group H such that H* = 0. [Hint: let H be slender such that H* = 0, and use (i).] 5. Use Corollary 2.4 to give another proof that a direct sum of slender groups is slender.
6. (i) A subgroup A of ZW is a countable product if and only if A is a closed subgroup of ZW (topology as in section 2) if and only if there exist elements {an: nEw} of ZW such that for all m, an E Urn for almost all n and A = U:=n rnan: rn E Z}. (ii) There is a closed subgroup A of zw such that ZW fA ~ Jp • [Hint: consider the function cp:Zw ---? J p defined by cp((xn)n) = LnXnpn.] This provides a counterexample to Lemma 95.1 of Fuchs 1973. 7. Prove the converse of 3.2: for any product C and group D both of non-w-measurable cardinality, there exist non-ce-measurable cardinals K and ,\ and a homomorphism f: Z>' ---? ZK such that ker(J) ~ D* and coker(J) ~ C EB Ext(D, Z). 8. Prove the following generalization of 3.1: let R be any ring, let H be any R-module, let M and N be reflexive R-modules and let f: M ---? N be a homomorphism such that f[M]l.. is a direct summand of N* and Ext(M*, H) = O. (Here M* = HomR(M, H).) Then there is a module D, which is a homomorphic image of M* such that ker(J) ~ D* and coker(J) ~ C EB Ext(D, H), where C is a direct summand of N.
298
IX. QUOTIENTS OF PRODUCTS OF THE INTEGERS
9. 1fT is a preradical, then for any groups Ai, T(EBl Ai) = EBl T(A) and T(Ill Ai) ~ III T(A i). [Hint: for ~ use 7rj: Il Ai --+ Aj; for ;2 use £j: A j --+ EBAd 10. A is cotorsion-free if and only if Hom(Zw, A) = O. [Hint: for ({=) use the structural description of Zw in Exercise VA]
11. Let X be a set of groups and let G = EBXEX X and H = IlXEX X , Show that Ra = Rx = R H and SG = S». 12. (i) An annihilator class of groups is a class of groups which is closed under direct products and subgroups. Show that the map which takes a radical R to {A: RA = O} is a one-one correspondence between annihilator classes and radicals. (ii) Formulate an analogous result for socles. 13. If G is torsion-free and R G commutes with countable products, then G is cotorsion-free. [Hint: to show that Jp is not contained in G, use the fact that Jp is a direct summand of a product of cyclic p-groups; to show that G is reduced, use the fact that !lQ is the torsion subgroup functor.]
14. Prove that the following are equivalent. (See Exercise 11.8 for the definition of 11:- torsionless.) (1) Rz = R~l; (2) for every group A, A x-torsionless =} A torsionless; (3) for every group A =I- 0, A x-torsionless =} A* =I- OJ (4) II: is LW1W-compact. [Hint: for the proof of (3) =} (4), use 111.3.2; cr. 4.9.] 15. There is a cotorsion-free group G such that RG does not commute with IlWl' [Hint: Let Ga(a E WI) be a rigid system of groups, i.e., Hom(Ga, G{3) = 0 if a =I- (3; let G = Il, Ga/Il~Wl Ga. Show that RaGa = Ga.] 16. The Chase radical v commutes with countable products. [Hint: cr. proof of 4.11.] For any radical R, define inductively Ra+l = R RJ.1. = na
0
Ra and
299
IX. Notes (i) JlOO is well-defined and is a radical which is a socle; (ii) Ra(A) = 2:{B ~ A: Hom(B, G) = O}; (iii) Hom(R(; A, G) = 0; (iv) if K, is regular, K, is LW1W-compaet if and only if R [Hint: show that (R'Z)[K] = (R~K])OO.]
z = R~[K].
NOTES 1.1 through 1.5 are based on Nunke 1962a. Parts (b) and (c) of Theorem 1.5 come from Dugas-Gobel 1979b. Theorem 1.9 and Corollary 1.11 are due to Huber 1983a. Section 2 is from Nunke 1962b, except for 2.6 which comes from Rychkov 1980 and Cobel-Richkov-Wald 1981. Theorem 3.1 and Corollaries 3.2 and 3.3 are due to Dugas-Gobel 1979b. For further generalizations see also Huber 1979, Dugas-Gobel 1981, and Huber-Warfield 1981. The notion of strongly cotorsion-free and Lemmas 3.6 and 3.7 are from Dugas-Gobel 1985a. For more on homomorphic images of ZK and Z