Rudiger Gobel and Saharon Shelah- Generalized E-Algebras via lambda-Calculus I

8/3/2019 Rudiger Gobel and Saharon Shelah- Generalized E-Algebras via lambda-Calculus I

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Generalized E-Algebras via -Calculus I

Rudiger Gobel and Saharon Shelah

Abstract

An R-algebra A is called an E(R)algebra if the canonical homomorphismfrom A to the endomorphism algebra EndRA of the R-module RA, taking anya A to the right multiplication ar EndRA by a, is an isomorphism ofalgebras. In this case RA is called an E(R)module. There is a proper class ofexamples constructed in [9]. E(R)-algebras arise naturally in various topics ofalgebra. So it is not surprising that they were investigated thoroughly in the lastdecade, see [8, 10, 13, 16, 21, 24, 25, 27, 32, 31]. Despite some efforts ([25, 10])it remained an open question whether proper generalized E(R)-algebras exist.These are Ralgebras A isomorphic to EndRA but not under the above canonicalisomorphism, so not E(R)algebras. This question was raised about 30 years ago

(for R = Z) by Schultz [34] (see also Vinsonhaler [37]). It originates from Problem45 in Fuchs [17], that asks one to characterize the rings A for which A = EndZA(as rings). We will answer Schultzs question, thus contributing a large class ofrings for Fuchs Problem 45 which are not E-rings. Let R be a commutative ringwith an element p R such that the additive group R+ is p-torsion-free andp-reduced (equivalently p is not a zero-divisor and

n p

nR = 0). As explainedin the introduction we assume that either |R | < 20 or that R+ is free, seeDefinition 1.1.

The main tool is an interesting connection between -calculus (used in the-oretical computer sciences) and algebra. It seems reasonable to divide the workinto two parts; in this paper we will work in V=L (Godels universe) where

stronger combinatorial methods make the final arguments more transparent. The

Supported by the project No. I-706-54.6/2001 of the German-Israeli Foundation forScientific Research & DevelopmentGbSh867 in Shelahs archivesubject classification (2000):primary: 20K20, 20K30;secondary: 16S60, 16W20;Key words and phrases: E-rings, endomorphism rings

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E(R)-algebras can also be defined dually, assuming that the homomorphism

EndR A A ( 1)

is an isomorphism. It is easy to see that E(R)algebras are necessarily commutative.For any S-ring (with S cyclically generated) that is S-incomplete we will construct

noncommutative Ralgebras with EndR A = A. Hence these As are generalized E(R)-algebras but not E(R)algebras. IfR = Z and RA is an abelian group, then we do notmention the ring Z: e.g. E(Z)modules are just Egroups. The existence of generalizedErings answers a problem in [34, 37].

If is a cardinal, then let o = { : cf() = , }. We will only need theexistence of a non-reflecting subset E o for some regular uncountable, not weaklycompact cardinal such that the diamond principle E holds. It is well-known (seee.g. Eklof, Mekler [12]) that E is a consequence for all non-reflecting subsets E ofregular uncountable, not weakly compact cardinals in Godels universe (V = L). Weindicate our (weaker) set theoretic hypothesis (which also holds in other universes) asE in our following main result.

Theorem 1.4 Let R be a S-incomplete S-ring for some cyclically generated S. If > | R | is a regular, uncountable cardinal and E o a non-reflecting subset withE, then there is anScotorsionfree, noncommutative Ralgebra A of cardinality|A| = with EndR A = A. Moreover any subset of cardinality < is contained in an

R-monoid-algebra of cardinality < .

A similar result without the set theoretic assumption will be shown in [23]:

Theorem 1.5 Let R be a S-incomplete S-ring for some cyclically generated S. Forany cardinal = + with| R | 0 = there is anScotorsionfree, noncommutativeRalgebra A of cardinality |A| = with EndR A = A.

It seems particularly interesting to note that the R-monoid A comes from (classical)-calculus taking into account that elements of an E(R)-algebra A are at the sametime endomorphisms ofA, thus the same phenomenon appears as known for computerscience and studied intensively in logic in the thirties of the last century. The problemsconcerning the semantics of computer science were solved four decades later by Scott[35, 36]. We will describe the construction of the underlying monoid M explicitly.Since this paper should be readable for algebraists with only basic background onmodel theory, we will also elaborate the needed details coming from model theory. Thebasic knowledge on model theory is in [33], for example. In Section 4 and 5 the monoidM will be completed and become the algebra A.

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2 Model theory of bodies and skeletons via-calculus

2.1 Discussion

We begin by defining terms for a skeleton and will establish a connection with -calculus. Let R be any commutative S-ring with S = p. (S-incompleteness will beadded in Section 5.)

By definition of generalized E(R)-algebras A, endomorphisms ofRA must be con-sidered as members of A. Hence they act on RA as endomorphisms while they areelements ofRA at the same time. Thus we will introduce the classical definitions from-calculus over an infinite set X of free variables and an infinite set Y of bound vari-ables to represent those maps. First note that we can restrict ourselves to unary, linear

functions because endomorphisms are of this kind. (The general argument to reduce-calculus to unary functions was observed by Schonfinkel, see [2, p. 6].) What are thetypical terms of our final objects, the bodies? If x1 and x2 are members of the gener-alized E(R)-algebras A and a, b R, then also polynomials like n(x1, x2) = ax

n1 + bx

32

belong to the algebra A, so there are legitimate functions pn(y) = y.n(y, x2) on Ataking y n(y, x2) and A must be closed under such generalized polynomials.This observation will be described in Definition 4.2 and taken care of in Proposition2.20 and in our Main Lemma 6.2. A first description of these generalized polynomialswill also be the starting point for our construction and we begin with its basic settings.

2.2 The notion of terms

Let be a vocabulary with no predicates; thus is a collection of function symbolswith an arity function defining the places of function symbols. Moreover letX be an infinite set of free variables. Then unspecified (, X)-terms (briefly calledterms) are defined inductively as the closure of the atomic terms under these functionsymbols (only), that is:

(i) Atomic terms are the 0-place functions: the individual constants (in our case 1)and members x from X.

(ii) The closure: If 0, . . . , n1 are terms and F is an n-place function symbol from, then F(0, . . . , n1) is a term.

We also define the (usual) length l() of a term inductively: Let l() = 0 if isatomic and l() = k + 1 if derives from (ii) with k = max{l(i) : i < n}.

If is an unspecified (, X)-term, then we define (also by induction on l()) a finitesubset F V() X of free variables of :

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(a) If is an individual constant, then F V() = and if X, then F V() = {}.

(b) If = F(0, . . . , n1) is defined as in (ii), then F V() = i


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(ii) (Step k = m+1) Suppose that skm is defined and (, x) t(skm, X) is a specified

term of length k with x = x0, . . . , xn1 and y = y0, . . . , yn1 suitable for x,then F(y0,...,yn1) is an nplace function symbol belonging to

skk (but not to

sk


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The axioms in Tsk


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(ii) If = and , are reduced, but is not of the form F0(y0,...,yt)(1, . . . , t),

then r = is reduced.

(iii) Suppose that and i (i t) are reduced and = F(y0,...,yt)(1, . . . , t), with(, x0, . . . xt) t(

sk


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(i) If1 is atomic, then 2 is atomic and 1 = 2.

(ii) If 1 = 1

1 such that

1,

1 are reduced, then 2 =

2

2 and

1

.= 2, 1

.= 2 .

(iii) If i = Fi(y0,...,ymi )

(i1, . . . , imi

) for i 2, then m1 = m2 and there is a permuta-tion of {1, . . . m1} with 1j

.= 2(j) for all j m1 and also

1

.= 2.

Observation 2.8 (i) The relation.

= is an equivalence relation on tr(sk


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(ii) r1.

= r2.

Proof. (ii) (i): From the Definition-Observation 2.6 follows Tsk 1 =r1, 2 =

r2 and by Observation 2.8 is T

sk r1 = r2 thus T

sk 1 = 2.(i) (ii): From (i) follows Tsk r1 =

r2. Thus

r1

.= r2 by Definition 2.7 and

Observation 2.8.

2.5 The skeleton freely generated by X

Next we construct and discuss free skeletons based on reduced terms. We will use theinfinite set X of free variables to construct a skeleton MX which is freely generated bya set which corresponds by a canonical bijection to X.

By Observation 2.8(i) we have an equivalence relation .= on sk := sk


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Proof. The axioms (2.1) are satisfied, e.g. the crucial condition (ii)(a) follows by

definition of [F].

Remark 2.13 In the construction of the free skeleton MX we also used an infinite setY of bound variables. However, it follows by induction that another infinite set Y ofbound variables leads to an isomorphic copy of MX . Thus we do not mention Y inTheorem 2.12.

Lemma 2.14 LetB be a subset of the Tsk


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Repeated application of (1) and (2) leads to finite sums liken

i=1 aiyi and we will

write bd := bd


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Proof. Generalized E(A)-algebras satisfy EndR A = A. Thus any function symbol

F of bd can be interpreted on A as a function and the axioms (2.1) and (2.3) hold.

But note that only free bodies arrive from skeletons, see Section 3.2.

2.7 Linearity of unary body functions from t(bd


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Proposition 2.20 The weak completeness of bodies. LetM be a body and (, x)

a polynomial (a term in t(bd, X)) with x = x0, . . . , xm and d1, . . . , dm M. Thenthere is (, x1, . . . , xn) t(

bd, X) and the following holds.

(i) M is an R-module.

(ii) The unary function .z(z, d1, . . . , dm) : M M (z (z, d1, . . . , dm)) is theR-endomorphism y.y(z, d1, . . . , dm) EndR(M) (y y (d1, . . . , dm)).

Remark. We will show here that there is a function symbol (, x) t(bd, X)such that (d, d1, . . . , dm) = d

(d1, . . . , dm) for all d M, see axioms (2.1)(ii)(a) ofthe skeletons.

Proof. By the axioms (2.3) of Tbd it is clear that M is an R-module. It remainsto show (ii). Let (, x) t(bd, X) with x = x0, . . . , xm and x = x1, . . . , xm. ByLemma 2.18 there are monomials (l, x

) t(sk, X) and al R such that

=l


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3 From the skeleton to the body

3.1 The monoid structure of skeletons

Recall from Theorem 2.12 that the skeleton on an infinite set X of free variables is theset MX = {[] : t

r(sk)} with n-place functions

[F] : (MX)n MX ([0], . . . , [n1]) [F(0, . . . , n1)]

for each n-place function symbol F = F(y0,...,yn1) with (, x) tr(sk) and F V() =

{x0, . . . , xn1}. For simplicity we will also write Roman letters for the members ofMX ,e.g. m = [] MX . The set MX has a distinguished element 1 = [1] and m1 = 1m = m

holds for all m MX (thus MX is an applicative structure with 1). In order to turnMX into a monoid, we first represent MX as a submonoid of Mono(MX), the injectivemaps on MX , say : MX Mono(MX):

Let a = [] MX and t(bd). We will use induction: If a = [1] then

[]a = , if a = [x], then []a = x and if a = [F(y0,...,yn1)x1, . . . , xn1] is a unaryfunction as above, then []a = (, x1, . . . , xn1) (so a = [y.y]). Thus a mapsany m = [] MX to m(a) = m(y.y) = [m] MX which can be representedby a reduced element using (2.2). If a = b MX , then 1(a) = a = b = 1(b) thus : MX Mono(MX) MX is an embedding. We define multiplication of elementsa, b MX as composition of functions (a)(b) = (ab). This is to say that from

a = [], b = [

] we get the product as the equivalence class of y.((y

)). We willwrite a b = ab and will often suppress the map . From Mono(MX) follows that alsoMX is a monoid. Also note that [x][x

] = [x][x] for any free variables x, x X. Weget an

Observation 3.1 The free skeleton(MX , , 1) with composition of functions as productis a non-commutative (associative) monoid with multiplication defined as above by theaction on MX : If [], [

] MX , then [] [] = [y.((y ))].

3.2 Free Bodies from skeletons

Finally we will associate with any skeleton M its (canonical) body BRM: Let BRMbe the R-monoid algebra RM of the monoid M. Moreover any n-place function F :Mn M extends uniquely by linearity to F : BRM

n BRM. We deduce a

Lemma 3.2 IfR is a commutative ring as above and M a skeleton, then the R-monoidalgebraBRM of the monoid M is a body. If the skeleton MX is freely generated by X,then also BRM is freely generated by X as a body. MoreoverRBRMX =

mMmR.

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Proof. It is easy to see that BRM (with the linear n-place functions) is a body.

We first claim that X, viewed as {[x] : x X} BRMX is a basis. First applyLemma 2.18 to the R-monoid BRMX : Any (, x) t(

bd, X) can be written as apolynomial =

l lal with monomial (l, x) t(

sk, X). Moreover, any l is viewedas an element of Mono(MX), so axiom (2.1)(ii)(a) applies and l becomes a product ofelements from X. Thus X generates BRMX . The monomials of the skeleton M extenduniquely by linearity to polynomials of the free R-module RBRMX =

mMmR from

its basis M.

We will also need the notion of an extension of bodies.

Definition 3.3 LetB and B be two bodies, thenB B (B extends B) if and onlyif B B as R-algebras and if (, x) t(bd, X) and F is a function symbol withcorresponding unary, R-linear function F ofB, then its natural restriction to B is thefunction forB corresponding to F.

Example 3.4 LetX X be sets of free variables andB,B be the free bodies generatedby the free skeletons obtained from X and X, respectively. ThenB B. In this casewe say thatB is free overB.

4 The technical tools for the main construction

The endomorphism ring EndR BRMX of the R-module RBRMX has natural elementsas endomorphisms, the linear maps, our (generalized) polynomials interpreted by theterms in t(bd, X) acting by scalar multiplication on BRMX as shown in Proposition2.20(ii). The closure under these polynomials is dictated by the properties of E(R)-algebras. Thus we would spoil our aim to construct generalized E(R)-algebras if welose these R-linear maps on the way.

Definition 4.1 Let (, x) t(bd, X) with x = x0, . . . , xn. If B is a body, d =d1, . . . , dn with d1, . . . , dn B, then we call sd(y) = y.(y, d) the (generalized) poly-nomial overB with coefficients d.

Note that d(y) is a sum of products of elements di and y. Here we must achieve(full) completeness of the final body, thus showing that any endomorphism is repre-sented. By a prediction principle we kill all endomorphisms that are not represented byt(bd, X) - thus the resulting structure will be complete: Any R-endomorphism of anextended body BRMX will be represented by a polynomial q(x) over BRMX , so BRMXis complete or equivalently an E(R)-algebra.

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The fact that BRMX is not just the R-linear closure (or A-linear closure for some

algebra A), makes this final task, to get rid of undesired endomorphisms harder thanin case of realizing algebras as endomorphism algebras (where the closure is not thatfloppy).

Definition 4.2 LetB be a body and G = RB. Then EndR G is called represented(by q(y)) if there is a generalized polynomial q(y) with coefficients in B such thatg = q(g) for all g G.

If all elements from EndR G are represented, then B is a generalized E(R)-algebra.As for other algebraic structures we have the

Lemma 4.3 Let R be anS-ring as above andB be a body generated by B, then B isa basis ofB if one of the following equivalent conditions holds.

(i) If B = BRMX is the body generated by the free skeleton MX and X B is abijection, then this map extends to an isomorphismB B

(ii) B is independent inB, i.e. if(1, x), (2, x) tr(bd, X) and the sequence b fromB is suitable for x such that 1(b) = 2(b), then T

bd 1(x) = 2(x).

(iii) For all bodies H and maps : B H there is an extension : BRMX Has Tbd-homomorphism.

Proof. The proof is well known for varieties (see Gratzer [26, p. 198, Theorem 3]or Bergman [3, Chapter 8]), so it follows from Observation 2.15.

Freeness Proposition 4.4 LetR be anS-ring as above and X X be sets of vari-ables and BRMX BRMX the corresponding free bodies. If u B := BRMX andv X \ X, then w := u + v B := BRMX is free overB, i.e. there is a basis X

ofB with w X X.

Proof. We will use Lemma 4.3 (c) to show that the set X := (X \ {v}) {w} is

a basis ofB

. First note that X

also generates B

, thus B

= BRMX .Given : X H for a body H, we must extend this map to : B H.

Let := (X \ {v}) and note that the set X \ {v} = X \ {w} is independent.Thus if B0 := BRMX\{v}, then

extends to : B0 H by freeness, and fromu B0 follows the existence of u H. We now define : if B0 := , then (X\{v}) = = (X\{v}). Thus it remains to extend to : B H in such away that w = w. Ifw =: h H, then we must have h = w = (u+v) = u+v.

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Hence put v := h u = h u. Now : B H exists, because X is free,

and w = (u + v) = u + h u = h = w, thus holds as required.

The following corollaries (used several times for exchanging basis elements) areimmediate consequences of the last proposition.

Corollary 4.5 Let X be a basis for the body B, v X and B be the subbody of Bgenerated by X\ {v} and w B, then X = X\ {v} {v + w} is another basis forB.

Corollary 4.6 If X X andBRMX BRMX , then any basis ofBRMX extends toa basis ofBRMX .

Proof. If X is a basis ofBRMX , then it is left as an exercise to show that ( X \X) X is a basis ofBRMX .

The last corollaries have another implication.

Corollary 4.7 Suppose thatB ( ) is an ascending, continuous chain of bodiessuch thatB+1 is free overB for all < . ThenB is free overB0 and ifB0 is free,thenB is free as well.

The proof of the next lemma is also obvious. It follows by application of the

distributive law in Tbd

and collection of summands with p.Lemma 4.8 Letq(y) be a generalized polynomial and r R. Then there is a polyno-mial q(y) such that q(y1 + ry2) = q(y1) + rq

(y1, y2).

Proof. By Lemma 2.18 we can write =

l


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Proof. (i) Write q(y) =n

i=1 mi(y) as a sum of minimal length of generalized R

monomials and suppose for contradiction that y appears n times in m1(y) with n > 1.Also let, without loss of generality, n be maximal for the chosen monomial m1(y).

By the distributive law the monomials of the polynomial q(g +v1+v2) include thosemonomials induced by m1(y) replacing all entries of the variable y by arbitrary choicesof v1 and v2. Let m

1 be one of these monomials. If there are further such monomials

mi (i k) alike m1 arriving from this substitution into monomials mi(y) of q(y) withk

i=1 mi = 0, then replacing all v1s and v2s by ys gives

ki=1 mi(y) = 0 contradicting

the minimality of the above sum. Thus m1 represents a true monomial (not canceledby others) of q(g + v1 + v2), and as n > 1 we may also assume that v1 and v2 bothappear in m1. This monomial does not exist on the righthand side of the equation in

the lemma a contradiction. Thus (a) holds.(ii) First substitute in the given equation v1 := y, v2 := 0 and v1 := 0, v2 := y,respectively. Thus q(g + y) = q1(y) + c and q(g + y) = q2(y) + c

, where c := q2(0), c :=

q1(0) B0. Subtraction now yields 0 = q1(y)q2(y)+(cc), thus q1(y)q2(y) = ccdoes not depend on y, as required.

In order to establish the Step Lemmas, we next prepare some preliminary results.Let X =

n Xn be a strictly increasing sequence of infinite sets Xn of variables

and fix a sequence vn Xn \ Xn1 of elements (n ). Moreover, let M = MX bethe skeleton and B := B(X) be the body generated by X for , respectively.

Note that by our identification B(X) is an R-algebra and restricting to the modulestructure G := RB(X) is an R-module, which is free by Lemma 3.2. Recall thatS = {pn : n } for some p R (with

n p

nR = 0) generates the S-topology onR-modules. Thus the S-topology is Hausdorff on G and G is naturally an S-pureR-submodule of its S-completion G; we write G G and pick particular elementswn G. If ai {0, 1} and ln N is increasing, then we define

wn(vn, ln, an) := wn :=kn

plklnakvk G (4.1)

and easily check that

wn pln+1lnwn+1 = anvn Gn for all n . (4.2)

Proposition 4.10 Letan {0, 1} and ln N be as above. If X+1 = X \ {vn : an =1, n > 0} W with W = {wn : n > 0} andB+1 := B(X+1), G+1 = RB+1, then thefollowing holds

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(i) G G+1

G

(ii) G+1/G is p-divisible, thus anS1R-module.

(iii) X+1 is a basis of the (free) skeleton M+1 = MX+1.

(iv) The R-algebraB+1 is freely generated by the skeletonM+1, thus B+1 = RM+1and G+1 =

mM+1

Rm.

(v) B(X+1) is free overB(Xn) (as body).

Proof. Claim (i): Clearly G G and G+1 G. From wn, wn+1 X+1,an = 1 and (4.2) follows vn G+1. Hence vn G+1 for all n and G B(X+1) = G+1 follows at once. Thus G+1/G G/G and purity (G+1 G)follows if G+1/G is p-divisible. This is our next

Claim (ii): By definition of the body B(X+1), any element g G+1 is the sumof monomials in X+1. If wn, wn+1 are involved in such a monomial, then we apply(4.2) and get wn = pwn+1 + anvn, which is wn pwn+1 mod G. Let m be the largestindex ofwns which contributes to g. We can remove all wi of smaller index i < m andalso write wm pwm+1 mod G. Thus g + G is divisible by p and G+1/G is anS1R-module.

Claim (iii): It is enough to show that X+1

is free, because X+1

generates M+1by definition of the skeleton. First we claim that

X = (X \ {vn}) {wn} is free.

We apply the characterization of a basis by Lemma 4.3 (ii). Let (1, x), (2, x) tr(bd, X) (x = x1, . . . , xk) be such that

1(y1, . . . , yk) = 2(y1, . . . , yk)

for some yi X and suppose that y1 = wn (there is nothing to show if wn does

not appear among the yis, because they are free; otherwise we relabel the yis such

that y1 = wn). Now we consider the above equation as an element in G and notethat the support [yi] X of the elements yi, (i > 1) is finite, while wn has infinitesupport {vk : k > n} [wn]. Thus we project i(y1, . . . , yk) onto a free summandfrom [wn] \

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Finally let y1, . . . , ym be any finite subset ofX+1. We may assume that y1, . . . , yk

Wn = {wi : i n} and yk+1, . . . , ym X+1 \ W. Hencey1, . . . , ym Wn X \ {v1, . . . , vn}

which is free by the last claim (4.3), thus X+1 is free and (iii) follows.

(iv) is a consequence of (iii) and the definitions.(v) We note that (by (iii)) the body B(X+1) is freely generated by X+1 and also

(using (4.2)) by the (free) set X = (X+1 \ {w1, . . . , wn}) {v1, . . . , vn}. HoweverXn X which generates B(Xn), hence (v) also follows.

Throughout the remaining part of this section and Section 5 we use the notationsfrom Proposition 4.10. Moreover we assume the following, where we view B as anR-algebra.

Let EndR G \ B, with Gn Gn for all n . (4.4)

Lemma 4.11 Let be as in (4.4). If w0 G+1, then the following holds.

(i) There exist m and a generalized polynomial q0(y) overB such that w0 =q0(wm).

(ii) There exists an n > m such that q0 is a polynomial overBn.

Proof. (i) If w0 G+1, then there exists some m such that w0 BRMX{w0,...,wm}. Using wi pwi+1 mod G it follows that

w0 BRMX{wm},

and there is a generalized polynomial q0(y) over B such that w0 = q(wm).(ii) The coefficients of q0 are in some Bn for some n > m.

5 The three Step (or Stop) Lemmas

We will use the notations from Proposition 4.10 and (4.4).We begin with our first Step Lemma, which will stop becoming an endomorphism

of our final module.

Step Lemma 5.1 Let EndR G be an endomorphism as in (4.4) such that for alln there is gn Xn+1 \ Gn with gn / BRMXn{gn}, and let G+1 be defined withwn = wn(gn, ln, 1) as in (4.1) for suitable elements ln . Then does not extend toan endomorphism in EndR G+1.

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Proof. We define inductively an ascending sequence ln :

IfC G is a submodule, then the p-closure ofC is defined by C = n(pnG + C).It is the closure ofC in the S-adic topology, which is Hausdorff on G (i.e.

n p

nG =

0). In particular C = C if C is a summand of G (e.g. C = 0 is closed).By hypothesis we have gn / BRMXn{gn} and BRMXn{gn} is a summand of G,

so it is closed in the S-topology. There is an l such that gn / BRMXn{gn} +plG.

Ifln1 is given, we may choose l = ln such that ln > 3ln1. We will ensure (just below)that G G+1 is S-pure, thus p

lnG+1 G plnG, and it follows:

There is a sequence ln with ln+1 > 3ln and gn / BRMXn{gn} +plnG+1. (5.1)

Hence G+1 is well defined and Proposition 4.10 holds; in particular G G+1and G is dense in G+1 (G = G+1). Suppose for contradiction that EndR Gextends to an endomorphism of G+1; this extension is unique, and we call it also EndR G+1. In particular w0 G+1; by Lemma 4.11(i)(ii) there is a polynomialq0(y) with coefficients in Bn for some n and with w0 = q0(wm) for some m .We choose n > max{n, m} and use (4.1) to compute w0:

w0 =ni=1

plil0gi +pln+1l0wn+1.

Application of gives

w0 n1i=1

plil0(gi) +plnl0(gn) mod p

ln+1l0G+1.

If i < n, then gi Gn and gi Gn by the choice of . The last equality becomesw0 pln

l0(gn) mod pln+1l0G+1 + Gn, hence

q0(w0) plnl0(gn) mod (p

ln+1l0G+1 + Gn).

Finally we determine plnl0(gn) in terms ofBRMXn{gn}:From w0 = q0(wm), n > m and the definition of wm in (4.2) we get

wm =n1i=m

plilmgi +plnlmgn +p

ln+1lmwn+1,

thus

q0(wm) q0(n1i=m

plilmgi +plnlmgn) mod p

ln+1lmG+1,

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and

plnl0(gn) q0(wm) q0(n1i=m

plilmgi +plnlmgn) mod (p

ln+1lmG+1 + Gn).

Now we use (again) that gi Gn for all i < n. The last equation reduces toplnl0(gn) BRMXn{gn} + p

ln+1lmG+1, hence gn B(Gn, gn) + pln+1lmlnG+1.

Note that ln+1 > 3ln by the choice of the lns, hence ln+1 lm ln > ln, so we get aformula

gn BRMXn{gn} +plnG+1

that contradicts (5.1) and the Step Lemma 5.1 follows.

Step Lemma 5.2 Let EndR G be an endomorphism as in (4.4). Moreover sup-pose there are elements un, gn Xn+1 \ Gn (for each n ) with un = q1n(un) andgn = q

2n(gn), where q

1n, q

2n are polynomials overB0 such that

q1n(y) q2n(y) / B0 i.e. y appears in the difference.

If G+1 is defined with wn = wn(gn + un, ln, 1) as in (4.1) for suitable elements ln ,then does not extend to an endomorphism in EndR G+1.

Proof. Let ck := gk + uk. The set X := (X \ {uk| k < }) {ck| k < } is now

a basis ofB by Corollary 4.5, thus Proposition 4.10 applies and G+1 is welldefined.By definition of wn and (4.2) we have pln+1lnwn+1 + cn = wn, and as in the proof of

Step Lemma 5.1 we get

w0 = q0(wm) =k0

plkl0(ck) = q0(km

plklmck)

for n, m , q0(y) as in Step Lemma 5.1. Furthermore,

ck = (gk + uk) = gk + uk = q1k(gk) + q

2k(uk).

Thus

k0

plkl0(q1k(gk) + q2k(uk)) = q0(km

plklm(gk + uk)),

where q1k(gk) RBX0{gk} and q2k(uk) RBX0{uk}, and arguments similar to Lemma

4.9 apply: in every monomial of q0(y) the variable y appears at most once as thereare no mixed monomials on the lefthand side, and the same holds for q1k(y), q

2k(y).

Furthermore, the variable y does not appear in q1k(y) q2k(y), which contradicts our

assumption on the qiks.

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The next lemma is the only place where we will use that R is S-incomplete in

order to find a sequence an {0, 1}, see Definition 1.1. Recall that this conditionfollows by Corollary 1.2 if the S-ring is a direct sum of S-invariant subgroups of size< 20. Hence it will be sufficient if R+ is free and S defines the usual p-adic topologyon R.

Step Lemma 5.3 LetR be a S-incomplete S-ring, let EndR G be an endomor-phism as in (4.4) and letq = q(y) be a polynomial in y with coefficients inB0 such thatg q(g) G0 for allg G. Moreover suppose that for all n there are elements

gn Xn+1 \ Gn such that gn q(gn) = 0.

If G+1 is defined with wn = wn(gn q(gn), ln, 1) as in (4.1) for suitable elementsln , then does not extend to an endomorphism in EndR G+1.

Proof. Choose gn Xn+1 as in the Lemma, and put hn = gn q(gn) = 0. Byassumption on and q it follows that hn G0. Let wn = wn(gn, n , an) be defined as in(4.1) for a suitable sequence of elements an {0, 1} and ln = n for all n . We defineagain G+1 as in Proposition 4.10 using the new choice of elements wn. Note that G0is a free R-module. By the assumption that R is S-incomplete there is a sequencean {0, 1} with

k p

kakhk / G0. However

k pkakhk G0 by the choice of

hn, hence

k pkakhk / G+1 by definition of G+1. Recall w0 =

k p

kakgk and

suppose that w0 G+1. We compute

w0 = (k

pkakgk) =k

pkak(gk)

andk

pkakhk =k

pkak(gk q(gk)) =k

pkakgk k

q(gk)pkak = w0

k

pkakq(gk).

From

k pkakhk / G+1 follows

k p

kakq(gk) / G+1. However, by the definitionof bodies B, the map taking g q(g) for any g G+1 is an endomorphism ofG+1,

and also w0 = k pkakgk G+1, hence k pkakq(gk) G+1 is a contradiction.We deduce w0 / G+1 and does not extend to an endomorphism of G+1.

6 Constructing Generalized E(R)-Algebras

Lemma 6.1 Let be a regular, uncountable cardinal andB = B a-filtration of

bodies. Also letG = RB and G = RB. Then the following holds for any EndR G.

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(i) If there is g G such that g / (B){g}, then there is also h G free overB

such that h / (B){h}.

(ii) If there are g G and a polynomial q(y) overB such that g q(g) / (B){g},then there is also h G free overB such that h q(h) / (B){h}.

Proof. If g G satisfies the requirements in (i) or (ii), respectively, then chooseany element h G which is free over B. If h

also satisfies the conclusion of thelemma, then let h = h and the proof is finished. Otherwise let h = h + g which is alsofree over B by Proposition 4.4. In this case h

(B){h} or h q(h) (B){h},

respectively. It follows h / (B){h} or h q(h) / (B){h}, respectively.

The next lemma is based on results of the last section concerning the Step Lemmasand Lemma 6.1. We will construct first the -filtration ofBs for application usingE for some non-reflecting subset E

o. Recall that E holds for all regular,uncountable, not weakly compact cardinals and non-reflecting subsets E in V = L.

Construction of a -filtration of free bodies: Let { : E} be the family of Jensenfunctions given by E. The body B and the R-module RB will be constructed as afiltration B =

B of bodies. We choose

|B| = || + |R| = |B+1 \ B|

and fix for each E a strictly increasing sequence

n \ E with supn

n = .

This is possible, because E consists of limit ordinals cofinal to only and we can pickn as a successor ordinal. We will use the same Greek letter for a converging sequenceand its limit, so the elements of the sequence only differ by the suffix.

As E is non-reflecting, we also may choose a strictly increasing, continuous sequence, cf() with

supcf()

= and \ E

if cf() > . This is crucial, because the body B of the (continuous) filtration ofB must be free in order to proceed by a transfinite construction. This case does notoccur for = 1.

Using Lemma 5.1, Lemma 5.2 and Lemma 5.3 inductively, we define the bodystructure on B. We begin with B0 = 0, and by continuity of the ascending chain the

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construction reduces to an inductive step passing from B to B+1. We will carry on

our induction hypothesis of the filtration at each step. In particular the following threeconditions must hold.

(i) B is a free body.

(ii) If \ E, then B is a free body over B.

(iii) If E then let n (n ) be the given sequence with supn n = . Supposethat the hypothesis of one of the three step lemmas, Lemma 5.1 or Lemma 5.2or Lemma 5.3 holds for Gn = RBn (n ). We identify B+1 with B+1 fromthe Step Lemmas (so does not extend to an endomorphism ofB+1).

Following these rules we step to +1: If the hypotheses of condition (iii) are violated,for instance, if / E we choose B+1 := (B){v} adding any new free variable v tothe body. However, next we must check that these conditions (i) to (iii) can be carriedover to + 1. If the hypotheses of condition (iii) are violated, this is obvious. In theother case the Step Lemmas are designed to guarantee:

Condition (i) is the freeness of B+1 in Proposition 4.10. Condition (ii) needsthat B+1 is a free body over B. However B Bn for a large enough n .Hence (ii) follows from freeness of Bn over B (inductively) and B+1 over Bn (byProposition 4.10 and Corollary 4.7).

In the limit case we have two possibilities: If cf() = then supn n = ,

hence B = nBn and B is a free body with the help of (i) and (ii) by induction,see Corollary 4.7. If cf() > , then by our set theoretic assumption (E is nonreflecting) we have a limit supcf() = of ordinals not in E. The union of thechain B =

cf()

B by (i) and (ii) is again a free body, see Corollary 4.7. Thus we

proceed and obtain B =

B which is a filtration of free bodies. It remains to

show the

Main Lemma 6.2 Assume E. Let be a regular, uncountable cardinal and B = B be the -filtration of bodies just constructed. Also let G = RB and G = RB.

Suppose that EndR G does not satisfy the following conditions (i) or (ii) for any and any polynomial q(y) overB:

(i) There is g G such that g / R(B){g}.

(ii) There is g G such that g q(g) / R(B){g}

Then is represented inB.

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Proof. Suppose for contradiction that is not represented in B. Let E o be

given from E, let { : E} be the family of Jensen functions and define astationary subset E = { E : G = }. Note that C = { : G G} is acub, thus E := E

C is also stationary.

As a consequence we see that there is E satisfying one of the following condi-tions.

(i) For every < E there is g G such that g / R(B){g}.

(ii) There is < E such that g R(B){g} for all g G (not case (i)) but forevery < and every polynomial q(y) over B represented by an endomorphismof G there is g G with g q(g) / G.

(iii) There is < E such that g R(B){g} and there is a polynomial q(y) overB with g q(g) G for all g G (neither (i) nor (ii) holds), but is notrepresented by B. Thus there are a sequence n < (n ) with supn n = and gn Gn such that gn q(gn) = 0 for all n .

By Lemma 6.1 we may assume that the elements g existing by (i) and (ii), respec-tively, are free over B. Moreover, if g G, then by cf > we can choose Esuch that g G.

If (i) holds, then we can choose a proper ascending sequence n E with supn n = and elements gn Gn+1 such that gn is free over Bn and

gn / R(Bn){gn} for all n .

We identify Gn with Gn in Step Lemma 5.1 and note that (by n E) Gnis an endomorphism with Gn Gn which is predicted as a Jensen function. Byconstruction of G+1 (as a copy of G+1 from Lemma 5.1), the endomorphism Gdoes not extend to EndR G+1. However End G, thus G+1 G for some < .Finally note that G+1 is the S-adic closure of G in G because G is S-dense in G+1and G+1 is a summand ofG hence S-closed in G. We derive the contradiction thatindeed G+1 EndR G+1. Hence case (i) is discarded.

Now we turn to case (ii). Suppose that (ii) holds (so condition (i) is not satisfied).In this case there is an ascending sequence n E with supn n = as above andthere are free elements gn, un Gn+1 (also free over Bn and polynomials q

1n, q

2n over

B0 such that un = q1n(un) = gn = q

2n(gn). Moreover, the polynomials q

1n(y) q

2n(y)

are not constant over B0 . Step Lemma 5.2 applies and we get a contradiction as incase (i). Thus also case (ii) is discarded.

Finally suppose for contradiction that (iii) holds (so (i) and (ii) are not satisfied).There are < and q(y) a polynomial over B such that g q(g) G for all g G.

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The polynomial q(y) is represented by an endomorphism of G. Moreover (from (iii))

we find gn Gn+1 free over Gn for a suitable sequence n with supn n = suchthat gn q(gn) = 0. We now apply Step Lemma 5.3; the argument from case (i) givesa final contradiction. Thus the Main Lemma holds.

Proof of Main Theorem 1.4. Let B be the body over the S-ring R constructedat the beginning of this section using E as in Theorem 1.4; moreover let G = RB.Thus |B | = and by the construction and Proposition 4.10 any subset of size < iscontained in an R-monoid-algebra of cardinality < ; the algebra B is the union of a-filtration of free bodies B. By Lemma 6.2 every element EndR G is represented(by a polynomial q(y) with coefficients from B); see Definition 4.2. Thus g = q(g)for all g G and = q(y) B. It follows that B = EndR G is the R-endomorphismalgebra of G.

Finally recall that there is B B which is an R-monoid-algebra over a non-commutative monoid from Observation 3.1. Thus B cannot be commutative either andTheorem 1.4 is shown.

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Rudiger GobelFachbereich MathematikUniversitat Duisburg-EssenD 45117 Essen, Germanye-mail: [email protected]

Saharon ShelahInstitute of Mathematics,Hebrew University, Jerusalem, Israel

and Rutgers University, New Brunswick, NJ, U.S.Ae-mail: [email protected]

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Rudiger Gobel and Saharon Shelah- Generalized E-Algebras via lambda-Calculus I