rudiger gobel and saharon shelah- indecomposable almost free modules - the local case

8/3/2019 Rudiger Gobel and Saharon Shelah- Indecomposable almost free modules - the local case

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Indecomposable almost free modules -

the local case

Rudiger Gobel and Saharon Shelah

Abstract

Let R be a countable, principal ideal domain which is not a field and A be

a countable R-algebra which is free as an R-module. Then we will construct

an 1-free R-module G of rank 1 with endomorphism algebra EndRG = A.

Clearly the result does not hold for fields. Recall that an R-module is 1-free if

all its countable submodules are free, a condition closely related to Pontryagins

theorem. This result has many consequences, depending on the algebra A in

use. For instance, if we choose A = R, then clearly G is an indecomposable

almost free module. The existence of such modules was unknown for rings with

only finitely many primes like R = Z(p), the integers localized at some prime

p. The result complements a classical realization theorem of Corners showing

that any such algebra is an endomorphism algebra of some torsion-free, reducedR-module G of countable rank. Its proof is based on new combinatorial-algebraic

techniques related with what we call rigid tree-elements coming from a module

generated over a forest of trees.

1 Introduction

Let R be a fixed countable, principal ideal domain which is not a field. An R-moduleA is reduced for if

sSsA = 0 where S = R \ {0} and A is torsions-free if sa = 0

(s S, a A) implies a = 0. Note that R is reduced because R is not a field. We

will consider R-algebras A which are torsion-free and reduced as R-modules AR. Inparticular this is the case when AR is free.

This work is supported by the project No. G-0294-081.06/93 of the German-Israeli Foundationfor Scientific Research & DevelopmentAMS subject classification: 20K20, 20K26, 20K30, 13C10Key words and phrases: indecomposable modules of local rings, 1-free modules of rank 1, realizingrings as endomorphism ringsSh 591 in Shelahs list of publications

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Let be infinite cardinals. We are interested in R-modules of size which are

free, which is the case when all its submodules of cardinality < are free R-modules.Can we find indecomposable free R-modules of cardinality ?We are mainly interested in the case when = and in particular when this

cardinal is 1.Such modules - by freeness - most likely want to decompose into non-trivial direct

sums and in fact, if is a singular cardinal, then by Shelahs [31] singular compactnesstheorem it follows that such R-modules are free (hence very decomposable), this holdsin particular for cardinals of cofinality , e.g. for , a result due to Hill [26], see Eklof,Mekler [14].

On the other hand, the existence of nonfree, 1free R-modules of cardinality 1

follows from Griffith [23], Hill [26], Eklof [11], Mekler [29] and a result of Shelahs inEklof [12, p.82, Theorem 8.8]. By an induction it can be shown, that there are nonfreenfree modules of cardinality n. A similar result for noncommutative groups is dueto Higman [24, 25]. The freeness-result at illustrates that induction breaks downat +1 and new techniques are needed to show that for certain cardinals only theexistence of nonfree free R-modules of cardinality follows, see Shelah, Magidor[28].

However only very little is known about algebraic properties of the nonfree freeR-modules of cardinality and this is also the case when = = 1, see Eklof [11]and Eklof, Mekler [14]. The following problem is immediate.

Investigate the algebraic properties of -free modules of cardinality .

The only earlier result known to us uses a construction from Shelah [33] of non-separable groups [12, 14] and is due to Eda [10]. He shows the existence of an 1freegroup G of cardinality 1 with Hom (G,Z) = 0. In this paper we want to present newtechniques which allow us to shed some more light on this problem.

In order to work exclusively in ZFC we restrict ourself to = 1 and 20.Recall that under negation of CH the cardinal can be quite arbitrary, see [27]. Wewill show the next corollary which will follow immediately from our Main Theorem inSection 3.

Main Corollary 3.1 Let A = 0 be a Rfree R-algebra over a countable, principalideal domain R which is not a field and let |A| < 20, then there exists an 1freeR-module G of cardinality with End G = A.

We will construct G as an A-module and A is identified with endomorphisms actingby scalar multiplication. If A = R, we derive the existence of 1free R-modules ofcardinality 1 with End G = R, a result about indecomposable R-modules known onlyin the case R = Z from our recent paper [21]. The main difficulty in passing from

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Z to R can be seen in the local case when R is a local ring with just one prime p,

e.g. if R = Z(p) is the ring of integers localized at p = 0. Infinitely many primes by arithmetic provide a rigid system (= modules with no homomorphisms = 0between them). Hence homomorphisms can be restricted in their activity on G bybuilding into G a rigid system in a suitable way [21]. Finally they calm down to scalarmultiplication on G. This is no longer possible in the local case. The only chance wehave is to utilize the existence of sufficiently many algebraically independent elementsin the p-adic completion ofZ(p) and this is hidden in our construction.

It may be interesting to see this result in the light of its predecessors. The firstexample of an 1free module which is not free is the BaerSpecker module R

, whichis the cartesian product of countably many copies of the ring R, known for sixty years;

cf. Baer [1] or [16, p.94]. Assuming CH, this module is an example of an R-moduleof cardinality 1 = 20. However, it is surely (by slenderness of R) a finite but not

an infinite direct sum of summands = 0. Under the same settheoretic assumptionof the continuum hypothesis it can be shown that A above can be realized as theendomorphism ring of an 1free R-module G of cardinality 1. The chronologicallyearlier realization theorem of this kind uses the weak diamond prediction principlewhich follows from 20 < 21. See Devlin and Shelah [5] for the weak diamond, Shelah[35] for the case End G = Z and Dugas, Gobel [7] for the case A = End G and extensionsto larger cardinals. Using, what is called Shelahs Black Box, the existence of 1freemodules G with |G| = 0 also follows from Corner, Gobel [4] using Dugas, Gobel[8] and combinatorial fine tuning from Shelah [36, 37], see also Shelah [41, ChapterVII] and [40]. Many of the older results however do not concentrate on the additionaldemand that the constructed modules with prescribed endomorphism algebra are 1-free, see [2, 18, 19, 20]. This of course was due to other difficulties that had to besettled first.

Assuming Martins axiom (MA) together with ZFC and 2 < 20 any 2free groupG of cardinality < 20 is separable and hence has endomorphism ring Z only in thetrivial case when G = Z, see [21].

Hence 1 in the Main Corollary can not be replaced by 2. This is in contrast tothe result [7] which holds in Godels universe: All algebras A as above are of the form

A= End G for all uncountable regular, not weakly compact cardinals = |G| > |A|such that G is a free R-module. A similar result was shown recently [22] using the

generalized continuum hypothesis G.C.H. only. In view of the theorem under Martinsaxiom,

the existence of indecomposable 2free R-modules of cardinality 2 and the exis-tence of such modules with endomorphism ring R, respectively, is undecidable.

Endomorphism ring results as discussed have well-known applications using theappropriate also well-known R-algebras A.

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If is any abelian semigroup, then we use Corners R-algebra A, implicitly dis-

cussed in Corner, Gobel [4], and constructed for particular s in [3] with special idem-potents (expressed below), with free R-module structure and |A| = max{||, 0}. If|| < 20, we may apply the main theorem and find a family of 1free R-modulesG( ) of cardinality 1 such that for , ,

G G = G+ and G = G if and only if = .

Observe that this induces all kinds of counterexamples to Kaplanskys test problemsfor suitable s. If we consider Corners algebra in [3], see Fuchs [17, p.145], then itis easy to see that AR is free and |A| = 0. The particular idempotents in A andour main theorem provide the existence of an 1free superdecomposable R-module of

cardinality 1, which seems to be new as well. Recall that a group is superdecomposableif any nontrivial summand decomposes into a proper direct sum.

Finally, we remark that, as the reader may suspect, it is easy to replace G inTheorem 3.1 by a rigid family of 2 such groups with only the trivial homomorphismbetween distinct members.

2 The Construction of 1free Modules

a. The Topology

Let R be a countable, principal ideal domain which is not a field, hence R is reduced.We consider any free R-algebra A of cardinality |A| < 20. In particular A is torsion-free and reduced as well. Enumerating S = R \ {0} = {sn : n } we obtain adescending chain of principal ideals qnA for

q0 = 1 and qn+1 = snq2n for all n (1)

with

n qnA = 0. The system qnA (n ) generates a Hausdorff topology, theR-topology on A.

b. The Geometry of a Tree and a Forest

Let T = >2 denote the tree of all finite branches v : n 2, n < , where (v) = ndenotes the length of the branch v. The branch of length 0 is denoted by = Tand we also write v = (v n 1)v(n 1). Moreover, 2 = Br (T) denotes all infinitebranches v : 2 and clearly v n T for all v Br (T), n . We often identifyinfinite branches v with their nodes v = {v n : n } which is a countable, maximal,

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linearly ordered subset of T. Following convention we will call a node v n finite

branch of length n of the tree Br (T). Ifv = w Br (T), then

br(v, w) = inf{n : v(n) = w(n)}

denotes the branch point of v and w. Hence m = br(v, w) is the largest ordinal withv m = w m.

If C , then we collect the subtree

TC = {v T : if e (v) \ C then v(e) = 0}.

Similarly

Br (TC) = {v Br (T) : if e \ C then v(e) = 0},

hence v n TC for all v Br (TC), n and as before we omit C if C = . Manyof our arguments use a finite trunk of these trees. If m < , then we define

TmC = { TC : () < m}. (2)

Finally let TC = TC Br (TC).

Next we use trees to build a forest.

Let 20 be two fixed infinite cardinals and let be regular and uncountable.

Then we choose a familyC

= {C : < } of pairwise almost disjoint, infinitesubsets of . Let T = {v : v T} be a disjoint copy of the tree T and letT = TC for < be the forest of trees (with finite branches), say

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Observation 2.1 Any perfect subtree of T has a subtree order isomorphic to T such

that for any ordinal n there is at most one finite branch v n of the subtree suchthat v (n + 1) = w (n + 1) for some branch w of this subtree.

Proof: Let T be a perfect tree. We will define a tree embedding p : T T suchthat p(T) has the desired branching property. The map p is defined as the union of achain of partial maps pi : T

i Tni. Let

B(T) = {v br(v, w) T : v = w T}

be the set of all branch points in T. Using that Br(T) has no isolated infinite branchesand Konigs Lemma, inductively we can choose a sequence of natural numbers ni (i )

such that n0 = 0 and ifni is given, then ni+1 is the least number x > ni such that for anyv Tx the set B(v, x) = {e : ni < e < x, v e B(T)} has cardinality |B(v, x)| > 2ni.Now we have enough room to extend a partial embedding pi : T

i Tni to pi+1 insuch a way that the branching condition of the Observation 2.1 holds when restrictedto pi+1(T

i+1). Hence p =

i pi is the desired tree embedding of T into that perfecttree.

The forest TC of pairwise disjoint perfect trees T and the sequence of sets infinitebranchesV from T which branch at almost disjoint sets will form our basic geometricalobjects for building modules. The geometry will help to distinguish elements and tocarry out calculation in the corresponding module. In view of Observation 2.1 we will

assume

If T TC and n there is at most one finite branch

v n such that v (n + 1) = w (n + 1) for some w T. (3)

c. The Base Module and its Completion

We consider the free A-moduleBC =

TC

A

which is a pure and dense submodule of its R-adic completion BC taken in the R-topology on BC. The A-module BC will be our base module and we will often omitC for convenience. The sequence V of infinite branches is used to identify certainelements in the completion BC. Any infinite branch v V, n < and any g BCgive rise to an element yvng in the completion BC. Note that

yvng =in

qiqn

(v i ) + gin

qiqn

v(i) (4)

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is a welldefined element of the A-module BC. The reader should keep in mind thatthe branch element yvng connects an infinite branch v Br (TC) Br (T) with finitebranches from the disjoint tree T. We will write yv0g = yvg and often omit the suffixg if this is clear from the context. Moreover

yvng snqnyvn+1g = v n + gv(n) (5)

follows from (1) and (4). We immediate obtain an equation concerning branchingbranches. For v = w V and the branch point m = br(v, w) we have v(i) = w(i) {0, 1} for i < m and v(m) w(m) = 1. Hence

yv =m

i=0 qi(v i ) + gm1

i=0 qiv(i) + qm+1yvm+1 + gqmv(m) and (6)mi=0

qi(v i ) + gm1i=0

qiv(i) =mi=0

qi(w i ) + gm1i=0

qiw(i) and

yv yw qmg = qm+1(yvm+1 ywm+1) for br(v, w) = m. (7)

The special form of branch elements allows us to recognize the geometry of the treesthrough the

d. Support and Norm of Elements and Subsets of BCElements g in BC have a natural support [g] TC, the at most countable set of finitebranches used in the sum-representation with respect to the R-adic completion of BCdefined by

g =[g]

g,

where g = 0 is a unique element in A. If TC and 1 A we identify with theelement 1 in B, hence TC BC and support is defined on TC as well.

Let v V and note that in particular [yvn0] = [vn ], where

[vn ] = {v j : j , j n} T.This infinite part of the branch v we also denote by [vn ] = [vn] because it is clearthat like v it comes from . The notion of support trivially extends to subsets X ofBC by taking unions [X] = xX [x], see also [4].

Each element g =

[g] g ofB also has a special possibly empty subset in [g],

theRsupport [g]R = { [g] : 0 = g R}.

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Branch elements g = yvn0 as in (4) have Rsupport [g]R = [g] = [vn]. The support of

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inductively A-submodules G BC subject to the following condition for any < .The sequence G is increasing, continuous withG0 = 0 and G =

must be a successorbecause E is a proper finite set, hence = 1 exists. Also note that G/G isa quotient of A-modules, hence an A-module. By induction it is enough to show that

E (U + G)/G G/G G+1/G for some free A-module (U + G)/G. (11)

First we want to find inductively an A-submodule U G. We note by (10) thatG+1 = G + F where

F = T {yvng : v V, n < }A (12)

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IfE G+1 is a set of representatives of the elements in E, then by (12) and (5) there

is a finite set F V, and a number m < such that

E U + G where U = Tm {yvmg : v F}A

Moreover we may assume that [vm] [wm] = for all v = w F. A support argumentshows that the defining generators ofU are A-independent modulo G, hence U+G/Gmust be A-free and G/G (U + G)/G = 0.

Now it is easy to show that U is R-pure in G which also implies the purity in (11).Ifh G \ G+1, then an easy support argument shows that G+1 is pure in G that is tosay that dh G+1 for any 0 = d R and in particular dh U. We may suppose thath G+1, and by the last considerations we find a finitely generated A-submodule

U = Tm

{yvmg : v F}A

for some number m m and finite set F F V with h U. We may assumethat m is chosen such that also

[vm] [wm ] = for all v = w F.

One more support argument now shows that U is a summand of U, we leave it to thereader to write down a complement of U in U. If dh U for some 0 = d R, thenh U follows from h U, which shows that U is pure in G.

3 The Constructed Modules And Their Endo-

morphism Algebras

The following Definition 3.2 rigid tree-elements is the critical tool of this paper. Theshort proof of our Main Theorem 3.1, following immediately below, is based on a Main-Lemma 3.3 which indicates our strategy. Moreover, the Definition 3.4 explains howto convert rigid tree-elements into algebraic content. We think that it may help thereader if we start at the end:

The main result of this paper is the following

Theorem 3.1 If A is a free R-algebra over a countable, principal ideal domain Rwhich is not a field and |A| < 20, then there exists an 1free R-module G ofcardinality with End G = A.

Remark: G will be the Amodule constructed in (2.3) and we have identified a Awith a idG.

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Proof: Let G be the A-module from the Construction 2.3. Clearly A End G by

scalar-multiplication because A acts faithfully on G, and G is an 1free R-module ofcardinality by Proposition 2.5. It remains to show that End G A.

Suppose End G\A. Recall from (2.3) that T G for all < , hence BC G.Inspection of (10) shows that G/BC is torsion-free divisible. This is needed to provethat there exists

g BC with BC/gA A-free and g Ag. (13)

Note that BC is a free A-module freely generated by some set J BC. If (13) doesnot hold, then e Ae for all e J, say e = aee. If also f J \ {e} then f = affand similarly e + f is another basic element and the negation of (13) would also give

(e + f) = ae+f(e + f) for some ae+f A, hence ae = ae+f = af by independence.The element a = ae does not depend on e J, and e = ae for all e J, hence BC = a idBC . The endomorphism extends uniquely to the A-module G by density,and = a idG A, which was excluded. Condition (13) is shown.

By Construction 2.3 we can find < such that the element g from (13) belongsto G , moreover we find

< < with g = g, hence G G and g = g G.In particular g = 0 by (13) and, since G is reduced, we find m0 such that

g = g, and g / qmG for all m m0.

We now apply (10)

G+1 = G T {yvn : v V, n }A where yvn = yvng (14)

and (7) implies for v = w V that

yv yw qmg = qm+1(yvm+1 ywm+1) if br(v, w) = m.

We may assume m m0 by assumption on V. Let tv = yv (v V V) and apply to the last equation. We derive the existence of a family of rigid tree-elements asdefined below:

The set of elements tv G (v T), T a subset of V satisfying the hypothesis of(3.3), constitute a family of rigid tree-elements for g, where

tv tw qmg qm+1G.

We now apply our Main-Lemma 3.3 and obtain

g Ag = Ag

which contradicts (13). Hence does not exist and the Main Theorem 3.1 follows.

We proceed with the definition of rigid tree-elements.

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Definition 3.2 LetG be the A-module and be the regular cardinal from (2.3) and

T V for some < be a set of cardinality . If , a family {tv G : v T}is called a family of rigid tree-elements for z G at a tree T, if

tv tw qmz qm+1G for all v = w T with br(v, w) = m. (t)

Main-Lemma 3.3 Let G be the A-module constructed in (2.3) and < , .If T is a set of branches from V branching above some m with |T| = and{tv G : v T} is a family of rigid tree-elements for z G \ qmG, then z Ag.

The proof of (3.3) follows after a number of steps where we replace T above by

equipotent subsets with stronger families of rigid tree-elements. Our final goal is afamily as in the following definition.

Definition 3.4 Let m0 be a natural number. We will say that the family {tv G :v T} as in (3.2) is an independent family of rigid tree-elements for z at some treeT over m0 if there are a sequence of ordinals 1 < < s < , m, j

> m0, a finiteset F with elements ax A for x F and an injective map

: T F is

Vi, (i)

such that any tv (v T) can be expressed as

tv =xF

axy(x,v)m. (ii)

Moreover,

is

[gi ]

[y(x,v)m] = , (iii)

the branches (x, v) j (x F) are all distinct and independent of v,

F =

is

Fi and (T Fi) Vi (i s). (iv)

For convenience we will some times omit above, writing Vi for Vi . Often we onlydeal with partial maps of say

v = {v} F and write v : F Fv = Im (v).

In order to find an independent family of rigid tree-elements we first concentrate onfinding a weaker family which satisfies (w) in (3.5) and comes from a given family ofrigid tree-elements at some fixed tree.

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Pigeon-Hole-Lemma 3.5 Let G be the A-module, be the cardinal given by (2.3),

and let < and . Assume that we also have a family of elements

tv G (v T) for some subset T V of cardinality

and elements h1,...,hk G with the property that

if v, w T and br(v, w) = m, then tv tw h1,...,hkA + qmG. ()

Then we find an equipotent subset T of T and ordinals 1 < < s < with

tw Ti , yvngi : v Vi, n < and i sA for all w T. (w)

Note that () is a weak form of the definition of a family of rigid tree-elements. Iffor the above family = , then by (w) we also find < such that tv G (v T)for an equipotent subfamily.

Proof: We must collect a small pigeon-hole the right-hand side of (w) andenough pigeons tw to land in (w). There are plenty of pigeons and we just discard alltrouble makers.

The proof is by induction on .If = 0, then G0 = 0 and it is nothing to show.Next we assume that is a limit ordinal. We will distinguish three cases (a),(b)

and (c) depending on the cofinality cf() of .(a) Suppose cf() > .Hence |T| < cf() and note that cf() is a regular cardinal. Then we can find < with h1,...,hk G and tv G for all v T. Condition () in (3.5) holds for < and the induction hypothesis applies to finish this case.(b) Suppose cf() < .Note that G =

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If Y, then we find a smallest G ( ) which contains tv . Since : and

G : are continuous sequences, must be a successor ordinal, say = + 1.We get a function g : Y ( g() = ) and note that g() < from tv G.Hence g is regressive and Fodors lemma applies, see [27, Theorem 22, p. 59]. There isa stationary subset X ofY - which must be stationary in by hypothesis (i) - on whichg is constant, taking some fixed value . Clearly |X| = and T = {tv : X}and G ( < ) satisfy the induction hypothesis () in (3.5). Again, the claimfollows in case (i) by induction.Finally we assume that(ii) Y is not a stationary subset of .In this case we have to work showing that (ii) can not occur. There is a cub C in

with Y C = . Inductively we may replace C by an equipotent subset, called Cagain, and replace the s by new ones such that

tv G+1 \ G for all C.

Note that T = {tv : C} still has cardinality , and apply Proposition 2.5 to notethat G+1/G is 1-free for all C . Moreover,

0 = tv + G G+1/G

and there are elements 0 = d R such that

tv + G d(G+1/G). (d)

From |R| < |C| we can find 0 = d R with C = { C : d = d} of cardinality .Pick any j with d | qj.

The set X = {tv : C} of cardinality must have elements tv , tv for < C

with branch point br(v, v) = m > j. In particular d | qm. We derive from thehypothesis () that

tv tv h1,...,hkA + qmG+1.

However h1,...,hk G0 G and also tv G from < . Hence

tv + G qm(G+1/G) d(G+1/G) = dv(G+1/G)

which contradicts (d) and case (ii) cannot come up. This finishes the case of limitordinals .

We may assume that = + 1 is a successor ordinal, and the lemma holds for < . We also have

G = G+1 = T, G, yvng : v V, n A.

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As in (10) and (12) we can write G+1 = G + F with

F = T {yvng : v V, n < }A.

Obviously G by construction (2.3) of G and Lemma 2.2 applies. We derive

Ag = G F

and G/Ag F/Ag is a direct sum.If h1,...,hk G and tv G (v T) are given by hypothesis, then we can

write tv = t0v + t

1v for all v T and similarly hi = h

0i + h

1i (i k) with t

0v, h

0i G

and t1v, h1i F. Moreover, if v, w T branch at br(v, w) = m, then by hypothesis

tv tw h1,...,hkA + qmG and

(t0v t0w) + (t

1v t

1w) =

ki=1

aih0i +

ki=1

aih1i + qmg

+ qmf

follows for some g G and f F. Hence

(t0v t0w

ki=1

aih0i ) qmg

= (t

1w t

1v +

ki=1

aih1i ) + qmf.

The left-hand side of the displayed equation is in G while the right-hand side is in F.The sum must be 0 modulo Ag by the direct sum above. In particular

t0v t0w h

01,...,h

0k, gA + qmG.

The induction applies for < . We find T T with |T| = and 1 < ... < k < such that

t0w Ti, yvngi : v Vi, i k, n A

for all w T. Finally let k+1 = and note that

tw = t0w + t

1w Ti, yvngi : v Vi, i k + 1, n A

for all w T. This completes the induction.

Next we will use the Pigeon-Hole-Lemma 3.5 to find an independent family of rigidtree-elements from an ordinary family of rigid tree-elements. Then we are ready toprove the Main-Lemma 3.3 which established already the Main-Theorem 3.1.

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Proposition 3.6 Let m0 be a natural number and G be the A-module constructed in

(2.3). Suppose there is a family of rigid tree-elements for 0 = z G at some tree.Then we can find an independent family of rigid tree-elements over m0 for z at thesame tree.

Remark: The new family need not be a subfamily of the old one.

Proof: Let {tv G : v T} be the given family of rigid tree-elements for z Gwhere T V for some < has cardinality .

Hence

tv tw Az + qm+1G for all v = w T with br(v, w) = m

follows from (t) in Definition 3.2. Shrinking T, we may assume m > m0. Then weapply the Pigeon-Hole-Lemma 3.5 for k = 1. There is an equipotent subset T replacingT and there are ordinals 1 < < s < with

tv s

i=1

Fi for all v T

where we write [as before in (12)]

Fi = Ti {ywngi : w Vi, n < }A.

We replace i by i and put

gi = gi, Ti = Ti, Vi = Vi and ywngi = ywn for w Vi.

Moreover let B = {1,...,s} and BB = Ti : i sA. For each v T we now can write

tv = bv +s

i=1

tl=1

m(vli)n=0

avlinyvlin (rp)

where bv BB, avlin A and yvlin are branch-elements with vli Vi depending onv T, l and i s.

Next we want to improve the representation (rp) by using relations in G and dis-carding some of the elements from T.

Note that w n Ti for w Vi and (5) can be applied for g = gi to replacem(vli)n=0 avlinyvlin by multiples of gi, an element in BB and of yvlim(v) where m(v) is

larger then the maximum of all the m(vli)s and m0, which is taken over a finite set of

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at most st + 1 numbers. We find new elements avlim(v), avi A, bv BB and m(v)

and new representations for all v T which are

tv = bv +s

i=1

(t

l=1

avlim(v)yvlim(v)) + avigi. (newrp)

Moreover we may assume, enlarging m(v) for each v T up to the supremum of allbranch points of distinct pairs {vli, vji} and the finite set

i=js Ci Cj where the Cis

are from C, that

[yvlim(v)] [yvnjm(v)] = for all (li) = (nj), 1 l, n t, 1 i, j s.

Also V = is[gi] is finite and V li[yvlim(v)] = can be obtained by enlarging m(v).This ensures the first part of (iii) in the Definition 3.4 of an independent family oftree-elements. Next we apply a pigeon-hole argument to simplify (newrp) even further.Recall that max{, |A|, |BB|} < = |T|. There is a subset of cardinality of T whichwe denote by T as well, with the following property.

There is a finite number of parameters for 1 i s, 1 l t(i) with elements

t(v, i) = t(i), m(v) = m > m0, avlim(v) = ali, avi = ai, bv = b

independent of v T. Equations (newrp) become

tv = b +s

i=1

(

t(i)l=1

aliyvlim) + aigi. (better)

Recall that vli Vi Ti , ali, ai A, b BB. Let

E = {(li) : 1 i s, 1 l t(i)}, Ev = {vx : x E} is

Vi

and it is easy to verify from (better) that v : E Ev(x vx) is a bijection.We can also choose jv large enough and > m0 such that the restriction map

(v jv) : E is

Ti : (x vx jv)

is injective with image (Ev jv) = {v(x) jv : x E}. Note that

is Ti is acountable set while T is uncountable. By a pigeon-hole argument we can shrink T suchthat jv = j

> m0 and (Ev j) = are constant for all v T, however the finite

branches in that is v(x) j (x E) are all distinct.

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In order to show that the total map is injective, we replace the old family of rigid

tree-elements tv by a new family tv b si=1 aigi (v T) and observe that the newfamily is a family of rigid tree-elements for z as well. The new family has a betterrepresentation, we can write

tv =xE

axyv(x)m.

The set {Ev : v T} of finite sets of infinite branches constitutes a -system and the-Lemma applies, see Jech[27, p. 225]. There is a new equipotent subset T replacingthe old T such that

Ew Ev = for all v = w T.

IfFv = Ev \ for (v T) and F = E\ 1v

, then the Fvs are pairwise disjoint, hence

is injective on T F and

d = dv =

xE\F

axyv(x)m (v T)

does not depend on v any more. Replacing tv again by tv d, we obtain a new familyof rigid tree-elements for z at the tree T with the best representation

tv =xF

axyv(x)m (v T). (best)

The new family is the desired independent family of rigid tree-elements. Recall that

Fv =is (Fv Vi) and the preimage of this decomposition is the decomposition of F

in (3.4). The proposition follows.

The ultimate step in proving the Main-Lemma 3.3 is the following proposition. TheMain-Lemma 3.3 is now immediate from

Proposition 3.7 LetG be the A-module constructed in (2.3). If there is an indepen-dent family of rigid tree-elements branching above some m0 for z G \ qm0G at thetree T, then z Ag.

Proof: We want to extract the arithmetical strength hidden in the given indepen-dent family of rigid tree-elements {tv G : v T} for z at some tree T with T Vof cardinality . By the last Proposition 3.6 and Definition 3.4 the elements can beexpressed in the form

tv =xF

axyv(x)m for ax A

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with pairwise disjoint sets Fv of infinite branches from si=1 Vi where

F =is

Fi, v : F Fv, v(Fi) = Fv Vi Vi (i s),

Ci = Ci C, Vi = Vi , Ti = Ti , 1 < < s < .

Let I = {i : i s} {}. Recall from (1) thatqn+1qn

= qnsn. Hence

z qj

qj1G = qj1sj1G qj1G qm0G for any j > m0 (15)

by assumption on m0. Moreover, note that Fv is a finite set of distinct branches,a=bI Ca Cb is a finite subset of ,

i qiG = 0 and

i qiA = 0. Also note that

Fv Fw = for distinct v, w T by Definition 3.4. We also have an element j > m0satisfying (3.4). All branches v(x) j

of length j for any x F are pairwise distinctbut independent of v. From these facts it is clear that the following combinatorialconditions hold.

v(y) j = v(x) j

(16)

br(v(y), w(y)) = br(v(x), w(x)) j for v = w T x = y F (17)

qj1z G \ qjG (18)

sup

a=bI

Ca Cb < j, (19)

and

ax = 0 ax qj1A for all x F. (20)

We also may assume

br(v, w) > j > m0 for all v = w T. (21)

Next we will show that the branch point of any two distinct branches v, w T isbounded by the branch point of some -pair of branches:

br(v, w) br(v(x), w(x)) for some x F (22)

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If

n = br(v, w), then tv tw qnz qn+1G (23)

by Definition 3.2 (t) of tree-elements and if also tv tw qn+1G, then qnz qn+1G.Hence z qnsnG qj1sj1G qj1G by (21), which contradicts (15). We have

tv tw qnG \ qn+1G for n = br(v, w). (24)

On the other hand tv =

axyv(x)m and tw =

axyw(x)m for ax A, hence

tv tw =

xFax(yv(x)m yw(x)m). (25)

Ifbr(v(x), w(x)) > n for all x F, then by (7) and the last expression tv tw qn+1Gcontradicts (24) and (22) follows.

We want to calculate (25) more accurate and define

Br (F) = {br(v(x), w(x)) : x F with ax = 0}.

Then min(Br (F)) = k = k(v, w) n follows from (22).Hence

F = {x F : br(v(x), w(x)) = k}

is a non-empty set.We suppose that k < n for contradiction. Then k + 1 n and qnG qk+1G and

tv tw qk+1G follows from (24).If x F \ F, then br(v(x), w(x)) > k, and ax(yv(x)m yw(x)m) qk+1G and

0 tv tw xF

ax(yv(x)m yw(x)m) xF

ax(yv(x)m yw(x)m) modqk+1qm

G. (26)

For any x F we let i(x) be the unique integer i s with x Fi. From (16) we infer

br(v(x), w(x)) = k j (27)

for x F and using (7) and (10) we can reduce (26) furtherxF

axqkgi(x) qk+1G.

If i(x) = i(y) for some x = y F, then x, y F Fi and all elements v(x), w(x),v(y), w(y) are branches of the same tree Ti and if x = y, then by (16) the pairs of

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branches (v(x), w(x)) and (v(y), w(y)) have two distinct branch points on Ti at the

same level k. This contradicts our assumption (3), which followed from Ti being aperfect tree, hence x = y, see Observation 2.1.

We have seen that |F Fi| 1 and note that |F Fi| = 1 for at least one of theis, because F = . We discard all other is and may assume

F Fi = {xi} for all i s.

If gi = gi(x) = gi and axi = ai, then the last displayed sum becomes

isaiqkgi qk+1G.

Recall from (27) that k = br(v(x), w(x)) j, hence k Ci Cj for any i = jand k can not be the splitting level of pairs of branches from two distinct trees Ti, Tj.Hence splitting of branches (v(x), w(x)) at this level k can only happen at one pair,say for the one with label i(x) = 1. We can reduce the last sum expression to

a1qkg1 qk+1G.

Hencea1g1

qk+1qk

G = skqkG qjG

by (1) and (27). However, g1 is pure in G, hence 0 = a1 qjA contradicts (20). Itfollows that k = n, F = F and

all pairs v(x), w(x) (x F) branch at level n = br(v, w). (28)

Note that elements ax(yv(x)m yw(x)m) such that v, w branches strictly above n areabsorbed into qn+1G by (7). As before, but now for n = k, it follows

tv tw ax(yv(x)m yw(x)m) mod qn+1G and F = {x}. (29)

If v(x) = v, w(x) = w

and ax = a then v, w V1 and v, w T V. The

independent family of tvs for z by (29) simply turns into

tv tw a(yvm ywm) mod qn+1G.

The pair v, w can not branch at level > n because 0 tv tw mod qn+1G by (24).Hence v, w and v, w branch at the same level n > j by (28). Either {v, w} = {v, w}or the pairs are different. In the second case branching of two distinct pairs of branches

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at such a high level n can only happen at the same tree. Hence in either case we must

have1 = and also g1 = g

by (20). Using (7), (10) and (t) we have

0 qnz aqng mod qn+1G.

As before, we derive z ag modqn+1qn

G = qnsnG from (1). The set T of infinitebranches has size and hence its branches split at arbitrarily large level. Choose anysequence of pairs (v, w) of branches from T with branch points converging to infinityand note that G is 1free by Proposition 2.5, hence G is reduced. We derive that

z ag n qnsnG = 0, hence z = ag Ag as required.Proof of the Main-Lemma 3.3: The family of rigid tree-elements given by (3.3) can

be traded into an independent family of rigid tree-elements over the given number mand the same tree by Proposition 3.6. Now the assumptions for Proposition 3.7 aresatisfied for m = m0 and that z. Hence the conclusion of (3.3) follows from (3.7).

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Rudiger GobelFachbereich 6, Mathematik und InformatikUniversitat Essen, 45117 Essen, Germanyemail: [email protected]

andSaharon ShelahDepartment of Mathematics

Hebrew University, Jerusalem, Israeland Rutgers University, Newbrunswick, NJ, U.S.Ae-mail: [email protected]

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rudiger gobel and saharon shelah- indecomposable almost free modules - the local case

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