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Every Coseparable Group May Be Free

arXiv:math/9305205v1 [math.LO] 15 May 1993Every Coseparable Group May Be FreeAlan H. Mekler1Department of Mathematics and StatisticsSimon Fraser UniversityBurnaby, B.C.,V5A 1S6CanadaSaharon Shelah2Institute of MathematicsThe Hebrew UniversityJerusalem, IsraelandDepartment of MathematicsRutgers UniversityNew Brunswick, New Jersey 08903USA1Research supported by NSERC. Research on this paper was begun while thefirst author was visiting the Hebrew University.2Publication #418.

Research supported by the BSF.

AbstractWe show that if 2ℵ0 Cohen reals are added to the universe, then for everyreduced non-free torsion-free abelian group A of cardinality less than thecontinuum, there is a prime p so that Extp(A, Z) ̸= 0. In particular if it isconsistent that there is a supercompact cardinal, then it is consistent (evenwith weak CH) that every coseparable group is free.

The use of some largecardinal hypothesis is needed.

1IntroductionThere are several approximations to being free for abelian groups. (Fromhere on by group we will mean abelian group.) For some of these notionssuch as being hereditarily separable or being a Whitehead group it is wellknown that it is independent from the usual axioms of set theory whetheror not every such group is free.

(A group is separable if every finite set iscontained in a free subgroup which is a direct summand and hereditarilyseparable if every subgroup is separable. A group A is a Whitehead groupif Ext(A, Z) = 0.Any Whitehead group is hereditarily separable.

)Onthe other hand it can be shown in ZFC that there are non-free groups ofcardinality ℵ1 which are ℵ1-separable (i.e., any countable subset is containedin free countable direct summand). A somewhat mysterious class is the classof coseparable groups (to be defined later).

Any hereditarily separable groupis coseparable while every coseparable group is separable. So this class liesbetween those which exhibit independence phenomena and those for whichwe have absolute existence results.Early on Chase [2] showed that CH implied the existence of a non-freecoseparable group of cardinality ℵ1.

Later Sageev-Shelah [7] and Eklof-Huber[3] elaborated this result to construct non-free groups of cardinality ℵ1 withspecified p-rank of Ext (as p ranges over the primes). Their proofs continuedto use CH.

This reliance on CH was in itself unusual. In the cases wherethe existence of a group was known to be independent, the existence wasalso independent of CH.

As well many results in abelian group theory thatfollow from CH can also be shown to follow from weak CH, i.e., 2ℵ0 < 2ℵ1.In this paper we will show that the use of CH is necessary. Namely we willshow that is consistent with weak CH (assuming as always the consistencyof ZFC) that every coseparable group of cardinality ℵ1 is free.

Furthermoreif it is consistent that a supercompact cardinal exists then it is consistent(with weak CH if desired) that every coseparable group is free. A referencefor the facts mentioned above and the ones we will quote below is [4].A group is ℵ1-free if every countable subgroup is free and, more generally,for an uncountable cardinal κ a group is κ-free if every subgroup of cardinalityless than κ is free.

A group A is coseparable if it is ℵ1-free and every subgroupB such that A/B is finitely generated contains a direct summand C of Aso that A/C is finitely generated.There are several equivalents to beingcoseparable. The one we will use throughout is that a group A is coseparable1

if and only if Ext(A, Z) is torsion-free or equivalently Extp(A, Z) = 0 for allprimes p. Here Extp(A, Z) denotes the subgroup of Ext(A, Z) consisting ofthose elements whose order is a power of p. There is a useful characterizationof when Extp(A, Z) = 0. For a group A and a prime p, Extp(A, Z) = 0 ifand only if for every homomorphism h: A →Z/pZ there is a homomorphismˆh: A →Z (called a lifting of h) so that ˆh/p = h. (If f is a homomorphism toZ, then f/p denotes the composition of f with the natural map to Z/pZ.

)Although (at least in some models of set theory) a subgroup of a cosepara-ble group is not necessarily coseparable, every pure subgroup of a coseparablegroup is coseparable. The following proposition sums up the discussion.Proposition 11.

If λ is the minimal cardinality of a non-free cosepa-rable group then every coseparable group is λ-free.2. Any homomorphism from a subgroup H of K to Z/pZ can be extendedto a homomorphism from K to Z/pZ provided that H is p-pure in K,i.e., for any a ∈H, a ∈pH if and only if a ∈pK.2ProofsMost of our set theoretic notation will be standard.

If κ is an infinite cardinalwe will use Fn(I, J, κ) to denote the set of partial functions from I to J whosedomain have cardinality less than κ. The poset Fn(κ, 2, ω) is usually calledthe poset (or forcing) for adding κ Cohen reals.

As is customary we will saythat (at least) κ Cohen reals are added to a set theoretic universe if we forcewith Fn(κ, 2, ω) (or Fn(µ, 2, ω) for some µ ≥κ) The principal lemma we willuse is the following.Lemma 2 Suppose κ is a cardinal and P = Fn(κ, 2, ω). Then P forces thatevery coseparable group of cardinality less than κ is free.This lemma has as immediate consequences our main theorems.Theorem 3 It is consistent with both 2ℵ0 < 2ℵ1 and 2ℵ0 = 2ℵ1, that everycoseparable group of cardinality ℵ1 is free.2

Theorem 4 If it is consistent that a supercompact cardinal exists then it isconsistent with both 2ℵ0 < 2ℵ1 and 2ℵ0 = 2ℵ1, that every coseparable group isfree.Proof.Shelah (see [1]) has shown that if κ is a supercompact cardinaland κ Cohen reals are added to the universe then every κ-free group ofcardinality κ is free. So after adding κ Cohen reals we have, by Lemma 2and Proposition 1, that every coseparable group is κ-free and hence free.

Aswell we have that 2ℵ0 = κ = 2ℵ1.For the other part of the theorem, it suffices to know that the consistencyof a supercompact cardinal implies the consistency of a cardinal κ such thatκ < 2ℵ1 and if κ Cohen reals are added to the universe then every κ-freegroup is free. This result is proved in Corollary 20.

✷We will also show that the use of some large cardinal is necessary. (Theconsistency of the statement “every coseparable group is free” implies theconsistency of the existence of measurable cardinals.) Also we will show bya more complicated forcing that assuming the consistency of a supercom-pact cardinal it is consistent that every coseparable group is free and thecontinuum is as small as possible, i.e., ℵ2.There are various cases to consider in the proof of the main lemma.

Atvarious times we will have to consider the case of a free group F and itsp-adic completion. Fix a set of free generators X for F. Then any elementy of the p-adic completion can be written uniquely as P pnyn where eachyn is a linear combination of elements of X such that all the coefficients arebetween 0 and p−1 (inclusive).

If we have fixed a set of free generators thengiven y, yn will always denote the element above. First we prove two easyand useful propositions.Proposition 5 Suppose A is contained in the p-adic completion of a freegroup F and X is a set of free generators for F. Assume h ∈Hom(A, Z/pZ)and ˆh is a lifting to Z.

If x, y ∈A, ˆh(x) = ˆh(y) and n is the least elementso that xn ̸= yn then h(xn) = h(yn).Proof. By subtracting Pi

So ˆh(xn) ≡ˆh(yn) mod p✷3

Proposition 6 Let A, F and X be as above. Assume that there is a finiteset Y of elements which are not divisible by p and a set B ⊆A so that|B| > |F| and for all x ̸= y ∈B, xn −yn ∈Y , where n is the least naturalnumber so that xn ̸= yn.

Then Extp(A, Z) ̸= 0.Proof. Suppose not.

For each a ∈Y , choose ha so that ha(a) ̸= 0. Let ˆhabe a lifting of ha.

Choose B′ ⊆B so that |B′| > |F| and for all x, y ∈B′ anda ∈Y , ˆha(x) = ˆha(y). By the last proposition for x ̸= y ∈B′ and a ∈Y ,xn −yn ̸= a, where n is the least natural number so that xn ̸= yn.

Thiscontradicts the choice of Y . ✷We will have occasion to use the following ad hoc definition.

If A is anabelian group a function h: A →Z/pZ is an counterexample to the vanishingof Extp(A, Z) if h has no lifting to a map to Z.Theorem 7 Suppose A is contained in the p-adic completion of a free groupF and |A| > |F|. Then after adding at least |F| Cohen reals to the universeExtp(A, Z) ̸= 0.Proof.

By splitting the forcing into two parts (if necessary) we can assumethat exactly |F| Cohen reals are added. Let X be a set of free generators ofF and let h be a Cohen generic function from X to p. More exactly, we canassume the Cohen reals are added by forcing with Fn(X, p, ω).

Let h be ageneric set for this forcing and we will also let h denote the unique extensionto a homomorphism from A to Z/pZ. (There are many other schemes toproduce a generic function.) Suppose ˆh is forced (by some condition) to be alifting of h. For each a ∈A, choose a condition qa and an integer ma so thatqa ⊩ˆh(a) = ma.

Since |A| > max{|X|, |Z|}, there is a condition q so that forsome m and some B of cardinality greater than |F|, q ⊩ˆh(x) = m, for allx ∈B. Let X0 be the domain of q and let Y be the set of linear combinationsof elements of X0 whose coefficients are between −p + 1 and p −1.Suppose first that for all x, y ∈B, xn −yn ∈Y where n is the leastnatural number such that xn ̸= yn.

Then we are done by Proposition 6. Sowe can choose x, y ∈B so that this is not the case.

Choose now q1 extendingq so that q1 ⊩h(xn −yn) ̸= 0. This contradicts Proposition 5.

So no suchlifting exists. ✷The non-existence of a group A which is coseparable and satisfies thehypotheses of the theorem is not a consequence of ZFC.

It is consistent that4

there is a Whitehead group of cardinality ℵ1 which is contained in the 2-adic completion of a countable set [9]. In fact the non-free coseparable groupconstructed by Chase [2], assuming CH, is contained in the Z-adic completionof a countable set.We can strengthen the the previous theorem to allow us to calculate thep-rank.Theorem 8 Suppose A is contained in the p-adic completion of a free groupF and |A| > |F|.Then if λ ≥|F| and λ Cohen reals are added to theuniverse, |Extp(A, Z)| ≥λ.Proof.There are two cases to consider.First assume that we have acollection {Yn: n < ω} of disjoint finite subsets of F so that for all n, noelement of Yn is divisible by p modulo the subgroup generated by Sm̸=n Ym,the subgroup generated by Sn<ω Yn is p-pure in F and for all n there is a setB of cardinality greater than the cardinality of F so that if x, y ∈B thenxi −yi ∈Yn where i is the least natural number so that xi ̸= yi.

In this caseby (the proof of) Proposition 6 we can find elements an ∈Yn such that ifh: A →Z/pZ is a homomorphism which is non-zero on an then h does notlift. There are 2ℵ0 homomorphisms {hi: A →Z/pZ: i < 2ℵ0} such that for alli ̸= j there is n such that hi −hj(an) ̸= 0.

So for any i ̸= j, hi −hj does notlift to a map to Z and hence does not represent 0 in Extp(A, Z). As hi andhj represent different elements of Extp(A, Z), |Extp(A, Z)| = 2ℵ0.For the second case we assume that the first does not hold.

In this casewe can find a pure finitely generated subgroup Y of F so that for all finitesubsets Z of F if every element of Z is not divisible by p modulo Y thenthere is no set B of cardinality greater than |F| so that for all x, y ∈B,xn −yn ∈Z where n is the least natural number so that xn ̸= yn.Weconsider the Cohen reals as giving λ generic functions {hα: α < λ} from Fto Z/pZ which are 0 on Y . In this case we can repeat the second part of theargument in Theorem 7, since for all α ̸= β, hα −hβ is generic.

✷We now turn to the proof of the main Lemma. Since if A is a non-freecoseparable group of minimal cardinality, A is |A|-free, it is enough to showthe result for all λ-free groups A of cardinality λ < 2ℵ0.

By Shelah’s singularcompactness theorem [8], λ is regular. We will consider a λ-filtration of A(i.e., write A as a union of a continuous chain (Aα: α < λ) of pure subgroupsof cardinality less than λ such that for all α, if A/Aα is not λ-free then5

Aα+1/Aα is not free). There will be two cases to consider.

The first andeasier case is dealt with by the following Theorem.Theorem 9 Suppose that |A| = λ and A is λ-free.Also suppose that(Aα: α < λ) is a λ-filtration of A and for a stationary set E, after adding atleast λ Cohen reals, Extp(Aα+1/Aα, Z) ̸= 0 for all α ∈E. Furthermore as-sume that for all α ∈E the non-vanishing of Extp(Aα+1/Aα, Z) is witnessedby a function in the ground model.

Then if we add at least λ Cohen reals,Extp(A, Z) has rank 2λ.Also if weak diamond of E holds then Extp(A, Z) has rank 2λ, providedthat for all α ∈E, Extp(Aα+1/Aα, Z) ̸= 0.Proof. This is a fairly routine weak diamond proof.

In the case where weadd Cohen reals the proof can be done using the definable weak diamond(see [5]).In order to pursue the main case we will give the proof usingthe Cohen reals directly. As well we will only show that Extp(A, Z) ̸= 0.

(The proof that Extp(A, Z) has rank 2λ is similar.) Choose a basis X =∪α<λXα of A/pA so that (under the obvious identifications) Xα is a basis of(Aα+1/Aα)/p(Aα+1/Aα).

For α ∈E, let h1α be a map, in the ground model,from Xα to Z/pZ which does not lift to a homomorphism from Aα+1/Aα toZ (even in the forcing extension). Let h0α denote the homomorphism whichis constantly 0 on Xα.

The important point to notice here is that if f is anyhomomorphism from Aα to Z then at most one of f/p + h0α and f/p + h1αlifts to a homomorphism of Aα+1 which agrees with f on Aα. Otherwise iff0, f1 were two such liftings then f1 −f0 would contradict the choice of h1α.For α /∈E, let h0α and h1α be any functions from Xα to Z/pZ.Now we can use λ Cohen reals to define a map which cannot be lifted.We can assume that the forcing is Fn(λ, 2, ω) and G is a generic functionfrom λ to 2.

Let h from X to Z/pZ be Pα<λ hG(α)α. Suppose ˆh is forced tobe a lifting of h (without loss of generality we can assume that it is forced tobe a lifting by the empty condition).

Since Cohen forcing is c.c.c. we can findα ∈E so that ˆh↾Aα is decided by G↾α.

So there is l = 0, 1 and a condition, q,whose domain is contained in α such that q forces that ˆh↾Aα does not extendto a function which lifts h↾Aα + hlα. If we choose r so that r extends q andr(α) = l then r forces that ˆh is not a lifting of h. This is a contradiction.

✷The last case of the main lemma is the following.6

Theorem 10 Suppose that λ is a regular cardinal and A is a group of car-dinality λ so that A is λ-free and there is a λ-filtration of A so that for astationary set E and all α ∈E there is a ∈Aα+1 \ Aα so that a is in the p-adic completion of Aα. Then if we add at least λ Cohen reals to the universe,Extp(A, Z) ̸= 0.

In fact, |Extp(A, Z)| ≥λ.Proof. In view of the Theorem 7 (or Theorem 8 for the stronger result) wecan assume that the p-adic closure in A of any set of cardinality < λ has size< λ.

Applying Fodor’s lemma we get that all but a non-stationary subset ofE consists of ordinals of cofinality ω. (Otherwise there would be some α sothat the p-adic closure in A of Aα has cardinality λ.) Also by Theorem 7,we can choose the filtration so that for all α the p-adic closure in A of Aα iscontained in Aα+1.Fix a basis X for A/pA.

We now redefine the filtration of A. We canassume A is the union of a λ-filtration where Aα = Nα∩A where (Nα: α < λ)is an increasing sequence of elementary submodels of cardinality less than λof some (H(κ), ∈) with the usual good properties (including X ∈N0).

Sincethe new filtration agrees with the old one on a closed unbounded set we canassume E consists only of ordinals of cofinality ω.Since the p-adic closure in A of any set of cardinality less than λ hascardinality less than λ, X has cardinality λ. We can assume that the forcingis Fn(X, p, ω).

Let h be the generic function from X to Z/pZ. Note that theobvious name for h is in N0.

Suppose that ˆh is a lifting of h.Choose a limit ordinal δ ∈E and let a be an element of Aδ+1 \ Aδ whichis in the p-adic closure in A of Aδ. Choose q which determines ˆh(a), sayit forces the value to be m. Choose α < δ so that q ∩Nδ = q ∩Nα.

Callthis restriction q1. Next take n maximal so that a is pn-divisible moduloAα.

(Since the p-adic closure in A of Aα is contained in Aδ such an n mustexist.) Choose y ∈Aα such that pn | a −y.

Let q2 ∈Nα be an extensionof q1 which determines ˆh(y). Now choose b ∈Aδ so that pn+1 | a −y −b.Notice that pn is the exact power of p which divides b (even modulo Aα).Since pn+1 ∤b + Aα, if we write b/pn as a (infinite) sum of elements of Xthere will be elements whose coefficients are not divisible by p which lieoutside Aα.

Hence there is a condition q3 ∈Nδ extending q2 so that q3 forcesh(b/pn) ̸≡(m −ˆh(y)/pn) mod p. Let r be a common extension of q3 and q.This gives a contradiction.7

To get the stronger statement on the cardinality of Extp(A, Z) it is enoughas we did in Theorem 8 to use the Cohen reals to code λ generic functionsand then note that for any two of them their difference is also generic. ✷Finally we can complete the proof of the main lemma.Proof.

(of Lemma 2) The proof is by induction on λ < κ. We show that anycoseparable group A of cardinality λ is free.

Suppose that A is a coseparablegroup, |A| = λ < κ and that the induction hypothesis is true for all ρ < λ.Assume, for the sake of contradiction that A is not free. By the inductionhypothesis, A is λ-free and by the singular compactness theorem [8] λ isregular.

Let (Aα: α < λ) be a λ-filtration of A. There are two main cases:either for a stationary set E ⊆λ, Aα+1/Aα contains a non-zero divisiblegroup or not.

In the first case we have by Theorem 10 that for every p thereis a counterexample to the vanishing of Extp(A, Z).The second case divides into two subcases. If there is a stationary set E sothat for all α ∈E, Aα+1/Aα is ℵ1-free then by the induction hypothesis andthe fact that there are only countably many primes we can choose a stationarysubset S of E and a prime p so that for all α ∈S there is a counterexampleto the vanishing of Extp(Aα+1/Aα, Z).

By absorbing at most λ of the Cohenreals into the ground model we can assume that A, (Aα: α < λ), E and thecounterexamples to the vanishing of Extp(Aα+1/Aα, Z) are all in the groundmodel. In this case we can apply Theorem 9.In the final subcase there is a stationary set E so that for all α ∈E,Aα+1/Aα is not ℵ1-free and contains no non-zero divisible subgroup.

So forall α ∈E there is a prime pα so that there is a counterexample to thevanishing of Extpα(Aα+1/Aα, Z) (see [4] XII.2.7 for a proof of this standardfact). If we choose a prime p and S a stationary subset of E so that pα = pfor all α ∈S, we can again apply Theorem 9 (as above).

✷Since we have been able to calculate the p-ranks of our groups in all caseswe have the following theorem on the structure of Ext.Theorem 11 Suppose it is consistent that a supercompact cardinal exists.Then it is consistent with either 2ℵ0 = 2ℵ1 or 2ℵ0 < 2ℵ1 that for any group Aeither Ext(A, Z) is finite or Ext(A, Z) has rank at least 2ℵ0.It is natural to ask whether the use of the large cardinals is necessary. Toshow that it is, we will give a construction of a coseparable group (which is8

similar to Chase’s) from principles whose negation implies consistency of theexistence of many measurable cardinals. By the principle, ∗(λ, ℵ0), we willmean that there exist a family {Si: i < λ+} of countable subsets of λ suchthat for any collection I of size λ there exists {S∗i : i ∈I} where for all i ∈ISi \S∗i is finite and for all i ̸= j, S∗i ∩S∗j = ∅.

(See page 157 of [4] for details. )Theorem 12 Suppose that λ is a cardinal, cf(λ) = ω, 2λ = λ+, ∗(λ, ℵ0)holds and for all µ < λ, 2µ < λ.

Then there is a non-free coseparable groupof cardinality λ.Proof. Let {Si: i < λ+} be as guaranteed by ∗(λ, ℵ0) and for each i let{sin: n < ω} be an enumeration of Si.

Let G0 be the group freely generatedby λ (= Si<λ+ Si). Let {fi: i < λ+} enumerate the homomorphisms fromG0 to Z/pZ as p varies over the primes.

We will inductively define a strictlyincreasing chain of free groups Gα and homomorphisms gα from Gα+1 →Zfor α < λ+ so that gα is a lifting of fα, Gα/G0 is divisible and for all β < α,gβ extends to a homomorphism from Gα to Z. (Since Gα/Gβ is divisiblethere is at most one extension.) If we can do this then G = Sα<λ+ Gα is thegroup required.

G is not free. Since G/G0 is divisible, for every prime p everyhomomorphism from G to Z/pZ is uniquely determined by its restriction toG0.

By the construction every homomorphism from G0 to Z/pZ lifts, so Gwill be as required.It remains to do the construction. We will work inside the Z-adic com-pletion of G0 so our use of infinite sums will be justified.

Choose a func-tion h: λ →λ so that the inverse image of any ν has cardinality λ. Foreach α we will choose xαn,0 ̸= xαn,1 so that h(xαn,l) = sαn and define xα =Pn<ω n! (xαn,0 −xαn,1).

The group Gα+1 will be the pure closure of Gα ∪{xα}.At limit ordinals we will take unions. If we do this then for each α the groupGα will be free but the group G will not be free (see [6] or [4] TheoremVII.2.13).Now we make the choices.

Suppose we are at stage α. We have alreadychosen {gβ: β < α} a set of at most λ functions.

Write α = Sn<ω wn, wherethe cardinality of each wn is less than λ and wn ⊆wn+1. Since 2|wn| < λ,for each n we can choose xαn,0 ̸= xαn,1 so that h(xαn,l) = sαn and for all β ∈wn,gβ(xαn,0) = gβ(xαn,1).

Then the homomorphisms extend to the infinite sum(namely xα) since each homomorphism is 0 on all but a finite number of theterms being summed. Finally we must choose gα.

Since Gα+1 is free there issome lifting of fα to Gα+1, so we can choose gα as required. ✷9

As we have mentioned the statement that the hypothesis of the theoremfails at all cardinals implies the consistency of many large cardinals. So theuse of some large cardinal hypothesis was necessary.

One might also wantto know if the continuum needs to be as large as we have made it. Theanswer to this question is no.

Assuming the consistency of the existenceof a supercompact cardinal it is possible to show that it is consistent with2λ = λ+ + ℵ2 that every coseparable group is free. To prove this result wewill divide our effort between a set-theoretic statement and a group theoreticone.

First we need to define a forcing axiom. By Ax+(ℵ1-complete forcing)we mean the statement “if Q is an ℵ1-complete forcing and ˜S is a Q-nameand ⊩Q “ ˜S is a stationary subset of ω1” then for some directed G ⊆Q, {α <ω1: there is q ∈G, q ⊩α ∈˜S} is stationary.

”The point of Ax+(ℵ1-complete forcing) is to give a reflection principlewhich shows that certain groups are not ℵ2-free.Lemma 13 Suppose that Ax+(ℵ1-complete forcing) holds and P is an ℵ2-complete forcing. (By ℵ2-complete we mean that any directed subset of cardi-nality ℵ1 has an upper bound.) If G is P-generic then the following statementis true in the generic extension.Suppose λ is a regular cardinal, A is a group of cardinality λ and{Aα: α < λ} is a λ-filtration of A such thatE = {α: Aα+1/Aα is not ℵ1-free}is stationary.

Then A is not ℵ2-free.Proof. Without loss of generality we can assume that the set underlyingA is λ and ˜A is a P-name for the group structure on λ which is forced bythe empty condition to be as in the hypothesis.

We will show that there is acondition which forces that ˜A is not ℵ2-free. Let Q be the product of P withthe forcing R which adds a generic function ˜f from ω1 onto λ by countableconditions.

Let ˜S be the name of a subset of ω1 such that q ⊩α ∈˜S if andonly if for some β > α, q determines ˜f↾β and the following holds. Let Xbe the determined value of the image of ˜f↾β.

Then q determines the groupstructure of ˜A on X and if Y is the determined image of ˜f↾α then X and Yare given group structures by q and X/Y contains a non-free group of finiterank.10

We need to see that ˜S is forced to be stationary. First recall that for anygroups C and D, where C ⊆D, D/C is not ℵ1-free if and only if there isa non-free subgroup of D/C of finite rank.

Let ˜T be the R-name for ˜SG× ˜H,where ˜H is the canonical name for an R-generic set. It is enough to see thatin V [G], ˜T is forced to be stationary.

This task is quite easy. Let ˜C bean R-name in V [G] for a closed unbounded subset of ω1.

By taking a newfiltration of A we can assume that E consists entirely of ordinals of cofinalityω. Let r be an element of R, i.e., r is a function from some countable ordinalµ to λ.

Choose N an elementary submodel of some appropriate (H(κ), ∈) sothat |N| < λ and N ∩A = Aα for some α ∈E also we can assume that rand everything else we have been talking about are elements of N. Choose acountable subgroup X0 of Aα+1 so that X0+Aα/Aα is not free. It is standardto choose an increasing sequence r = r0, r1, .

. .

of conditions in N ∩R wherethe domain of rn is µn so that there is an increasing sequence (νn: n < ω) ofordinals such that for all n, rn ⊩νn ∈˜C, µn < νn+1 and Sn<ω rge(rn) is asubgroup of Aα which contains X0 ∩Aα. Let rω denote Sn<ω rn.

Finally letrω+1 be an extension of rω so that the range of rω+1 is X0 + rge(rω). Thenrω+1 forces that Sn<ω νn ∈˜C ∩˜T.So by Ax+(ℵ1-complete forcing) there is a directed set H of cardinalityℵ1 such that {α: there is q ∈H, q ⊩α ∈˜S} is stationary.

Let p be an upperbound to the first coordinates of {q: q ∈H}.By the definition of ˜S, Hdetermines a function, f from ω1 to λ and a group structure on the range off which p forces to coincide with a subgroup of ˜A. Further the definition of˜S and the choice of H guarantees that the group structure on the range of fis not free.

✷Lemma 14 Suppose that λ is a regular cardinal, A is not coseparable and|A| = λ. If Q is a λ-complete notion of forcing then Q ⊩A is not coseparable.Proof.

Without loss of generality we can assume that the set underlying Ais λ. Suppose p is a prime and h: A →Z/pZ is a function in the ground modelwhich does not lift.

Suppose that ˜g is forced to be a lifting of h. Choosean increasing sequence of conditions (qα: α < λ) so that qα determines ˜g(α).Then f defined by f(α) = n if qα ⊩˜g(α) = n is a lifting of h. ✷Theorem 15 Suppose that the following statements are true: for all infinitecardinals λ, 2λ = λ+ + ℵ2; every coseparable group of power less than 2ℵ0 is11

free; and Ax+(ℵ1-complete forcing). Then there is a generic extension suchthat(1) cardinalities and cofinalities are preserved,(2) for all infinite cardinals λ, 2λ = λ+ + ℵ2,(3) every coseparable group is free,(4) suppose λ is a regular cardinal, A is a group of cardinality λ and {Aα: α <λ} is λ-filtration of A such that E = {α: Aα+1/Aα is not ℵ1-free} is station-ary.

Then A is not ℵ2-free.Proof. The forcing is an iteration over all regular cardinals greater thanor equal ℵ2, where the support is an initial segment.

The forcing ˜Qλ is thePλ-name for adding a Cohen subset of λ. The result of this forcing we willuse is that ♦(S) holds for all regular λ > ℵ1 and S a stationary subsetof λ.

Properties (1) and (2) are standard. Property (4) has been proved inLemma 13.

It remains to verify that every coseparable group is free. Since theforcing is ω2-complete, we have, by Lemma 14, that every coseparable groupof cardinality at most ℵ1 is free.

Suppose that A is a non-free coseparablegroup of cardinality λ where λ is minimal. (Note that λ > ℵ1.) We canassume that the set underlying A is λ.

By forcing with the initial segmentwhich adds the subsets to µ for µ ≤λ, we can assume we are working in auniverse satisfying (1), (2), (4) where every coseparable group of cardinalityless than λ is free, A is a non-free λ-free group and ♦(S) holds for everystationary subset of λ. Choose a λ-filtration (Aα: α < λ) of A.

The filtrationshould be chosen so that Aα+1/Aα is not free if A/Aα is not λ-free.Let E = {α: Aα+1/Aα is not ℵ1-free}. Then E is not stationary, since by(4), if E were stationary A would not be ℵ2-free.

By the inductive hypothesis,{α: Extp(Aα+1/Aα, Z) ̸= 0} is stationary for some p. (In fact, for all p.) Henceby Theorem 9 and the fact that ♦(S) holds for all stationary subsets of λ weare done. ✷It remains to show the hypothesis is consistent.Theorem 16 Suppose it is consistent that a supercompact cardinal existsthen it is consistent that: for all infinite cardinals λ, 2λ = λ+ + ℵ2; everycoseparable group of power less than 2ℵ0 is free; and Ax+(ℵ1-complete forcing).12

Proof. Since for any ℵ1-complete forcing P, there is some cardinal λ so thatP×Fn(ω1, λ, ω1) is equivalent to Fn(ω1, λ, ω1) it is enough to consider posetsof the form Fn(ω1, λ, ω1) in the verification of Ax+(ℵ1-complete forcing).

Wewill call Fn(ω1, λ, ω1) the forcing for collapsing λ to ω1. Suppose that κ issupercompact.

We can assume that GCH holds in the ground model. Letf : κ →κ be a function such that for any µ and λ there is some M andan elementary embedding of j: V →M so that j(f)(κ) = µ and λM ⊆M.The forcing is an iteration (Pi, ˜Qi: i < κ), where if i = 2j, ˜Qi is the Pi-namefor the forcing collapsing f(j) to ω1 and for i = 2j + 1, Qi is the forcing foradding ℵ1 Cohen reals.

A function p is in Pκ if for all i, p↾i ⊩p(i) ∈˜Qi,the support of p intersect the even ordinals is countable and the support ofp intersect the odd ordinals is finite.It is standard to verify that Ax+(ℵ1-complete forcing) is forced to hold.As well κ is forced to be either ℵ1 or ℵ2. We will show that ℵ1 is preservedby this forcing and that every coseparable group of cardinality ω1 is freein the forcing extension.

(Notice that the first statement is implied by thesecond since otherwise CH would hold and so there would be a non-freecoseparable group of cardinality ℵ1. In fact we will verify that ω1 is preservedas part of the verification that all coseparable groups of cardinality ℵ1 arefree.) We will only sketch the proof and leave details to the reader.

We canreconstruct the iteration differently. For any i, let Ri be the finite supportiteration of ( ˜Q2j+1: 2j + 1 < i).

The forcing Pκ can be rewritten as Rκ ∗˜Sκ,where ˜S is the countable support iteration of ( ˜Q2j: 2j < i). We claim thatif H is an Rκ-generic set and G is Pκ-generic then V [H] and V [G] havethe same countable sets of ordinals.To show this we show by inductionon i, that V [Hi] and V [Gi] have the same countable sets of ordinals.

HereHi (respectively, Gi) denotes the restriction of H (respectively, G) to Ri(respectively, Pi). In particular we have from the induction hypothesis thatfor any λ, Fn(ω1, λ, ω1)V [Hi] = Fn(ω1, λ, ω1)V [Gi].

So we can in the forcingreplace ˜Q2j by the R2j-name for the forcing for collapsing f(j) to ω1, callthis name ˜Q′2j.Any element of ˜Q′2jH2j of the form {{(ˇα, ˇβ)}×Aαβ: α < γ, β < f(j)} whereγ < ω1, for all α and β Aαβ ⊆R2j, for all α, β, τ, if p ∈Aαβ and q ∈Aατ thenp and q are incompatible, and for all α, S β < f(j)Aαβ contains a maximalantichain. Call such names pleasant.

Notice that any pleasant name is forcedby the empty condition to be a function with countable domain from ω1 to13

λ. As well, if q is any name which is forced by the empty condition to bea model then there is a pleasant name which the empty condition forces toextend q.

The important property that we will use is that the union of acountable chain of pleasant names which is forced by the empty condition tobe increasing in stength is again a pleasant name. With this observation inhand, we can prove the claim.

There are two cases to consider, the successorcase and the limit case. We will just do the limit case as the successor one issimilar.

Suppose i is a limit ordinal and ˜g is a Pi-name for a function fromω to the ordinals. It is enough to show that there is a name ˜s in ˜Si so thatthe empty condition in Ri forces that ˜s determines the value of ˜g.

By thediscussion there is a sequence (˜sn: n < ω) such that for all n and 2j in thesupport of ˜sn, ˜sn(2j) is a pleasant name, the empty condition forces that ˜sndetermines the value of ˜g(n) and for all n and 2j in the support of ˜sn, theempty condition (in R2j) forces that ˜sn+1(2j) extends ˜sn(2j). Then ˜s can betaken to be the coordinatewise union of the ˜sn.Similarly, we can prove the following fact.Suppose ˜g is an Pκ-name for a function from ω1 to ω1.

Then thereis a sequence (˜sα: α < ω1) of elements of ˜Sκ, so that the emptycondition in Rκ forces that the sequence is increasing and that ˜sαdetermines ˜g up to α.With this fact in hand we can prove that all coseparable groups of cardi-nality ℵ1 are free. Suppose A is a non-free coseparable group in the genericextension.

By doing an initial segment of the forcing we can assume that Ais in the ground model. By Lemma 2, we know that A is not coseparable inthe generic extension by Rκ.

Let h : A →Z/pZ be a function which does notlift. By adding some of the Cohen reals to the ground model we can assumethat h is in the ground model.

Without loss of generality, we can assumethat the set underlying A is ω1. Suppose that ˜g is forced to be a lifting ofh to Z.

Let (˜sα: α < ω1) be as above. Then if H is Rκ-generic, H togetherwith the sequence determines a function, ϕ from A to Z.

Since the valueof any 3 elements is determined by some condition, this function must be ahomomorphism which lifts h. But ϕ is in V [H] which contradicts the choiceof h.14

32ℵ0-free may imply freeIn [1] it is shown that if κ is a supercompact cardinal and κ Cohen reals areadded to the universe then 2ℵ0-free implies free. In order to prove Theorem 4,we need to know that if we begin with a supercompact cardinal κ and firstadd κ+ Cohen subsets of ω1 and then add κ subsets of ω, then 2ℵ0-free impliesfree.

The proof is relatively standard but we will present it below since weneed the corollary.Lemma 17 Suppose A is a non-free group. Then in any extension of theuniverse which preserves cofinalities and stationary sets, A is non-free.Proof.Consider such an extension and suppose that A is any non-freegroup.

By restricting to a non-free subgroup of A we can assume that Ais λ-free and the cardinality of A is λ or A is countable.By the singu-lar compactness theorem λ is regular. The theorem is by induction on λ.By Pontryagin’s criterion, being non-free is absolute upwards for countablegroups.

(More generally since satisfaction of sentences in Lω1ω is absolutebeing non-free is absolute upwards for countable structures in any variety ina countable language. )Suppose we know the result for all κ < λ and A is λ-free, non-free andof cardinality λ.Then there is a λ-filtration (Aα: α < λ) so that for astationary set E, Aα+1/Aα is not free for all α ∈E.

By induction we knowin the extension Aα+1/Aα is not free. Hence A is not free in the extension.✷The lemma above applies in the more general situation of a variety in acountable language.Lemma 18 Suppose that GCH is true in the ground model and let P =Fn(I, 2, ω1)×Fn(J, 2, ω).

Suppose Q is Fn(µ, 2, ω1)×Fn(ρ, 2, ω) where µ, ρ >ℵ1. If G is Q generic and K = V [G], then in K, P is ℵ2-c.c.

and preservesstationary sets.Proof.It is easy to show that P is ℵ2-c.c. in K, since Q × P is ℵ2-c.c.over the ground model.

For amusement, we will give the proof only withthe assumption that K is an extension by an ℵ2-c.c. notion of forcing.

SinceK is an extension by an ℵ2-c.c. poset, for any large enough regular κ there15

is a club C of subsets of H(κ) such that for all N ∈C, G is generic overN.Suppose now that A is a maximal antichain in P (in K) and ˜A is aname for A. Choose N ≺H(κ) of cardinality ℵ1 so that N is closed undersequences of length ω, ˜A ∈N and G is generic over N. We will show thatany element of P is compatible with an element of N[G] ∩A.

Since N[G]has cardinality ℵ1, this suffices. Consider any element (p1, p2) ∈P.

SinceN is closed under countable sequences, (p1↾(I ∩N), p2↾(J ∩N)) = q is inN. Hence q is compatible with some element r of N[G] ∩A.

Finally r and(p1, p2) are compatible.All that remains to see is that P doesn’t destroy any stationary subsets ofω1 (in K). Let S be any subset of ω1 in K which is forced by some conditionin P to be non-stationary.

Since Q is ℵ2-c.c. we can choose a name ˜S for Swhich uses only Fn((L, 2, ω1) × Fn(T, 2, ω)) where the cardinality of L, T isℵ1.

Finally the homogeneity of Fn((L, 2, ω1) × Fn(T, 2, ω)) × Fn((I, 2, ω1) ×Fn(J, 2, ω)) implies that S is not stationary in K. ✷Theorem 19 Suppose κ is a supercompact cardinal, V satisfies CH andP = Fn(µ, 2, ω1) × Fn(ρ, 2, ω), where µ, ρ > ℵ1. Then P forces that everyκ-free group is free.Proof.

Suppose that G is P-generic and A in V [G] is κ-free. Let λ = |A|.Let ˜A be a name for A.

Choose an embedding j : V →M so that j(κ) > λand M is closed under sequences of length max{κ, µ, λ, ρ}.Let H be aj(P)-generic set which contains j”G.Notice that A is isomorphic to theinterpretation of j” ˜A in M[j”G] and M[H]. (We denote this image as j”A.

)If A is not free then j”A is not free in M[j”G] and hence by the two lemmasabove also not in M[H]. However j extends to an elementary embedding ofV [G] into M[H].

So j(A) is j(κ)-free and hence in M[H], j”A is free. ✷Corollary 20 If it is consistent with ZFC that a supercompact cardinalexists then both of the statements “every 2ℵ0-free group is free and 2ℵ0 < 2ℵ1”and “every 2ℵ0-free group is free and 2ℵ0 = 2ℵ1” are consistent with ZFC.Furthermore, if it is consistent that there is a supercompact cardinal then itis consistent that there is a cardinal κ < 2ℵ1 so that if κ Cohen reals areadded to the universe then every κ-free group is free.16

References[1] Ben David, S. On Shelah’s compactness of cardinals, Israel J. Math.31(1978), 34–56. [2] Chase, S. On group extensions and a problem of J. H. C. Whitehead, inTopics in Abelian Groups, Scott, Foresman and Co., 1963, 173–197.

[3] Eklof, P. and Huber, M. On the p-ranks of Ext(A, G), assuming CH,in Abelian Group Theory, Lecture Notes in Mathematics No. 874,Springer-Verlag 1981, 93–108.

[4] Eklof, P. and Mekler, A. Almost Free Modules:set-theoreticmethods, North-Holland 1990.

[5] Mekler, A. and Shelah, S. Diamond and λ-systems, Fund. Math.131(1988) 45–51.

[6] Mekler, A. and Shelah, S. When κ-free implies strongly κ-free, inAbelian Group Theory (ed. by R. G¨obel & E.A.

Walker), Gordonand Breach, 1987, 137–148. [7] Sageev, G. and Shelah, S. On the structure of Ext(A, Z) in ZFC+, J.Symbolic Logic 50(1985), 302–315.

[8] Shelah, S. A compactness theorem for singular cardinals, free algebras,Whitehead problem and transversals, Israel J. Math.

21, 319–349. [9] Shelah, S. Whitehead groups may not be free even assuming and CH, II,Israel J.

Math. 35(1980), 257–285.17

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