saharon shelah and lutz strungmann- on the p-rank of extz(g,z) in certain models of zfc

8/3/2019 Saharon Shelah and Lutz Strungmann- On the p-rank of ExtZ(G,Z) in certain models of ZFC

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On the p-rank of Ext Z (G, Z ) in certain models of ZF C

Saharon Shelah and Lutz Str ungmann

Abstract. We show that if the existence of a supercompact cardinal is consis-tent with ZF C , then it is consistent with ZF C that the p-rank of Ext Z (G, Z )is as large as possible for every prime p and any torsion-free abelian group G .Moreover, given an uncountable strong limit cardinal of countable conalityand a partition of (the set of primes) into two disjoint subsets 0 and 1 ,we show that in some model which is very close to ZF C there is an almost-freeabelian group G of size 2 = + such that the p-rank of Ext Z (G, Z ) equals2 = + for every p 0 and 0 otherwise, i.e. for p 1 .

1. Introduction

In 1977 the rst named author solved the well-known Whitehead problem by show-ing that it is undecidable in ordinary set-theory ZF C wether or not every abelian

group G satisfying Ext Z (G, Z ) = {0} has to be free (see [Sh2 ], [Sh3 ]). However,this did not clarify the structure of Ext Z (G, Z ) for torsion-free abelian groups - aproblem which has received much attention since then. Easy arguments show thatExt Z (G, Z ) is always a divisible group for every torsion-free group G. Hence it isof the form

Ext Z (G, Z ) = p

Z ( p )( p ) Q ( 0 )

for some cardinals p , 0 ( p ) which are uniquely determined. The obviousquestion that arises is which sequences ( 0 , p : p ) of cardinals can appear asthe cardinal invariants of Ext Z (G, Z ) for some (which) torsion-free abelian group?Obviously, the trivial sequence consisting of zero entries only can be realized by anyfree abelian group. However, the solution of the Whitehead problem shows that itis undecidable in ZF C if these are the only ones. There are a few results aboutpossible sequences ( 0 , p : p ) provable in ZF C . On the other hand, assuming

2000 Mathematics Subject Classication. Primary 20K15, 20K20, 20K35, 20K40; Secondary18E99, 20J05.

Number 874 in Shelahs list of publications. The rst author was supported by project No.I-706-54.6/2001 of the German-Israeli Foundation for Scientic Research & Development .The second author was supported by a grant from the German Research Foundation DFG.

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2 SAHARON SHELAH AND LUTZ STR UNGMANN

Godels constructible universe ( V = L) plus there is no weakly compact cardinala complete characterization of the cardinal invariants of Ext Z (G, Z ) for torsion-free abelian groups G has recently been completed by the authors (see [ EkHu ],[EkSh1 ], [GrSh1 ], [GrSh2 ], [HiHuSh ], [MeRoSh ], [SaSh1 ], [SaSh2 ], [ShSt ]and [Sh4 ] for references). In fact, it turned out that almost all divisible groups Dmay be realized as Ext Z (G, Z ) for some torsion-free abelian group G of almost anygiven size.In this paper we shall take the opposite point of view. It is a theorem of ZF C thatevery sequence ( 0 , p : p ) of cardinals such that 0 = 2 0 for some innite0 and p 0 is either nite or of the form 2 p for some innite p can arise asthe cardinal invariants of Ext Z (G, Z ) for some torsion-free G. The rst purpose of this paper is to show that this result is as best as possible by constructing a model

of ZF C in which the only realizable sequences are of this kind. We shall assumetherefore the consistency of the existence of a supercompact cardinal (see [ MeSh ]).This is a strong additional set-theoretic assumption which makes the model we areworking in be far from ZF C .On the other hand, we will also work in models very close to ZF C assuming onlythe existence of certain ladder systems on successor of strong limit cardinals of co-nality 0 . Although this model is close to ZF C it allows us to construct almost-freetorsion-free abelian groups G such that for instance Ext Z (G, Z ) is torsion-free, i.e.G is coseparable. Also this can be considered as a result at the borderline of what isprovable in models close to ZF C since the existence of non-free coseparable groupsis independent of ZF C (see [EkMe , Chapter XII] and [ MeSh ]).

Our notation is standard and we write maps from the right. All groups underconsideration are abelian and written additively. We shall abbreviate Ext Z ( , )by Ext( , ) and will denote the set of all primes. A Whitehead group isa torsion-free group G such that Ext Z (G, Z ) = 0. If H is a pure subgroup of theabelian group G, then we shall write H G. We shall assume sufficient knowledgeabout forcing, large cardinals and prediction principles like weak diamond etc. asfor example in [EkMe ], [Ku ] or [Sh5 ]. Also reasonable knowledge about abeliangroups as for instance in [ Fu ] is assumed. However, the authors have tried to makethe paper as accessible as possible to both algebraists and set theorists.

2. The structure of Ext( G, Z )

In this section we recall the basic results on the structure of Ext( G, Z ) for torsion-free groups G. Therefore let G be a torsion-free abelian group. It is easy to seethat Ext( G, Z ) is divisible, hence it is of the form

Ext( G, Z ) = p

Z ( p )( p ) Q ( 0 )


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ON THE p-RANK OF Ext Z (G, Z ) IN CERTAIN MODELS OF ZF C 3

for certain cardinals p , 0 ( p ). Since the cardinals p ( p ) and 0 completelydetermine the structure of Ext( G, Z ) we introduce the following terminology. LetExt p(G, Z ) be the p-torsion part of Ext( G, Z ) for p . We denote by r e0 (G) thetorsion-free rank 0 of Ext( G, Z ) which is the dimension of Q Ext( G, Z ) and byr e p(G) the p-rank p of Ext( G, Z ) which is the dimension of Ext( G, Z )[ p] as a vectorspace over Z /p Z for any prime number p . There are only a few results provablein ZF C when G is uncountable, but assuming additional set-theoretic assumptionsa better understanding of the structure of Ext( G, Z ) is obtained. For instance, inGodels universe and assuming that there is no weakly compact cardinal a completecharacterization is known. The aim of this paper is to go to the borderline of thecharacterization. On the one hand we shall show that one can make the p-ranksof Ext( G, Z ) as large as possible for every torsion-free abelian group G by working

in a model of ZF C which assumes strong additional axioms (the existence of largecardinals). On the other hand we shall work in a model which is very close to ZF C but still allows to construct uncountable torsion-free groups G such that Ext( G, Z )is torsion-free.We rst justify our restriction to torsion-free G. Let A be any abelian group andt(A) its torsion subgroup. Then Hom( t(A), Z ) = 0 and hence we obtain the shortexact sequence

0 Ext( A/t (A), Z ) Ext( A, Z ) Ext( t(A), Z ) 0

which must split since Ext( A/t (A), Z ) is divisible. Thus

Ext( A, Z ) = Ext( A/t (A), Z ) Ext( t(A), Z ).

Since the structure of Ext( t(A), Z ) = p Hom( A, Z ( p )) is well-known in ZF C

(see [Fu ]) it is reasonable to assume that A is torsion-free and, of course, non-free.Using Pontryagins theorem one proves

Lemma 2.1. Suppose G is a countable torsion-free group which is not free. Then r e0(G) = 2 0 .

Proof. See [EkMe , Theorem XII 4.1].

Similarly, we have for the p-ranks of G the followingLemma 2.2. If G is a countable torsion-free group, then for any prime p, either r e p(G) is nite or 2 0 .

Proof. See [EkMe , Theorem XII 4.7].

This claries the structure of Ext( G, Z ) for countable torsion-free groups G in ZF C .We now turn our attention to uncountable groups. There is a useful characterization


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of r e p(G) using the exact sequence

0 Z p Z Z /pZ 0.

The induced sequence

Hom(G, Z )p

Hom(G, Z /p Z ) Ext( G, Z ) p Ext( G, Z )

shows that the dimension of

Hom(G, Z /p Z )/ Hom(G, Z ) p

as a vector space over Z /p Z is exactly r e p(G).

The following result due to Hiller, Huber and Shelah deals with the case whenHom(G, Z ) = 0.

Lemma 2.3. For any cardinal 0 of the form 0 = 2 0 for some innite 0 and any sequence of cardinals ( p : p ) less than or equal to 0 such that each pis either nite or of the form 2 p for some innite p there is a torsion-free groupG of cardinality 0 such that Hom( G, Z ) = 0 and r e0 (G) = 0 , r e p(G) = p for all primes p .

Proof. See [HiHuSh , Theorem 3(b)].

Together with the following lemma we have reached the borderline of what is prov-able in ZF C .

Lemma 2.4. If G is torsion-free such that Hom(G, Z ) = 0 , then for all primes p,r e p(G) is either nite or of the form 2 p for some innite p | G|.

Proof. See [EkMe , Lemma XII 5.2].

Assuming G odels axiom of constructibility one even knows a complete characteri-zation in the case when Hom( G, Z ) = 0.

Lemma 2.5 (V=L) . Suppose G is a torsion-free non-free group and let B be a subgroup of A of minimum cardinality such that A/B is free. Then r e0 (G) = 2 .In particular, r e0 (G)is uncountable and r e0(G) = 2 | G | if Hom( G, Z ) = 0 .

Proof. See [EkMe , Theorem XII 4.4, Corollary XII 4.5].

Note that the above lemma is not true in ZF C since for any countable divisiblegroup D it is consistent that there exists an uncountable torsion-free group G withExt( G, Z ) = D , hence r e0 (G) = 1 is possible taking D = Q (see [Sh4 ]).

The following result is a collection of theorems due to Grossberg, Mekler, Roslanowski,Sageev and the authors. It shows that under the assumption of ( V = L) almost all


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possibilities for r e p(G) can appear if the group is not of weakly compact cardinalityor singular cardinality of conality 0 .

Lemma 2.6 (V=L) . Let be an uncountable cardinal and suppose that ( p : p )is a sequence of cardinals such that for each p, 0 p 2 . Moreover, let H be a torsion-free group of cardinality . Then the following hold.

(i) If is regular and less than the rst weakly compact cardinal, then thereis an almost-free group G of cardinality such that r e0 (G) = 2 and for all primes p, r e p(G) = p ;

(ii) If is a singular strong limit cardinal of conality , then there is notorsion-free group G of cardinality such that r e p(G) = for any prime p;

(iii) If is weakly compact and re p(H ) for some prime p, then r

e p(H ) = 2

;(iv) If is singular less than the rst weakly compact cardinal and of conality cf( ) > 0 , then there is a torsion-free group G of cardinality such that r e0 (G) = 2 and for all primes p, r e p(G) = p .

Proof. For (i) see [MeRoSh , Theorem 3.7], for (ii) we refer to [ GrSh1 , The-orem 1.0], for (iii) see [SaSh1 , Main Theorem] and (iv) is contained in [ ShSt ].

The above results show that under the assumption of ( V = L) and the non-existenceof weakly compact cardinals, the structure of Ext( G, Z ) for torsion-free groups G of cardinality is claried for all cardinals and almost all sequences ( 0 , p : p )

can be realized as the cardinal invariants of some torsion-free abelian group inalmost every cardinality. However, if we weaken the set-theoretic assumptions toGCH (the generalized continuum hypothesis), then even more is possible whichwas excluded by ( V = L) before (see Lemma 2.5).

Lemma 2.7. The following hold.

(i) Assume GCH . For any torsion-free group A of uncountable cardinality , if Hom( A, Z ) = 0 and r e0(A) < 2 , then for each prime p, r e p(A) = 2 ;

(ii) It is consisitent with ZF C and GCH that for any cardinal 1 , thereis a torsion-free group G such that Hom(G , Z ) = 0 , r e0 (G ) = and for all primes p, r e

p(G ) = 2 1 .

Proof. See [EkMe , Theorem XII 5.3] and [ EkMe , Theorem XII 5.49].

It is our aim in the next section to show that this rich structure of Ext( G, Z ) (Gtorsion-free) which exists in ( V = L) does not appear in other models of ZF C .As a motivation we state two results from [ MeSh ] which show that using Cohenforcing we may enlarge the p-rank of Ext( G, Z ) for torsion-free groups G.


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Lemma 2.8. Suppose G is contained in the p-adic completion of a free group F and |G| > |F |. Then, if | F | and Cohen reals are added to the universe,| Ext p(G, Z )| . In particular, adding 2 0 Cohen reals to the universe impliesthat for every torsion-free reduced non-free abelian group G of cardinality less than the continuum, there is a prime p such that r e p(G) > 0.

Proof. See [MeSh , Theorem 8].

Assuming the consistency of large cardinals we even get more. Recall that a cardinal is compact if it is uncountable regular and satises the condition that for everyset S , every -complete lter on S can be extended to a -complete ultralter onS . This is equivalent to saying that for any set A such that |A| , there exists ane measure on P (A) (the set of all subsets of A of size less than or equal to ).If we require the measure to satisfy a normality condition, then we get a strongernotion. A ne measure U on P (A) is called normal if whenever f : P (A) Ais such that f (P ) P for almost all P P (A), then f is constant on a set inU . A cardinal is called supercompact if for every set A such that |A| , thereexists a normal measure on P (A) (see [EkMe , Chapter II.2] or [ Je , Chapter 6,33. Compact cardinals] for further details on supercompact cardinals).

Lemma 2.9. Suppose that it is consistent that a supercompact cardinal exists. Then it is consistent with either 2 0 = 2 1 or 2 0 < 2 1 that for any group G either Ext( G, Z ) is nite or r e0(G) 2 0 .


3. The free (p-)rank

In this section we introduce the free (p-)rank of a torsion-free group G ( p a prime)which will induce upper bounds for the cardinal invariants of Ext( G, Z ).

Definition 3.1. For a prime p let K p be the class of all torsion-free groups Gsuch that G/p G is free. Moreover, let K 0 be the class of all free groups.

Note, that for G K p ( p a prime) we have G = p G F for some free group F andhence Ext( G, Z )[ p] = 0 since p G is p-divisible. Note that p G is a pure subgroupof G. Thus r e p(G) = 0 for G K p and any prime p. Clearly, also r e0 (G) = 0 for allG K 0 .

Definition 3.2. Let G be a torsion-free group. We call

fr rk0 (G) = min {rk( H ) : H G such that G/H K 0}

the free rank of G and similarly we call

fr rk p(G) = min {fr rk( H ) : H G/p G such that (G/p G)/H K p}


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the free p-rank of G for any prime p .

We have a rst easy lemma.Lemma 3.3. Let G be a torsion-free group and p a prime. Then the following hold.

(i) r e p(G) = r e p(G/p G);(ii) If H is a pure subgroup of G, then r e p(H ) r e p(G) and r e0 (H ) r e0 (G);

(iii) fr rk0(G) fr rk0 (G/p G);(iv) fr rk p(G) = fr rk p(G/p G);(v) fr rk p(G) fr rk0 (G).

Proof. We rst show (i) and let p be a prime. Since p G is pure in G we have

that p

G is p-divisible, hence0 p G G G/p G 0

induces the exact sequence

0 Hom(G/p G, Z /p Z ) Hom(G, Z /p Z ) Hom( p G, Z /p Z ) = 0

the latter being trivial because p G is p-divisible. Thus we have

Hom(G, Z /p Z ) = Hom(G/p G, Z /p Z )

and it follows easily that

Hom(G, Z /p Z )/ Hom( G, Z ) p

= Hom(G/p

G, Z /p Z )/ Hom( G/p

G, Z ) p

.Therefore, r e p(G) = r e p(G/p G).In order to show (ii) we consider the exact sequence

0 H G G/H 0

which implies the exact sequence

Ext( G/H, Z ) Ext( G, Z ) Ext( H, Z ) 0.

Since G/H is torsion-free we conclude that Ext( G/H, Z ) is divisible and henceIm( ) is divisible. Thus Ext( G, Z ) = Ext( H, Z ) Im( ) and therefore r e0 (H )

r e0(G) and r e p(H ) r e p(G) for every prime p.Claim (iii) is easily proved noting that, whenever G = H F for some free groupF , then p G H for every prime p, hence G/p G = H/p G F .To show (iv) note that p (G/p G) = {0} and hence fr rk p(G) = fr rk p(G/p G)easily follows.Finally, (v) follows from (iii), (iv) and the denition of fr rk p(G) and fr rk0 (G)since the class K 0 is contained in the class K p for every prime p.


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Remark 3.4. If G is a torsion-free group and p a prime, then Lemma 3.3 (i)and (iv) imply that, regarding the free p-rank of G, we may assume without loss of generality that G is p-reduced. This is also justied by the fact that

fr rk p(G) = {H G such that G/H K p}

is easily proven.

To simplify notations we let 0 = { 0} in the sequel.

Lemma 3.5. Let G be a torsion-free group. Then the following hold.

(i) r e0 (G) 2 where = max { 0 , fr rk0(G)}; In particular, r e0(G) 2fr rk 0 (G ) if fr rk0(G) is innite;

(ii) r e p(G) pfr rk p (G ) for all p .

Proof. In order to prove (i) choose a subgroup H G such that rk( H ) =fr rk0(G) and G/H K 0 . Hence G = H F for some free group F andso Ext( G, Z ) = Ext( H, Z ) which implies that r e0(G) = r e0 (H ) 2 where =max{ 0 , rk( H )} = max { 0 , fr rk0(G)}.We now prove (ii). Let p be a prime, then r e p(G) = r e p(G/p G) and fr rk p(G) =fr rk p(G/p H ) by Lemma 3.3 (i) and (iv). Hence we may assume that p G = {0}without loss of generality. Let H G be such that fr rk0(H ) = fr rk p(G) andG/H K p. Then G/H = D F for some free group F and some p-divisible groupD . As in the proof of Lemma 3.3 (i) it follows that r e p(G) = r e p(H ). Now, we letH = H F for some free group F such that rk( H ) = fr rk0(H ) = fr rk p(G).

Hence Ext( H, Z ) = Ext( H , Z ) and therefore r e p(G) = r e p(H ) = r e p(H ). Conse-quently, r e p(G) = r e p(H ) prk( H ) = pfr rk p (G ) .

Note, that for instance in ( V = L) for any torsion-free group G, 2fr rk 0 (G ) is theactual value of r e0(G) by Lemma 2.5. The following lemma justies that, as far asit concerns the free p-rank of a torsion-free group, one may also assume withoutloss of generality that fr rk p(G) = rk( G) if p 0 .

Lemma 3.6. Let G be a torsion-free group, p 0 and H G such that

(i) G/ (H F ) K p for some free group F ;(ii) rk( H ) = fr rk p(G).

Then fr rk p(H ) = rk( H ) and r e p(G) = r e p(H ).

Proof. Let G, H and p be given. If p = 0, then the claim is trivially true.Hence assume that p and that rk( H ) = fr rk p(G). Then there is a free groupF such that H = H F is a pure subgroup of G satisfying G/H K p. Thusfr rk p(G) = rk( H ) = fr rk0(H ). Without loss of generality we may assumethat G/H is p-divisible by splitting of the free part. By way of contradictionassume that fr rk p(H ) < rk( H ). Let H 1 H such that H/H 1 K p and


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fr rk0(H 1) = fr rk p(H ) < rk( H ). Then there are a free group F 1 and a p-divisible group D such

H/H 1 = D F 1 .

Choose a pure subgroup H 1 H 2 H such that H 2 /H 1 = D . Thus H/H 2 = F 1and so H = H 2 F 1 and without loss of generality H = H 2 F 1 . Consequently,rk( H 2) = rk( H ) since rk( H ) = fr rk p(G). Let H 3 = H 1 F 1 F . Then

fr rk0(H 3 ) = fr rk0(H 1) < rk( H ) = fr rk p(G).

Moreover,

G/H = (G/H 3) / (H /H 3)

is p-divisible. Since also H /H 3 = H 2 /H 1 is p-divisible and all groups under consid-

eration are torsion-free we conclude that G/H 3 is p-divisible. Hence fr rk p(G) fr rk0(H 3) < fr rk p(G) - a contradiction. Finally, r e p(G) = r e p(H ) follows as inthe proof of Lemma 3.5.

We now show how to calculate explicitly fr rk p(G) for torsion-free groups G of nite rank and p (note that fr rk0(G) can be easily calculated). Recall that atorsion-free group G of nite rank is almost-free if every subgroup H of G of smallerrank than the rank of G is free.

Lemma 3.7. Let G be a non-free torsion-free group of nite rank n and p .Then we can calculate fr rk p(G) as follows.

(i) If G is almost-free, then let H Q be the outer type of G. Then (a) fr rk p(G) = n if H is not p-divisible;(b) fr rk p(G) = 0 if H is p-divisible.

(ii) If G is not almost-free, then choose a ltration {0} = G0 G1 Gm G with Gk+1 /G k almost-free. Then

fr rk p(G) =k 0 whenever we add |G| Cohen reals to theuniverse;


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(iii) If rk( G) 0 , then adding |G| Cohen reals to the universe adds a new member to Ext p(G, Z ) preserving the old ones.

Proof. Let G and p be as stated. Part (i) is an easy application of the rstauthors Singular Compactness Theorem from [ Sh1 ].One implication of (ii) is trivial, hence assume that G K p. By Lemma 3.3 (ii)we may assume that G does not have any pure subgroup of smaller rank than Gsatisfying (ii). It is easily seen that the rank = rk( G) of G must be uncountable.Thus > 0 must be regular by (i). Let G =


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and hence x kyn pn G with kyn H . Therefore x cl p(G, H ) and (ii) holds.Finally, (iii) follows easily from (ii).

4. Supercompact cardinals and large p-ranks

In this section we shall assume that the existence of a supercompact cardinal is con-sistent with ZF C . We shall then determine the cardinal invariants ( r e0 (G), r e p(G) : p ) of Ext( G, Z ) for every torsion-free abelian group in this model and showthat they are as large as possible. We start with a theorem from [ MeSh ] (see also[Da ]). Recall that for cardinals , and we can dene a partially ordered setF n (,, ) by putting

F n (,, ) = {f : dom(f ) : dom(f ) , |dom(f ) | < }.

The partrial order is given by f g if and only if g f as functions.

Lemma 4.1. Suppose is a supercompact cardinal, V is a model of ZF C which satises 2 0 = 1 and P = F n (, 2, 1) F n (, 2, 0), where , > 1 . Then P forces that every -free group is free.


As a corollary one obtains

Lemma 4.2. If it is consistent with ZF C that a supercompact cardinal exists then

both of the statements every 2 0 -free group is free and 2 0 < 2 1 and every 2 0 -free group is free and 2 0 = 2 1 are consistent with ZF C . Furthermore, if it isconsistent that there is a supercompact cardinal then it is consistent that there isa cardinal < 2 1 so that if Cohen reals are added to the universe then every -free group is free.

Proof. See [MeSh , Corollary 20].

We are now ready to prove our main theorem of this section working in the modelfrom Lemma 4.2. Assume that the existence of a supercompact cardinal is consis-tent with ZF C . Let V be any model in which there exists a supercompact cardinal such that the weak diamond principle 3 + holds for all regular cardinals .Now, we use Cohen forcing to add Cohen reals to V to obtain a new model V .Thus, in V we still have 3 + for all regular and also 3 holds. Moreover, wehave 2 0 = and every -free group (of arbitrary cardinality) is free by [ MeSh ].

Theorem 4.3. In any model V as described above, the following is true for every non-free torsion-free abelian group G and prime p .

(i) r e0 (G) = 2 max { 0 ,fr rk 0 (G )} ;


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(ii) If fr rk p(G) is nite, then r e p(G) = fr rk p(G);(iii) If fr rk p(G) is innite, then r e p(G) = 2 fr rk p (G ) .

We would like to remark rst that the above theorem shows that r e p(G) ( p 0 ) isas large as possible for every torsion-free abelian group G in the model describedabove. Moreover, Lemma 2.3 shows that every sequence of cardinals ( p : p 0)not excluded by Theorem 4.3 may be realized as the cardinal invariants of Ext( H, Z )for some torsion-free group H .

Proof. Let p 0 be xed. By Lemma 3.3 we may assume that p G = 0if p . Moreover, Lemma 3.6 shows that also fr rk p(G) = rk( G) holdswithout loss of generality. We now prove the claim by induction on the rank = rk( G) = fr rk p(G).

Case A: is nite.In this case we may assume without loss of generality that Hom( G, Z ) = {0} sinceG is of nite rank. If p = 0, then r e0 (G) = 2 0 = follows from Lemma 2.1 since Gis not free. Thus assume that p > 0. Then r e p(G) is the dimension of Hom( G, Z /p Z )as a vectorspace over Z /p Z . Since Hom(G, Z /p Z ) is the vectorspace dual of G/pG itfollows that r e p(G) = rk( G) = fr rk p(G). Note that G is p-reduced by assumption.

Case B: = 0 .Since G is of countable rank it is well-known that there exists a decomposition G =G F of G where F is a free group and G satises Hom( G , Z ) = {0}. Moreover, byassumption fr rk p(G) = = 0 , hence we obtain that rk( G ) = fr rk p(G ) = 0 .Therefore, r e0 (G) = r e0 (G ) = 2 0 = follows from Lemma 2.1. If p > 0 we concludethat Hom( G , Z /p Z ) has cardinality 2 0 . Since Hom(G , Z ) = {0} it follows byLemma 3.5 (ii) that

2 0 r e p(G ) = r e p(G) 2 0

and thus r e p(G) = 2 0 = for p 0 .

Case C: 0 < < .Note that = 2 0 = 2 , hence r e0 (G) = 2 0 = follows from Lemma 2.5. In fact,by induction hypothesis every Whitehead group H of size less than has to be free(because r e p(H ) = fr rk0(H )) which suffices for Lemma 2.5. Now, assume that p . By Lemma 3.5 (ii) we deduce

r e p(G) 2fr rk p (G ) 2 = 2 0

and hence it remains to prove that r e p(G) 2 0 . The proof is very similar tothe proof of [MeSh , Theorem 8] (see also Lemma 2.8), hence we shall recall itonly briey. Let V be the ground model and P be the Cohen forcing, i.e. P =P ( , 2, ) = {h : Dom(h) {0, 1} : Dom(h) is a nite subset of }. If G


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is a P -generic lter over V let h =g G

g. For notational reasons we may also write

V = V [G ] = V [h] for the extension model determined by the generic lter G . LetA [] such that G belongs to V [h A ]. Without loss of generality we may

assume that A if and only if A whenever + = + . We shall provethe claim by splitting the forcing. For each such that \ A let

f V [A [, + ) ]

be a member of Hom( G, Z /p Z ) computed by h [, + ) . Note that f ex-ists by Proposition 3.8 (iii). Then f is also computed from h [, + ) overV [h \ [, + ) ] and hence f is not equivalent to any f V [h \ [, + ) ]modulo Hom( G, Z ) p . Thus the set of homomorphisms {f : < , A}exemplies that r e p(G) = 2 0 and hence

2 = 2 0 = r e p(G) 2

which shows r e p(G) = 2 = 2 fr rk p (G ) .

Case D: .Let p 0 . We distinguish two subcases. Note that for p our assumptionfr rk p(G) = rk( G) also implies that fr rk0(G) = rk( G) by Lemma 3.3 (v).

Case D1: is regular.The case p = 0 follows as in [EkMe , Theorem XII 4.4]. Let G : < be a

ltration of G into pure subgroups G ( < ) so that if G/G is not -free, thenG +1 /G is not free. Choose by [EkHu , Lemma 2.4] an associate free resolutionof G, i.e. a free resolution

0 K F G 0

of G such that F =


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that Ext( H , Z ) = 0 and hence also Ext( G(E ), Z ) = 0. As in [EkHu , Lemma 1.1](see also [EkMe , Lemma XII 4.2]) there is an epimorphism

Ext( G, Z ) 0. Again, let G : < be a ltration of G into pure sub-groups G ( < ) such that Ext p(G +1 /G , Z ) = 0 if and only if Ext p(G /G , Z ) =0 for some > . Fix < . We claim that G/ cl p(G, G ) is not free. By wayof contradiction assume that G/ cl p(G, G ) is free. Hence G = cl p(G, G ) F for

some free group F . Therefore, G/G = (cl p(G, G ) F ) /G = cl p(G, G )/G F is a direct sum of a p-divisible group and a free group by Lemma 3.10 (iii). It fol-lows that fr rk p(G) rk( G ) < contradicting the fact that fr rk p(G) = .By [MeSh ] we conclude that G/ cl p(G, G ) is not -free since we are working inthe model V . Let G / cl p(G, G ) G/ cl p(G, G ) be a non-free pure subgroup of G/ cl p(G, G ) of size less than . Then there exists < such that G G .By purity it follows that cl p(G, G ) G = cl p(G , G ). Hence

G / cl p(G , G ) = ( G + cl p(G, G )) / cl p(G, G )

is torsion-free but not free. Without loss of generality we may assume that = +1.

Hence we may assume that for all < the quotient G +1 / cl p(G +1 , G ) is atorsion-free non-free group. Note that G +1 / cl p(G +1 , G ) is also p-reduced sincecl p(G +1 , G ) is the p-closure of G inside G +1 .Since the cardinality of G +1 / cl p(G +1 , G ) is less than the induction hypothe-sis applies. Hence r e p(G +1 / cl p(G +1 , G )) = 2 fr rk p (G +1 / clp (G +1 ,G )) . We claimthat Ext p(G +1 / cl p(G +1 , G ), Z ) = {0} stationarily often. If not, then there is acub C such that for all < C we have Ext p(G +1 / cl p(G +1 , G ), Z ) = {0}and equivalently Ext p(G /G , Z ) = 0, hence fr rk p(G ) = fr rk p(G ) by induc-tion hypothesis. As in [ EkMe , Proposition XII 1.5] it follows that Ext p(G/G , Z ) =0 for all C . This easily contradicts the fact that fr rk p(G) = . It fol-lows that without loss of generality for every < there exist homomorphismsh0 , h1 Hom( G +1 , Z /p Z ) such that

(1) h0 clp (G +1 ,G ) = h1 clp (G +1 ,G ) ;(2) There are no homomorphisms g0 , g1 Hom( G +1 , Z ) such that

(a) g0 clp (G +1 ,G ) = g1 clp (G +1 ,G ) (or equivalently g0 G = g1 G );(b) h0 = g0 p and h1 = g1 p.

To see this note that there is a homomorphism : G +1 / cl p(G +1 , G ) Z /p Zwhich can not be factored by p since Ext p(G +1 / cl p(G +1 , G ), Z ) = {0}. Let


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h0 = 0 and h1 be given by

h1 : G +1 G +1 / cl p(G +1 , G ) Z /p Z .

Then it is easy to check that h0 and h1 are as required. In particular we mayassume that h0 = 0 for every < . An immediate consequence is the followingproperty (U).

Let f : G Z /p Z and g : G Z such that g p = f . Then there exists(U) f : G +1 Z /p Z such that f G = f and there is no homomorphism

g : G +1 Z satisfying both g G = g and g p = f .

To see this, letf : G +1

Z/p

Zbe any extension of f which exists by the pureinjectivity of Z /p Z . If f is as required let f = f . Otherwise let g : G +1 Z be

such that g G = g and g p = f . Put f = f h1 . Then f G = f . Assume thatthere exists g : G +1 Z such that g G = g and g p = f . Choosing g1 = g gwe conclude g1 p = h1 contradicting (2). Note that h0 = 0.We now proceed exactly as in [ HiHuSh , Proposition 1] to show that r e p(G) =2fr rk p (G ) = 2 . We therefore recall the proof only briey and for simplicity we evenshall assume that 3 holds. It is an easy exercise (and therefore left to the reader)to prove the result assuming the weak diamond principle only. Assume that r e p(G) = < 2 and let L = {f : < } be a complete list of representatives of elements inHom(G, Z /p Z )/ Hom( G, Z ) p . Without loss of generality let {g : G Z : since = 2 0 is regular. By induction on < wechoose subgroups K of G such that the following hold.

(1) K is a pure non-free subgroup of G;(2) |K | < ;(3) K


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(iii) p is either nite or of the form 2 p for some innite cardinal p .

Proof.The proof follows easily from Lemma 2.3 and Theorem 4.3.

Corollary 4.5. In the model V let p : p 0 be a sequence of cardinals. Then there exists a non-free 1-free abelian group G such that r e p(G) = p for all p 0if and only if p 0 and p = 2 p for some innite cardinal p for every p 0 .

Proof. By Theorem 4.3 we only have to prove the existence claim of thecorollary. It suffices to construct 1-free groups G p for p 0 such that r e p(G) =r e0(G) = 2 0 = and r eq (G) = 0 for all p = q . Then B = G

( 0 )0

p G( p ) p

will be as required (see for instance the proof of [ HiHuSh , Theorem 3(b)]). Fix

p 0 . From [EkMe , Theorem XII 4.10] or [ SaSh2 ] it follows that there existsan 1-free non-free group G p of size 2 0 such that r e p(G p) = 2 0 = if p . In[EkMe , Theorem XII 4.10] it is then assumed that 2 0 = 1 to show that alsor e0(G p) = . However, since we work in the model V and G p is not free it followsfrom Theorem 4.3 that fr rk0(G p) 1 and hence r 0 p(G p) = 2 fr rk 0 (G p ) = .

Recall that a reduced torsion-free group G is called coseparable if Ext( G, Z ) istorsion-free. By [ MeSh ] it is consistent that all coseparable groups are free. How-ever, by [EkMe , Theorem XII 4.10] there exist non-free coseparable groups assum-ing 2 0 = 1 . Note that the groups constructed in Lemma 2.3 are not reduced,hence do not provide examples of coseparable groups.

Corollary 4.6. In the model V there exist non-free coseparable groups.

Proof. Follows from Corollary 4.5 letting 0 = 2 0 and p = 0 for all p.

5. A model close to ZF C

In this section we shall construct a coseparable group which is not free in a modelof ZF C which is very close to ZF C . As mentioned in the previous Section 4 it isundecidable in ZF C if all coseparable groups are free.Let 0 < be a regular cardinal and S a stationary subset of consisting of limit

ordinals of conality . We recall the denition of a ladder system on S (see forinstance [ EkMe , page 405]).

Definition 5.1. A ladder system on S is a family of functions = : S such that : is strictly increasing with sup(rg( )) = , where rg( )denotes the range of . We call the ladder system tree-like if for all , S andevery n, n , (n) = (m) implies n = m and (k) = (k) for all k n.

In order to construct almost-free groups one method is to use -free ladder systems.


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Definition 5.2. Let be an uncountable regular cardinal. The ladder system iscalled -free if for every subset X S of cardinality less than there is a sequenceof natural numbers n : X such that

{ l : n < l < } : X

is a sequence of pairwise disjoint sets.

Finally, recall that a stationary set S with uncountable regular is callednon-reecting if S is not stationary in for every < with cf( ) > 0 .

Theorem 5.3. Let be an uncountable strong limit cardinal such that cf() = and 2 = + . Put = + and assume that there exists a -free tree-like ladder

system on a non-reecting stationary subset S . If = 0 1 is a partition of into disjoint subsets 0 and 1 , then there exists an almost-free group G of size such that

(i) r e0 (G) = 2 ;(ii) r e p(G) = 2 if p 0 ;

(iii) r e p(G) = 0 if p 1 .

Proof. Let = : S be the -free ladder system where S is a stationarynon-reecting subset of consisting of ordinals less than of conality . Withoutloss of generality we may assume that S = . Let pr : 2 be a pairing function,hence pr is bijective and if then we shall denote by (pr 1(), pr 2( )) the uniquepair ( , ) 2 such that pr(pr 1(), pr 2()) = . Let L be the free abelian group

L =


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Fix < and assume that we have constructed ( g , , ) for all < . Choosea function h : such that h () > p for all < and

(5.1) { l : l [h (), )} : <

is a sequence of pairwise disjoint sets. Note that such a choice is possible since theladder system is -free by assumption. Moreover, by (3) the pairing function primplies that also

(5.2) { l , l : l [h (), )} : <

is a sequence of pairwise disjoint sets. Now, we choose the function g such that(2) and (5) hold. For < let n = n(, ) = h (). Since L is free we may chooserst g (x ) satisfying g (x ) + p Z = f (x ) for every such that pr 1() = (l)for all < and l n(, ), that is to say for those such that x does not appearin (5). Secondly, for < , we choose by induction on m n(, ) integers b,m +1such that

0 + p Z = b,m +1 + f (x (m ) ) f (x (m ) ) + p Z

and then choose g (x (m ) ) and g (x (m ) ) such that (5) holds for . Note that thisinductive process is possible by the choice of h and condition (5.1).Finally, let =

n An be the union of an increasing chain of sets An such that

|An | < (recall that we have assumed without loss of generality that S = , so is of conality ). By induction on n < we now may choose (n) and (n) asdistinct ordinals such that

(n), (n) (n), (n) { (m), (m) : m < n } pr 1( (n)) = pr 1( (n)) = (n); g(x (n ) ) : An = g (x (n ) ) : An .

Hence (3), (4) and (6) hold and we have carried on the induction. Now, let G befreely generated by L and {y,n : < ,n } subject to the following relationsfor < and n .

p 1 n

p y,n +1 = y,n + x (n ) x (n ) .

Then G is a torsion-free abelian group of size . Moreover, since the ladder system is -free and S is stationary but not reecting it follows by standard calculationsusing (5.2) that G is almost-free but not free (see for instance [ EkSh2 ]). It remainsto prove that (i), (ii) and (iii) of the Theorem hold. For < let

G = L, y ,n : < , n G

so that G =


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f Hom( G, Z /p Z ). By assumption there is < such that ( p,f L ) = ( p , f ).Inductively we shall dene an increasing sequence of homomorphisms g, : G Z for such that g, p = f G . For = we choose n(, ) andb,m : m [n(, ), ) as in (5) for < . We let g, L = g where g is chosen

as in (1). Moreover, put g, (y,m ) = b,m for m [n(, ), ) and < . Bydownwards induction we chose g, (y,m ) for m < n ( ), < . It is easily seenthat g, is as required, i.e. satises g, p = f G . Now, assume that > . If is a limit ordinal, then let g, =

2 . Consequently, r e p(G) = 2 . Similarly, it follows thatr e0(G) = 2 which nishes the proof.

Corollary 5.4. Let be an uncountable strong limit cardinal such that cf() = and 2 = + . Put = + and assume that there exists a -free ladder system on a stationary subset S . Then there exists an almost-free non-free coseparablegroup of size .

Proof. Follows from Theorem 5.3 letting 0 = and 1 = .

References

[Da] S. Ben David , On Shelahs compactness of cardinals , Israel J. Math. 31 (1978), 3456.[EkHu] P.C. Eklof and M. Huber , On the rank of Ext, Math. Zeit. 174 (1980), 159185.[EkMe] P.C. Eklof and A. Mekler , Almost Free Modules, Set-Theoretic Methods (revised edi-

tion) , Amsterdam, New York, North-Holland, Math. Library.[EkSh1] P.C. Eklof and S. Shelah , The structure of Ext( A, Z ) and GCH : possible co-Moore

spaces , Math. Zeit. 239 (2002), 143157.[EkSh2] P.C. Eklof and S. Shelah , On Whitehead modules , J. Algebra 142 (1991), 492510.[Fu] L. Fuchs , Innite Abelian Groups, Vol. I and II , Academic Press (1970 and 1973).[GrSh1] R. Grossberg and S. Shelah , On the structure of Ext p (G, Z ), J. Algebra 121 (1989),

117128.[GrSh2] R. Grossberg and S. Shelah , On cardinalities in quotients of inverse limits of groups ,

Math Japonica 47 (1998), 189-197.


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[HiHuSh] H. Hiller, M. Huber and S. Shelah , The structure of Ext( A, Z ) and V = L , Math.Zeit. 162 (1978), 3950.

[Je] T. Jech , Set Theory , Academic Press, New York (1973).[Ku] K. Kunen , Set Theory - An Introduction to Independent Proofs , Studies in Logic and the

Foundations of Mathematics, North Holland, 102 (1980).[MeSh] A. Mekler and S. Shelah , Every coseparable group may be free , Israel J. Math. 81

(1993), 161178.[MeRoSh] A. Mekler, A. Roslanowski and S. Shelah , On the p-rank of Ext, Israel J. Math.

112 (1999), 137156.[SaSh1] G. Sageev and S. Shelah , Weak compactness and the structure of Ext( G, Z ), Abelian

group theory (Oberwolfach, 1981), ed. R. G obel and A.E. Walker, Lecture Notes in Mathe-matics 874 Springer Verlag (1981), 8792.

[SaSh2] G. Sageev and S. Shelah , On the structure of Ext( A, Z ) in ZF C + , J. of SymbolicLogic 50 (1985), 302315.

[Sh1] S. Shelah , A compactness theorem for singular cardinals, free algebras, Whitehead problem and transversals , Israel J. Math. 21 (1975), 319349.

[Sh2] S. Shelah , Whitehead groups may not be free even assuming CH, I , Israel J. Math. 28(1977), 193203.

[Sh3] S. Shelah , Whitehead groups may not be free even assuming CH, II , Israel J. Math. 35(1980), 257285.

[Sh4] S. Shelah , The consistency of Ext( G, Z ) = Q , Israel J. Math. 39 (1981), 7482.[Sh5] S. Shelah , Proper and improper forcing , Perspectives in Mathematical Logic, Springer

Verlag (1998).[ShSt] S. Shelah and L. Str ungmann , A characterization of Ext( G, Z ) assuming (V = L ),

submitted.[Tr] J. Trlifaj , Whitehead test modules , Trans. Amer. Math. Soc. 348 (1996), 15211554.

Department of Mathematics, The Hebrew University of Jerusalem, Israel, and Rutgers

University, New Brunswick, NJ U.S.A.

E-mail address : [email protected]

Department of Mathematics, University of Duisburg-Essen, 45117 Essen, Germany


Current address : Department of Mathematics, University of Hawaii, 2565 McCarthyMall, Honolulu, HI 96822-2273, USA


saharon shelah and lutz strungmann- on the p-rank of extz(g,z) in certain models of zfc

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