Saharon Shelah- ℵ-N Free Abelian Group with No Non-Zero Homomorphism to Z

8/3/2019 Saharon Shelah- -N Free Abelian Group with No Non-Zero Homomorphism to Z

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N-FREE ABELIAN GROUP WITH NONON-ZERO HOMOMORPHISM TO Z

SH883

Saharon Shelah

The Hebrew University of JerusalemEinstein Institute of Mathematics

Edmond J. Safra Campus, Givat Ram

Jerusalem 91904, IsraelDepartment of MathematicsHill Center-Busch Campus

Rutgers, The State University of New Jersey110 Frelinghuysen Road

Piscataway, NJ 08854-8019 USA

Abstract. We, for any natural n, construct an n-free abelian groups which havefew homomorphisms to Z. For this we use n-free (n +1)-dimensional black boxes.The method is hopefully relevant to other constructions of n-free abelian groups.

I would like to thank Alice Leonhardt for the beautiful typing.This research was supported by German-Israeli Foundation for Scientific Research and Develop-mentFirst Typed - 06/Mar/6Latest Revision - 09/Mar/5

Typeset by AMS-TEX

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2 SAHARON SHELAH

Annotated Content

1 Constructing k()+1-free Abelian group

[We introduce x is a combinatorial k()-parameter. We also give a shortcut for getting only there is a non-Whitehead k()+1-free non-free abeliangroup (this is from 1.6 on). This is similar to [Sh 771, 5], so proofs areput in an appendix, except 1.14, note that 1.14(3) really belongs to 3.]

2 Black boxes

[We prove that we have black boxes in this context, see 2.1; it is based onthe simple black box. Now 2.3 belongs to the short cut.]

3 Constructing abelian groups from combinatorial parameter

[For x Kcbk()+1 we define a class Gx of abelian groups constructed from it

and a black box. We prove they are all k()+1-free of cardinality ||x + 0

and some G Gx satisfies Hom(G,Z) = {0}.]

4 Appendix 1

[We give adaptation of the proofs from [Sh 771, 5] with the relevantchanges.]


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n-FREE ABELIAN GROUP, ETC. 3

0 Introduction

For regular = n we look for a -free abelian group G with Hom(G,Z) = {0}.

We first construct G and a pure subgroup Zz G which is not a direct summand.If instead not direct product we ask not free so naturally of cardinality , weknow much: see [EM02].

We can ask further questions on abelian groups, their endormorphism rings,similarly on modules; naturally questions whose answer is known when we demand1-free instead n-free; see [GbTl06] . But we feel those two cases can serve asa base for significant number of such problems (or we can immitate the proofs).Also these cases are reasonable for sorting out the set theoretical situation. Whynot = and higher cardinals? (there are more reasonable cardinals for whichsuch results are not excluded), we do not fully know: note that also in previous

questions historically this was harder.Note that there is such an abelian group of cardinality 1, by [Sh:98, 4] and seemore in Gobel-Shelah-Struungman [GShS 785]. However, if MA then 2 < 20 any 2-free abelian group of cardinality < 20 fail the question.

The groups we construct are in a sense complete, like Z. They are close tothe ones from [Sh 771, 5] but there S = {0, 1} as there we are interested in Borelabelian groups. See earlier [Sh 161], see representations of [Sh 161] in [Sh 523, 3],[EM02].

However we still like to have = , i.e. -free abelian groups. Concerningthis we continue in [Sh 898].

We thank Ester Sternfield and Rudiger Gobel for corrections.

We shall use freely the well known theorem saying

0.1 Theorem. A subgroup of a free abelian group is a free abelian group.

0.2 Definition. 1) Pr(, ): means that for some G we have:

(a) G = G : + 1

(b) G is an increasing continuous sequence of free abelian groups

(c) |G+1| ,

(d) G+1/G is free for < ,

(e) G0 = {0}

(f) G/G is free if

(g) some h Hom(G;Z) cannot be extended to h Hom(G+1,Z).

2) We let Pr(, ,) be defined as above, only replacing G+1/G is free for < by G+1/G is -free.


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1 Constructing k()+1-free abelian groups

1.1 Definition. 1) We say x is a combinatorial parameter if x = (k,S, ) =(kx, Sx, x) and they satisfy clauses (a)-(c)

(a) k <

(b) S is a set (in [Sh 771], S = {0, 1}),

(c) k+1(S) and for simplicity || 0 if not said otherwise.

1A) We say x is an abelian group k-parameter when x = (k,S, , a) such that(a),(b),(c) from part (1) and:

(d) a is a function from to Z.

2) Let x = (kx, Sx, x) or x = (kx, Sx, x, ax). A parameter is a k-parameter forsome k and Kcbk()/K

grk() is the class of combinatorial/abelian group k()-parameters.

3) We may write ax,n instead ax(, n). Let wk,m = w(k, m) = { k : = m}.

4) We say x is full when x = k()(S).5) If x let x be (kx, Sx, ) or (kx, Sx, , a ( )) as suitable. We maywrite x = (y, a) if a = ax, y = (kx, Sx, x).

1.2 Convention. Ifx is clear from the context we may write k or k(), S, , a insteadof kx, Ss, x, ax.

A variant of the above is

1.3 Definition. 1) For S = Sm : m k we define when x is a S-parameter: x m kx m

(Sm).2) We say is a (x, )-black box or witness Qr(x, ) when:

(a) = m : m kx

(b) = :

x

(c) = ,m,n : m kx, n < and ,m,n < m

(d) if hm : x

m m for m kx then for some x we have: m kx n S and k() =

m S and for some we have n < , = m, n} where = m, n means m = m n and k() = m =

}

(b) xk() is {x

m : m k()}

(c) m() = m if xm.

1.5 Definition. 1) We say a combinatorial k()-parameter x is free when there isa list : < () of x such that for every for some m k() and somen < we have

() m m, n / {m m, n : < }.

2) We say a combinatorial k-parameter x is -free when x = (k, Sx, ) is freefor every x of cardinality < .

Remark. 1) We can require in () even (n)[m(n) / { (n

) : k , S) freely as an abelian group except the equations (n!)y,n+1 = y,n + xn.Note that if K is the countable subgroup generated by {x :

>2} then G/Kis a divisible group of cardinality continuum hence G is not free. So G is 1-freebut not free.

Now we have the abelian group version of freeness, the positive results in 1.12, 1.13and the negative results in 1.13.

1.12 The Freeness Claim. Let x Kk().1) The abelian group GUu/GU,u is free if U

S, u S\U and |u| k k()and |U| k()k.2) If U S and |U| k(), then GU is free.

1.13 Claim. 1) If x is a combinatorial k()-parameter then x is k()+1-free.2) If x is an abelian group k()-parameter and (kx, Sx, x) is free, then Gx is free.

Proof. 1) Easily follows by (2).2) Similar and follows from 3.2 as easily G belongs to G(k(),Sx,x), see Definition3.3.


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1.14 Claim. Assume x Kcbk() is full (i.e. x = k()+1((Sx))).

1) If U S and |U| (|S| + 0)+(k()+1), the (k() + 1 )-th successor of |S| + 0.Then GxU is not free.

2) If |Sx| k()+1 then Gx is not free.3) Assume x Kcbk(), |S

x

| + < +1 for < k() and |x| k() and G Gx

(see Definition 3.3) then G is not free.

Proof. 1) Let = |S|. Assume toward contradiction that GU is free and let be large enough; for notational simplicity assume |U| = +k()+1, this is O.K.as a subgroup of a free abelian group is a free abelian group. We choose N bydownward induction on k() such that

(a) N is an elementary submodel2

of (H(), , = N |N U|.]

()3 for m < k() the set U3m, =: U2

1r=m

Nr has cardinality +m

[Why? By downward induction on m. For m = 1 as U2 Nm and|U2 | = ++1 and clause (b). For m < 1 similarly.]

Now for = 0 choose U2 , possible by ()2 above. Then for > 0, k()

choose U30,. This is possible by ()3. So clearly

()4 U and Nm U = m for , m k().

[Why? If = 0, then by its choice, U2 , hence by the definition ofU

2 in

()2 we have / N, and

U

1+1 hence

N+1 . . . Nk() by ()1

so ()4 holds for = 0. If > 0 then by its choice, U30, but U3m, U2by ()3 so U

2 hence as before

N+1 . . . Nk() and

/ N.

Also by ()3 we have

1r=0

N so ()4 really holds.]

Let = : k() and let G be the subgroup ofGU generated by {x :

m k() and k()+1U and n < }{y,n : k()+1U but = and n < }.

Easily G G recalling G = NGU hence {G : k()} G, but y,0 / G

hence

()5 y,0 /

{G : k()}.

But for every n

()6 n!y,n+1 y,n = {x : m k()} {G : k()}.[Why? x Gm as

(k()) + 1\{m}) Nm by ()4.]

We can conclude that in GU/

{G : k()}, the element y,0 +

{G : k()} is not zero (by ()5) but is divisible by every natural number by ()6.This contradicts ()0 so we are done.2),3) Left to the reader. 1.14


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2 Black Boxes

2.1 Claim. 1) Assume k() < , = 0 , = k()() andS = , = k()+1(S)or just S = = ,() =

0 = for k() and =

k()

(S) and

x = (k(), , ) so x is a full combinatorial S : k()-parameter. Then hasa -black box, i.e. Qr(

xk() , ), see Definition 1.3.2) Moreover, x has the : k()-black box, i.e. for every B = B : xk() satisfying clause (c) below we can find h : such that:

(a) h is a function with domain { m, n : m k(), 2 n < }

(b) h

( m, n) B

(c) Bm,n is a set of cardinality m

(d) if h is a function with domain xk(), see Definition 1.4 such that h(

m, n) B() for , m k(), n < and < for k()then for some x, h h and (0) = for k().

3) Assume = 0 , +1 =

+1 for k(). If S = for simplicity, for

k(), x is a full combinatorial (S, k())-parameter, and |B| k() for x then we can find h : x as in part (2), moreover such that:

(e) if then is increasing

(f) if is regular then we can in clause (d) above add: if E is a club of for k() then we can demand: if x then for each for some < we have (E {})

(g) if is singular of uncountable cofinality, = {,i : i < cf()},cf(i,) = i, increasing with i we can add: if u cf() is unbounded,E,i a club of ,i then (Ei, {}) for some i u.

Proof. Part (1) follows form part (2) which follows from part (3), so let us provepart (3). To uniformize the notation in 2.1(1), i.e. 1.3(2) and 2.1(2),(3) we shalldenote:

1 h( m, n) = k(),m,n.


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Note that without loss of generality3 xk() B = |B|, i.e. without loss

of generality x n < m k() B = n and we use k(),m,n =

h( m, n) for x, m k() and n < . We prove part (3) by induction on

k(). Let k = x and without loss of generality S = .

Case 1: k() = 0.By the simple black box, see [Sh 300, III,4], or better [Sh:e, VI,2], see below

for details on such a proof.

Case 2: k() = k + 1.Let

2 k = k,m,n : k, n < ,m k witness parts (2),(3) for k, i.e. for x

k,

hence no need to assume xk is full.

So = k(), = k() and let H = {h : h is a function from k to }. So

|H| ()0k = . By the simple black box, see below, we can find h :

such that

3 () h = h,n : n < and h,n H for

() if f = f : > and f H for every such and < and > is increasing then for some increasing we have and n < h,n = fn

() if cf() > 0 and E is a club of then we can add {(n) : n < } E.

[Why? First assume = . Let g = g, : < n : < enumerate >Hsuch that for each g >H the set { < : g = g} is unbounded in . Nowfor and n < let h,n = g(k),n for every k large enough if well definedand g(n+1),n otherwise. So clause () of 3 holds and as for clause () of 3, let

f = f : > be given, f H.

Assume > is increasing. We choose n by induction on n < such that:

4 () n = (n) if n < g()

3Why? (As doubts were cast we shall elaborate.) For k() let B = {i : i < |B|} for

k() and let g be a one-to-one function from B onto B . Now assume that h : k()is as required in the claim for B : and define a fucntion h with domain Dom(h

) =

{ m, n : m k() and n < } such that h( m, n) = g(h( m, n)) B for

, m k(), n < . Define the function h with domain k() by h() = g1 h, so h

is well

defined with domain k() such that h() B . By the choice ofh

: k() there is

such that m k()n < h( m, n) = h( m, n). But by the choice of h , h we have

m k() n < h( m, n) = g1 (h

( m, n)) = g

1 (h

( m, n) = h( m, n)as required.


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() n < and n > m if n = m + 1

() if n g() then n satisfies gn = f: min(E\m).

Second, if > but still 0 , let g : < 0 list>H, and let hn :

for n < be such4 that < < (n)(hn() = hn()) and let cd: >

be one to one onto. Now for and n < let h,n be g where is the uniqueordinal < such that for every k < large enough (cd((k)))(n) = hn() so inparticular g(cd((k)) : k < is going to infinity or h,n is not well defined; infact, we can use only the case g(cd((k)) = k; stipulating h,n

{0} when notdefined. So we have defined h,n : , n < . Now we immitate the previousargument: clause () of2 holds.

Next we shall define k()

= k()

,m,n : k+1, m k(), n < as required;so let = : k() k() we define

k() =

k(),m,n : m k(), n < as

follows:

5 if k() and 0, . . . , k()1 k then for m k() and n <

() if m = k() then k(),m,n = hk(),n(0, . . . , k()1) < m

() if m < k(), i.e. m k then k(),m,n =

kk(),m,n < m.

Clearly k()

,m,n < m in all cases, as required, (in clause (a),(b),(c) of 2.1(2) and(e) of 2.1(3). But we still have to prove that

k(),m,n :

k+1, m k(), n let f : k = k() be defined by: f( : k) =: h( :

k). So by 3 above for some increasing k() we have k()(0) =

k()

and

6 n < fk()n = hk(),n.

Now substituting the definition of f we have

7 0, . . . , k k n < hk()

,n(0, . . . , k) = h(0, . . . , k, () n).

4recall (N) means for every large enough n <


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Substituting the definition of k we have

8 if 0, . . . , k k and n < then k() = h(0, . . . , k,

k()

n).

Now we define a function h with domain xk

k by: if xk

k then h() =

h(k()).

So by the choice of k in 2 we can find 0 , . . . , k k with no repetitions

such that (0) = for k and in 2

9 m k n < k0 ,...,

k,m,n = h

(0 , . . . , k (m, n)).

Let = 0 , . . . , k,

k+1,

= 0 , . . . , i .

Note that

10 if m k, n < then h( k, m) = h(( m, n)k()) = h(

m, n).

Now by 9 + 10 and 5() this means

11 if m k and n < then k(),m,n = h(

k, m).

So by putting together 8 + 11 we are clearly done, i.e. we can check that0 , . . . ,

k,

k() is as required. 2.1

2.2 Conclusion. For every k < there is an k+1-free abelian group G of cardinalityk+1 and pure (non-zero) subgroup Zz G such that Zz is not a direct summandof G.

Proof. Let = 20 and x be a combinatorial k-parmeter as guaranteed by 2.1.Now by 2.3(2) below we can expand x to an abelian group k-parameter, so Gx isas required.

2.3 Claim. 1) If x is a combinatorial k-parameter such that Qr(x, 20) then forsome a, y := (x, a) is an abelian group k-parameter such that h Hom(Gy,Z) h(z) = 0.2) For every k there is an k+1-free abelian group G of cardinalityk+1 and z Ga pure z G as above.

Proof. 1) Let witness Qr(x, 20). We define a function :Ord Z by: () in if < , is n if = + n < + and is zero otherwise. For each x we


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14 SAHARON SHELAH

shall choose a sequence a,n : n < of integers such that for any b Z\{0} forno c Z do we have:

for each n < we have

n!cn+1 = cn + a,nb + {(,m,n) : m k()}.

This is easy: for each pair (b, c0) Z Z the set of an : n < Z such that

there is at least one sequence (and always at most one sequence) c0, c1, c2, . . . ofintegers such that holds for them, is meagre, even no-where dense so the choiceof a,n : n < is possible.

Now toward contradiction assume that h is a homomorphism from Gx to zZsuch that h(z) = bz,b Z\{0}. We define h : xk by h

() = n if n < and

h(x) = nz and h() = + n if n < and h(x) = (n)z.By the choice of , for some x we have: m k n < h( m, n) =

,m,n. Hence h(x(m,n)) = (,m,n)z for m k, n < .Let cn Z be such that h(y,n) = cnz. Now the equation ,n in Definition 1.6

is mapped to the n-th equation in , so an obvious contradiction.2) By part (1) and 2.2. 2.3

2.4 Remark. 1) We can replace by a set of cardinality in Definition 1.3. UsingZz instead of simplify the notation in the proof of 2.3.2) We have not tried to save in the cardinality of G in 2.3(2), using as basic of theinduction the abelian group of cardinality 0 or 1.

2.5 Claim. 1) If 0 = 00 , m+1 = 2

m and m = m for m k for the -fullcombinatorial k-parameter x, the (x, )-black box exist.

2.6 Conclusion. Assume 0 < . . . < k() are strong limit of cofinality 0 (or

0 = 0), = + , = 2

.Then in 2.1 for x we can let h,m has domain { xm : [ = for

= m + 1, . . . , k()}.


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3 Constructing abelian groups from combinatorial parameters

3.1 Definition. 1) We say F is a -regressive function on a combinatorial param-eter x Kcbk() when S

x is a set of ordinals and:

(a) Dom(F) is x

(b) Rang(F) [x xk()]0

(c) for every x and m k() we5 have sup Rang(m) > sup({Rang(m) : F()}); note

x or xk() as F() is a set of such objects.

1A) We say F is finitary when F() is finite for every .1B) We say F is simple if k()(0) determined F() for

x.2) For x, F as above and x we say that is free for (x, F) when: x andthere is a sequence : < () listing =

{F() : } and sequence

: < () such that

(a) k()

(b) if < () and then F() { : < } { m, n : < is such that x and n < , m k()}

(c) if < () and then for some n < we have , n / { , n : < ,

} { : < }.

3) We say x is -free for F is (x, F) is -free when x, F are as in part (1) and every

x of cardinality < is free for (x, F).

3.2 Claim. 1) If x Kcbk() and F is a regressive function on x then (x, F) is

k()+1-free provided that F is finitary or simple.2) In addition: if k k(), x has cardinality k and u = u : satisfies u {0, . . . , k()}, |u| > k, then we can find

: < k, : j,m+1 (both are ordinals!).

Why () is true? As by the definition of j,m+1 for some i < and wj,m+2\wj,m+1 we have ,0(i) = j,m+1. By the definition of wj,m+2 as /wj,m+1, there is wj,m+1 such that

F() .As / wj,m+1 necessarily / wj,m hence by the definition of j,m we know

that (i < )(,0(i) < j,m). By clause (c) of Definition 3.1(1) as F() we

know that ,0(i) < sup{,0(i) : i < }. By the last two sentences we are done

proving (), so we are done defining wj+1 hence we finish justifying .Now let (j,i) : i < ij list wj+1\wj such that: if i1, i2 < i

j and

(j,i1)

F((j,i2)) then i1 < i2; we prove existence by F being regressive. Let j,i : i < ij list {F() : wj+1\wj}\

x\{F() : wj}.Let j = {i

j(1) + i

j(1) : j(1) < j}. Now we choose for <

j for j < j() as

follows:

(a) j

+i = j,i if i < ij

(b) j

+ij

+i = (j,i) if i < ij .

Lastly, we choose nj+i < for i < ij as in case 1A.

Now check.

Subcase 1C: F is simple.Note that F() when defined is determined by k()(0) and is included in {

xk() x : sup Rang(k()) < k()(0)}. So let u = {k()(0) : } and

u = u {sup(u) + 1} and for u let = { : k()(0) = } and for u


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let


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{x : F()\x} {y,n : F()

x and

(m k())( m, n) F()}.

3.4 Claim. If x Kcbk() and G Gx (i.e. G is an abelian group derived from x),then G is k()+1-free.

Proof. We use claim 3.2. So let H be a subgroup of G of cardinality k(). Wecan find such that

() (a) x has cardinality k()

(b) every equation which X = {x, y,n : m k(), n < , }satisfies in G, is implied by the equations from = {,n : }

(c) H G = x, y,n : , m k(), n < G

(d) if then F() is included in { (, n) : , k() andn < }.

So it sufices to prove that G is a free (abelian) group.

Let the sequence (

, ) : < () be as proved to exist in 3.2. LetU

={ < () : } {()} and for () let X0 = {x : U , m k() and n < } and X1 = X

0 {

: \U}. So for each Uthere is n = n, : v such that: v {0, . . . , k()}, n, < andX1+1\X

1 = {x : v and n [n,, )}.

For () let G, = {y,n, x : U and X1}G . Clearly

G, : () is purely increasing continuous with union G, and G,0 = {0}.So it suffices to prove that G,+1/G, is free. If / U the quotient is trivially afree group, and if U we can use v to prove that it is free giving a basis.

3.4

3.5 Conclusion. For every k() < there is an k()+1-free abelian group G ofcardinality = k()+1 such that Hom(G,Z) = {0}.

Proof. We use x and h : x from 2.1(3), and we shall choose G Gx. So Gis k()+1-free by 3.4.


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Let S = {(ai, i) : i < i1(bj, j, nj) : j < j1 : i1 < , ai Z, i x

k()

and j1 < , bj Z, j x, nj < } (actually S =

x

k() suffice noting j =

j, : k()).

So |S| = k() and let p be such that:

(a) p = p : <

(b) p lists S

(c) p = (ai , i ) : i < i(b

j ,

j , n

j ) : j < j so

j =

j, : k()

(d) sup Rang(i,k()) < , sup Rang(j,k()) < if i < i, j < j.

Now to apply Definition 3.3 we have to choose z (for Definition 3.3) as {ai xi :i < i} + {bj yj ,nj : j < j} and z = z,n = zk()(0) for

x, n < then for

x

we choose b,n : n <

Z such that:

there is no function h from {z}{y,n : n < }{x : m k(), n 2}

Y2 =

xm,, :m = k(),

k()U and for no t < t do we have

= t k() & {t,k() : st < }

Y3 = {yt,n : t < t and n [st, )}.

Now

()1 Y1 Y2 Y3 {z} generates GU.

[Why? Let G be the subgroup ofGU which Y1 Y2 Y3 generates. First we proveby induction on n < that for k()U and nS we have xk(),, G

. Ifxk(),, Y2 this is clear; otherwise, by the definition ofY2 for some < (in fact = n) and t < such that st we have = t k(), = t,k() .

Now

(a) yt,+1 , yt, are in Y3 G.

Hence by the equation ,n in Definition 1.6, clearly xk(),, G. So as Y1

G GU, all the generators of the form xk(),, with each U are in G.

Next note that

(b) xm,t{ik():i=m}, belong to Y1 G if m < k().


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Now for each t < we prove that all the generators yt,n are in G. If n st

then clearly yt,n Y3 G. So it suffices to prove this for n st by downward

induction on n; for n = st by an earlier sentence, for n < st by ,n. Together all

the generators are in this subgroup so we are done.]

()2 Y1 Y2 Y3 {z} generates GU freely.

[Why? Translate the equations, see more in [Sh 771, 5].]1.8

Proof of 1.10. 0), 1) Obvious.2),3),4) Follows.5) Let : < m() list u, U = U (u\{}) so GU,u = GU+0

. . . + GUm()1 .

First, GU,u GUu follows by the definitions. Second, we deal with provingGU,u pr GUu. So assume z G, a Z and az belongs to GU0 + . . . + GUm()so it has the form {bix : i < i()} + {cjyj,nj : j < j()} + az withi() < , j() < and a, bi, cj Z and i, i, j are suitable sequences of membersofU(i), U(i), Uk(j) respectively where (i), k(j) < m(). We continue as in [Sh 771].6) Easy.7) Clearly U1 v = U2 u hence GU1u GU1v = GU2u hence GU,u + GU1u isa subgroup ofGU,u + GU2u, so the first quotient makes sense.

Hence (GU,u+ GU2u)/(GU,u+ GU1u) is isomorphic to GU2u/(GU2u (GU,u+GU1u)). Now GU1,v GU1v = GU2v GU,u + GU2,u and GU1,v GU,v =GU,v\U = GU,u GU,u + GU2,u. Together GU1,v is included in their intersection,

i.e. GU2u (GU,u + GU1u) include GU1,v and using part (1) both has the samedivisible hull inside G+. But as GU1,v is a pure subgroup of G by part (5) hence ofGU1v . So necessarily GU1u (GU,u + GU1,u) = GU1,v, so as GU2u = GU1v weare done.8) See [Sh 771, 5]. 1.10

Proof of 1.12. 1) We prove this by induction on |U|; without loss of generality |u| =k as also k = |u| satisfies the requirements.

Case 1: U is countable.So let { : < k} list u be with no repetitions, now if k = 0, i.e. u = then

GUu = GU = GU,u so the conclusion is trivial. Hence we assume u = , and letu := u\{ } for < k.

Let t : t < t list with no repetitions the set U,u := {

x k()+1(Uu): for no < k does k()+1(U u)}. Now comes a crucial point: let t < t,for each < k for some rt, k() we have t,rt, =

by the definition of U,u, so


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|{rt, : < k}| = k < k() + 1 hence for some mt k() we have < k rt, = mtso for each < k the sequence t (k() + 1\{mt}) is not from {s : s k() ands = mt : s (U u) for every s k() such that s = mt}.

For each t < t

we define J(t) = {m k(): the set {t,s : s k() & s = m}is included in U u for no k}. So mt J(t) {0, . . . , k()} and m J(t) t {j k() : j = m} / k()+1\{m}(U u) for every k. For m k() lett,m := t {j k() : j = m} and

t :=

t,mt

. Now we can choose st < byinduction on t < t such that

() if t1 < t, m k() and t1,m

= t,m, thent,m st / {t1,m : < }.

Let Y = {xm,, GUu : xm,, / GUu for < k} {y,n GUu : y,n /GUu for < k}.

Let

Y1 = {xm,, Y: for no t < t do we have m = mt & = t}.

Y2 = {xm,, Y : xm,, / Y1 but for no

t < t do we have m = mt & = t &

t,mt st t,mt}

Y3 = {y,n : y,n Y and n [st, ) for the t < t such that = t}.

Now the desired conclusion follows from

()1 {y + GU,u : y Y1 Y2 Y3} generates GUu/GU,u

()2 {y + GU,u : y Y1 Y2 Y3} generates GUu/GU,u freely.

Proof of()1. It suffices to check that all the generators ofGUu belong to GUu =:Y1 Y2 Y3 GU,uG.

First consider x = xm,, where k()+1(U u), m k() and nS for

some n < . If x / Y then x GU,u for some < k but GUu GU,u GUu

so we are done, hence assume x Y

. If x Y1 Y2 Y3 we are done so assumex / Y1 Y2 Y3. As x / Y1 for some t < t we have m = mt & = t. As x / Y2,clearly for some t as above we have t,mt st t,mt . Hence by Definition1.6 the equation t,n from Definition 1.6 holds, now yt,n, yt,n+1 Y3 G

Uu.

So in order to deduce from the equation that x = xt belongs to GUu, it

suffices to show that xt,j

GUu for each j k(), j = mt. But each such

xt,j

belong to GUu as it belongs to Y1 Y2.


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[Why? Otherwise necessarily for some r < t we have j = mr, t,j =

r,mr

andr,mr sr t n r,mr so n sr and as said above n st. Clearly r = t asmr = j = mt, now as t,mr =

r,mr

and t = r (as t = r) clearly t,mr = r,mr .

Also (r < t) by () above applied with r, t here standing for t1, t there as r,mr sr t,j n r,mr . Lastly for ift < r, again () applied with t, r here standingfor t1, t there as n mt gives contradiction.]So indeed x GUu.

Second consider y = y,n GUu, if y / Y then y GU,u GUu, so assume

y Y. If y Y3 we are done, so assume y / Y3, so for some t, = t andn < st. We prove by downward induction on s st that y,s GUu, this clearlysuffices. For s = st we have y,s Y3 G

Uu; and if y,s+1 G

Uu use the

equation t,s from 1.6, in the equation y,s+1 GUu and the xs appearing in

the equation belong to GUu by the earlier part of the proof (of ()1) so necessarilyy,s

GUu

, so we are done.

Proof of ()2. We rewrite the equations in the new variables recalling that GUuis generated by the relevant variables freely except the equations of ,n fromDefinition 1.6. After rewriting, all the equations disappear.

Case 2: U is uncountable.As 1 |U| k()k, necessarily k < k().Let U = { : < } where = |U|, list U with no repetitions. Now for each

|U| let U := { : < } and if < |U| then u = u {}. Now

1 (GU,u + GUu)/GU,u : < |U| is an increasing continuous sequence ofsubgroups of GUu/GU,u.[Why? By 1.10(6).]

2 GU,u + GU0u/GU,u is free.[Why? This is (GU,u + Gu)/GU,u = (GU,u + Gu)/GU,u which by 1.10(8)is isomorphic to Gu/G,u which is free by Case 1.]

Hence it suffices to prove that for each < |U| the group (GU,u+GU+1u)/(GU,u+GUu) is free. But easily

3 this group is isomorphic to GUu/GU,u .[Why? By 1.10(7) with U, U+1, U , , u here standing for U1, U2, U , , uthere.]

4 GUu/GU,u is free.[Why? By the induction hypothesis, as 0 + |U| < |U| k()(k+1) and|u| = k + 1 k().]


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2) If k() = 0 just use 1.8, so assume k() 1. Now the proof is similar to (buteasier than) the proof of case (2) inside the proof of part (1) above.

1.12


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REFERENCES.

[EM02] Paul C. Eklof and Alan Mekler. Almost free modules: Set theoreticmethods, volume 65 of NorthHolland Mathematical Library. NorthHolland Publishing Co., Amsterdam, 2002. Revised Edition.

[GbTl06] Rudiger Gobel and Jan Trlifaj. Approximations and endomorphismalgebras of modules, volume 41 of de Gruyter Expositions in Mathe-matics. Walter de Gruyter, Berlin, 2006.

[GShS 785] Rudiger Gobel, Saharon Shelah, and Lutz Strungmann. Almost-FreeE-Rings of Cardinality 1. Canadian Journal of Mathematics, 55:750765, 2003. math.LO/0112214.

[Sh:e] Saharon Shelah. Nonstructure theory, accepted. Oxford University

Press.[Sh 898] Saharon Shelah. pcf and abelian groups. Forum Mathematicum,

preprint. 0710.0157.

[Sh 52] Saharon Shelah. A compactness theorem for singular cardinals, freealgebras, Whitehead problem and transversals. Israel Journal of Math-ematics, 21:319349, 1975.

[Sh:98] Saharon Shelah. Whitehead groups may not be free, even assumingCH. II. Israel Journal of Mathematics, 35:257285, 1980.

[Sh 161] Saharon Shelah. Incompactness in regular cardinals. Notre Dame Jour-nal of Formal Logic, 26:195228, 1985.

[Sh 300] Saharon Shelah. Universal classes. In Classification theory (Chicago,IL, 1985), volume 1292 of Lecture Notes in Mathematics, pages 264418. Springer, Berlin, 1987. Proceedings of the USAIsrael Conferenceon Classification Theory, Chicago, December 1985; ed. Baldwin, J.T.

[Sh 523] Saharon Shelah. Existence of Almost Free Abelian groups and re-flection of stationary set. Mathematica Japonica, 45:114, 1997.math.LO/9606229.

[Sh 771] Saharon Shelah. Polish Algebras, shy from freedom. Israel Journal ofMathematics, 181:477507, 2011. math.LO/0212250.

Saharon Shelah- ℵ-N Free Abelian Group with No Non-Zero Homomorphism to Z

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