saharon shelah- non-existence of universal members in classes of abelian groups

8/3/2019 Saharon Shelah- Non-Existence of Universal Members in Classes of Abelian Groups

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NON-EXISTENCE OF UNIVERSAL MEMBERS

IN CLASSES OF ABELIAN GROUPS

Saharon Shelah

Institute of MathematicsThe Hebrew University

Jerusalem, Israel

Rutgers University

Mathematics DepartmentNew Brunswick, NJ USA

The Fields InstituteToronto, Ontario

Canada

Abstract. We prove that if + < = cf() < 0 , then there is no universalreduced torsion free abelian group of cardinality . Similarly if0 < < 20 . Wealso prove that if < + < = cf() < 0 , then there is no universal reducedseparable abelian p-group in . We also deal with the class of1-free abelian group.

(Note: both results fail if (a) = 0

or if (b) is strong limit, cf() =0 < ).

2000 Mathematics Subject Classification. 20K20, 03E04, 20K10

Key words and phrases: Abelian groups, Universal, pcf, 1-free

This research was partially supported by the German-Israel Foundation for ScienceI would like to thank Alice Leonhardt for the beautiful typing.Latest Revision - 00/Aug/21Publ. No. 622

Typeset by AMS-TEX

1


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2 SAHARON SHELAH

0 Introduction

We deal with the problem of the existence of a universal member in K for K a

class of abelian groups, where K is the class of G K of cardinality ; universalmeans that every other member can be embedded into it. We are concerned mainlywith the class of reduced torsion free groups. Generally, on the history of the exis-tence of universal members see Kojman-Shelah [KjSh 409]. From previous works,a natural division of the possible cardinals for such problems is:

Case 0: = 0.

Case 1: = 0 .

Case 2: 0 < < 20

Case 3: 20 + + < = cf() < 0 .

Case 4: 20 + + + cf() < < 0 .

Case 5: = +, cf() = 0, ( < )(0 < ).

Case 6: cf() = 0, ( < )(0 < ).

Subcase 6a: is strong limit.

Subcase 6b: Case 6 but not 6a.

Our main interest was in Case 3, originally for K = Krtf, the class of torsion freereduced abelian groups. Note that if we omit the reduced then divisible torsionfree abelian groups of cardinality are universal. A second class is Krs(p), the class

of reduced separable p-groups (see Definition 2.3(4), more in Fuchs [Fu]) but we areinterested in having methods and in the class of 1-free abelian groups. Kojman-Shelah [KjSh 455] show that for K = Krtf, Krs(p) in Case 3 there is no universalmember if we restrict the possible embeddings to pure embeddings. This stressesthat universality depends not only on the class of structures but also on the kind ofembeddings. In [Sh 456] we allow any embeddings, but restrict the class of abeliangroups to (< )-stable ones. In [Sh 552, 1,5] we allow any embedding and allG K but there is a further restriction on related to the pcf theory (see [Sh:g]).This restriction is weak in the following sense: it is not clear if there is any cardinal(in any possible universe of set theory) not satisfying it. We here prove the full

theorem for > with no further restrictions:

() for > in Case 3, K= Krtf, Krs(p) there is no universal member in K

(where we define inductively 0 = 0,n+1 = 2n , =

n


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NON-EXISTENCE OF UNIVERSAL MEMBERS 3

1 deals with Krtf using mainly type theory. In 2, we apply combinatorial idealswhose definition has some built-in algebra and purely combinatorial ones to getresults on Krs(p); there is more interaction between algebra and combinatorics than

in [Sh 552]. Similarly in 3 we work on the class of 1-free abelian groups.What about the other cases? Case 4 (which is like case 3 but singular) for Krtfand pure embedding, was solved showing non-existence of universals in [KjSh 455]provided that some weak pcf assumption holds and in [Sh 552] this was done forembeddings under slightly stronger pcf assumptions. For both assumptions, it is notclear if they may fail. Note that the results on consistency of existence of universalsin this case cannot be attacked as long as more basic pcf problems remain open.

Concerning Case 5 - if we try to prove the consistency of the existence of universals,it is natural first to prove the existence of the relevant club guessing; here we expectconsistency results. (Of course, consistently there is club guessing(by C = C : S, S , otp(C) = ) and then there is no universal.) Also wewere first of all interested in the existence of universal reduced torsion free groupsunder embeddings, but later we also looked into some of the other cases here. Seemore in [Sh:F319].

Case 1 ( = 0). By subsequent work there is a universal member of Krtf , and

(see Fuchs [Fu]) in Krs(p) there is a universal member, but in K

1-free there is no

universal member (see forthcoming work).

Case 0 ( = 0). In Krtf there is no universal member (see above or 3.17) and in

Krs(p)

there is a universal member (see Fuchs [Fu]).

Case 2 (0 < < 20). For Krtf we prove here that there is no universal member

(by 1.2), whereas for Krs(p) this is consistent with and independent of ZFC (see [Sh

550, 4]).We also deal with Case 6 (( < )0 < , > cf() = 0). There is a universal

member for Ktrf and also for Krs(p) . See [Sh:F319].

We thank two referees and Mirna Dzamonja and Noam Greenberg for many cor-rections.

Notation: The cardinality of a set A is|A

|, the cardinality of a structure G is

G

.

H(+) is the set of sets whose transitive closure has cardinality and


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4 SAHARON SHELAH

1 Non-Existence of Universals AmongReduced Torsion Free Abelian Groups

The first result (1.2) deals with satisfying 0 < < 20 and show the non-existence of universal members in Ktrf which improves [Sh 552]. The proof isstraightforward by analyzing subgroups and comparing Bauers types.

Then we deal with 20 + + < = cf() < 0 . We add witnesses to bar theway against undesirable extensions (see [Sh:F319] on classes of modules) whichis a critical new point compared to [Sh 552].

1.1 Definition. Let Krtf denote the class of torsion free reduced abelian groupsG where torsion free means that nx = 0, n Z, x G n = 0 x = 0 andreduced means that (Q, +) cannot be embedded into G. The subclass of G Krtf

of cardinality is denoted by Krtf . Moreover, Ktf is the class of torsion free abeliangroups.

1.2 Claim. 1) If 0 < < 20 then Krtf has no universal member.

2) Moreover, there is no member ofKrtf universal forKrtf1

.

Proof. Let P be the set of all primes and let {Qi : i < 20} be a family of infinite

subsets ofP, pairwise with finite intersection. Let 2 for < 1 be pairwise

distinct. Let H

be the divisible torsion free abelian group with {x : < 1} amaximal independent subset. For i < 20 let Hi be the subgroup of H

generatedby

{x : < 1} {pnx : p P

\Qi, < 1 and n < }

{pn(x x) : , < 1 and p P and

p = p and n < }.

Clearly Hi Krtf and Hi = 1 . Let G Krtf , and we shall prove that atmost of the groups Hi are embeddable into G.

So assume Y 20 , |Y| > and for i Y we have hi, an embedding ofHi into G

and we shall derive that G is not reduced; a contradiction. We choose by inductionon n a set n

n and pure abelian subgroups G of G for n, as follows.For n = 0 we let 0 = {} and let G = G. For n + 1, for n such thatG > 0 we let n, = { : < G}, and let G = G : < G be


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an increasing continuous sequence of subgroups of G of cardinality < G withunion G such that:

() for < G we haveG = G (the Skolem Hull of G in (H(+), , 0 with < G} and =n|

1 1 (as >). So for some , , , , y and i0 < i1 from Y we

have (for = 0, 1)

i = , i = , i = , i = , hi(x) = y.

Now as hi embeds Hi

into G and hi(x) = y, necessarily

() if p P\Qi and n < then in G, pn divides y.

So this holds for every p (P\Qi0) (P\Qi1) = P

\(Qi0 Qi1).Now Qi0 Qi1 is finite so let p

P be above its supremum. As { : Xi0

}is a dense subset of { : Xi0}, there is X

i0

such that p = p

(=i0 p

). Let hi0(x) = y, it is in G.

So in (H(+), ,


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6 SAHARON SHELAH

Hence by (), i.e. by the choice of G : < G, necessarily for somey G we have (y

, y). Now y = y as y G, y / G. Also y y is

divisible by pn for p P, n < .

[Why? If p p

because both y and y

are divisible by pn

and if p < p

becausey y = (y y) (y y) and both y y and y y are divisible by pn.]As G is torsion free, the pure closure in G of {y y}G is isomorphic to Q, acontradiction to G is reduced. 1.2

1.3 Definition. 1) Let P be the set of primes.2) For G Krtf and x G\{0} let

(a) P(x, G) = {p P : x n


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For proving () for a given G, x let G be the divisible hull of G and let

G0 = {y G : for some n > 0, pny G},

G1 = {y G : for some b Z, b > 0 not divisible by p we have by G}. Clearly

G = G0 G1. We define the following subsets of G Q:

H0 = {(y, 0) : y G} (so G is isomorphic to H0)

H1 = {(pnbx,pnb) : b, n Z}

H2 = {(0, c1/c2) : c1, c2 Z and c2 not divisible by p}.

Easily all three are additive subgroups of G Q and H2 = Z(p). Let G = H0 +

H1 + H2, a subgroup of G Q.We claim that G (G {0}) = H0. The inclusion should be clear. For the

other direction let z G (G {0}); as z G there are (y, 0) H0, (soy G), (pnbx,pnb) H1 (so b Z, n Z and x G is the constant from()) and (0, c1/c2) H2 (so c1, c2 Z and p does not divide c2) and integersa0, a1, a2 such that z = a0(y, 0) + a1(p

nbx,pnb) + a2(0, c1/c2) which means z =(a0y + a1p

nbx,a1pnb + a2c1/c2).

As z G {0} clearly a1pnb + a2c1/c2 = 0, so as p does not divide c2, necessarilya1p

nb is an integer, hence a1pnbx G, hence as y G clearly a0y + a1p

nbx G

and hence z G {0} = H0 as required.It is easy to check now that H0 is a pure subgroup of G.

Also letting y = (0, 1) clearly (x, 0) y is divisible by pk for every k < (as(pkx, pk) H1 G

for every k Z) and y is divisible by any integer b when b isnot divisible by p (as 1

by = (0, 1/b) H2 G

).

Identifying y G with (y, 0) G we are done: G is as required in (), with y

witnessing p P(x, G). 1.4

1.5 Claim. If G1 Krtf is full and G2 K

rtf and h is an embedding of G1 intoG2 then:

for x G1\{0}, P(x, G1) = P(h(x), G2).

Proof. Without loss of generality h is the identity, now reflect using 1.4(1), 1.4(2)and the definition of full. 1.5


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8 SAHARON SHELAH

1.6 Conclusion. Assume

() 20 < + < = cf() < 0 .

Then there is no universal member in Krtf .

Proof. Let S { < : cf() = 0 and 2 divides } be stationary and =

: S where each is an increasing -sequence of ordinals < with limit such that (n) n is well defined and divisible by ; so 1 = 2 Rang(1) Rang(2) is finite. Let {p

n : n < } list the primes in the increasing order. Let

G0 be the abelian group generated by {x : < } {y : S} {z,n, : n, . Note that K

rs(p) is defined in 2.3(4).In the first section we have eliminated the very weak pcf assumptions for the

theorem concerning Krtf (though the = cf() > + remains, i.e. we assume we

are in Case 3). This was done using the infinitely many primes, so in the languageof e.g. [KjSh 455] the invariant refers to one element x. This cannot be generalized

to Krs(p) . However, in [Sh 552, 5] we use an invariant on e.g. suitable groups

and related stronger combinatorial ideals. We continue this, using combinatorialideals closer to the algebraic ones to show that the algebraic is non-trivial.

We rely on the GCH right version provable from ZFC, see [Sh 460] hence the

condition > is used.

2.1 Definition. 1) For = : < and t = t : < (with 1 < t < ) wedefine J4

t,.

It is the family of subsets A of


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10 SAHARON SHELAH

2) Let

J4t,,


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(a) T m


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12 SAHARON SHELAH

(cB

is defined in part 3) below).

When = + instead of< we may write . If n = , we may write instead of.

3) For X B, recall XB is the subgroup of B which X generates and

cB

(X) =

x : (n)(y X)(x y pnB)

.

4) Let Krs(p) be the family of pure subgroups of some B.

5) Ifp is not clear from the context, we may write B[p], B[p], etc.

2.4 Claim. Assume = n : n < , t = t : < , t = p and the ideal J4t,,

is proper (so n p1()+ is enough by 2.2(2)). Then the ideal I1, is proper

(and I1

,

is a +-complete ideal).

Proof. We define a function h from


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For m (n, k] and

[n,m)

[]t we let n,k be n,k1

m whenever

1 [n,k)

[]t (by clause (b) in () of 2.1 it is well defined). Fix n and

k [n, ) for awhile. Let A = An,k [m]

0 be such that {n,ku

(m) : u

[]tm} = [A]tm and without loss of generality (otp stands for the order type)

() otp(A) = and n,k

u(m) = OPA,(u)

(where OPA,(i) = iff i = otp(A )).

Now for m (n, k] and

[n,m)

[]t we define

y = y

n,k

=

h(

n,k

) :

[n,k)[]

t

and ()[g() < k () [0, t]

where denotes being an initial segment. So y B and we shall prove by

downward induction on m (n, k] that for every

[n,m)

[]t we have (


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14 SAHARON SHELAH

so the equality is trivial.

Case 2: n < m < k.

Here (with equalities in the equation being in B, modulo pk B), we have:

y =

[by the definition of y, yu]

=

{yu : u [{0, . . . , tm}]tm} =

[by the induction hypothesis]

= {(k1

=m+1

(t + 1))(


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+ k1

=m+1

(t + 1)

(pm xm ) ((tm + 1) 1) : A, | A| tm

=

[remember tm = p and pm+1

xm = 0]

= (tm + 1) k1=m+1

(t + 1)


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16 SAHARON SHELAH

2.6 Claim. 1) For every , for some < for every (p1(),) wehave (letting n = ) UI1

,() = (hence UI0() = ).

2) If cf() > 0, then for some < , for every (p1(),) and < we

have UI1,() < .

Proof. By 2.4, I, is a -complete proper ideal on a set of cardinality 0 , for any

, as in the assumptions. By [Sh 460] for each for some = [] < for every (,) we have cov(

, +, +, ) = , i.e. there is P [] of

cardinality such that: if Y [] then Y is included in the union of < members ofP. As I

1, is a

+-complete ideal on a set of cardinality it follows

that UI1,() |P| =

(and trivially UI1,() ). This proves part (1).

For part (2) we are assuming cf() > 0 so for some < , for arbitrarily large

< , [] ; clearly we are done. 2.6

2.7 Conclusion. If + < = cf() < 0 , then in K

rs(p) there is no universal

member.

Proof. By 2.6 and [Sh 552, 5.9].

Moreover

2.8 Claim. Assume

(a)


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Moreover, each G is slender, i.e. there is no homomorphism from Z into G

with range of infinite rank.

Proof. Same proof as that of [Sh 552, 5.9], [Sh 552, 7.5].


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18 SAHARON SHELAH

3 Non-existence of universals for 1-free abelian groups

The first section dealt withKrtf

improving [Sh 552]. But the groups used thereare almost divisible. So what occurs if we replace Krtf by a variant avoiding this?We suggest to consider the 1-free abelian groups where type arguments like thosein 1 break down. So the proof of [Sh 552] becomes relevant and it is natural toimprove it as in 2 (which deals with Krs(p)), for diversity we use a stronger ideal.We have not looked at the problem for 1-free abelian groups of cardinality when0 < < 2

0.So we concentrate here on torsion free (abelian) groups.

3.1 Definition. 1) Let t = t : < , 2 t < . For abelian group H, thet-valuation is

xt = Inf{2m :

0, nx

belongs to xH}. Moreover P Cp

H(X) is the p-adic closure in H of the subgroup ofH which X generates.3) Let Krtf[t] be the class of t-reduced torsion free abelian groups, i.e. the G Krtf

such thatn


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3.3 Proposition. Suppose that

() 0

() G is an 1-free abelian group, |G| = () t is as in 3.1 such that ()(m > ) (t divides tm).

Then there is an 1-free group H such that G H, |H| = and H has t-density .

Proof. Choose n < for n < such thatn


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20 SAHARON SHELAH

Moreover, the equations holding among those elements are deducible from the equa-tions of the form

(a)

equations of K1(b) y0i = xi for i I

(c) tnyn+1i + x

ni(n)

= yni for i I, n < .

We can find ki : i < such that [i = j & i I & j I & n ki & n kj & i = j i(n) = j(n)].

Now we know that K1 is free (being a countable subgroup of G), and it sufficesto prove that K/K1 is free. But K/K1 is freely generated by{yni : i I and n > ki} {x

n : (n, ) J but for no i I do we have

n > ki, i(n) = }. So K is free.

Fact C: H0 = xni : n < , i <

nH satisfies:

(a) i < dt(xi, H0) = inf{dt(xi, z) : z H0} = 0

(b) x G dt(x, H0) = 0

(c) x H dt(x, H0) = 0.

Proof. First note that

()1 Y = {x H : dt(x, H0) = 0} is a subgroup of H.

Also for every i < and n

()2 yni = x

ni(n)

+ tnyn+1i = x

ni(n)

+ tnxn+1i(n+1)

+ tntn+1yn+2i

=

mk=n

k1=n

t

xki(k) + k=n

t

yk+1i

(prove by induction on m n), and note that as ()(m > )(ti divides tm)

necessarily ()(m)(t divides tm) hence (k)(m)(

i

t dividesmi=k

ti). Now

()2 implies

()3 yni Y.


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But xi = y0i and hence clause (a) holds, so as {xi : i < } is dense in G also clause

(b) holds. So G Y (by clause (b)), and xn Y (as H0 Y and the choice ofH0) and y

ni Y (by ()3).

By ()1 clearly Y = H, as required in clause (c).

Fact D: |H| = .

Fact E: The t-density of H is .

Proof. It is as H0 has cardinality and is t-dense in H, it is , as we nowshow.

Define a function h with domain the generators of H listed above, into H. Let

h(x) = 0 ifx G; h(xm ) = 0 ifm > 0 < 0; h(x

m ) = x

m if m = 0&0 0

I5,tH,H,


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Then the ideal Ii,rtft,,

is proper.

Proof. 1) If not, we can find X L =: Grtft,

for < such that Grtft,

=


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26 SAHARON SHELAH

j m = , j(m) = {j , j+1}. By clause (e) above we know that there are j

such that j j


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(i


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28 SAHARON SHELAH

(c) = UI6t,

() < 0 where tm =


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REFERENCES.

[Fu] Laszlo Fuchs. Infinite Abelian Groups, volume I, II. Academic Press,New York, 1970, 1973.

[KjSh 409] Menachem Kojman and Saharon Shelah. Non-existence of UniversalOrders in Many Cardinals. Journal of Symbolic Logic, 57:875891,1992. math.LO/9209201.

[KjSh 455] Menachem Kojman and Saharon Shelah. Universal Abelian Groups.Israel Journal of Mathematics, 92:113124, 1995. math.LO/9409207.

[RuSh 117] Matatyahu Rubin and Saharon Shelah. Combinatorial problems ontrees: partitions, -systems and large free subtrees. Annals of Pureand Applied Logic, 33:4381, 1987.

[Sh 550] Saharon Shelah. 01 laws. Preprint. math.LO/9804154.

[Sh:F319] Saharon Shelah. Notes on Abelian Groups.

[Sh:g] Saharon Shelah. Cardinal Arithmetic, volume 29 of Oxford LogicGuides. Oxford University Press, 1994.

[Sh 456] Saharon Shelah. Universal in (< )-stable abelian group. MathematicaJaponica, 43:111, 1996. math.LO/9509225.

[Sh 552] Saharon Shelah. Non-existence of universals for classes like reducedtorsion free abelian groups under embeddings which are not necessar-

ily pure. In Advances in Algebra and Model Theory. Editors: ManfredDroste and Ruediger Goebel, volume 9 of Algebra, Logic and Applica-tions, pages 229286. Gordon and Breach, 1997. math.LO/9609217.

[Sh:f] Saharon Shelah. Proper and improper forcing. Perspectives in Mathe-matical Logic. Springer, 1998.

[Sh 620] Saharon Shelah. Special Subsets of cf(), Boolean Algebras andMaharam measure Algebras. Topology and its Applications, 99:135235, 1999. 8th Prague Topological Symposium on General Topologyand its Relations to Modern Analysis and Algebra, Part II (1996).

math.LO/9804156.[Sh 460] Saharon Shelah. The Generalized Continuum Hypothesis revisited. Is-

rael Journal of Mathematics, 116:285321, 2000. math.LO/9809200.

saharon shelah- non-existence of universal members in classes of abelian groups

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