P. C. Eklof, A. H. Mekler and S. Shelah- Hereditarily Separable Groups and Monochromatic Uniformization

8/3/2019 P. C. Eklof, A. H. Mekler and S. Shelah- Hereditarily Separable Groups and Monochromatic Uniformization

1/23

442

re

vision:2000-10-27

modified:2000-10-30

Hereditarily Separable Groups and

Monochromatic Uniformization

P. C. Eklof

University of California, IrvineA. H. Mekler

Simon Fraser University

S. ShelahHebrew University and

Rutgers University

Abstract

We give a combinatorial equivalent to the existence of a non-free

hereditarily separable group of cardinality 1. This can be used, to-

gether with a known combinatorial equivalent of the existence of a

non-free Whitehead group, to prove that it is consistent that everyWhitehead group is free but not every hereditarily separable group is

free. We also show that the fact that Z is a p.i.d. with infinitely many

primes is essential for this result.

Introduction

An abelian group G is said to be separable if every finite rank pure subgroupis a free direct summand ofG; G is hereditarily separable if every subgroupof G is separable. It is well-known that a Whitehead group is hereditarily

separable. In fact, we have the following implications:

free W-group hereditarily separable

The authors thank Rutgers University for its support.Research partially supported by NSERC grant #9848Research partially supported by the BSF. Publication #442

1


2/23

442

re

vision:2000-10-27

modified:2000-10-30

2

Whiteheads Problem asks if the first arrow is reversible. For each of thetwo arrows, it has been proved by the third author that it is independent ofordinary set theory whether the arrow reverses. (See [7], [9] and [10], or theaccount in [3].)

Now if we consider the two arrows together, there are four possible cases,three of which have already been shown to be consistent:

1. Both arrows reverse. That is, every hereditarily separable group is free.This is true in a model of V = L. (See [3, VII.4.9].)

2. Neither arrow reverses. That is, there are Whitehead groups which are

not free, and hereditarily separable groups which are not Whitehead.This is true in a model of MA + CH. (See [3, VII.4.5, VII.4.6 andXII.1.11].)

3. The second arrow reverses but not the first. That is, every hereditarilyseparable group is Whitehead and there are Whitehead groups whichare not free. This is true in a model of Ax(S) + (1 \ S) plus (E)for every regular > 1 and every stationary subset E of . (See [3,Exer. XII.16].)

4. The first arrow reverses but not the second. That, is every Whitehead

group is free, but there are non-free hereditarily separable groups. Itis an application of the main theorem of this paper that this case isconsistent. (See section 3.)

We also give additional information about the circumstances under whichCases 2 and 3 can occur. (See section 4.) Finally, we show that Case 4 isimpossible for modules over a p.i.d. with only finitely many (but at leasttwo) primes. (See section 5.)

Our methods involve the use of notions of uniformization, which haveplayed an important role in this subject since [8]. (See, for example, [3] and

the recent [4].) It has been proved that there exists a non-free Whiteheadgroup of cardinality 1 if and only if there is a ladder system on a stationarysubset of 1 which satisfies 2-uniformization. (Definitions are given in detailin the next section.) Our main theorem here is the following:


3/23

442

re

vision:2000-10-27

modified:2000-10-30

3

Theorem 1A necessary and sufficient condition for the existence of a non- free hereditarily separable group of cardinality 1 is the existence of a lad-

der system on a stationary subset of 1 which satisfies monochromatic uni-formization for colours.

A ladder system = {: S} on S is said to satisfy monochromaticuniformization for colours if for every function c: S , there is a functionf: 1 such that for every S, f((n)) = c() for all but finitely manyn .

We believe the main theorem is of independent interest aside from its usein Case 4. We will prove sufficiency in section 1 and necessity in section 2.

We will then derive the consistency of Case 4 by standard forcing techniqueslike those used in [10]. (Actually, we need only the sufficiency part of themain theorem for this.) A knowledge of forcing is required only for sections 3and 4.

We would like to thank Bill Wickless for his help in answering a questionabout finite rank torsion-free groups.

Preliminaries

We will always be dealing with abelian groups or Z-modules; we shall simplysay group. A group G is said to be 1-free if every countable subgroup ofG is free, or equivalently, every finite rank subgroup is free. (See [3, IV.2.3];throughout the paper we will usually cite [3] for results we need, rather thanthe original source.) An 1-free group G is separable if and only if everypure subgroup H of finite rank is a direct summand of G, i.e., there is aprojection h: G H (a homomorphism such that hH is the identity on H).The following are two useful facts (cf. [3, IV.2.7 and VII.4.2]):

Lemma 2 (i) An 1-free group G is separable if every pure cyclic subgroupof G is a direct summand of G.

(ii) An1-free group G is hereditarily separable if B is separable wheneverB is a subgroup of G such that G/B is isomorphic to a subgroup ofQ/Z andthere is a finite set P of primes such that the order of every element of G/Bis divisible only by primes in P.


4/23

442

re

vision:2000-10-27

modified:2000-10-30

4

A group G is said to be a Whitehead group if Ext(G,Z) = 0. EveryWhitehead group is separable ([3, XII.1.3]). Since a subgroup of a White-head group is also a Whitehead group, every Whitehead group is hereditarilyseparable.

A group is said to be a Shelah group if it has cardinality 1, is 1-free,and for every countable subgroup B there is a countable subgroup B Bsuch that for every countable subgroup C satisfying C B = B, C/B is free.In that case, we say that B has the Shelah property over B. In [7] and [9]it is proved a consequence of Martins Axiom plus CH that the Whiteheadgroups of cardinality 1 are precisely the Shelah groups.

Notions of uniformization (in our sense) were first defined in [2] and [8].Let S be a subset of lim(1). If S, a ladder on is a function : which is strictly increasing and has range cofinal in . A ladder system onS is an indexed family = {: S} such that each is a ladder on .The ladder system is tree-like if whenever (n) = (m), then n = mand (k) = (k) for all k < n.

For a cardinal 2, a -coloring of a ladder system on S is an indexedfamily c = {c: S} such that c: . A uniformization of a coloringc of a ladder system on S is a pair g, g where g: 1 , g

: S andfor all S and all n g(), g((n)) = c(n). If such a pair exists, we saythat c can be uniformized. We say that (, )-uniformization holds or that

satisfies -uniformization if every -coloring of can be uniformized.A monochromatic colouring c of a ladder system is one such that for

each S, c is a constant function. We shall, from now on, consider amonochromatic colouring with colours to be a function c: S (whichgives the constant value, c(), of the colouring of ). Then a uniformizationof a monochromatic colouring c is a pair f, f where for all S and alln f(), f((n)) = c(). If every monochromatic -colouring of canbe uniformized we say satisfies monochromatic uniformization for colours.

Define a ladder system based on a countable set to be an indexed

family = {: S} such that each is a function from to a fixed count-able set I. We can define notions of colouring and uniformization analogousto those above. (See [3, pp. 367369].) The following two results, thoughstated and proved for ladder systems on a stationary subset of 1, are truealso for ladder systems based on a countable set.

For ease of reference, we include the following result (compare [3, XII.3.3]).


5/23

442

re

vision:2000-10-27

modified:2000-10-30

5

Lemma 3If there is a ladder system on a stationary subset of 1 which sat-isfies -uniformization (resp. monochromatic uniformization for colours),

then there is a tree-like ladder system on a stationary subset of 1 which sat-isfies -uniformization (resp. monochromatic uniformization for colours).

Proof. Suppose {: S} satisfies -uniformization (resp. monochro-matic uniformization for colours). Choose a one-one map from


6/23

442

re

vision:2000-10-27

modified:2000-10-30

6

is the sequence 0c(), 1, 1, 1, . . . , 1

i.e., c() zeroes followed by 2k c() ones. Let g, g be a uniformizationof c. To define f, f, let f() be the least n so that n = 2k+1 where2k > max{c(), g()}. We define f so that for all S and m f(),f((m)) = c(). To see that f is well-defined, consider the case when(m) = (m) and m f

(), f(). Since is tree-like, (j) = (j)for all j m. By definition of f, there is a k such that 2k+1 m, and2k > max{c(), c(), g(), g()}. But then the values of g((j)) for j =2k + 1, . . . , 2k+1 code c() and also c(), so c() = c().

Remark. The proof actually shows that if is tree-like and satisfies 2-uniformization, then satisfies monochromatic uniformization for colours.

1 Sufficiency

Theorem 5 If there is a ladder system on a stationary subset of 1 whichsatisfies monochromatic uniformization for colours, then there is a non-freegroup of cardinality 1 which is hereditarily separable.

Proof.

By hypothesis there is a stationary subset S of 1 and a laddersystem = {: S} such that every monochromatic colouring with colours can be uniformized. By Lemma 3, without loss of generality we canassume that is tree-like.

We begin by defining the group. Let pn (n < ) be an enumeration of theprimes. The group G will be generated by {x: < 1}{y,n: S, n < },subject to the relations

pny,n+1 = y,n + x(n)

For any , we let G denote the subgroup ofG generated by {x, yn: , (n) > }. There is a surjective map : F B whichis the identity on B G and such that (u) = x g(x) and (w,n) =y,n g(y,n). The kernel K of is generated by elements of the form (w,n +

u(n)pnw,n+1)+(g(y,n)+g(x(n))png(y,n+1)). Let h: F Z be defined

so that hB G = h1, h(u) = h(xg(x)) and h(w,n) = h(y,ng(y,n)).

Since h is constantly 0 on K, it induces a homomorphism from B to Z whichagrees with h on the generators of B.

Remark. The same proof works with any tree-like ladder system basedon a countable set. (The assumption that the ladder system is tree-like is

necessary, as witnessed by Hausdorff gaps). In particular, if there is a set of1 branches through the binary tree of height which satisfies monochro-matic uniformization for colours, then the group built from these branchesis hereditarily separable. This group is just the group constructed in [3,VII.4.3]; there it is shown that MA + CH implies this group is hereditarilyseparable. Given these comments, one might expect that it is possible toshow that MA + CH implies that any system of 1 branches through thebinary tree satisfies monochromatic uniformization for colours. Indeed,this is the case: given a set of 1 branches and a monochromatic colouringc by colours, let the poset, P, consist of pairs (s, B) where s is a functionfrom n2 and B is a finite subset of the branches such that for all b B,s(bn) = c(b). If (t, c) P and dom(t) = m2, we define (t, C) (s, B) iffs t, B C and for all b B and n k m, t(bk) = s(bn) = c(b). Theproof that for each n, {(s, B): n2 dom(s)} is dense uses the fact that thecolouring is monochromatic. On the other hand the poset is c.c.c., since anytwo conditions with the same first element are compatible.


9/23

442

re

vision:2000-10-27

modified:2000-10-30

9

2 NecessityThe following lemma can be derived as a consequence of the fact that theRichman type of a finite rank torsion free group is well-defined (see [6] or[5]); but for the convenience of the reader we give a self-contained proof.

Lemma 7 Suppose A is a torsion free group of rank r + 1 and every rank rsubgroup is free. If B and C are pure rank r subgroups, then the type of A/Bis the same as the type of A/C.

Proof. The proof is by induction on r. We can assume that r 1 and

B = C. Consider first the case r = 1; then B C = 0. If b B and c Care generating elements, then A Qb Qc and it is enough to prove that mdivides b (mod C) if and only if m divides c (mod B). Now if m dividesb (mod C), then b = ma + nc for some a A and n Z. Since B = b ispure in A, m and n must be relatively prime. Hence there exist s, t Z suchthat ns + mt = 1. But then c = m(tc sa) + sb, so m divides c (mod B).

Now suppose r > 1. Consider B C; since r + 1 = rk(B + C) = rk(B) +rk(C) rk(B C) and 2r > r + 1, we have that rk(B C) 1. SinceB C is a pure free subgroup of A we can find x B C which is a puresubgroup of A. Note that A/x has the property that every subgroup of

rank r 1 is free. Now apply the induction hypothesis to A/x, B/x andC/x.

Theorem 8 If there is a non-free hereditarily separable group G of cardi-nality 1, then there is a ladder system on a stationary subset of 1 whichsatisfies monochromatic uniformization for colours.

Proof. We can write G =


10/23

442

re

vision:2000-10-27

modified:2000-10-30

10

p and an element y

+ F of M which is not divisible by p. Without lossof generality (again using [3, II.4.5]), we can assume that there is a prime psuch that p = p for all S such that M is not divisible.

For each S, let {y : 0 r} G+1 be such that {y + G: 0

r 1} is a basis of F/G, yr / F and y

r + F is not divisible by p in M if

M is not divisible. Then G+1 is generated by G {y : r} {z

n: n },

where the zn satisfy equations

rnzn =

r

s,n y + g

n

where gn G and rn, s,n Z.Define functions on for each S by:

(n) = gm, s

,m , r

m: r, m n.

Notice that (n) determines the isomorphism type of the finitely generatedsubgroup ofG+1/G generated by (the cosets of) {y

: r} {z

m: m n}.

As in [3, XII.3] see especially Theorem XII.3.3 and the beginning ofthe proof of XII.3.1 (p. 381) if we show that = {: S} satisfiesmonochromatic uniformization for colours, then there is a ladder system

on a stationary subset of 1 with the same property.So fix a monochromatic colouring c: S of . We are going to use c todefine a subgroup B of G with a pure cyclic subgroup Z. By the hypothesison G, there will be a projection h: B Z. Because of the way we define Bwe will be able to use h to define f: {(n): S, n } such that foreach S, f((n)) = c() for all but finitely many n .

We will define a continuous chain of subgroups B of G by induction on and let B = 1B. To begin, let {xn: n } be a basis of G0, and letB0 be the subgroup of G0 generated by {px0} {pxn+1 xn: n }. Thus

G0/B0 = Z(p) and Z

def= Zpx0 is a pure subgroup of B0.

Let A = {tn: n } G0 be a complete set of representatives of G0/B0such that t0 = 0. For each pair (d, a) where d > 0 and a A, fix an element[d, a] such that dt[d,a] + B0 = a + B0.

We will define the B so that for all

1. B + G0 = G and


11/23

442

re

vision:2000-10-27

modified:2000-10-30

11

2. for all < , B G = B.

Notice then that G/B = G0/B0, and Z is pure in each B.The crucial case is when we have defined B already and S. We will

define B,m by induction on m and then let B+1 = mB,m. Let

B,0 = B {y : < r} {y

r tc()}.

Then B,0 G = B since {y : r} is independent mod G. Suppose B,m

has been defined so that B,m G = B. Thus (B,m + G)/B,m = G0/B0.Let dm > 0 be minimal such that dmz

m B,m + G. If dmz

m am A

(mod B,m), let B,m+1 = B,m + Z(zm t[dm,am]).

Then we will have B,m+1 G = B. So, in the end, B+1 G = B.Moreover, B+1 + G0 = G+1, because, by construction, every generator ofG+1 belongs to B+1 + G0.

If / S, the construction of B+1 is essentially the same, except that thecolouring c plays no role; we begin with a set Y G+1 which is maximalindependent mod G and let B,0 = B Y; then define B,m by inductionas before (using a well-ordering of type of a set of generators of G+1 modG). This completes the description of the construction of B.

Now fix a projection h: B Z and fix a well-ordering, , of Zr+1 of

order type . We are going to define the uniformizing function f. We mustdefine f() for each of the form (n). (Note that there may be many such that = (n).) Suppose

= gm, sm , rm: r, m n.

Let = () be minimal such that gm G for all m n. For each k

we can construct a group B(k) just as in the construction of B+1, which is

generated by B {y: < r} {yr tk} together with elements of the formzm t[dm,am] (m n) where the zm satisfy the relations

() rmzm =

r sm y + gm.

This is an abstract group, which can be regarded as a subgroup of the freegroup on G {y: r} {zm: m n} modulo the relations in G and


12/23

442

re

vision:2000-10-27

modified:2000-10-30

12

the relations given by (i.e., the equations ()). If is such that (n) = and c() = k, then there is an embedding of B

(k) into B,n+1 which fixes B

(and is an isomorphism if = ). As before, Z = px0 is a pure subgroup

of B(k) . Since B

(k) is isomorphic to a subgroup of G, it is separable.

Since h exists, there is a -least tuple w: rk in Zr+1 for

which there is a projection h: B(k) Z with hB = hB, h

(y) = wfor < r and h(yr tk) = wr. Note that this tuple determines h

on B(k) .Define f() = k.

We have to show that this definition works, that is, for each S, f((n))equals c() for sufficiently large n . Fix S. With respect to the well-

ordering , there are only finitely many wrong guesses which come beforethe right answer h(y): < rh(y

r tc())c(). So we just have to

show that no wrong guess can work for all n if it involves a k = c(). If therewere a wrong guess that worked for all n for some k = c(), then there wouldbe a projection h onto Z whose domain, B, contains {y : < r} {y

r tk},

elements of the form zn a,n for all n (with a,n A), and B where is minimal such that gn G for all n .

Let G denote

G0 + B = G + {y

: r} {z

n: n }.

Notice that for each g G0, there is a j such that pjg B0 B, which isa subset of dom(h) and dom(h). Hence we can extend h and h uniquely tohomomorphisms from G into Q(p) Z. (Here Q(p) is the group of rationalswhose denominators are powers of p.) Denote the extension by h (resp. h).We claim that h = h.

Assume for the moment that this is true. Then h(yr) = h(yr). Now

h(yr tc()) Z and h(yr tk) Z. So

h(tc() tk) = h(yr tc()) h

(yr tk) Z.

Since k = c(), there is an s Z so that s(tc() tk) x0 (mod B0), soh(x0) Z. But p(h(x0)) = h(px0) = px0; this contradicts the fact that px0generates Z.

It remains to prove the claim. Let H = G/G, which is isomorphic toG+1/G. Now h h

induces a homomorphism from H into Q(p) since hand h agree on B, hence on G0 (since G0/B0 = Z(p

)) and so on G. So


13/23

442

re

vision:2000-10-27

modified:2000-10-30

13

it suffices to prove that Hom(H,Q(p)

) = 0. Assume, to the contrary, thatthere is a non-zero : H Q(p). Let K = ker(); then the rank of K is r.Now H/K is isomorphic to a subgroup ofQ(p) and hence is not divisible. ByLemma 7, H/K is isomorphic to M = G+1/F. So by the choice of S andp, H/K is not p-divisible; since H/K is isomorphic to a subgroup ofQ(p),this implies H/K is free. But this is impossible, since H is not free and Kis a subgroup of rank r, and hence free.

Corollary 9 If there is an hereditarily separable group of cardinality 1which is not free, then there exist 21 different 1-separable groups of cardi-

nality 1 which are hereditarily separable.

Proof. By the theorem, the given hypothesis implies that there is a laddersystem on a stationary subset of 1 which satisfies monochromatic uni-formization for colours. Using this ladder system, we can construct an1-separable group which is hereditarily separable as in the proof of Theo-rem 5. By a standard trick we can, in fact, construct such groups with 21

different -invariants. (Compare [3, VII.1.5].)

Similarly to the proof of Theorem 8 we can prove the following:

Theorem 10 If there is an hereditarily separable group G of cardinality 1which is not a Shelah group, then there is a ladder system based on a countableset which satisfies monochromatic uniformization for colours.

3 Consistency of Case 4

The consistency of Case 4 in the Introduction will now follow from Theorem 5and the following set-theoretic result.

Theorem 11 It is consistent with ZFC + GCH that the following all hold:

(i) there is a ladder system on a stationary subset of 1 which satisfiesmonochromatic uniformization for colours;

(ii) there is no ladder system on a stationary subset of 1 which satisfies2-uniformization;

(iii) (E) holds for every stationary subset, E, of every regular cardinal > 1.


14/23

442

re

vision:2000-10-27

modified:2000-10-30

14

Proof.We assume familiarity with the methods of [10]. For simplicitylet our ground model be L; fix a stationary, co-stationary subset S of 1

and a ladder system on S. Our forcing P will be an iterated forcing withcountable support using two types of posets: R, which adds a Cohen subsetof 1, and Q(c) which is the poset uniformizing a monochromatic colouringc: S of with countable conditions, i.e.,

Q(c) = {f: f: for some successor < 1 and for all S ,f((n)) = c() for almost all n }.

In the iteration P we force with R at successors of even ordinal stages andforce with Q(c) at the successors of odd ordinal stages, where, as usual, thenames c are chosen so that all possibilities occur. The posets R and Q(c) areproper, so stationary sets are preserved by P. Also, P is (1 \ S)-closed andof cardinality 2, so GCH holds in the generic extension as well as (E) forevery stationary subset of every regular cardinal > 1.

It remains to show that in the generic extension 2-uniformization fails forevery stationary subset E of 1 and every ladder system = {: E}.By doing an initial segment of the forcing we can assume that E and areboth in the ground model. Let X be the generic set for the first copy of R

in the iteration ofP. Consider the 2-colouring {c: E} of defined asfollows: c(n) = 0 if and only if + n X. The proof that this colouring isnot uniformized now follows along the same lines as that in [10].

Corollary 12 It is consistent with ZFC + GCH that there is an hereditar-ily separable group of cardinality 1 which is not free, and every Whiteheadgroup (of arbitrary cardinality) is free.

Proof. We use the model of ZFC + GCH constructed in Theorem 11.Clause (i) in Theorem 11 together with Theorem 5 imply that there is a non-

free hereditarily separable group of cardinality 1. Clause (ii) implies thatthere is no non-free Whitehead group of cardinality 1. (See [3, XII.3.1(i)].)Finally clause (iii) enables one to do an inductive proof that there is no non-free Whitehead group of any cardinality (as, for example, in [3, XII.1.6]).


15/23

442

re

vision:2000-10-27

modified:2000-10-30

15

4 Cases 2 and 3In Cases 2 and 3 of the Introduction we are in the situation where there isa Whitehead group which is not free; here we shall consider two hypotheseswhich are stronger than this hypothesis: first, that there is a Whiteheadgroup which is not a Shelah group; and, second, that every Shelah group isa Whitehead group.

The following theorem says that the hypothesis that there is a Whiteheadgroup which is not a Shelah group is not consistent with Case 3. It also givesanother consistency proof for Case 2 since it is known that it is consistent thatthere are Whitehead groups which are not Shelah groups (see [3, XII.3.11]).

Theorem 13 If there is a Whitehead group of cardinality 1 which is nota Shelah group, then there is an hereditarily separable group which is not aWhitehead group.

Proof. By [3, XII.3.19] there is a ladder system = {: 1} based on acountable set I which satisfies 2-uniformization. Without loss of generality wecan assume that I = and each : is strictly increasing. Moreover,as in the proof of Lemma 3, we can assume that is tree-like, and hence,as in the proof of Lemma 4, satisfies monochromatic uniformization for

colours.For each 1 and n let

k,n = ((n) + 1)!

andk,n = (n)!

Let G be the group generated by {xn: n < } {y,n: 1, n < }, subjectto the relations

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

As in the proof of Theorem 5, G is hereditarily separable.It remains to show that G is not a Whitehead group. For this we shalldefine an epimorphism : H G with kernel Z which does not split. Let Hbe the group generated by {xn: n < } {y

,n: 1, n < } {z}, subject

to the relationsk,n+1y

,n+1 = y

,n + x

(n)

+ k,n+1z. (2)


16/23

442

re

vision:2000-10-27

modified:2000-10-30

16

There is an epimorphism taking y

,n to y,n, x

m to xm, and z to 0; the kernelof is the pure subgroup of H generated by z. Aiming for a contradiction,assume there is a splitting of , i.e., a homomorphism : G H such that = 1G. So (y,n) y

,n ker() for all < 1, n . Since a

countable union of countable sets is countable, there exists = such that(0) = (0), (1) = (1) and (y,0) y

,0 = (y,0) y

,0. Let m ( 2)

be minimal such that (m) = (m).We claim that (y,n) y

,n = (y,n) y

,n if n < m. The proof is

by induction on n < m; the initial case n = 0 is by choice of and . Sosupposing the result is true for n < m1, we will prove it for n+1. Applyingthe homomorphism to equation (1) for as well as and subtracting weget that (in H)

k,n+1(y,n+1) k,n+1(y,n+1) = (y,n) (y,n) (3)

since x(n) = x(n) because n < m. But then by induction

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

,n. (4)

Now by equation (2), since x(n) = x(n) and k,n+1 = k

,n+1 (the latter

because n < m 1), we have

k,n+1y

,n+1 k,n+1y

,n+1 = y

,n y

,n. (5)so by equations (4) and (5) we have

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

,n+1). (6)

Since n < m 1, k,n+1 = k,n+1, so cancelling k,n+1 from equation (6), weobtain the desired result, and the claim is proved.

Now equation (4) holds for n = m 1 so

k,m(y,m) k,m(y,m) = y,m1 y

,m1. (7)

In this case, instead of (5) we have

k,my,m k,my

,m (k

,m k

,m)z = y

,m1 y

,m1. (8)

so combining (7) and (8) we have

k,m(y,m) k,m(y,m) = k,my,m k,my

,m (k

,m k

,m)z. (9)


17/23

442

re

vision:2000-10-27

modified:2000-10-30

17

Say (m) < (m). Then k,m, k

,m and k,m are all divisible by k,m =((m) + 1)! so equation (9) implies that ((m)+1)! divides k,mz = (m)!z

in H which is a contradiction, since z generates a pure subgroup of H.

Now we consider the hypothesis that every Shelah group is a Whiteheadgroup. This is true in a model of Martins Axiom, in which case there arehereditarily separable groups which are not Whitehead groups, i.e., Case 2holds. Here we show that it is consistent that every Shelah group is a White-head group but every hereditarily separable group is a Whitehead group,i.e., there is a model for Case 3 in which every Shelah group is a Whiteheadgroup. For this purpose we use the notion of stable forcing. A poset, P, is

stable if for every countable subset P0 there is a countable subset P1 so thatfor every p P there is an extension p ofp and an element p P1 so that p

and p are compatible with exactly the same elements of P0. In [1] the basicfacts about c.c.c. stable forcings are proved. There are a few basic facts thatwe will use:

Proposition 14 1. [1] Any iteration of c.c.c. stable forcings with finitesupport is c.c.c. and stable.

2. The forcing adding any number of Cohen reals is stable.

3. IfA is a Shelah group and

0 ZB

A 0

is a short exact sequence, then the finite forcing, Q(), constructing thesplitting of is (c.c.c. and) stable.

Proof. We will prove only the last of the statements. Write A as


18/23

442

re

vision:2000-10-27

modified:2000-10-30

18

for some y0, . . . , ym since G A+n+1 = G A+n and A+n+1 has the Shelahproperty over A+n. Extend p to p Q() such that

dom(p) = M y0, . . . , ym

where M G A+n is a finite rank pure subgroup of G A+n such thatdom(p) M y0, . . . , ym. Let p

= pM P1.It suffices to prove that ifq P0 is compatible with p

, then q is compatiblewith p. So suppose that r Q() such that r q, p. Without loss ofgenerality dom(r) G A+n. Define r

with

dom(r

) = dom(r) y0, . . . , ymby: rdom(r) = r and ry0, . . . , ym = p

y0, . . . , ym. Clearly r q, p.

Moreover, dom(r) is pure in A since G is pure in A and dom(r) is pure inG A+n; so r

Q().

Theorem 15 It is consistent that every Shelah group is a Whitehead groupand every hereditarily separable group is a Whitehead group.

Proof. We do our forcing over L by iteratively adding subsets of 1 byfinite conditions and adding splittings for Shelah groups. More precisely, our

forcing P will be an iterated forcing with finite support and of length 2 usingtwo types of posets: R, the finite functions from 1 to 2, and Q() which isthe finite forcing splitting as in Proposition 14(3). If we choose the iterantscorrectly, then in the generic extension every Shelah group of cardinality 1will be a Whitehead group and (E) will hold for every stationary subset ofevery regular cardinal greater than 1. It will suffice then to show that everyhereditarily separable group of cardinality 1 is a Shelah group (because wehave all instances of diamond above 1: cf. [3, Exer. XII.16(ii)].)

By Theorem 10 it is enough to show that, in the generic extension, if = {: < 1} is a ladder system based on , then does not satisfy

monochrome uniformization for colours. In fact, we will show that doesnot satisfy monochrome uniformization for 2 colours. By absorbing an initialsegment of the forcing into the ground model we can assume that is inthe ground model and the forcing P is first R, the finite functions from 1 to2, followed by a name T for a c.c.c. stable forcing. We define the colouringc: 1 2 to be the generic set for R; let c be a name for c.


19/23

442

re

vision:2000-10-27

modified:2000-10-30

19

In order to obtain a contradiction, assume that this colouring can be uni-formized. Then there is a pair f , f of names for functions and there isa p P such that p f , f uniformizes c. Now let P0 be a countablesubset of P containing p as well as for every n < , a maximal antichainwhich determines the value of f(n). Let P1 be as given by the definitionof a stable poset for this P0. For each 1 choose p p

so that pdetermines the values of f() and c() and there exists p P1 so that pand p are compatible with exactly the same elements of P0. Say

p f() = m c() = e.

By the pigeon-hole principle, there exists an uncountable set E 1 andp P1 so that for all E, p

= p

. Since p is compatible with p, thereexists q1 p

, p. By the definition ofR, there exists 0 E and q2 P suchthat q1 q2 and q2 c(0) = e0 . So q2 k > m0 s.t. f((k)) = e0.Thus there exists q3 q2 and k0 > m0 such that q3 f((k0)) = e0 .But there is a maximal antichain in P0 of conditions forcing the value off((k0)). Hence there exists r P0 and q4 such that r q4, q3 q4 andr f((k0)) = 1 e0 . Then r is compatible with p

= p0 and hence with

p0 . But this is a contradiction since p0 c() = e0 f((k0)) = c()since k0 > m0 .

5 Finitely many primes

The proof of Theorem 5 uses infinitely many primes. Otherwise said, the typeof the (torsion-free rank one) non-free quotients G+1/G in that constructionis (1, 1, 1, . . .). We may ask what happens if we are allowed only finitely manyprimes. For example, we may consider modules over Z(P) (where P is a set ofprimes and Z(P) denotes the rationals whose denominators in reduced formare not divisible by an element of P) and ask whether the main theorem,Theorem 1, holds. If P is infinite, i.e., Z(P) has infinitely many primes,

then our proofs apply and there is a non-free hereditarily separable Z(P)-module of cardinality 1 if and only if there is a ladder system on 1 whichsatisfies monochrome uniformization for colours. On the other hand if thecardinality of P is finite but at least two, we can show that Theorem 1 doesnot hold, and Case 4 in the Introduction is impossible. In fact, this sectionis devoted to proving the following result:


20/23

442

re

vision:2000-10-27

modified:2000-10-30

20

Theorem 16Suppose R is a countable p.i.d. with only finitely many but atleast 2 primes. If there is an hereditarily separable R-module of cardinality

1 which is not free, then there is a Whitehead R-module of cardinality 1which is not free.

Proof. The method of proof is to show that if there is an hereditarilyseparable R-module of cardinality 1 which is not free, then there is a laddersystem on a stationary subset of1 which satisfies 2-uniformization. We firstprove that

()

there is a tree-like ladder system = {: S} on

a stationary subset S of lim(1) such that for every2-colouring c = {c: S} of , there is a functionf: 1 2 such that for all S there existsm such that f((n), m) = c(n) for all n .

Let N be an hereditarily separable R-module of cardinality 1. As in theproof of Theorem 8, we write N =


21/23

442

re

vision:2000-10-27

modified:2000-10-30

21

Let c be a 2-colouring of = {: S}. Following the pattern of theproof of Theorem 8, we will use c to define a subgroup B of N. We begin byletting {xn: n } be a basis of N0, and letting B0 be the subgroup of N0generated by {px0} {pxn+1 xn: n }. Also, let A = {tn: n } N0be a complete set of representatives of N0/B0 such that t0 = 0 and for eacha A, fix an element [p,a] such that pt[p,a] + B0 = a + B0.

Assume we have defined B so that B + N0 = N and for all < ,B N = B. We now define B,m inductively so that z

m B,m + N0. Let

B,0 be generated by B {y0, . . . , y

r}. If B,m1 has been defined, we have

pzm B,m1 + N0, so pzm a

m (mod B,m1) for some a

m A. Let

B,m = B,m1 + R(zm t[p,am] c(m)x0).

Having defined B,m for all m, we can extend

m B,m to B+1 such thatB+1 + N0 = N+1 and B+1 N = B.

Finally, let B =


22/23

442

re

vision:2000-10-27

modified:2000-10-30

22

belongs to Rpx0, and we know h(z

k t[p,a

k]) by induction (because we knowhN0). Now h(z

k t[p,a

k]) x0 and h(z

k t[p,a

k]) cannot both belong to Rpx0.

Ifh(zk t[p,ak]) belongs to Rpx0, c(k) must equal 0; otherwise let c(k) = 1.

(If the latter value does not make h(zkt[p,ak]c(k)x0) belong to Rpx0, then

our supposition must have been wrong, and we can let f(, m) be arbitrary.)This completes the proof of (). At this point we use the assumption that

there are at least two primes. Then the proof of necessity, i.e. of Theorem 8,is still valid. (Referring to the last paragraph of that proof, we use thefact that there are two primes when we assert that H/K is not divisible.)Moreover, there is a single ladder system which satisfies the property of ()

as well as monochromatic uniformization for colours. (Indeed, by reducingto a smaller set we can assume that the same set S is used in both the proof of() and the proof of Theorem 8; then we can let be a ladder system derivedfrom functions which give combined information about the equations usedin the proof of () and the equations used in the proof of Theorem 8.)

Given a 2-colouring c of , let f be as in (). Define a monochromaticcolouring c of by: c() = m where m is such that f((n), m) = c(n)for all n . There is a uniformization g, g of c. Define h: 1 2 by:h() = f(, g()). Then for all S for sufficiently large n,

h((n)) = f((n), g((n))) = f((n), m) = c(n).

The third author, in [9, Thm. 3.6], claimed to prove that if the non-freenessofG involves only finitely many primes, then G is hereditarily separable if andonly if G is Whitehead. However, the proof given seems to be irredeemablydefective. We do not know if the result claimed is true. Thus we still havethe following open questions:

If R is a countable p.i.d. with exactly one prime, does Theo-rem 16 hold? IfR has finitely many primes, is every hereditarily

separable R-module of cardinality 1 a Whitehead module? Ifnot, find a combinatorial equivalent, analogous to Theorem 1, tothe existence of a hereritarily-separable R-module which is not aWhitehead module.


23/23

2

re

vision:2000-10-27

modified:2000-10-30

23

References[1] U. Avraham and S. Shelah, Forcing with stable posets, J. Symbolic

Logic 47, 3742 (1982).

[2] K. Devlin and S. Shelah, A weak version of which follows from20 < 21, Israel J. Math. 29, 239247 (1978).

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

[4] P. C. Eklof, A. H. Mekler and S. Shelah, Uniformization and the

diversity of Whitehead groups, Israel J. Math. 80, 301321 (1992).

[5] H. P. Goeters and W. J. Wickless, Hyper- groups, Comm. in Algebra,17, 12751290 (1989).

[6] F. Richman, A class of rank-2 torsion free groups, in Studies onAbelian Groups (B. Charles, ed.), Springer-Verlag, 327334, (1968).

[7] S. Shelah, Infinite abelian groups, Whitehead problem and some con-structions, Israel J. Math 18, 243256 (1974).

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

[9] S. Shelah, On uncountable abelian groups, Israel J. Math. 32, 311330(1979).

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

Documents

P. C. Eklof, A. H. Mekler and S. Shelah- Hereditarily Separable Groups and Monochromatic Uniformization