Rudiger Gobel and Saharon Shelah- ℵn Free Modules with Trivial Duals

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Result.Math. 99 (9999), 1121422-6383/990001-12, DOI 10.1007/s00025-006-xxxx-yc 2009 Birkhauser Verlag Basel/Switzerland Results in Mathematics

n-Free Modules With Trivial Duals

Rudiger Gobel and Saharon Shelah

Abstract. In the first part of this paper we introduce a simplified version of anew Black Box from Shelah [11] which can be used to construct complicatedn-free abelian groups for any natural number n N. In the second partwe apply this prediction principle to derive for many commutative rings Rthe existence ofn-free R-modules M with trivial dual M

= 0, where M =

Hom(M,R). The minimal size of the n-free abelian groups constructed belowis n, and this lower bound is also necessary as can be seen immediately ifwe apply GCH.

Mathematics Subject Classification (2000). Primary 13C05, 13C10, 13C13,20K20, 20K25, 20K30; Secondary 03E05, 03E35.

Keywords. prediction principles, almost free abelian groups, dual groups.

1. Introduction

The existence of almost free R-modules M over countable principal ideal domains

(but not fields) with trivial dual M := Hom(M, R) = 0 is a well-known factby results using strong prediction principles like diamonds in V = L. Only notethat any (non-trivial) R-module with endomorphism ring End M = R is such anexample. For every regular cardinal > | R | (which is not weakly compact) wecan find (strongly) -free R-modules M of size with End M = R. But fromthe singular compactness theorem follows, that such modules M do not exist forsingular cardinals, see e.g. [4] or [8]. Thus we want to get rid of additional settheoretic restrictions and work exclusively with ZFC:

If we restrict to 1-free R-modules (meaning that all countable submodulesare free) and do not care about the size of M, then we have an abundance of suchmodules and those of minimal cardinal 20 can be constructed by applications of

This is GbSh 920 in the second authors list of publications.The collaboration was supported by an NSF-Grant DMS-0100794 and by the project No. I-963-98.6/2007 of the German-Israeli Foundation for Scientific Research & Development and theMinerva Foundation.


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2 Rudiger Gob el and Saharon Shelah Result.Math.

the (ordinary) Black Box, see for example [8]. However, it is much harder to findexamples like this of size 1 (recall that 1 may be much smaller than 2

0 and hencewe do not have that many possible extension of a module to eliminate unwantedhomomorphisms. A first example of an 1-free R-module M of size 1 with trivial

dual was given in Eda [3]. Some years later we improved this result showing theexistence of such modules with endomorphism ring R, see [7, 2] or [8]. If we want toreplace 1 by 2 or any higher cardinal, then we necessarily encounter additionalset theoretic restriction, see [6]. If we require only the existence of indecomposableabelian groups, then their -freeness is restricted to small cardinals; see [10]. Thus,in order to construct n-free R-modules M with trivial dual we must relax therestriction on the size of M. Clearly (note that GCH is not excluded, in whichcase n = n), the size of M must be at least n. These cardinals are definedinductively as 0 = 2

0 and n+1 = 2n ; see Jech [9]. Using a recent refined

Black Box from Shelah [11] which takes care of additional freeness of the module,we can give a reasonably short proof of the existence of n-free R-modules Mwith trivial dual M = 0 of cardinality | M| = n for any natural number n;

see Theorem 4.3 and Corollary 4.4. It remains an open question if we can go anyfurther and pass or possibly replace M = 0 by End M = R. In this contextit is also worthwhile to recall (from [10]) that there are models of ZFC in which2+1-free implies 2+2-free.

We would like to thank Daniel Herden for several very useful suggestions andsome corrections.

2. The Combinatorial Black Box for

The new Black Box depends on a finite sequence of cardinals satisfying some

cardinal conditions. Thus let k < and = 1, . . . , k be a sequence ofcardinals such that for l :=

0l (l k) the following -conditions holds.

ll+1 = l+1 (l < k). (2.1)

We will also say that is a -sequence and note that l = 0l < l+1.

Condition (2.1) is used to enumerate all maps which we want to predict beforeconstructing the modules. If is any cardinal, then we can define inductively a-sequence: Let 1 =

0 and if l is defined for l < k, then choose a suitablel+1 > l with (2.1), e.g. put l+1 =

ll . The sequence 1, . . . ,k is an example

of such a -sequence.

If is a cardinal, then will denote all order preserving maps : (which we also call infinite branches) on , while > denotes the family of all

order preserving finite branches : n on , where the natural number n, and (the first infinite ordinal) are considered as sets, e.g. n = {0, . . . , n 1},thus the finite branch has length n.


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Vol. 99 (9999) n-free modules with trivial duals 3

For the sake of generality we first consider any sequence = 1, . . . , kof cardinals such that 0l = l (l k) (which later will be strengthened to a

-sequence). Moreover, we associate with two sets and . Thus let

=

1

k .For the second set we replace the m-th (and only the m-th) coordinate m bythe finite branches >m, thus

m =1

>m k for m k and let =

mk

m. (2.2)

The elements of , will be written as sequences = (1, . . . , k) withl or l >, respectively. Using these s as support of elements ofthe module will make enough room for linear independence which will then given-freeness.

With each member of we can associate a subset of :

Definition 2.1. If = (1, . . . , k) and m k, n < , then let m, n be

the following element in m (thus in )

( m, n)l =

l if m = l k

m n if l = m.

We associate with its support [] = { m, n | m k, n < } which is acountable subset of . Similarly, for m k also let [ m] = { m, n | n }, and >, then there is such that 0 = , and n = f n for all n < .

Proof. Since |P | = = 0 = | |, we can fix an embedding

: P .

And since | > | = there is also a list > = | < with enoughrepetitions for each >:

{ < | = } is unbounded.

Moreover we define for each n < a coding map

n :nP n

2

>


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= 0, . . . , n1 n = (0 n) . . . (n1 n).

Finally let X be the collection of all order preserving maps : suchthat the following holds.

= i | i <

P and < with ( n)n = n for all n > k. (2.3)By definition of n it follows that is uniquely determined by (2.3). (Just

note that, n determines m n for all n > k, m.)We now prove the two statements of the lemma. For (a) we consider any

. If / X, then we can choose arbitrary members n P, and if X,then choose the uniquely determined sequence from (2.3) and let n = n, so = .

For (b) we consider some f = {f | f P, >} and >. Inthis case we must define an extension = n | n < of . Thus put0 = , n = n for n < lg(). And if n lg(), then using the above andthat the list of s is unbounded, we can choose inductively n > n1 withf m | m < nn = n .

Finally we check statement (b). Using (2.3) it will follow that the sequence belongs to X:

If = f i | i < P and k = lg(), then we have

( n)n = f m | m < nn = n = n for all n > k

and n = n = f n for all n < is immediate.

The -Black Box 2.4. Let 1, . . . , k be a -sequence satisfying (2.1), , asabove and C as in Definition 2.2. Then there is a family of set-traps | satisfying the following

prediction principle: If : C is any map with the trap-condition m Cm (m k) and k, then for some there is a set-trap with and 0k = .

Proof. The proof of Theorem 2.4 will follow by induction on k. Temporarily

we will attach parameters k to the above symbols like k , Ck

, k , k , . . . .

The first step is k = 1. In this case the claim is a special case of Lemma 2.3.Indeed, we have k = k and

k =

>k and m, n = k n holds. Weput P = Ck = Ckk and note that |P | k =

0k

. The trap functions aredefined by

( m, n) = kn

and with f = condition (b) of Lemma 2.3 reads as

( m, n) = kn = fk n = ( m, n)

and the prediction principle in Theorem 2.4 is clear.

The induction step k = k + 1:Suppose that Theorem 2.4 is shown for k. We must find a family of traps {k |


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3. The Black Box for n-free modules

Let R be a commutative ring with S a countable multiplicatively closed subsetsuch that the following holds.

(i) The elements ofS are not zero-divisors, i.e. if s S, r R and sr = 0, thenr = 0.

(ii)sS sR = 0.

We also say that R is an S-ring. If (i) holds, then R is S-torsion-free and if (ii) holds,then R is S-reduced, see [8]. To ease notations we use the letter S only if we wantto emphasize that the argument depends on it. If M is an R-module, then thesedefinitions naturally carry over to M. Finally we enumerate S = {sn | n < }, lets0 = 1 and put qn =

in si, thus qn+1 = qnsn+1.

Similar to the Black Box in [1], we first define the basic R-module B, whichis

B =

Re.

Definition 3.1. If U , then we get a canonical summand BU =

URe ofB, and in particular, let B = B[] andB m = B[ m] be the canonical summandof and m ( ), respectively.

We have several Rfree summands

B m =n


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Definition 3.2. LetG be any R-module. A trap (for B, G) is a partial R-homomor-phism of B into G with a label , say : Dom() G, such that B Dom() B.

The -Black Box 3.3. Given a -sequence = 1, . . . , k with (2.1) and anR-module G of size | G | 1, let , be as above. Then there is a family oftraps ( ) with the following property:The prediction: If : B G is an R-homomorphism and k, then thereis with 0k = and .

Proof. The theorem is an immediate consequence of Theorem 2.4. We viewthe set maps in Theorem 2.4 as the restrictions of the R-homomorphisms in The-orem 3.3 to the canonical R-basis {e | } of B. There is a one-to-onecorrespondence between these maps and thus Theorem 3.3 follows.

4. The R-modules

Let R be an S-torsion-free and S-reduced commutative ring of size | R | < 20 ,k =

0k

= k be as before, B =

Re the R-module freely generated by

{e | } and

=

[] with [] = { m, n | m k, n < }.

We also choose any bijection

: k .

Thus we can write the basis elements of B in the form e() for any k .

From [5] follows that the S-adic completion

R of R has 20 algebraically

independent elements over R, and in particular R = 20 .Next we define particular elements in B. If , then let

yk =nk

qnqk

(

km=1

e m,n + bne(0k ))

where bn R. Moreover let y = y0. We will choose R and write =n


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and yk sk+1yk+1 =k

m=1 e m,k + bke(0k ), follows

sk+1yk+1 = yk k

m=1e m,k bke(0k ). (4.1)

We want to define an R-module M with B M B which is S-pure inB. Thus M/B is S-torsion-free and S-divisible. It follows that for any non-trivialhomomorphism : M R there is with e = 0. If , then we willadjoin to B for some suitable R the element y = n


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10 Rudiger Gobel and Saharon Shelah Result.Math.

Proof. Besides the -support [g] (discussed at the beginning of the lastsection) any element g of the module M = B, y | has a refined naturalfinite support [g] arriving from the definition (4.3). It consists of all those elementsof and contributing to g. We observe that g is generated by elements yand e m,n and simply collect the s and m, n needed. Clearly [g] is a finitesubset of . Hence any submodule H of M has a natural support [H] takingthe union of supports of its elements and if | H| < for any cardinal > | R |,then there is a subset of size | | < such that H is a submodule of thepure R-submodule

M = e m,n, e(0k ), y | , m k, n < B,

which also has size < . Thus, in order to show k-freeness of M, we only mustconsider any of size | | < k and show the freeness of the module M.We may assume that | | = k1. Let F : be the map which assignes to the element F = (0k) .

By Proposition 2.6 we can express the generators of M of the form

M = e m,n, eF, yn | < k1, m k, n <

and find a sequence of pairs (, n) (k + 1) such that for n n

, n / { , n | < } {

F | < }. (4.4)

Let M = e m,n, eF, yn | < , m k, n < for any < k1;thus

M+1 = M + e m,n, eF, yn | m k, n <

= M + e ,n | n < n + yn | n n

+eF, e m,n | = m k, n < .

Hence any element in M+1/M can be represented in M+1 modulo M ofthe form

nn

rnyn +n


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Vol. 99 (9999) 11

Theorem 4.3. Let R be an S-torsion-free and S-reduced commutative ring of size< 20. Then for any -sequence = 1, . . . , k with (2.1) there exists an k-free R-module M of size k with trivial dual Hom(M, R) = 0. In particular, if Ris a principal ideal domain but not a field of size < 20, then there is an k-free

R-module M of size k with trivial dual.

Proof. If M is the R-module above, then M is k -free by Proposition 4.2.Obviously M has size k . If : M R is a non-trivial homomorphism, thenthere is such that for some basis element e = 0. By the Black Box 3.3there is with (0k) = and B[] = . We apply Proposition 4.1 tosee that this is a contradiction. Hence Hom(M, R) = 0 follows.

Corollary 4.4. If n is a natural number, then we find n-free abelian groups of sizen with trivial dual.

References

[1] A. L. S. Corner and R. Gobel, Prescribing endomorphism algebras A unified treat-ment, Proc. London Math. Soc. (3) 50 (1985), 447 479.

[2] A. L. S. Corner and R. Gobel, Small almost free modules with prescribed topologicalendomorphism rings, Rendiconti Sem. Mat. Univ. Padova 109 (2003) 217 234.

[3] K. Eda, Cardinality restrictions on preradicals. pp. 277283 in Abelian group theory(Perth, 1987), Contemp. Math. 87, Amer. Math. Soc., Providence, RI, 1989.

[4] P. Eklof and A. Mekler, Almost Free Modules, Set-theoretic Methods, North-Holland,2002.

[5] R. Gobel and W. May, Independence in completions and endomorphism algebras,

Forum mathematicum 1 (1989), 215 226.

[6] R. Gobel and S. Shelah, G.C.H. implies the existence of many rigid almost freeabelian groups, pp. 253 271 in Abelian Groups and Modules, Marcel Dekker NewYork 1996.

[7] R. Gobel and S. Shelah, Indecomposable almost free modules the local case, Cana-dian J. Math. 50 (1998), 719 738.

[8] R. Gobel and J. Trlifaj, Endomorphism Algebras and Approximations of Modules,Expositions in Mathematics 41, Walter de Gruyter Verlag, Berlin (2006).

[9] T. Jech, Set Theory, Monographs in Mathematics, Springer, Berlin Heidelberg(2002).

[10] M. Magidor and S. Shelah, When does almost free imply free? (For groups, transver-

sals, etc.), J. Amer. Math. Soc. 7 (1994), 769830.[11] S. Shelah, n-free abelian groups with no non-zero homomorphisms to Z, Cubo - A

Mathematical Journal 9 (2007), 59 79.


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12 Rudiger Gobel and Saharon Shelah Result.Math.

Rudiger GobelFachbereich Mathematik,Universitat Duisburg EssenCampus Essen, D 45117 Essen, Germany

e-mail: [email protected]

Saharon ShelahInstitute of Mathematics,Hebrew University, Einstein Institute of Mathematics,Givat Ram, Jerusalem 91904, Israel andRutgers University,Department of Mathematics, Hill Center, Busch CampusPiscataway, NJ 08854 8019, U.S.Ae-mail: [email protected]

Received: September 12th, 2007

Rudiger Gobel and Saharon Shelah- ℵn Free Modules with Trivial Duals

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