Saharon Shelah- A Combinatorial Principle Equivalent to the Existence of Non-free Whitehead Groups

8/3/2019 Saharon Shelah- A Combinatorial Principle Equivalent to the Existence of Non-free Whitehead Groups

1/23

505

re

vision:1994-03-04

modified:1996-03-19

address: Department of Mathematics, University of California, Irvine, Irvine,CA 92717

email: pceklof@@uci.edu1 2 address: Institute of Mathematics, Hebrew University, Jerusalem 91904,

Israelemail: shelah@@math.huji.ac.ilA Combinatorial Principle]A Combinatorial Principle Equivalent to the Ex-

istence of Non-free Whitehead GroupsAms classification: Primary 20K20, 20K35; Secondary 03E05

1


2/23

505

re

vision:1994-03-04

modified:1996-03-19

[

Saharon Shelah

Abstract

As a consequence of identifying the principle described in the title,

we prove that for any uncountable cardinal , if there is a -free White-

head group of cardinality which is not free, then there are many nice

Whitehead groups of cardinality which are not free.

1 Introduction

Throughout, group will mean abelian group; in particular, free group willmean free abelian group.

Two problems which have been shown to be undecidable in ZFC (ordinaryset theory) for some uncountable are the following:

Is there a group of cardinality which is -free (that is, every subgroupof cardinality < is free), but is not free?

Is there a Whitehead group G (that is, Ext(G,Z) = 0) of cardinality which is not free?

(See [6] for the first, [7] for the second; also [2] is a general reference for unex-plained terminology and further information.)

The second author has shown that the first problem is equivalent to a prob-lem in pure combinatorial set theory (involving the important notion of a -system; see Theorem 3.) This not only makes it easier to prove independenceresults (as in [6]), but also allows one to prove (in ZFC!) group-theoretic resultssuch as:

() if there is a -free group of cardinality which is not free, thenthere is a strongly -free group of cardinality which is not free.

Thanks to Rutgers University for funding the authors visits to Rutgers.The second author was partially supported by the Basic Research Foundation adminis-

tered by The Israel Academy of Sciences and Humanities. Pub. No. 505. Final version.

2


3/23

505

re

vision:1994-03-04

modified:1996-03-19

(See [10] or [2, Chap. VII].) A group G is said to be strongly -free if everysubset of G of cardinality < is contained in a free subgroup H of cardinality< such that G/H is -free. One reason for interest in this class of groupsis that they are precisely the groups which are equivalent to a free group withrespect to the infinitary language L (see [1]).

There is no known way to prove () except to go through the combinatorialequivalent.

As for the second problem, the second author has shown that for = 1,there is a combinatorial characterization of the problem:

there is a non-free Whitehead group of cardinality 1 if and only ifthere is a ladder system on a stationary subset of 1 which satisfies2-uniformization.

(See the Appendix to this paper; a knowledge of the undefined terminology inthis characterization is not needed for the body of this paper.) Again, there aregroup-theoretic consequences which are provable in ZFC:

() if there is a non-free Whitehead group of cardinality 1, thenthere is a strongly 1-free, non-free Whitehead group of cardinality1.

(See [2, XII.3]. It is consistent with ZFC that there are Whitehead groups ofcardinality 1 which are not strongly 1-free.)

Our aim in this paper is to generalize () to cardinals > 1 by combiningthe two methods used to prove () and (). Since the existence of a non-free

Whitehead group G of cardinality 1 implies that for every uncountable cardinal there exist non-free Whitehead groups of cardinality (e.g. the direct sum of copies of G which is not -free), the appropriate hypothesis to consider is:

there is a -free Whitehead group of cardinality which is not free.

By the Singular Compactness Theorem (see [8]), must be regular. It canbe proved consistent that there are uncountable > 1 such that the hypothesisholds. (See [4] or [11].) In particular, for many (for example, = n+1, n )it can be proved consistent with ZFC that is the smallest cardinality of a non-free Whitehead group; hence there is a -free Whitehead group of cardinality which is not free. (See [4].)

Our main theorem is then:

Theorem 1 If there is a -free Whitehead group of cardinality which is notfree, then there are 2 different strongly -free Whitehead groups of cardinality.

Our proof will proceed in three steps. First, assuming the hypothesis callit (A) of the Theorem, we will prove

3


4/23

505

re

vision:1994-03-04

modified:1996-03-19

(B) there is a combinatorial object, consisting of a -system with atype of uniformization property.

Second, we will show that the combinatorial property (B) can be improvedto a stronger combinatorial property (B+), which includes the reshuffling prop-erty. Finally, we prove that (B+) implies

(A+): the existence of many strongly -free Whitehead groups.

Note that we have found, in (B) or (B+), a combinatorial property which isequivalent to the existence of a non-free -free Whitehead group of cardinality, answering an open problem in [2, p. 453]. Certainly this combinatorialcharacterization is more complicated than the one for groups of cardinality 1cited above. This is not unexpected; indeed the criterion for the existenceof -free groups in Theorem 3 implies that the solution to the open problem isinevitably going to involve the notion of a -system. A good reason for assertingthe interest of the solution is that it makes possible the proof of Theorem 1.

In an Appendix we provide a simpler proof than the previously publishedone of the fact that the existence of a non-free Whitehead group of cardinality1 implies the existence of a ladder system on a stationary subset of 1 whichsatisfies 2-uniformization.

We thank Michael OLeary for his careful reading of and comments on thispaper.

2 Preliminaries

The following notion, of a -set, may be regarded as a generalization of thenotion of a stationary set.

Definition 2 (1) The set of all functions

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

(n ) is denoted


5/23

505

re

vision:1994-03-04

modified:1996-03-19

(a) for all S, is a final node of S if and only if

= 0

;(b) if S \ Sf, then S implies , < and

E = { < : S} is stationary in .

(3) A -system is a -set together with a set B for each S such thatB = , and for all S\ Sf:

(a) for all E, |B| < ;(b) {B: E} is a continuous chain of sets, i.e. if

are inE , then B B, and if is a limit point of E, then B ={B: < , E};

(4) For any -system = (S, , B: S), and any S, let B ={Bm: m ()}. Say that a family S = {s: Sf} of countable sets isbased on if S is indexed by Sf and for every Sf, s B .

(5) A family S = {si : i I} is said to be free if it has a transversal, thatis, a one-one function T : I S such that for all i I, T(i) si. We sayS is -free if every subset of S of cardinality < has a transversal.

It can be proved that a family S which is based on a -system is not free.(See [2, Lemma VII.3.6].)

The following theorem now gives combinatorial equivalents to the existenceof a -free group of cardinality which is not free. (See [10] or [2, VII.3].)

Theorem 3 For any uncountable cardinal , the following are equivalent:

1. there is a -free group of cardinality which is not free;

2. there is a family S of countable sets such that S has cardinality and is-free but not free;

3. there is a familyS of countable sets such thatS has cardinality, is -free,and is based on a -system.

Definition 4 A subtree S of


6/23

505

re

vision:1994-03-04

modified:1996-03-19

If = (S,

, B

: S) is a -system of height n, and S = {s

: Sf

}is a family of countable sets based on , let sk = s Bk for 1 k n.

The following is useful in carrying out an induction on -systems.

Definition 6 Given a -system = (S, , B: S) and a node of S, letS = { S: }. We will denote by the -system which is naturallyisomorphic to (S, , B : S

) where B = and B = B if = .

(That is, we replace the initial node, , of S by , and translate the othernodes accordingly.)

If S = {s : Sf} is a family of countable sets based on and Sf, let

s = {sk : k > ()}. LetS

= {s : Sf}; it is a family of countable sets

based on .

In order to construct a (strongly) -free group from a family of countablesets based on a -system, we need that the family have an additional property:

Definition 7 A family S of countable sets based on a -system is said tohave the reshuffling property if for every < and every subset I of Sf suchthat |I| < , there is a well-ordering


7/23

505

re

vision:1994-03-04

modified:1996-03-19

is a pure subgroup of H, and that H/L is a rank one group which is not free(because zr + L is non-zero and divisible by q0q1 qm for all m ) andhence not finitely-generated. But this is impossible if H is free.

Conversely, let C be as stated, and let L be a pure subgroup of rank r. ThenL is free (say with basis z0,...,zr1) and C/L is a non-free torsion-free groupof rank 1. Thus C/L contains a subgroup with a non-zero element zr + L suchthat either: zr + L is divisible by all powers of p for some prime p (in whichcase we let qm = p for all m); or zr + L is divisible by infinitely many primes(in which case we let {qm : m } be an infinite set of primes dividing zr). Itis then easy to see that H exists as desired. 2

3 (A) implies (B)

Theorem 9 For any regular uncountable cardinal , if(A) there is a Whitehead group of cardinality which is -free but not free,then(B) there exist integers n > 0 and r 0, and:

1. a -system = (S, , B : S) of height n;

2. one-one functions k ( Sf, 1 k n) with dom(k ) = ;

3. primes q,m ( Sf, m ); and

4. integers d,m ( Sf, m , < r)

such that

(a) if we define s =n

k=1 rge(k ), then S = {s : Sf} is a

-free family of countable sets based on ; in particular, rge(k ) Bk;

and

(b) for any functions c : Z ( Sf), there is a functionf :

S Z such that for all Sf there are integers a,j (j )such that for all m ,

c (m) = q,ma,m+r+1 a,m+r


8/23

505

re

vision:1994-03-04

modified:1996-03-19

-system following the procedure given in [2, VII.3.4]; we review that procedurehere.

Choose a -filtration ofG, that is, write G as the union of a continuous chain

G =


9/23

505

re

vision:1994-03-04

modified:1996-03-19

is not free. We can assume that for each S\ Sf

and each E

,

B,+1 + B = G, + B, + B .

We can also assume that for all Sf, G has been chosen so that G +B /B has finite rank r + 1 for some r such that every subgroup of rank r is free. By restricting to a sub--set, we can assume that there is an rsuch that r + 1 = r + 1 for all Sf and that there is an n such that hasheight n (cf. Lemma 5). Moreover, we can assume (easing the purity condition,if necessary) that G + B/B is as described in Lemma 8, that is, it isgenerated modulo B by the cosets of elements z,j which satisfy precisely therelations which are consequences of relations

q,mz,m+r+1 = z,m+r +


10/23

505

re

vision:1994-03-04

modified:1996-03-19

The kernel of will be generated by an element e M. We will define tobe the union of a chain of homomorphisms : M A 0 with kernel Ze.The will be defined by induction on . At the same time, we will also define,as we go along, a chain of set functions : A M such that = idA .Let 0 be the zero homomorphism : Ze A0 = {0}.

Suppose that and have been defined for all < for some < ;say () = where =, for some S, E . Suppose first that isa limit ordinal. Let : M


11/23

505

re

vision:1994-03-04

modified:1996-03-19

that

is the identity. We extend

to

: M

A

= B,

by usingthe surjectivity of Ext(A ,Z) Ext(B,1+1 + B ,Z); finally we extend

.

This completes the definition of : M G and of the set map : G M (= the direct limit of the ). Since G is a Whitehead group, there is ahomomorphism : G M such that is the identity on G. In order todefine f, consider an element x of S; x is an ordered pair equal to k (m)(possibly for many different ( , k , m)). Ifg is the second coordinate of x, letf(x) be the unique integer such that

(g) (g) = f(x)e.

Also for any Sf and j define a,j such that

z

,j (z,j ) = a,j e.Then applying to the equation (3) and subtracting the result from equation(3), we obtain

q,m(z,m+r+1 (z,m+r+1)) = (z,m+r (z,m+r))+


12/23

505

re

vision:1994-03-04

modified:1996-03-19

satisfying (a) and (b) as in (B), with the additional properties:

S = {s : Sf} has the reshuffling property; and

for all Sf and k, i , rge(i ) rge(k ) = if i = k.

proof. We shall refer to the data in (B+), with the given properties, as astrong Whitehead -system.

Suppose that = (S, , B : S

), k , q,m , and d

,m is a Whitehead

-system (as in (B)); in particular, S = {s : Sf} is a family of countable

sets based on , where s =n

k=1 rge(k ). In [2, VII.3A] is contained a

proof that if there exists a family S of countable sets based on a -system

which is -free, then there is a family S of countable sets based on a -system

which has the reshuffling property. Our task is to examine the proof and showhow the transformations carried out in the proof can be done in such a way thatthe additional data and properties given in (B) namely the existence of thefunctions k , primes q,m, and integers d

,m satisfying (b) continue to hold.

The transformations in question change the given S and into S and whichare beautiful, that is, they satisfy the following six properties:

1. for , S, if B B = , then there are S and , so that = and = ;

2. for , Sf and k, i , if sk s

i = then k = i, () = n = () for

some n and for all j = k 1,1 (j) = (j);

3. for each k and , sk

is infinite and has a tree structure; that is, for each

there is an enumeration tk0 , tk1 , . . . of s

k so that for all , Sf and

n , if tkn+1 sk , then t

kn s

k ;

4. S is -free;

5. for all E, and S are beautiful; and

6. one of the following three possibilities holds:

(a) every E has cofinality and there is an increasing sequence ofordinals {n: n } approaching such that for all Sf if (0) = then s1 = {n, tn: n } for some tns; moreover, these enumerations

of the s1 satisfy the tree property of (iii);

(b) there is an uncountable cardinal and an integer m > 0 so that for all E the cofinality of is and for all Sf, m = ; moreover, foreach E there is a strictly increasing continuous sequence {: < }

1Note that this corrects an error in [2]. A list of errata for [2] is available from the firstauthor.

12


13/23

505

re

vision:1994-03-04

modified:1996-03-19

cofinal in such that for all Sf

if (0) = then s1

= {(m)

} Xfor some X;

(c) each E is a regular cardinal and = ; moreover, for every Sf, s1 = {(1)} X for some X .

By [2, Thm. VII.3A.6], if S and are beautiful, then S has the reshufflingproperty. Thus it is enough to show that we can transform S and into abeautiful S and and at the same time preserve the additional structure of(B).

Let us begin with property (i). We do not change S (the tree), but for every S \ Sf and E, we replace B

, with B

, {}. Define

k (m) = k (m), k 1 B

k { k 1}.

The definitions of the rest of the data are unchanged. Then (B)(b) continuesto hold since given the c , define f by f(

k (m)) = f

(k (m)), where f is the

function associated with the original data (and the same c ). The function fis well-defined because if k1(m1) =

k2

(m2), then k1

(m1) = k2

(m2). Note

that property (i) implies that rge(i ) rge(k ) = if i = k.

Property (ii) of the definition of beautiful is handled similarly.To obtain property (iii), we do not change S, but for all S we replace

B with


14/23

505

re

vision:1994-03-04

modified:1996-03-19

the original data satisfy (B), f : S Z and a,j

( Sf

, j ) exist.

Let f(k (m)) = f(k()(m)); the (contrapositive of the) final hypothesis on

implies that f is well-defined. Let a,m = a(),m. Then for each Sf, theequation

q(),ma(),m+r+1 = a

(),m+r +


15/23

505

re

vision:1994-03-04

modified:1996-03-19

Define : F Z by setting S = f and (z,j

) = a,j

. We just need tocheck that extends . But for all Sf and m , we have

(w,m ) = q,ma,m+r+1 a,m+r

1 and the result is proved for n 1. Again, let C E

such that is a limit point ofE. Again we have that 1(m) G for all m

when (0) = . We will consider the -system . (See Definition 6.) Note

that has height n 1, and the group G+1/G is defined as in (5) and (5)relative to this -system. Hence by induction G+1/G is not free.

Finally, we will use the reshuffling property given by (B+) to prove that Gis strongly -free. As in the proof of [2, VII.3.11], we will prove that for all {1} and all > , G/G+1 is free. Let I = { Sf: (0) < }, andlet


16/23

505

re

vision:1994-03-04

modified:1996-03-19

To see that the elements ofZ,

generate G

/G+1

, we proceed by inductionwith respect to


17/23

505

re

vision:1994-03-04

modified:1996-03-19

a ladder system on a stationary subset of 1

which has the 2-uniformizationproperty. This was left to the reader in the original paper by the second author[9, Thm. 3.9, p. 277]. The only published proof is a rather complicated one in [2,XII.3]; so considering the importance of this result, it seems to us worthwhileto give another proof which is conceptually and technically simpler than thatone. The proof given here resembles the original proof found by the secondauthor, which was also the basis of the proofs in [3] and in this paper.

Our goal is to prove the following.

Theorem 12 If there is a non-free Whitehead group A of cardinality 1, thenthere is a ladder system { : E} on a stationary set E which has the2-uniformization property.

We begin with an observation. It is sufficient to show that the hypothesis ofTheorem 12 implies that there is a family { : E} of functions which hasthe 2-uniformization property and is based on an 1-filtration, that is, indexedby a stationary subset E of 1 and such that there is a continuous ascendingchain {B : } of countable sets such that for all E, : B.(Note that what we are talking about, in the language of the preceding sections,is a family of countable sets based on an 1-system.) Indeed, by a suitablecoding we can assume that B = and if the range of is not cofinal in, we can choose a ladder on , replace (n) by (n),

(n), and re-

code, to obtain a ladder system on E C, (for some cub C) which has the2-uniformization property.

Fromnow on, let A denote a non-free Whitehead group of cardinality 1.Then we can write A as the union, A =


18/23

505

re

vision:1994-03-04

modified:1996-03-19

2. there is a prime p such that for all E, the type ofA+1

/A

is(0, 0,...0, , 0,...) where the occurs in the pth place.

(See [5, pp. 107ff].) We next give the easy combinatorial lemmas needed for thefirst, and simplest, subcase.

Lemma 13 Suppose Y and Y are finite subsets of an abelian group G suchthat |Y|2 < |Y|. Then there exists b Y such that Y and b + Y are disjoint.[Here b + Y = {b + y : y Y}.]

proof. Choose b Y \ {x y : x, y Y}. 2

Lemma 14 For any positive integer p > 1 there are integers a0 and a1 and afunction Fp : Z/pZ 2 = {0, 1} such that for all m Z with (2|m| + 1)2 < p,

Fp(m + a +pZ) = , for = 0, 1.

proof. Let a0 = 0 and let a1 = b as in Lemma 13, where G = Z/pZ = Y

and Y = {m + pZ : (2|m| + 1)2 < p}. Then since Y = a0 + Y and a1 + Y aredisjoint, we can define Fp. (Note that Fp is a set function, not a homomorphism.)2

proof of theorem 12 (in special subcase (i)): For all E there is aninfinite set P of primes such that

A+1/A = {m

n Q : n is a product of distinct primes from P}.

Then if P = {p,n : n }, A+1 is generated over A by a subset {y,n : n

} satisfying the relations (and only the relations)

() p,ny,n+1 = y,0 g,n

for some g,n A. We define (n) = p,n, g,n. Then { : E} isbased on an 1-filtration, in fact on the chain {Z A : < 1}.

Given functions c : 2, we are going to define a homomorphism :A A with kernel Ze and then use the splitting : A A to define theuniformizing function H.

We define : A A inductively along with a set function : A A

such that = 1A . The crucial case is when and have been defined

and E. (When / E we can use the fact that Ext(A+1,Z) Ext(A,Z)is onto.) We define A+1 by generators {y

,n : n } over A

satisfying

relations() p,ny

,n+1 = y

,0 (g,n) + ae

18


19/23

505

re

vision:1994-03-04

modified:1996-03-19

where a

is as in Lemma 14 for p = p,n

and = c

(n).In the end we let = : A

= A A and = . Then since

A is a Whitehead group, there exists a homomorphism such that = 1A.For any g A, (g) (g) ker() = Ze; we will abuse notation and identify(g) (g) with the unique integer k such that (g) (g) = ke. For anyw E rge(), if w = p,g, let H(w) = Fp((g) (g) +pZ).

Note that w may equal (n) (= p,n, g,n) for many pairs (, n). Tosee that this definition of H works, fix E. For any n , applying toequation () and subtracting from equation (), we have

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

,0 (y,0) ((g,n) (g,n)) + a

so that (g,n) (g,n) is congruent to y,0 (y,0) + a mod p,n. Then if

n is large enough, (2| y

,0 (y,0)| + 1)2

< p,n so by choice of Fp,n and a,

H((n)) = Fp,n((g,n) (g,n) +p,nZ) =Fp,n(y

,0 (y,0) + a +p,nZ) = = c(n).

This completes the proof in the first special subcase.For the purposes of the second special subcase we need another combinatorial

lemma.

Lemma 15 Fix a positive integer p > 1. Define a strictly increasing sequenceof positive integers ti inductively, as follows. Let t0 = 0. If ti1 has been definedfor some i 1, let ti = ti1 + di where di is the least positive integer such that(2pti1 + 1)2p2ti1 < pdi . Then for every i 1 there exists a function

Fi : Z/ptiZ 2

and integers an {0,...,p 1} (ti1 n < ti, = 0, 1) such that whenever|m0| p

ti1 and aj {0,...,p 1} for j < ti1, then for = 0, 1

Fi(m0 +

j


20/23

505

re

vision:1994-03-04

modified:1996-03-19

Then A+1

is generated over A

by a subset {y,n

: n } satisfying therelations (and only the relations)

() py,n+1 = y,n g,n

for some g,n A. Let (m) = g,j : j < tm+1 for all m . Givenfunctions c : 2, we define : A A with kernel Ze inductively alongwith a set function : A A

such that = 1A . The crucial case

is when and have been defined and E. Then we define A+1 bygenerators {y,n : n } over A

satisfying relations

() py,n+1 = y,n (g,n) + a

(n)n e

where (n) is taken to be c(i 1) when ti1 n < ti. In the end we let = : A = A A and = . Then since A is a Whitehead group,there exists a homomorphism such that = 1A. For any w Erge(),ifw = gj : j < ti, let H(w) = Fi(

n


21/23

505

re

vision:1994-03-04

modified:1996-03-19

In other words, A+1

is generated by A

and a subset {z,k

: k = 1,...,r}{y,n : n } modulo (only) the relations in A plus relations:

1. () p,ny,n+1 = y,0 +r

k=1 ,k(n)z,k g,n for some family of distinctprimes p,n and ,k(n) Z , g,n A; or

2. () py,n+1 = y,n +r

k=1 ,k(n)z,k g,n for some ,k(n) Z andg,n A for each n .

For use in (the harder) subcase (ii), define a strictly increasing sequence ofpositive integers ti inductively, as follows. Let t0 = 0. If ti1 has been definedfor some i 1, let ti = ti1 + di where di is the least positive integer such that

(2pti1 + 1)2r+2p2ti1 < pdi .

Then we have the following generalization of Lemma 15. (Note that when r = 0the sequence is empty.)

Lemma 16 Fix p > 1 and r 0. For every sequence of functions =1,...,r, where k : Z and every i 1 there exists a function

Fi, : Z/ptiZ 2

and integers an, {0,...,p 1} (ti1 n < ti, = 0, 1) such that Fi, and

an, depend only on ti (= 1 ti,...,r ti) and are such that wheneverm0,...,mr are integers with |mk| p

ti1 for allk r and aj {0,...,p 1} for

j < ti1, then

Fi,(m0 +r

k=1

(

j


22/23

505

re

vision:1994-03-04

modified:1996-03-19

Lemma 17 Givenp > 1 andr 0, and a sequence of integers = 1

,...,r

,let tp be maximal such that (2tp + 1)2

r+2 < p. Then there exists a function

Fp, : Z/pZ 2

and integers ap, {0,...,p 1} ( = 0, 1) such that whenever m0,...,mr areintegers such that |mk| tp for all k r, then

Fp,(m0 +r

k=1

kmk + a

p, +pZ) = .

2

Now define the function with domain by letting

1. (m) = ,k(m) : k = 1,...,r , p,m, g,m; or

2. (m) = ,k(n) : k = 1,...,r , g,n : n < tm+1.

Given functions c : 2, we define : A A inductively along witha set function : A A such that = 1A . The crucial case is when and have been defined and E. Then we define A+1 by generators{z,k : k = 1,...,r} {y

,n : n } over A

satisfying relations

1. () p,ny,n+1 = y,0 +

rk=1 ,k(n)z

,k (g,n) + a

p,n,

e; or

2. () py,n+1 = y,n +

rk=1 ,k(n)z

,k (g,n) + a

,n,e

where ap,n, (respectively, a,n,) is as in Lemma 17 (respectively, Lemma 16)

for = c(n) (respectively, = c(i 1) if ti1 n < ti ) (and the appropriateprime or primes are used).

In the end we use a splitting of = : A = A

A to define

H(w) as follows:

1. ifw = k : k = 1,...,r , p , g, let H(w) = Fp,((g) (g) +pZ); or

2. ifw = k(n) : k = 1,...,r , gn : n < ti, let H(w) =Fi,(

n


23/23

5

re

vision:1994-03-04

modified:1996-03-19

[2] P. C. Eklof and A. H. Mekler, Almost Free Modules, North-Holland(1990).

[3] P. C. Eklof, A. H. Mekler and S. Shelah, Uniformization and the diversityof Whitehead groups, Israel J. Math 80 (1992), 301-321.

[4] P. C. Eklof and S. Shelah, On Whitehead Modules, J. Algebra 142 (1991),492510.

[5] L. Fuchs, Infinite Abelian Groups, vol II, Academic Press (1973).

[6] M. Magidor and S. Shelah, When does almost free imply free? (For groups,transversal etc.), to appear in Journal Amer. Math. Soc.

[7] S. Shelah, Infinite abelian groups, Whitehead problem and some construc-tions, Israel J. Math 18 (1974), 24325.

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

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

[10] S. Shelah, Incompactness in regular cardinals, Notre Dame J. Formal Logic26 (1985), 195228.

[11] J. Trlifaj, Non-perfect rings and a theorem of Eklof and Shelah, Comment.Math. Univ. Carolinae 32 (1991), 2732.

23

Saharon Shelah- A Combinatorial Principle Equivalent to the Existence of Non-free Whitehead Groups

Documents

Transcript of Saharon Shelah- A Combinatorial Principle Equivalent to the Existence of Non-free Whitehead Groups