Rudiger Gobel and Saharon Shelah- Almost free splitters

8/3/2019 Rudiger Gobel and Saharon Shelah- Almost free splitters

1/35

arXiv:math/9

910161v1

[math.L

O]28Oct1999

ALMOST FREE SPLITTERS

Rudiger Gobel and Saharon Shelah

Abstract

Let R be a subring of the rationals. We want to investigate self splittingR-modules G that is Ext R(G,G) = 0 holds. For simplicity we will call such

modules splitters, see [16]. Also other names like stones are used, see a dictio-nary in Ringels paper [14]. Our investigation continues [10]. In [10] we answeredan open problem by constructing a large class of splitters. Classical splitters arefree modules and torsion-free, algebraically compact ones. In [10] we concen-trated on splitters which are larger then the continuum and such that countablesubmodules are not necessarily free. The opposite case of 1-free splitters ofcardinality less or equal to 1 was singled out because of basically different tech-niques. This is the target of the present paper. If the splitter is countable, thenit must be free over some subring of the rationals by Hausen [12]. Contrary tothe results in [10] and in accordance to [12] we can show that all 1-free splittersof cardinality 1 are free indeed.

1 Introduction

Throughout this paper R will denote a subring of the rationals Q and we will considerR-modules in order to find out when they are splitters. Splitters were introduced inSchultz [16]. They also come up under different names as mentioned in the abstract.

Definition 1.1 An R-module G is a splitter if and only if Ext R(G, G) = 0 or equiva-lently if Ext Z(G, G) = 0 which is the case if and only if any R-module sequence

0 G

X

G 0

This work is supported by the project No. G-0294-081.06/93 of the German-Israeli Foundationfor Scientific Research & DevelopmentAMS subject classification:primary 13C05, 18E40, 18G05, 20K20, 20K35, 20K40secondary: 13D30, 18G25, 20K25, 20K30, 13C10Key words and phrases: self-splitting modules, criteria for freeness of modulesGbSh 682 in Shelahs list of publications

1
http://arxiv.org/abs/math/9910161v1http://arxiv.org/abs/math/9910161v1http://arxiv.org/abs/math/9910161v1http://arxiv.org/abs/math/9910161v1http://arxiv.org/abs/math/9910161v1http://arxiv.org/abs/math/9910161v1http://arxiv.org/abs/math/9910161v1http://arxiv.org/abs/math/9910161v1http://arxiv.org/abs/math/9910161v1http://arxiv.org/abs/math/9910161v1http://arxiv.org/abs/math/9910161v1http://arxiv.org/abs/math/9910161v1http://arxiv.org/abs/math/9910161v1http://arxiv.org/abs/math/9910161v1http://arxiv.org/abs/math/9910161v1http://arxiv.org/abs/math/9910161v1http://arxiv.org/abs/math/9910161v1http://arxiv.org/abs/math/9910161v1http://arxiv.org/abs/math/9910161v1http://arxiv.org/abs/math/9910161v1http://arxiv.org/abs/math/9910161v1http://arxiv.org/abs/math/9910161v1http://arxiv.org/abs/math/9910161v1http://arxiv.org/abs/math/9910161v1http://arxiv.org/abs/math/9910161v1http://arxiv.org/abs/math/9910161v1http://arxiv.org/abs/math/9910161v1http://arxiv.org/abs/math/9910161v1http://arxiv.org/abs/math/9910161v1http://arxiv.org/abs/math/9910161v1http://arxiv.org/abs/math/9910161v1http://arxiv.org/abs/math/9910161v1http://arxiv.org/abs/math/9910161v1http://arxiv.org/abs/math/9910161v1http://arxiv.org/abs/math/9910161v1http://arxiv.org/abs/math/9910161v1http://arxiv.org/abs/math/9910161v1http://arxiv.org/abs/math/9910161v1http://arxiv.org/abs/math/9910161v1


2/35

splits.

A short exact sequence

0 B

C

A 0

represents 0 in Ext (A, B) if and only if there is a splitting map : A C such that = idA. Here maps are acting on the right.

Recall an easy basic observation, see [5]:

If Ext (A, B) = 0, A A and B B, then Ext(A,B/B) = 0 as well.

The first result showing freeness of splitters is much older then the notion of splittersand is due to Hausen [12]. It says that any countable, torsion-free abelian group is a

splitter if and only if it is free over its nucleus. The nucleus is the largest subring R ofQ which makes the abelian group canonically into an R-module. More precisely

Definition 1.2 The nucleus R of a torsion-free abelian group G = 0 is the subring RofQ generated by all 1

p(p any prime) for which G is p-divisible, i.e. pG = G.

The fixed ring R mentioned at the beginning will be the nucleus R = nuc G of theassociated abelian group G.

The following result reduces the study of splitters among abelian groups to those whichare torsion-free and reduced modules over their nuclei.

Theorem 1.3 ([16]) Let G be any abelian group and G = D C a decompositionof G into the maximal divisible subgroup D and a reduced complement C. Then the

following conditions are equivalent.

(i) G is a splitter.

(ii)

(a) D is torsion (possibly 0), C is a torsion-free (reduced) splitter with

pC = C for all p primary components Dp = 0 of D.

(b) D is not torsion and C is cotorsion.

Many splitters are constructed in [10], in fact we are also able to prescribe theirendomorphism rings. This shows that uncountable splitters are not classifiable in anyreasonable way, a result very much in contrast to classical well-known (uncountable)splitters which are the torsion-free algebraically compact (or cotorsion) groups.

2


3/35

The classical splitters come up naturally among many others when considering

Salces work [15] on cotorsion theories: A cotorsion theory is a pair of classes of R-modules (F,C) which are maximal, closed under extensions such that the torsion-freeclass F is closed under subgroups and the cotorsion class C is closed under epimorphicimages and Ext (F, C) = 0 for all F F and C C. The elements in F C aresplitters and in case of Harrisons classical cotorsion theory these are the torsion-free,algebraically compact groups. For the trivial cotorsion theory these are free R-modules.

Hausens [12] theorem mentioned above can be slightly extended without mucheffort, see [10].

Theorem 1.4 If R = nuc G is the nucleus of the torsion-free group G and G is asplitter of cardinality < 20, then G is an 1-free R-module.

Recall that G is an 1-free R-module if any countably generated R-submodule is free.The algebraic key tool of this paper can be found in Section 2. We consider torsion-freeR-modules M of finite rank which are minimal in rank and non-free. They are (bydefinition) n-free-by-1 R-modules if rk M = n + 1; the name is self explaining: Theyare pure extensions of a free R-module of rank n by an R-module of rank 1. Similarto simply presented groups, n-free-by-1 groups are easy represented by free generatorsand relations. Using these minimal R-modules we will show the following

Main Theorem 1.5 Any 1-free splitter of cardinality 1 is free over its nucleus.

The proof will depend on the existence of particular chains of 1-free R-module ofcardinality 1 which we use to divide 1-free R-modules of cardinality 1 into threetypes (I,II,III). This may be interesting independently and we would like to drawattention to Section 3. In Sections 4-7 we use our knowledge about these chains toshow freeness of splitter. The proof is divided into two main cases depending on thecontinuum hypothesis CH (Section 5) and its negation (Section 4). In the appendixSection 8 we present a proof of the main result of Section 5 under the weaker settheoretic assumption WCH 20 < 21, a weak form of CH which will be interesting(only) for splitters of cardinality > 1. The results in Section 6 and 7 on splitters oftype II and III do not use the case distinction by additional axioms of set theory.

2 Solving Linear Equations

Let R be a subring ofQ. Then R-modules of minimal finite rank which are not freewill lead to particular infinite systems of linear equations. Consider the Baer-SpeckerR-module R of all R-valued functions f : R on , also denoted by f = (fm)m.

3


4/35

Lemma 2.1 Let p = (pm)m, ki = (kim)m R (i < n) where each pm is not

a unit of R. Then we can find a sequence s = (sm)m R

such that the followingsystem of equations (s) has no solution x = (x0, . . . , xn) Rn+1, ym R with

y0 = xn, ym+1pm = ym +i


5/35

If G G is a pure R-submodule of some R-module G which is of finite rank, not

a free R-module such that all pure R-submodules of G

of smaller rank are free, thenwe will say that G is minimal non-free. Such minimal non-free modules are simplypresented in the sense that there are xi, ym G

(i < n, m ) such that

G = B, ymR : m with B =i


6/35

be a short exact sequence. Then we can find an R-module

H = H B, ym : m

with B =i


7/35

We get x = l

m=0(zm+ gm)km andl

m=0(ym+1pm ymi


8/35

Since G/G is also 1-free, the Chase radical is the smallest submodule with 1-free

quotient. If U is a submodule of G we write

UG = G for the Chase radical of G over U which is defined by (G/U) = G/U.

Given any 1-free R-module G of cardinality |G| = 1, we fix an 1-filtration

G =


9/35

Otherwise G/Gi is not 1-free, and by Pontryagins theorem we can find a finite

rank minimal non-free pure R-submodule M/Gi of G/Gi . Since G = i1 G0i and = M \ Gi G, there is also a least ordinal = (M) = (M/Gi) < 1, such that

(M \ Gi) (G0+1 \ G

0) = . (3.5)

Among the candidates M we choose one with the smallest (M) and take it forM = Gi+1. This completes the construction of the Gi

s. Notice that either theconstruction of G stops as in case (3.4) or we arrive at the second possibility:

Gi+1/Gi is minimal non-free for each i < 1 and |G/G1| = 1. (3.6)

It remains to show that in case (3.6) the following holds.G0G = G or equivalently G/G is 1-free. (3.7)

Suppose that G/G is not 1-free and let X be a non-free submodule of minimal finiterank in G/G which exists by Pontryagins theorem. Representing X in G we have

G = xi, ym, G : i < n, m with G/G = X,

see also Gobel, Shelah [10]. There are elements gm G (m ) such that

ym+1pm = ym +i


10/35

By minimality of j =: (Gj+1/Gj) we must have j j < and

(Gj+1 \ Gj) (G0j+1

\ G0j) =

and (Gj G0) (j 1) is a strictly increasing chain of length 1 of the countable

module G0 , which is impossible. Hence G/G is 1-free and (3.7) is shown. We havea useful additional property of the constructed chain which reflects (3.7).

Corollary 3.1 If 0 = 1 is not a limit ordinal, then G = G0G.

Proof: We concentrate on the case (3.6) and only note that the case (3.4) is similar.Recall from (3.7), that G/G is 1-free, hence the claim of the corollary is equivalent

to say that any submodule X of G must be G if only G0 X with G/X 1-free.

Let > 0 and suppose G0 X G and 0 = G/X is 1-free. First we claimthat

G X for all < .

If this is not the case, then let < be minimal with G X. Recall that can notbe a limit ordinal and we can write = + 1 for some < . We have G =

j1

Gj ,

hencei = min {j 1 : Gj X} 1

exists. If i = 0, then G0 G

0 X from > and G0 X. We get G0 =

G, G0 X and G

0 X requires G X, contradicting minimality of . Hence

i > 0 and i = j + 1. We have Gi X and Gj X from j < i and minimalityof i. However Gj+1/Gj is minimal non-free, and 0 = Gj+1 + X/X G/X is anepimorphic image, hence non-free as well. Therefore G/X is not 1-free, a contradictionshowing our first claim.From the first claim we derive

0, hence i = j + 1. We findGj X, Gj+1 X and G/X cannot be 1-free, a final contradiction.

We now distinguish cases for G depending on the existence of particular filtrations.Let G =


11/35

the same notation) {G : 1} of countable, pure and free R-submodules ofG such

that G0 = C and each G+1/G ( > 0) is minimal non-free. In this case we say thatG and the filtration are of type I.In the opposite case the chain only terminates at the limit ordinal 1 i.e. G = G

for all < 1. We have a proper filtration G =


12/35

Corollary 4.2 (1 < 20). If G is a splitter of cardinality 1 and nuc G = R,

then there is an 1-filtration G = 1 G of pure and free R-submodules G such thatnuc G = R for all 1.

Proof: From 1 < 20 and Gobel and Shelah [10], see Theorem 1.4, follows that G

is an 1-free R-module and G has an 1-filtration as in the hypothesis of Proposition4.1. Hence Corollary 4.2 follows by the Proposition 4.1.

Definition 4.3 Let G be a torsion-free abelian group with nuc G = R and X an R-submodule of G. ThenX is contra-Whitehead in G if the following holds.There are zm G, pm, kim R (i < n, m ) such that the system of equations

Ym+1pm Ym +i


13/35

According to the above definition we also say that a W is contra-Whitehead if (4.2)

has no solutions ym (m ) in G (hence in Ga) for some z and Xi = ai. Otherwisewe say that a is pro-Whitehead. IfG = 1

G is an 1-filtration ofG, then we define

W for X = G and let S = { 1: there exists a W contra-Whitehead }.

Proposition 4.4 If G =

1

G and S as above is stationary in 1, then G is not a

splitter.

Before proving this proposition we simplify our notation. If S we choose

zm G, a = (a0 , a

1 , . . . , a

n), pam = pm, gam = g

m, kaim = kim, yam = y

m

so that equations (4.1) and (4.2) become for X = G

ym+1pm = ym +i


14/35

which are increasing continuously. Let

(0) 0 H0id

H0h0 0 0

be defined for H0 = H0 with h0 the zero-map and suppose () is defined for all

< < 1 with a limit ordinal. We take unions and () is defined. If 1 \ S,we extend () trivially to get ( + 1) and if S we must work for ( + 1):We apply Proposition 2.2 to find H H+1 with

H+1 = H, em, xm : m , i < n.

and relations

em+1 pm = em +i


15/35

Put

fm = e

m g

m, vi = xi a

i , wm = ym g

m and note that

fmh = emh gmh = g

m g

m = 0

hence fm ker h = H0. Similarly wm, vm H

0. The last equation turns into

fm+1 pm = fm +i


16/35

If 2, then let

z

= (e)ze : e = (z

e ).Recall that X is pro-Whitehead in G, hence the systems of equations

ym+1pm ym +i 1, which is a contradition.So we find pm R and zm G such that pm1 does not devide zm + X in G for all

m . The above argument applies again for n = 0 and leads to a contradiction.

Corollary 4.6 Any splitter of cardinality at most 1 < 20 is free over its nucleus.

16


17/35

Proof: Let

G = 1

G

be an 1-filtration of the splitter G. By Corollary 4.2 we may assume that each G isa pure and free R-submodule of G with R = nuc G. If S denotes the set

{ 1 : G is contra-Whitehead in G},

then S is not stationary in 1 by the last Proposition 4.4. We may assume that all Gare pro-Whitehead in G and each G+1/G is countable, hence free by Theorem 4.5.We see that G must be free as well.

5 Splitters Of Type I Under CH

In view of Section 4 we may assume CH to derive a theorem in ZFC showing freenessfor 1-free splitters of cardinality 1 of type I. The advantage of the set theoreticalassumption is - compared with the proof based on the weak continuum hypothesisWCH in Section 8 - that the proof given here by no means is technical. Recall that Gis of type I ifG =

1

G for some 1-filtration {G : 1} of pure submodules G

such that each G+1/G ( > 0) is a minimal non-free R-module. In this section wewant to show the following

Proposition 5.1 (ZFC + CH) Modules of type I are not splitters.

Combining Proposition 5.1 and Corollary 4.6 we have can remove CH and have theimmediate consequence which holds in ZFC.

Corollary 5.2 Any 1-free splitter of type I (and cardinality 1) is free over its nu-cleus.

The proof of Proposition 5.1 is based on an observation strongly related to typeI concerning splitting maps. Then we want to prove a step lemma for applications

of CH. Finally we use CH to show Ext (G, G) = 0 in Theorem 5.1. In Section 1 wenoticed that if

0 B

C

A 0

is a short exact sequence, hence representing an element in Ext (A, B), then this elementis 0 if and only if there is a splitting map : A C such that = idA. This simplefact is the key for the next two results.

17


18/35

Observation 5.3 LetG =

1G be a filtration of type I. For 1, let

0 H0 H G 0 ()

be a continuous, increasing chain of short exact sequences with union

0 H0 H G 0 (1)

and let H0 = G be 1-free. Then any splitting map of (1) has at most one extensionto a splitting map of (1).

Proof: We may assume that the splitting map : G1 H1 of (1) has two extensions

,

: G H which split. Since 1 is a limit ordinal, there is some < 1 minimalwith ( ) G = 0. Clearly is not a limit ordinal and induces a non-trivial

map : G/G1 H. The domain of this map is minimal non-free, while its rangeis 1-free, hence must be 0, a contradiction.

Step-Lemma 5.4 LetG =

1

G be a filtration of type I and let

0 H0 Hh

G 0

be a short exact sequence with H0 = G. If : G H is a splitting map, then thereis an extension of this sequence such that does not extend to a splitting map of thenew short exact sequence:

0 H0 Hh

G 0

0 H0 Hh

G+1 0

Moreover, the vertical maps in the diagram are inclusions and if G+1/G is n-free-by-1, then B+1 is a free R-module of rank n and

H = H B+1, y

m : m

and B+1 is mapped under h mod G onto a free maximal pure R-submodule of

G+1/G.

18


19/35

Proof of the Step-Lemma 5.4: We will use special elements sm R (m ) to

kill extensions. It will help the reader to pose precise conditions on the choice of thesms only when needed, which will be at the end of the proof. Readers familiar withsuch proofs will know that we are working to produce a p-adic catastrophe.

First we use the fact that G = G+1/G is minimal non-free, say n-free-by-1. By(2.3) and (2.4) we have

G = i


20/35

Suppose that : G+1 H is an extensions of : G H such that

h = idG+1.

Now we want to derive a contradiction when choosing the sms accordingly (inde-pendent of !) We apply to () and get the equations in H:

() ym+1pm = ym

+i


21/35

particular choices of the extensions of as required in the Lemma.

Proof of Proposition 5.1: Let

H0 = G =1

G

be the module of type I. We must show that Ext (G, H0) = 0 and need a non-splittingshort exact sequence

0 H0 Hh

G 0 (5.3)

which we construct inductively as an ascending, continuous chain of short exact se-quences

0 H0 Hh G 0

with union (5.3). Let

0 H0 H1h1 G1 0

be the first step with G1 a free R-module of countable rank. By Observation 5.3 andCH we can enumerate all possible splitting maps : G H of extensions h as in(5.3) of all h1s by 1, and let { : G H, 1} be such a list. Using theStep-Lemma 5.4 and the uniqueness in Observation 5.3 we can discard any at stage

when constructing

0 H0 H+1h+1 G+1 0.

The resulting extension (5.3) can not split.

6 Splitters Of Type II

An R-module G is of type II if G has an 1-filtration G = 1 G of pure submodulesG such that G/G is 1-free for all non-limit ordinals 1, see Section 3. In thissection we want to show our second main

Theorem 6.1 If G is of type II, then G is a splitter if and only if G is free over itsnucleus R.

21


22/35

Remark Theorem 6.1 includes that strongly 1-free R-modules are never splitters,

except if trivially the module is free. This was very surprising to us.

Proof: If S = { 1 : G/G is not 1-free}, then S is a set of limit ordinals by(3.1), and if S we also may assume that G+1/G is minimal non-free, compare3.We get a - invariant (G) defined by S modulo the ideal of thin sets, see e.g. [4]. If(G) = 0, then we find a cub C 1, with CS = and G =

C

G. Let 0 = min C.

Then G = G0 F for some free R-module F, and G0 is a countable submodule ofGwhich must be free over R by Hausens [12] result, see also [10]. Hence G is free. Notethat the hypothesis ofG being 1-free is not used in this case! If (G) = 0 we want toshow that Ext (G, G) = 0. Theorem 6.1 can be rephrased as

If G is of type II, then G is a splitter if and only if (G) = 0. (6.1)

Now assume that S is stationary in 1. We want to construct some Hh

G 0with kernel ker h = H0, G isomorphic to H0 by , which does not split. If G =H ( 1), then

H0 =1

H

is a (canonical) 1-filtration of H0 copied from G. First we pick elements z H0

such that zR = R and H0/zR is 1-free, e.g. take any basis element from a layerH+2 \ H

+1 of the filtration of H

0. Then we define inductively a continuous chain ofshort exact sequences ( 1).

()

0 H0 H

h G 0

countable, free submodules H H, and ordinals < 1

subject to various conditions. At the end we want in particular H =

1

H =

1

H.

If = 0, then G0 = 0 and we take the zero map h0 : H0 G0 0 with kernel

H0.

Suppose () is constructed for all < . If is a limit, we take unions h =


23/35

Put H+1 = H F with F a free R-module of the same rank as the free R-module

G+1/G. As G+1 = F

G, we may choose an isomorphism h

: F F

and extend h to h+1 by h+1 = h h

. Clearly ker h+1 = ker h = H0 and

Im h+1 = G+1.If S, then we must work. We have G+1/G = B

+1, y

m : m from (2.3)

and (2.4). Hence

G+1 = G, B+1, ymR : m , B+1 =i


24/35

as required in (b). Similarly, by Proposition 2.2 the map h extends to an epimorphism

h+1 : H+1

G+1. It is now easy to check that (c) holds and it is also easy to seethat ker h+1 = H0. Next we extend H H+1 carefully such that ( + 1), (a), (b),

(d), (e) and (f) hold.

Im h+1 = G+1 is a countable module of the 1- free R-module G, hence free and h+1must split. There is a splitting map

: G+1 H+1 such that h+1 = idG+1,

hence H+1 = H0 G+1. Let

0 : H+1 H0, 1 : H

+1 G+1

be the canonical projections with 0 + 1 = idH+1. Recall that B+1 H+1. Choose

1 large enough such that S, , + 1 and (B+1 + H)0 H

. This

is easy because 1 \ S is unbounded and (B+1 + H)0 is countable. Put H+1 =H (G+1) and = ( + 1)

. Note that

G+1 = H+1h+1 H+1h+1 G+1h+1 = G+1

and (a) follows.

If (+1), then H+1+H = H+1 and if (+1), then H+1+H

= H

G+1

and (d) follows. We see immediately H1 G+1 and H0 H(+1) , hence

H H0 + H1 H+1

and ( + 1) holds; similarly B+1 H+1 for (c). From H(+1) H

0 and the modular

law we have H+1 H0 = H(+1) (G+1 H

0) = H(+1) and (f) holds.

Finally we choose H =

1

H, h =

1

h and

0 H0 Hh

G 0 (6.5)

is established and it remains to show that (6.5) does not split. Suppose for contradictionthat : G H is a splitting map for h. We have H =

1

H and G =

1

G

1-filtration. Using the above properties of the Hs, it follows by a back and forthargument that

E = { 1 : H H0 = H, G H}

24


25/35

is a cub. On the other hand S is stationary in 1 and we find

S E.

From (6.3) and (6.4) we have

ym+1pm = ym +i


26/35

From Theorem 6.1 we see that non-free but strongly 1-free abelian groups are never

splitters. We find this very surprising. Particular groups like the Griffith-group Gbelow which is a Whitehead group (Ext (G,Z) = 0) under Martins axiom and CH isnot a splitter. Recall a nice and easy construction ofG which is sometimes Whiteheadbut always fails to be a splitter in general.

Let P = Z1 =1

Z the cartesian product ofZ. If 1 is a limit ordinal choose

an order preserving map : with sup() = . Then, along this ladder systemwe define branch-elements

cn =in

(i)i!

n!

which are a divisibility chain of c0 modulo 1 Z, henceG =

1

Z, cn : 1, a limit ordinal, n

is a pure subgroup ofP. We see that |G| = 1 and G is 1- free by 1-freeness ofP; see[5] (Vol 1, p 94, Theorem 19.2). Moreover G = 0 because G = G


27/35

By a basic observation from Section 1 it is enough to show that Ext (G1, G) = 0.

Inductively we will construct a non-trivial element in Ext (G1, G). We consider thefollowing diagram

(0) 0 G H0h0 G00 = 0 0

() 0 G Hh

G0 0

(1) 0 G Hh

G1 0

The first row is the trivial extension with G = H0 and h0 = 0. Vertical maps andmaps between G and Hs are inclusions. The sequences () are increasing continuousand suppose () is constructed for all < . Then h =


28/35

where the sm R will be specified later on, and gm H.

Suppose that (1) splits and consequently : G1 H is a splitting map for h.Then letdm = ym ym, ei = xi xi and fm = gm gm.

From splitting we get again

dm, ei, fm G ( 1, m , i < n).

Using G =

1

G for 1 we find 1 such that

dm, ei, fm G for all i < n, m .

Consider a map : 1 1 taking any 1 to

() = min { 1; a limit ordinal, dm, ei, fm G, , m , i < n}

and note that C = { 1 : () = } is a cub in 1 and a subset of

E = { : a limit ordinal, dm, ei, fm G for all < ,i < n, m }.

Hence E is a cub in 1. Next we apply to (7.2) and subtract (7.3). Hence we get asystem of equations in H+1.

dm+1pm = dm +i


29/35

8 Appendix: Splitters Of Type I Under 20 < 21

In Section 5 we have seen a proof that CH implies modules of type I are never splitters.A slight variation but some what technical modification of the proof, shows that thisresult can be extended to WCH that is 20 < 21 . Due to Section 5 this is not neededfor the main result of this paper dealing with modules of cardinality 1 but it will beinteresting when passing to cardinals > 1. We outline the main steps, their proofs aresuggested by the proofs in Section 5.

Theorem 8.1 ( ZFC + 20 < 21) Modules of type I are not splitters.

Step-Lemma 8.2 LetG = 1 G be a filtration of type I and let0 K H

h G 0

be a short exact sequence with some z K such thatzR = R andK,K/zR are 1-free.Then there are two commuting diagrams ( = 0, 1)

0 K Hh

G 0

0 K Hh G+1 0

with vertical maps inclusions such that any third row with H 1-free,

0 K H

h G 0

and any splitting map of h cannot have two splitting extensions of h :

0 K Hh

G 0

0 K Hh G+1 0

0 K Hh

G 0

Moreover H = H

B

+1, ym : m and B

+1 is mapped under h mod Gonto a free maximal R-submodule of G+1/G cf. (2.3).

29


30/35

Definition 8.3 If such an extension as in (8.2) exists for some {0, 1} we say

that splits over (H, h).

Proof of Lemma 8.2: Compare the proof of the Step-Lemma 5.4 but note that atthe end you must take once more differences of the elements dm, ei for = 1 and = 0 respectively. Then we are able to apply Lemma 2.1 to get a contradiction fromsplitting.

We then apply the Step-Lemma and weak diamond 1 to construct a short exactsequence

0 H0 Hh

G 0.

Let : G H0 be a fixed isomorphism. Later we will use consequences of 1to show that h does not split.

Proof of Theorem 8.1: If H = G, then H0 =

1

H is an 1-filtration if

G =

1

G is the given filtration of type I.

Let T = 1>2 be the tree of all branches : 2 for some 1. We call = l()the length of . Branches are ordered as usually, hence < if Dom = . Theempty set is the bottom element of the tree. If T, then we construct triples

(H, H, h)

of R-modules H H with H free of countable rank and a homomorphism

h : H G

subject to various natural conditions.

(i)

H = H0, H = 0 and h = 0, hence

0 H0 Hh G0 = 0 0 is short exact.

(ii)

If < , then (H, H

, h) (H , H

, h), i.e.

H H , H H

and h h .

(iii) H0 = ker h and Im h = Gl() = Hh.

30


31/35

(iv) If 1 is a limit and 2, then we take unions

(H, H, h) =


32/35

Similarly h() =


33/35

and (ii) holds. Similarly B+1 H and (S i) is shown. From H H

0 and the

modular law we have

H H0 = H (G+1 H

0) = H

and (S iii) holds.

The construction of the tree with triples is complete. We are ready to use the weakdiamond 1(S) to show that G is not a splitter.

We will use 1(S) as stated in Eklof, Mekler [4, p. 143, Lemma 1.7] and note thatG =


34/35

If () = P() = 0, then (H0, h0) is part of the construction of

0 H0 H+1 G+1 0

and does not split over (H0, h0), but is a global splitting map, hence splits at over (H0, h0), a contradiction.Necessarily () = P() = 1 and by (a) does not split over (H1, h1), but this

time (H1, h1) was used in the construction of Hh

G and a contradiction follows.This shows that is no splitting map, and G is not a splitter.

References

[1] T. Becker, L. Fuchs, S. Shelah, Whitehead modules over domains, ForumMathematicum 1 (1989), 5368.

[2] A.L.S. Corner, R. Gobel, Prescribing endomorphism algebras - A unified treat-ment, Proceed. London Math. Soc. (3) 50 (1985), 471 483.

[3] P. EklofSet Theoretic Methods in Homological Algebra and Abelian Groups, LesPresses de l Universite de Montreal 1980.

[4] P. Eklof, A. Mekler Almost free modules, Set-theoretic methods, North-Holland, Amsterdam 1990.

[5] L. Fuchs, Infinite abelian groups - Volume 1, 2 Academic Press, New York (1970,1973).

[6] P. A. Griffith, A solution of the splitting mixed problem of Baer, Trans. Amer-ican Math. Soc. 139 (1969), 261269.

[7] R. Gobel, R. Prelle, Solution of two problems on cotorsion abelian groups,Archiv der Math. 31 (1978), 423431.

[8] R. Gobel, Cotorsion-free abelian groups with only small cotorsion images, Aus-tral. Math. Soc. (Ser. A) 50 (1991), 243247.

[9] R. Gobel, New aspects for two classical theorems on torsion splitting, Comm.Algebra 15 (1987), 24732495.

34


35/35

[10] R. Gobel, S. Shelah, Cotorsion theories and splitters, to appear in Trans. Amer.

Math. Soc. (1999)[11] R. Gobel, J. Trlifaj, Cotilting and a hierarchy of almost cotorsion groups, to

appear in Journal of Algebra (1999)

[12] J. Hausen, Automorphismen gesattigte Klassen abzahlbaren abelscher Gruppen,Studies on Abelian Groups, Springer, Berlin (1968), 147181.

[13] C. M. Ringel, The braid group action on the set of exceptional sequences of ahereditary artin algebra, pp. 339 352 in Abelian group theory and related topics,Contemporary Math. 171 American Math. Soc., Providence, R.I. 1994.

[14] C. M. Ringel, Bricks in hereditary length categories, Resultate der Mathematik6 (1983), 64 70.

[15] L. Salce, Cotorsion theories for abelian groups, Symposia Math. 23 (1979), 1132.

[16] P. Schultz, Self-splitting groups, Preprint series of the University of WesternAustralia at Perth (1980)

[17] S. Shelah Infinite abelian groups, Whitehead problem and some constructions,Israel Journal Math. 18 (1974), 243 256.

[18] S. Shelah On uncountable abelian groups, Israel Journal Math. 32 (1979), 311

330.

[19] S. Shelah, A combinatorial theorem and endomorphism rings of abelian groupsII, pp. 37 86, in Abelian groups and modules, CISM Courses and Lectures, 287,Springer, Wien 1984.

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

and

Saharon ShelahDepartment of MathematicsHebrew University, Jerusalem, Israeland Rutgers University, Newbrunswick, NJ, U.S.Ae-mail: [email protected]

Rudiger Gobel and Saharon Shelah- Almost free splitters

Documents

Transcript of Rudiger Gobel and Saharon Shelah- Almost free splitters