Rudiger Gobel and Saharon Shelah- G.C.H. implies existence of many rigid almost free abelian groups

8/3/2019 Rudiger Gobel and Saharon Shelah- G.C.H. implies existence of many rigid almost free abelian groups

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G.C.H. implies existence of many rigid almost

free abelian groups

Rudiger Gobel

Fachbereich 6

Mathematik

Universitat Essen

45117 Essen, Germany

e-mail:

[email protected]

Saharon Shelah

Department of Mathematics

Hebrew University

Jerusalem, Israel

and Rutgers University

New Brunswick, N. Y., U.S.A.

e-mail:

[email protected]

October 6, 2003

1 Introduction

More than thirty years ago A. L. S. Corner [2] proved a very remarkable theo-

rem concerning the existence of pathological abelian groups; it is formulatedin terms of endomorphism rings.

If A is a countable ring with reduced and torsion-free additive group AZ, thenthere exists a reduced and countable abelian group G with endomorphism ringEndZG = A.

This result has been generalized in various directions, e.g. replacing theground ring Z by a more general ring and dropping the cardinal restriction (inCorners result this is actually |A| < 20), see [2], [20], [8], [9], [4], [23].

Other extensions include abelian groups which are not necessarily torsion-free, see Corner, Gobel [4] for references.

Here we are interested in a different strengthening of Corners important re-sult: As a byproduct of older constructions of abelian groups G with prescribedendomorphism ring A we have obtained (in a strong sense) almost-free groupsprovided AZ is free and provided we are working in G odels universe V = L.

This work is supported by GIF project No. G-0294-081.06/93 of the German-IsraeliFoundation for Scientific Research & Development

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Let us recall the appropriate definition. An R-module G (of cardinality ) with|R| < |G| is strongly -free if G has a -filtration (a continuous chain of sub-modules {G : < } with |G | < and


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additional set theoretic conditions as possible. From the result mentioned abovewe know that Martins axiom must be excluded. Our results discussed beloware surprising in two ways. First of all, we will only need a special case ofG.C.H., which ensures the existence of the desired modules. In another paper[43], Shelah and Spasojevic will show by proper forcing arguments that evenG.C.H. will not be strong enough for End G = Z, |G| = and G -free if istoo large, e.g. if is any strongly inaccessible cardinal. Hence we are limitedto small large cardinals. Secondly it turns out that the restriction to G.C.H.forces us to develop novel ideas to get the desired modules. A new Step-Lemmais needed! Here we will also apply combinatorial tools fortunately developed byShelah [35], [41] a decade ago. The main results of this paper can be summarizedas follows.

In order to construct pathological groups we begin with the existence of

groups with trivial duals for cardinals n (n ). Then we derive resultsabout strongly n-free abelian groups of cardinality n (n ) with prescribedfree, countable endomorphism ring. Finally we use combinatorial results [35],[41] to give similar answers for cardinals > . As in Magidor and Shelah[31], a paper concerned with the existence of -free, non-free abelian groups ofcardinality , the induction argument breaks down at . Recall that isthe first singular cardinal and such groups of cardinality do not exist by thewell-known Singular Compactness Theorem [34], see also Hill [28] for cardinalscofinal to and, for a survey, Eklof and Mekler [15, p. 107].

We will fix a countable commutative ring R with multiplicatively closed sub-sets S of regular elements containing 1. An R-module G will be called torsion-free (with respect to S) if sg = 0 (s S, g G) implies g = 0. Moreover G

is reduced (with respect to S) ifsS sG = 0. If A is an R-algebra, and theR-module AR is R-free (torsion-free or reduced) we will say that the algebra A

is free (torsion-free or reduced). We will also fix a countable, free R-algebra Aand derive the following results.

Theorem 5.1: Let (R, S) be as above. If H is a strongly -free R-module ofregular cardinality with trivial dual H = 0 and = 2 = + (the successorcardinal of ), then there exists a strongly -free R-module of cardinality withtrivial dual.

As a consequence of (5.1) we obtain the existence of such R-modules ofcardinality n (n ) for the n-th successor n of provided 2i = i+1 for

all i < n. The existence of H as in (5.1) for |H| = 1 under 20

= 1 is well-known, see Dugas [7], Shelah [38] or Dugas, Gobel [8], cf. also Eklof, Mekler[15, p. 391]. Hence the existence of the derived strongly n-free R-modules ofcardinality n with trivial dual is immediate (see also (5.2)).

In 6 we will work with a stronger algebraic hypothesis, which again issatisfied automatically for 1 = 2

0 by the papers [8],[38] above. We derive the

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Theorem 6.1: Let (R, S), = 2 = + and regular as above. If H is astrongly -free R-module of cardinality with EndRH = A for some countable,free R-algebra A, then there exists a strongly -free R-module G of cardinality with EndRG = A.

Now the existence of strongly n-free R-modules G of cardinality n withEndRG = A follows for each countable, free R-algebra A, provided i+1 = 2

i

for each i < n. As one of many consequences we see that under the mild restric-tion i+1 = 2

i (i < n) there are counterexamples to Kaplanskys test problemsfor each cardinal n (n ) even ones which are strongly n-free (see [3] for therequired rings). Similarly, we obtain indecomposable, strongly n-free groups

of cardinality n. By the result of Shelah, Spasojevic [43] we know that a real-ization theorem similar to (6.1) and even the existence of modules with trivialdual (5.1) does not follow from G.C.H. for large enough cardinals, e.g. for thefirst strongly inaccessible cardinal . What happens at smaller cardinals, forinstance at +1? The Theorems 6.1 and (5.1) can be extended to +1, henceto +n (n ) by (5.1) and (6.1), and to certain larger cardinals < . Wemust strengthen the methods from 3,4 and 5, and this procedure is sketchedin 7.

2 Representation of strongly -free modules and

the notion of a type of freenessWe fix some notations throughout this paper.(2.1) Let R be a commutative ring and S R a countable, multiplicativelyclosed subset of regular elements such that RR (as R-modules) is reduced andtorsion-free (with respect to S), compare 1. We will also assume 1 S and

enumerate S = {sn : n } and set qn =ni=1

si.

Our results concern almost free R-modules (mainly for |R| = 0), hence wewill refine the notion of almost free and strongly -free R-modules from 1:

We say that a strongly -free R-module H has freeness type f tH = u ifu = 1, , n is a descending sequence of regular cardinals i with 1 =

, n = 0 and if H =


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type 2

, , n

and set S1

= { 1

: H+1

/H

is not free }. By inserting< 1 members between H and H+1 we may assume that H+1/H = R if S1 . Moreover we assume S1 E2 := { 1, cf = 2}. If S1 ,then H+1/H is not free and we proceed similarly. We can find a continuouschain

H = H0 ... H ... H2 = H+1 ( < 2)

with quotients H+1/H either free or of type 3, , n. Recursively weunravel all of 1, , n and let u

i = 1 i (i = 1, , n). Weobtain a refined, continuous chain {H, ui} of free R-modules H withH/H+1 free for + 1. Moreover H+1/H is either = R or of typei+1, , n. The set u

i is ordered lexicographically and each ui is of theform = 1

i with j < j . As usual, + 1 will denote the successor of

in u

i

which is + 1 =

1

i + 1, and 0 denotes the initial element of u

i

.Moreover, again refining the chain {H : un}, we may assume thatH+1/H S

1R for each un and H0 = 0. We will identify un with in

3 6 and write for S= { : H+1/H is not free }, hence

(2.2)

either H+1 = H t0R, if S or H+1 = H, tk : k is a free R-module with qktk+1 = tk +

j


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3 A Step-Lemma and a Freeness-PropositionThe -0-Step-Lemma 3.1: Let H be a strongly -free R-module of cardi-nality and H = 0. Moreover, let F = nKn be a direct sum of freemodules Kn of rank . Choose a partial basis xn ( < ) for each Kn, letKn = xn : < and let f K0 be any pure element with K0/f R free.Then we can find two extensions Fi F (i = 0, 1) with the following properties.

(1) If Kn = xi : i < n or < = i


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where 11

= 1 and 10

= 0.In order to show (2), we consider the maps i : F

i H defined by i|F = 0and sini = tn for all S, n , s

ii = t for \S and i = 0, 1.

These maps preserve the relations (ri) and (i), hence i can be extended to anR-homomorphism i : F

i H. Its kernel is clearly F, is epic and (2) follows.We have a decomposition into free summands Kn Ln = Kn where K

n =

xn : < , and Ln comes from the complement of the partial basis, hence

(ii)

F = m


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Clearly C is generated by Cn and C

n

.Any linear combination s =

rks

0k C

nmay be multiplied by a large

enough q S such that the summand qrks0k by (r0) becomes a sum with

summand qrkxk. Now s Cn is only possible if all the independent termsqrkxk ( , k n) vanish, hence rk = 0 and s = 0. The sum (iv) must

be direct. The freeness of Cn

follows similarly, again using independence ofthe xk under , and (iv) follows.

The quotient F0

/Cn =mn Lm C

nis free by (iv) and F

0is free by

(iii) and (1) is shown for i = 0.In order to show (3) we consider any F extending to i Fi for

i = 0, 1 and recall that f F0 F is a prescribed basic element by hypothesis.First we consider the pushout D of F Fi, which is

D = F0 F1/ with = {(g, g) : g F} = F.

We also have a pushout extension i D (i = 0, 1) by the pushoutproperties and i.

If g = (g, 0) + , g = (0, g) + for g F0 and g F1 respectively, thenwe have canonical identifications F0 + F1 = D withF = F0 F1 = {(g, 0) + : g F} = {(0, g) + : g F}.We omit in the following. Let U =F, dn, d : S, n < , \S D for dn = s0n s

1n, d = s

0 s

1.

Observe that

U + Fi = F0 + F1 = D and U Fi = F0 F1 = F.

The desired relations from (ri) are

(v)

qndn+1 = dn +i


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The following Freeness-Proposition is a special case of a more general Freeness-Proposition. In order to make the idea of the construction more transparent, wepostpone again the extension of (3.2) to Section 7. The more general Freeness-Proposition 7.2 involves a few more technical difficulties, due to more com-plicated notation. The Proposition will ensure that continuous chains of freesubmodules in our construction remain free at limit ordinals below +.

The -Freeness-Proposition 3.2: Let be a regular cardinal and < +

be an ordinal with {Gi : i } an ascending, continuous chain of submoduleswith the following properties for all i < .

(a) Gi is free of rank .

(b) Gi+1/Gi is strongly -free and letS

= {i < : Gi+1/Gi is not free }(c) There exist Gi Hi Gi+1 and H0i a free summand of Gi with free

quotient such that Hi = Gi H0i

H1i is a pushout with H0i H

1i free

modules of rank .

(d) Gi+1 = Hi Ki with Ki freely generated by xi ( < ).

(e) If i S, then Gi = Hi and H1i = H0i (hence Gi+1 = Gi Ki).

(f) If i S, then cf(i) = . Moreover

(i) there exists ai i of order type with ai S= and sup ai = i.

(ii) there exist Aij for all j ai such that Aij Aij = for j = j

and Aij Ai

j

is bounded in for (ij) = (i

j

).(iii) Let Kj = xj : Aij, H

0i =

jai

Kj . If k ai, < and

Kk =j


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If i S, choose any fixed i

< such that i

< i

and observe thatH0i = K

i Ci by (f)(iii) for k = min ai, = i and

Kk = Ki := xj : j ai, < i.

Now we are able to collect a basis of G : Choose all basis elements xj notinvolved in extension of Gi at i S, and we have the set

B = {xj : j < , there is no i S with j ai, Aij and > i}.

Moreover C =iSCi is direct because of (+), and each Ci is free by

(f)(iii). We may collect its basis BC. Finally choose all basis elements xj( < i) which are used for the distinct K

i , say B

. Then B BC B is a

basis for G .

4 The prediction principles

We will use prediction principles that hold under G.C.H. First recall some well-known definitions, tailored for our applications. Let be a regular cardinal and = {A : < } be a -filtration of A =


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In the first application we will use

(E0

), a weak consequence of (4.2).Later we will replace = 0 by some regular > 0 (cf. 7). Moreover we willuse

Proposition 4.3: If = + and E is non-small, then we can decomposeE =


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(1)

Gi, Gj, Gi/Gjare free R-modules for j aiand g Gj0 , with Gj0/gRfree, say j0 = min ai.Moreover Gj+1 = Gj Kjwith Kjfreely generated by some xj ( < ).

(2)

Xis (as a graph) a homomorphism hX Gi .If K

j = xj : Aij

(j ai) and if we identify (Gi, j


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We conclude the following results.

Theorem 5.1: Let R be a countable commutative ring as in 2. If H is astrongly -free R-module of regular cardinality with trivial dual H = 0 and = 2 = +, then there exists a strongly -free R-module of cardinality withtrivial dual.

Corollary 5.2: If 2i = i+1 for i = 0, , n 1, then there exists a stronglyn-free abelian group of cardinality n with trivial dual.

The corollary follows by induction. Ifi = 0, then any subgroup ofQ differentfrom Z serves as H in (5.1). From (5.1) we obtain a strongly 1-free abelian

group of cardinality 1 with trivial dual, and so on.

6 Endo-rigid modules

The following result establishes the existence of R-modules with endomorphismring R, the so-called rigid R-modules. Such modules are known in ZFC, see e.g.Corner [2] or [20], [21], [40]. However we will also assume that the R-modulesin question are almost free. A result in ZF C + MA (= Martins axiom) showsthat extra axioms are needed, because such modules turn out to be separableunder ZFC + MA, see Gobel, Shelah [22]. Assuming V = L, almost free mod-ules were constructed in Dugas [7], Shelah [38] and Dugas, G obel [8] severalyears ago. Here we have to work harder to derive the existence under ZFC and

a weak form of G.C.H.

Theorem 6.1: LetR be a countable, commutative ring as in (2.1) and assume = 2 = + for some regular cardinal . Suppose there exists a strongly -freeR-module H of cardinality with End H = R. Then we can find a strongly-free R-module G of cardinality with End G = R.

The proof is as in (5.1) divided into three parts, a Step-Lemma similarto (3.1), however with crucial, not-so-obvious changes. The construction of Gand the actual proof of the claims in (6.1) are similar to 5. We will keep theproof short. However, the reader familiar with 3-5 will follow the argumentseasily. We begin with the Step-Lemma.

The -0-Step-Lemma 6.2: LetR be as in 2 and let H be a strongly -freeR-module of cardinality with End H = R. Moreover, let F = nKn be adirect sum of free modules Kn of rank . Choose a partial basis xn ( < ) foreach Kn, let f K0 and h H such that K0/f R is free and H/hR is strongly

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-free. Then we can find two extensions F Fi (i = 0, 1) with the followingproperties.

(1) If Kn = xi : i < n or < , then Fi/Kn is a free R-module foreach n , < .

(2) Fi/F = H/hR is strongly -free.

(3) If End F extends to both i End Fi (i = 0, 1), then f f R.

Proof. Let H = H/hR and choose a -filtration H =i


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by (). We find a projection : D V, because - as in (3.1) - f R has acomplement L1 in K0 F (cf. 3.1) and we find complements Ln of eachKn in Kn+1, hence D V

n1 Ln gives . The pushout D provides

End D extending i (i = 0, 1) and = |V is an endomorphism of V.Finally End H = R forces f = f f R.

Proof (6.1): The proof is similar to 5, however (3.1) is replaced by (6.2). Weonly indicate the changes. Assuming = + = 2 we prepare the applicationof (E) for E = { , cf = }. Condition (2) in 5 must be replaced bya new condition

(2)

Xis an endomorphism hX End Gi. If Kj = xj : Aij

(j ai) and if we identify (Gi, j


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Theorem 6.1: Let R be a countable, commutative ring which is S-reducedand S-torsion-free for some multiplicatively closed set S of regular elementsin R. Assume = 2 = + for some regular cardinal and suppose thereexists a strongly -free A-module H of cardinality with EndRH = A and|A| < , AR free. Then we can find a strongly -free A-module G of cardinality with EndRG = A.

Hence we have a corollary which ensures counterexamples for Kaplanskystest problems, cf. Corner [3] and Corner, Gobel [4]. The next corollary strength-ens (6.3).

Corollary 6.3: LetR be as in (6.1) and assume 2i = i+1 fori = 0, , n

1. If A is a countable, free R-algebra, then there exists a strongly n-free R-module G with |G| = n and End G = A.

7 Almost free abelian groups which are of car-

dinality >

In order to derive results like the existence of strongly |G|-free groups G with|G| = +1 and End G = Z under G.C.H., the induction used in 3, 4 and5 breaks down at and we must extend the basic results of the earlier sec-tions in order to overcome this difficulty. The changes will be sketched below.If is a regular cardinal, then let r() be the set of all decreasing sequences

u = 1, , n of regular cardinals 1 > > n. Recall the notionof a u-resolution from (2.3) and of the freeness type of an R-module H from2. In case End H = A we assume that H is a strongly -free A-module andreplace R by A in the above definitions. If H0/hA is A-free for H0 as in (2.2)and h H0, then f tH = f t(H/hA) follows. Observe that we showed in (3.1)(1) and (2) that a strongly -free R-module H of freeness type u = , 0 hasa (0)-resolution. This implication does not depend on additional properties ofH as H = 0 or End H = A.

Moreover, we must refine the use of -sequences converging to particularordinals. If u = 1, , n r(), then let u = 1 n be equippedwith the lexicographical order. We consider strictly order preserving monomor-

phisms : u and let a = Im() and i = sup a which is cofinal to 1.We say that X a is bounded (with respect to ) if there is = (1, , n) u such that X Im(|).We may view as an n-ladder at a. A bounded set X is bounded on eachladder in an obvious sense. If n = 1, then a is a 1-sequence representingcf(i) = 1. The bounded subsets of a constitute an ideal of the set of subsets

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P(a) of a. The set a i will be used to find an extension on Gi

Gi+1

whichwill lead to the desired module G =

i


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qsj+1 = s

j + j


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Corollary 7.4 (ZFC + G.C.H.): If (R, S) is as in 2 and A is a countable,free R-algebra, then we can find strongly +1-free R-modules G of cardinality+1 with EndRG = A and of freeness type +1, 1.

We would like to remark that a further extension of (7.5) would providemodules as in (7.4) for cardinals k+1 (k ). Clearly (7.4) in conjunctionwith (5.1) and (6.1), respectively, provides examples for all +n (n 1). Asimilar result then follows for all k+n (k, n 1). Magidor and Shelah [31]show, however, that there are models of ZFC + G.C.H. in which 2+1-freeabelian groups of cardinality 2+1 are free. This shows that the results abovecannot be extended any further.

Theorem 7.5 (ZFC + G.C.H.): Suppose that the following conditions hold.

(1) Let(R, S) be as in 2 and A a countable, free R-algebra.

(2) There exists a strongly -free A-module H of cardinality withEndRH =A of freeness type u = 1, , n.

(3) Let be a regular, uncountable cardinal with a stationary, non-small subsetS of ordinals cofinal to some cardinal 1.

(4) There exists an n-ladder system ai i (i S) with (+) and sup ai =i, ai S= .

If = +, then there exists a strongly -free R-module G with |G| =

, ft(G) =

u and EndRG = A.

Remark: A similar result holds for modules with trivial dual.

Proof. We indicate the changes in the construction of G, which are based on(2.2) using un = u and the nladder i : u i with Im(i) = ai. Recallthat cf(i) = . From ai S= and the inductive construction of F = j


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References[1] R. Baer: Abelian groups without elements of finite order, Duke Math. J.

3 (1937), 68122.

[2] A.L.S. Corner: Every countable reduced torsionfree ring is an endomor-phism ring, Proc. London Math. Soc. (3) 13 (1963), 687710.

[3] A.L.S. Corner: Additive categories and a theorem of W.G. Leavitt, Bull.Amer. Math. Soc. 75 (1969), 7882.

[4] A.L.S. Corner and R. Gobel: Prescribing endomorphism algebras, aunified treatment, Proc. London Math. Soc. (3) 50(1985), 447479.

[5]A.L.S. Corner

andR. Gobel:

On the existence of an 1free abeliangroup of cardinality 1 with no free summands (unpublished).

[6] K. Devlin and S. Shelah: A weak version of3 which follows from 20

Rudiger Gobel and Saharon Shelah- G.C.H. implies existence of many rigid almost free abelian groups

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