aa r X i v : . [ m a t h . L O ] N ov Generalized E -Algebras via λ -Calculus I R¨udiger G¨obel and Saharon Shelah
Abstract An R -algebra A is called an E ( R )–algebra if the canonical homomorphismfrom A to the endomorphism algebra End R A of the R -module R A , taking any a ∈ A to the right multiplication a r ∈ End R A by a , is an isomorphism ofalgebras. In this case R A 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 R –algebras A isomorphic to End R A 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 ∼ = End Z A (as rings). We will answer Schultz’s 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 and p -reduced (equivalently p is not a zero-divisor and T n ∈ ω p n R = 0). As explainedin the introduction we assume that either | R | < ℵ or that R + is free, seeDefinition 1.1.The main tool is an interesting connection between λ -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 (G¨odels universe) wherestronger 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 Shelah’s archivesubject classification (2000):primary: 20K20, 20K30;secondary: 16S60, 16W20;Key words and phrases: E -rings, endomorphism rings roof based entirely on ordinary set theory (the axioms of ZFC) will appear ina subsequent paper [23]. However the general strategy will be the same, but thecombinatorial arguments will utilize a prediction principle that holds under ZFC. E ( R )-Algebras Let S be a countable, multiplicatively closed subset of a commutative ring R with 1.An R –module M is S –reduced if T s ∈ S sM = 0 and it is S –torsion–free if sm = 0 , m ∈ M, s ∈ S implies m = 0. Suppose that R (as an R -module) is S –reduced and S -torsion-free. Then R is called an S -ring, see [28]. In order to avoid zero–divisors as in thecase of Z –adic completion Q p J p of Z we also assume that S is cyclically generated,i.e. S = h p i := { p n : n ∈ ω } for some p ∈ R . We will concentrate on S –cotorsion–free modules. An S –torsion-free and S –reduced R -module M is S –cotorsion–free ifHom( b R, M ) = 0, where b R denotes the S -completion of R . A submodule U ⊆ M is S –pure (we also write U ⊆ ∗ M ) if sM ∩ U ⊆ sU for all s ∈ S . Note that R , being S –reduced, is Hausdorff in the S –topology. In the proof of Step Lemma 5.3 we willalso use the following condition on the additive group R + of R which implies that R is S -cotorsion-free. Definition 1.1 An R -module M is Σ S -incomplete if for any sequence = m n ∈ M ( n ∈ ω ) there are a n ∈ { , } with P n ∈ ω p n a n m n / ∈ M . If M = R + we say that R is Σ S -incomplete. All S -rings of size < ℵ are Σ S -incomplete as shown in [22]. Thus it follows easilythat any S -ring which is a direct sum of S -invariant subgroups of size < ℵ is Σ S -incomplete as well. So we deduce from [22] a Corollary 1.2
If an S -ring R is a direct sum of S -invariant subgroups of size < ℵ ,then R is Σ S -incomplete. In particular, if S generates the ordinary p -adic topology(i.e. for ∈ R there is p ∈ h i and S = h p i ) and the additive group R + is free, then R is Σ S -incomplete. We recall the main definition.
Definition 1.3 If A is an R –algebra, then δ : A −→ End R A denotes the homomor-phism which takes any a ∈ A to the R –endomorphism aδ = a r which is multiplicationby a on the right. If this homomorphism is an isomorphism, then A is called an E ( R ) –algebra and R A is called an E ( R ) –module. By R A we denote the R -module structureof an R -algebra A . ( R )-algebras can also be defined dually, assuming that the homomorphismEnd R A −→ A ( ϕ −→ ϕ )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 Σ S -incomplete we will constructnon–commutative R –algebras with End R A ∼ = A . Hence these A s are generalized E ( R )-algebras but not E ( R )–algebras. If R = Z and R A is an abelian group, then we do notmention the ring Z : e.g. E ( Z )–modules are just E –groups. The existence of generalized E –rings answers a problem in [34, 37].If κ is a cardinal, then let κ o = { α : cf( α ) = ω, α ∈ κ } . We will only need theexistence of a non-reflecting subset E ⊆ κ o for some regular uncountable, not weaklycompact cardinal κ such that the diamond principle ♦ κ E holds. It is well-known (seee.g. Eklof, Mekler [12]) that ♦ κ E is a consequence for all non-reflecting subsets E ofregular uncountable, not weakly compact cardinals κ in G¨odel’s universe (V = L). Weindicate our (weaker) set theoretic hypothesis (which also holds in other universes) as ♦ κ E in our following main result. Theorem 1.4
Let R be a Σ S -incomplete S -ring for some cyclically generated S . If κ > | R | is a regular, uncountable cardinal and E ⊆ κ o a non-reflecting subset with ♦ κ E , then there is an S –cotorsion–free, non–commutative R –algebra A of cardinality | A | = κ with End R A ∼ = A . Moreover any subset of cardinality < κ is contained in an R -monoid-algebra of cardinality < κ . A similar result without the set theoretic assumption will be shown in [23]:
Theorem 1.5
Let R be a Σ S -incomplete S -ring for some cyclically generated S . Forany cardinal κ = µ + with | R | ≤ µ ℵ = µ there is an S –cotorsion–free, non–commutative R –algebra A of cardinality | A | = κ with End R A ∼ = A . It seems particularly interesting to note that the R -monoid A comes from (classical) λ -calculus taking into account that elements of an E ( R )-algebra A are at the sametime endomorphisms of A , 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 monoid M will be completed and become the algebra A .3 Model theory of bodies and skeletons via λ -calculus We begin by defining terms for a skeleton and will establish a connection with λ -calculus. Let R be any commutative S -ring with S = h p i . (Σ S -incompleteness will beadded in Section 5.)By definition of generalized E ( R )-algebras A , endomorphisms of R A must be con-sidered as members of A . Hence they act on R A as endomorphisms while they areelements of R A at the same time. Thus we will introduce the classical definitions from λ -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, linearfunctions because endomorphisms are of this kind. (The general argument to reduce λ -calculus to unary functions was observed by Sch¨onfinkel, see [2, p. 6].) What are thetypical terms of our final objects, the bodies? If x and x are members of the gener-alized E ( R )-algebras A and a, b ∈ R , then also polynomials like σ n ( x , x ) = ax n + bx belong to the algebra A , so there are legitimate functions p n ( y ) = λy.σ n ( y, x ) on A taking y −→ σ n ( y, x ) 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. Let τ be a vocabulary with no predicates; thus τ is a collection of function symbolswith an arity function τ −→ ω defining the places of function symbols. Moreover let X be an infinite set of free variables. Then unspecified ( τ, X )-terms (briefly called‘terms’) 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 σ , . . . , σ n − are terms and F is an n -place function symbol from τ , then F ( σ , . . . , σ n − ) is a term.We also define the (usual) length l ( σ ) of a term σ inductively: Let l ( σ ) = 0 if σ isatomic and l ( σ ) = k + 1 if σ derives from (ii) with k = max { l ( σ i ) : i < n } .If σ is an unspecified ( τ, X )-term, then we define (also by induction on l ( σ )) a finitesubset F V ( σ ) ⊂ X of free variables of σ : 4a) If σ is an individual constant, then F V ( σ ) = ∅ and if σ ∈ X , then F V ( σ ) = { σ } .(b) If σ = F ( σ , . . . , σ n − ) is defined as in (ii), then F V ( σ ) = S i The theories T sk See Gr¨atzer [26, p. 167] or Bergman [3, Chapter 8].We immediately derive one of our central definitions. Definition 2.3 Let T sk := T sk<ω and τ sk = τ sk<ω be as in Observation 2.2. Any T sk -model (an algebra satisfying T sk ) is called a skeleton and two skeletons are called iso-morphic if they are isomorphic as T sk -models; see e.g. [3, p. 262] or [33, p. 5]. For applications it is useful to recall the following Definition 2.4 of (free) generators of a T sk An unspecified term σ ∈ t ( τ sk Every term ( σ, x ) can be reduced to a (normalized) re-duced term red ( σ, x ) with T sk Using induction, we say when two reduced elements σ , σ ∈ t r ( τ sk This is immediate by induction using (2.1).Using normalization of x and b from Definition-Observation 2.6 we can deduce a Proposition 2.9 If M is a T sk Let h ≤ k ≤ ω and ( σ, x ) ∈ t ( τ sk For σ , σ ∈ τ sk the following are equivalent(i) T sk ⊢ σ = σ . ii) σ r . = σ r . Proof. ( ii ) −→ ( i ): From the Definition-Observation 2.6 follows T sk ⊢ σ = σ r , σ = σ r and by Observation 2.8 is T sk ⊢ σ r = σ r thus T sk ⊢ σ = σ .( i ) −→ ( ii ): From (i) follows T sk ⊢ σ r = σ r . Thus σ r . = σ r by Definition 2.7 andObservation 2.8. Next we construct and discuss free skeletons based on reduced terms. We will use theinfinite set X of free variables to construct a skeleton M X 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 τ sk := τ sk<ω with equiv-alence classes [ σ ] for any σ ∈ τ sk . Let M X = { [ σ ] : σ ∈ t( τ sk ) } . The equivalence classes [ σ ] of atoms σ are singletons by Definition 2.7(i). If B = { [ x ] : x ∈ X } , then ι : X −→ B ( x [ x ]) is a bijection and we see that M X is a skeleton withbasis B ⊆ M X . Moreover, [ ] is compatible with the application of function symbols:If F = F σ ∈ τ sk with ( σ, x ) ∈ t ( τ sk , X ) and x = h x , . . . , x n − i is an n -place functionsymbol and σ i . = σ ′ i ( i < n ), then F ( σ , . . . , σ n − ) r . = F ( σ ′ , . . . , σ ′ n − ) r by Observation2.8(ii), thus F ([ σ ] , . . . , [ σ n − ]) = [( F ( σ , . . . , σ n − )] ( σ i ∈ t r ( τ sk ))is well-defined as follows from Observation 2.8(ii).We have the following Theorem 2.12 If X is an infinite set (of free variables) and M X is defined as above,then the following holds.(a) M X = { [ σ ] : σ ∈ t r ( τ sk ) } .(b) M X is a skeleton with n -place functions [ F ] : ( M X ) n −→ M X ([ σ ] , . . . , [ σ n − ]) [ F ( σ , . . . , σ n − )] for each n -place function symbol F = F σ ( y ,...,y n − ) for ( σ, x ) ∈ t r ( τ sk , X ) with F V ( σ ) = { x , . . . , x n − } .(c) M X is freely generated by B = { [ x ] , x ∈ X } , called the free skeleton over X .Using ι above we identify B and X . roof. The axioms (2.1) are satisfied, e.g. the crucial condition (ii)(a) follows bydefinition of [ F ]. Remark 2.13 In the construction of the free skeleton M X we also used an infinite set Y of bound variables. However, it follows by induction that another infinite set Y ′ ofbound variables leads to an isomorphic copy of M X . Thus we do not mention Y inTheorem 2.12. Lemma 2.14 Let B be a subset of the T sk The theories T bd See Gr¨atzer [26, p. 198, Theorem 3] or Bergman [3, Chapter 8]. Definition 2.16 Let T bd := T bd<ω and τ bd := τ bd<ω be as in Observation 2.15. Any T bd -model (an algebra satisfying T bd ) is called a body and two bodies are isomorphic if theyare isomorphic as T bd -models; see e.g. [3, p.262] or [33]. Observation 2.17 Any (generalized) E ( A ) -algebra is a body. roof. Generalized E ( A )-algebras satisfy End R A = A . Thus any function symbol F σ of τ 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. t ( τ bd Let x = h x , . . . x m − i and ( σ, x ) ∈ t ( τ bd This is an easy induction on the length of σ :Case 1: If σ = 1 and σ = x , then the claim holds trivially.Case 2: If σ = F ( σ , . . . , σ m ), then the claim follows from the axioms (2.3)(ii) of T bd . Similarly, if σ = F + ( σ , σ ) and σ = F a ( σ ), then the linearity follows by definitionof these functions and induction hypothesis.Recall the notion from λ -calculus in Remark 2.1(ii).13 roposition 2.20 The weak completeness of bodies. Let M be a body and ( σ, x ) a polynomial (a term in t ( τ bd , X ) ) with x = h x , . . . , x m i and d , . . . , d m ∈ M . Thenthere is ( σ ′ , h x , . . . , x n i ) ∈ t ( τ bd , X ) and the following holds.(i) M is an R -module.(ii) The unary function λ.zσ ( z, d , . . . , d m ) : M −→ M ( z σ ( z, d , . . . , d m )) is the R -endomorphism λy.yσ ′ ( z, d , . . . , d m ) ∈ End R ( M ) ( y −→ yσ ′ ( d , . . . , d m )) . Remark. We will show here that there is a function symbol ( σ ′ , x ) ∈ t ( τ bd , X )such that σ ( d, d , . . . , d m ) = dσ ′ ( d , . . . , d m ) for all d ∈ M , see axioms (2.1)(ii)(a) ofthe skeletons. Proof. By the axioms (2.3) of T bd it is clear that M is an R -module. It remainsto show (ii). Let ( σ, x ′ ) ∈ t ( τ bd , X ) with x ′ = h x , . . . , x m i and x = h x , . . . , x m i . ByLemma 2.18 there are monomials ( σ l , x ′ ) ∈ t ( τ sk , X ) and a l ∈ R such that σ = X l From the skeleton to the body Recall from Theorem 2.12 that the skeleton on an infinite set X of free variables is theset M X = { [ σ ] : σ ∈ t r ( τ sk ) } with n -place functions[ F ] : ( M X ) n −→ M X ([ σ ] , . . . , [ σ n − ]) [ F ( σ , . . . , σ n − )]for each n -place function symbol F = F σ ( y ,...,y n − ) with ( σ, x ) ∈ t r ( τ sk ) and F V ( σ ) = { x , . . . , x n − } . For simplicity we will also write Roman letters for the members of M X ,e.g. m = [ σ ] ∈ M X . The set M X has a distinguished element 1 = [1] and m m = m holds for all m ∈ M X (thus M X is an applicative structure with 1). In order to turn M X into a monoid, we first represent M X as a submonoid of Mono( M X ), the injectivemaps on M X , say ι : M X −→ Mono( M X ):Let a = [ σ ] ∈ M X and σ ′ ∈ t ( τ bd ). We will use induction: If a = [1] then[ σ ′ ] a = σ ′ , if a = [ x ], then [ σ ′ ] a = σ ′ x and if a = [ F σ ( y ,...,y n − ) h x , . . . , x n − i ] is a unaryfunction as above, then [ σ ′ ] a = σ ( σ ′ , x , . . . , x n − ) (so aι = [ λy.yσ ]). Thus aι mapsany m = [ σ ′ ] ∈ M X to m ( aι ) = m ( λy.yσ ) = [ mσ ] ∈ M X which can be representedby a reduced element using (2.2). If a = b ∈ M X , then 1( aι ) = a = b = 1( bι ) thus ι : M X −→ Mono( M X ) ⊆ M X is an embedding. We define multiplication of elements a, b ∈ M X as composition of functions ( aι )( bι ) = ( ab ) ι . This is to say that from a = [ σ ] , b = [ σ ′ ] we get the product as the equivalence class of λy. (( yσ ′ ) σ ). We willwrite a · b = ab and will often suppress the map ι . From Mono( M X ) follows that also M X 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 ( M X , · , with composition of functions as productis a non-commutative (associative) monoid with multiplication defined as above by theaction on M X : If [ σ ] , [ σ ′ ] ∈ M X , then [ σ ′ ] · [ σ ] = [ λy. (( yσ ′ ) σ )] . Finally we will associate with any skeleton M its (canonical) body B R M : Let B R M be the R -monoid algebra RM of the monoid M . Moreover any n -place function F : M n −→ M extends uniquely by linearity to F : B R M n −→ B R M . We deduce a Lemma 3.2 If R is a commutative ring as above and M a skeleton, then the R -monoidalgebra B R M of the monoid M is a body. If the skeleton M X is freely generated by X ,then also B R M is freely generated by X as a body. Moreover R B R M X = L m ∈ M mR . roof. It is easy to see that B R M (with the linear n -place functions) is a body.We first claim that X , viewed as { [ x ] : x ∈ X } ⊆ B R M X is a basis. First applyLemma 2.18 to the R -monoid B R M X : Any ( σ, x ) ∈ t ( τ bd , X ) can be written as apolynomial σ = P l σ l a l with monomial ( σ l , x ) ∈ t ( τ sk , X ). Moreover, any σ l is viewedas an element of Mono( M X ), so axiom (2.1)(ii)(a) applies and σ l becomes a product ofelements from X . Thus X generates B R M X . The monomials of the skeleton M extenduniquely by linearity to polynomials of the free R -module R B R M X = L m ∈ M mR fromits basis M .We will also need the notion of an extension of bodies. Definition 3.3 Let B and B ′ be two bodies, then B ≤ B ′ ( B ′ extends B ) if and onlyif B ⊆ B ′ as R -algebras and if ( σ, x ) ∈ t ( τ bd , X ) and F σ is a function symbol withcorresponding unary, R -linear function F of B ′ , then its natural restriction to B is thefunction for B corresponding to F σ . Example 3.4 Let X ⊆ X ′ be sets of free variables and B , B ′ be the free bodies generatedby the free skeletons obtained from X and X ′ , respectively. Then B ≤ B ′ . In this casewe say that B ′ is free over B . The endomorphism ring End R B R M X of the R -module R B R M X has natural elementsas endomorphisms, the linear maps, our (generalized) polynomials interpreted by theterms in σ ∈ t ( τ bd , X ) acting by scalar multiplication on B R M X 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 we‘lose these R -linear maps’ on the way. Definition 4.1 Let ( σ, x ) ∈ t ( τ bd , X ) with x = h x , . . . , x n i . If B is a body, d = h d , . . . , d n i with d , . . . , d n ∈ B , then we call s d ( y ) = λy.σ ( y, d ) the (generalized) poly-nomial over B with coefficients d . Note that σ d ( y ) is a sum of products of elements d i 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 by t ( τ bd , X ) - thus the resulting structure will be complete: Any R -endomorphism of anextended body B R M X will be represented by a polynomial q ( x ) over B R M X , so B R M X is complete or equivalently an E ( R )-algebra.16he fact that B R M X is not just the R -linear closure (or A -linear closure for somealgebra 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 Let B be a body and G = R B . Then ϕ ∈ End R G is called represented(by q ( y )) if there is a generalized polynomial q ( y ) with coefficients in B such that gϕ = q ( g ) for all g ∈ G .If all elements from End R 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 an S -ring as above and B be a body generated by B , then B isa basis of B if one of the following equivalent conditions holds.(i) If B ′ = B R M X is the body generated by the free skeleton M X and X −→ B is abijection, then this map extends to an isomorphism B ′ −→ B (ii) B is independent in B , i.e. if ( σ , x ) , ( σ , x ) ∈ t r ( τ bd , X ) and the sequence b from B is suitable for x such that σ ( b ) = σ ( b ) , then T bd ⊢ σ ( x ) = σ ( x ) .(iii) For all bodies H and maps ϕ : B −→ H there is an extension ϕ : B R M X −→ H as T bd -homomorphism. Proof. The proof is well known for varieties (see Gr¨atzer [26, p. 198, Theorem 3]or Bergman [3, Chapter 8]), so it follows from Observation 2.15. Freeness Proposition 4.4 Let R be an S -ring as above and X ⊆ X ′ be sets of vari-ables and B R M X ⊆ B R M X ′ the corresponding free bodies. If u ∈ B := B R M X and v ∈ X ′ \ X , then w := u + v ∈ B ′ := B R M X ′ is free over B , i.e. there is a basis X ′′ of B ′ with w ∈ X ′′ ⊇ X . Proof. We will use Lemma 4.3 (c) to show that the set X ′′ := ( X ′ \ { v } ) ∪ { w } isa basis of B ′ . First note that X ′′ also generates B ′ , thus B ′ = B R M X ′′ .Given ϕ : X ′′ → H for a body H , we must extend this map to ϕ : B ′ → H .Let ϕ ′ := ϕ ↾ ( X ′ \ { v } ) and note that the set X ′ \ { v } = X ′′ \ { w } is independent.Thus if B := B R M X ′ \{ v } , then ϕ ′ extends to ϕ ′ : B → H by freeness, and from u ∈ B follows the existence of uϕ ′ ∈ H . We now define ϕ : if ϕ ↾ B := ϕ ′ , then ϕ ↾ ( X ′ \{ v } ) = ϕ ′ = ϕ ↾ ( X ′ \{ v } ). Thus it remains to extend ϕ ′ to ϕ : B ′ → H in such away that wϕ = wϕ . If wϕ =: h ∈ H , then we must have h = wϕ = ( u + v ) ϕ = uϕ + vϕ .17ence put vϕ := h − uϕ = h − uϕ ′ . Now ϕ : B ′ → H exists, because X ′ is free, ϕ ′ ⊆ ϕ and wϕ = ( u + v ) ϕ = uϕ + h − uϕ = h = wϕ , thus ϕ ⊆ ϕ 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 B generated by X \ { v } and w ∈ B ′ , then X ′ = X \ { v } ∪ { v + w } is another basis for B . Corollary 4.6 If X ⊆ X ′ and B R M X ⊆ B R M X ′ , then any basis of B R M X extends toa basis of B R M X ′ . Proof. If X ′′ is a basis of B R M X , then it is left as an exercise to show that ( X ′ \ X ) ∪ X ′′ is a basis of B R M X ′ .The last corollaries have another implication. Corollary 4.7 Suppose that B α ( α ≤ δ ) is an ascending, continuous chain of bodiessuch that B α +1 is free over B α for all α < δ . Then B δ is free over B and if B is free,then B δ is free as well. The proof of the next lemma is also obvious. It follows by application of thedistributive law in T bd and collection of summands with p . Lemma 4.8 Let q ( y ) be a generalized polynomial and r ∈ R . Then there is a polyno-mial q ′ ( y ) such that q ( y + ry ) = q ( y ) + rq ′ ( y , y ) . Proof. By Lemma 2.18 we can write σ = P l Let X , X , X be pairwise disjoint infinite sets, B := B ( X ) ⊆ B := B ( X ∪ X ∪ X ) and q ( y ) , q ( y ) , q ( y ) polynomials over B such that q ( g + v + v ) = q ( v ) + q ( v ) for some g ∈ B , v ∈ X , v ∈ X . Then the following holds.(i) q ( y ) is a linear polynomial in y , i.e. y appears at most once in every monomial.(ii) q ( y ) − q ( y ) does not depend on y . roof. (i) Write q ( y ) = P ni =1 m i ( y ) as a sum of minimal length of generalized R –monomials and suppose for contradiction that y appears n times in m ( y ) with n > n be maximal for the chosen monomial m ( y ).By the distributive law the monomials of the polynomial q ( g + v + v ) include thosemonomials induced by m ( y ) replacing all entries of the variable y by arbitrary choicesof v and v . Let m ′ be one of these monomials. If there are further such monomials m ′ i ( i ≤ k ) alike m ′ arriving from this substitution into monomials m i ( y ) of q ( y ) with P ki =1 m ′ i = 0, then replacing all v s and v s by y s gives P ki =1 m i ( y ) = 0 contradictingthe minimality of the above sum. Thus m ′ represents a true monomial (not canceledby others) of q ( g + v + v ), and as n > v and v bothappear in m ′ . This monomial does not exist on the right–hand side of the equation inthe lemma – a contradiction. Thus (a) holds.(ii) First substitute in the given equation v := y, v := 0 and v := 0 , v := y ,respectively. Thus q ( g + y ) = q ( y ) + c and q ( g + y ) = q ( y ) + c ′ , where c := q (0) , c ′ := q (0) ∈ B . Subtraction now yields 0 = q ( y ) − q ( y )+( c − c ′ ), thus q ( y ) − q ( y ) = c ′ − c does not depend on y , as required.In order to establish the Step Lemmas, we next prepare some preliminary results.Let X ω = S n ∈ ω X n be a strictly increasing sequence of infinite sets X n of variablesand fix a sequence v n ∈ X n \ X n − of elements ( n ∈ ω ). Moreover, let M α = M X α bethe skeleton and B α := B ( X α ) be the body generated by X α for α ≤ ω , respectively.Note that by our identification B ( X α ) is an R -algebra and restricting to the modulestructure G α := R B ( X α ) is an R -module, which is free by Lemma 3.2. Recall that S = { p n : n ∈ ω } for some p ∈ R (with T n ∈ ω p n R = 0) generates the S -topology on R -modules. Thus the S -topology is Hausdorff on G α and G α is naturally an S -pure R -submodule of its S -completion b G α ; we write G α ⊆ ∗ b G α and pick particular elements w n ∈ b G ω . If a i ∈ { , } and l n ∈ N is increasing, then we define w n ( v n , l n , a n ) := w n := X k ≥ n p l k − l n a k v k ∈ b G ω (4.1)and easily check that w n − p l n +1 − l n w n +1 = a n v n ∈ G n for all n ∈ ω. (4.2) Proposition 4.10 Let a n ∈ { , } and l n ∈ N be as above. If X ω +1 = X ω \ { v n : a n =1 , n > } ∪ W with W = { w n : n > } and B ω +1 := B ( X ω +1 ) , G ω +1 = R B ω +1 , then thefollowing holds i) G ω ⊆ ∗ G ω +1 ⊆ ∗ b G ω (ii) G ω +1 /G ω is p -divisible, thus an S − R -module.(iii) X ω +1 is a basis of the (free) skeleton M ω +1 = M X ω +1 .(iv) The R -algebra B ω +1 is freely generated by the skeleton M ω +1 , thus B ω +1 = RM ω +1 and G ω +1 = L m ∈ M ω +1 Rm .(v) B ( X ω +1 ) is free over B ( X n ) (as body). Proof. Claim (i): Clearly G ω ⊆ ∗ b G ω and G ω +1 ⊆ b G ω . From w n , w n +1 ∈ X ω +1 , a n = 1 and (4.2) follows v n ∈ G ω +1 . Hence v n ∈ G ω +1 for all n ∈ ω and G ω ⊆ B ( X ω +1 ) = G ω +1 follows at once. Thus G ω +1 /G ω ⊆ b G ω /G ω and purity ( G ω +1 ⊆ ∗ b G ω )follows if G ω +1 /G ω is p -divisible. This is our nextClaim (ii): By definition of the body B ( X ω +1 ), any element g ∈ G ω +1 is the sumof monomials in X ω +1 . If w n , w n +1 are involved in such a monomial, then we apply(4.2) and get w n = pw n +1 + a n v n , which is w n ≡ pw n +1 mod G ω . Let m be the largestindex of w n s which contributes to g . We can remove all w i of smaller index i < m andalso write w m ≡ pw m +1 mod G ω . Thus g + G ω is divisible by p and G ω +1 /G ω is an S − R -module.Claim (iii): It is enough to show that X ω +1 is free, because X ω +1 generates M ω +1 by definition of the skeleton. First we claim that X ′ = ( X ω \ { v n } ) ∪ { w n } is free.We apply the characterization of a basis by Lemma 4.3 (ii). Let ( σ , x ) , ( σ , x ) ∈ t r ( τ bd , X ) ( x = h x , . . . , x k i ) be such that σ ( y , . . . , y k ) = σ ( y , . . . , y k )for some y i ∈ X ′ and suppose that y = w n (there is nothing to show if w n doesnot appear among the y i s, because they are free; otherwise we relabel the y i s suchthat y = w n ). Now we consider the above equation as an element in b G ω and notethat the support [ y i ] ⊆ X ω of the elements y i , ( i > 1) is finite, while w n has infinitesupport { v k : k > n } ⊆ [ w n ]. Thus we project σ i ( y , . . . , y k ) onto a free summandfrom [ w n ] \ S
Let ϕ be as in (4.4). If w ϕ ∈ G ω +1 , then the following holds.(i) There exist m ∈ ω and a generalized polynomial q ( y ) over B ω such that w ϕ = q ( w m ) .(ii) There exists an n ∗ > m such that q is a polynomial over B n ∗ . Proof. (i) If w ϕ ∈ G ω +1 , then there exists some m ∈ ω such that w ϕ ∈ B R M X ω ∪{ w ,...,w m } . Using w i ≡ pw i +1 mod G ω it follows that w ϕ ∈ B R M X ω ∪{ w m } , and there is a generalized polynomial q ( y ) over B ω such that w ϕ = q ( w m ).(ii) The coefficients of q are in some B n ∗ for some n ∗ > m . We will use the notations from Proposition 4.10 and (4.4).We begin with our first Step Lemma, which will stop ϕ becoming an endomorphismof our final module. Step Lemma 5.1 Let ϕ ∈ End R G be an endomorphism as in (4.4) such that for all n ∈ ω there is g n ∈ X n +1 \ G n with g n ϕ / ∈ B R M X n ∪{ g n } , and let G ω +1 be defined with w n = w n ( g n , l n , as in (4.1) for suitable elements l n ∈ ω . Then ϕ does not extend toan endomorphism in End R G ω +1 . roof. We define inductively an ascending sequence l n ∈ ω :If C ⊆ G ω is a submodule, then the p -closure of C is defined by C = T n ∈ ω ( p n G ω + C ).It is the closure of C in the S -adic topology, which is Hausdorff on G ω (i.e. T n ∈ ω p n G ω =0). In particular C = C if C is a summand of G ω (e.g. C = 0 is closed).By hypothesis we have g n ϕ / ∈ B R M X n ∪{ g n } and B R M X n ∪{ g n } is a summand of G ω ,so it is closed in the S -topology. There is an l ∈ ω such that g n ϕ / ∈ B R M X n ∪{ g n } + p l G ω .If l n − is given, we may choose l = l n such that l n > l n − . We will ensure (just below)that G ω ⊆ ∗ G ω +1 is S -pure, thus p l n G ω +1 ∩ G ω ⊆ p l n G ω , and it follows:There is a sequence l n ∈ ω with l n +1 > l n and g n ϕ / ∈ B R M X n ∪{ g n } + p l n G ω +1 . (5.1)Hence G ω +1 is well defined and Proposition 4.10 holds; in particular G ω ⊆ ∗ G ω +1 and G ω is dense in G ω +1 ( G ω = G ω +1 ). Suppose for contradiction that ϕ ∈ End R G ω extends to an endomorphism of G ω +1 ; this extension is unique, and we call it also ϕ ∈ End R G ω +1 . In particular w ϕ ∈ G ω +1 ; by Lemma 4.11(i)(ii) there is a polynomial q ( y ) with coefficients in B n ∗ for some n ∗ ∈ ω and with w ϕ = q ( w m ) for some m ∈ ω .We choose n > max { n ∗ , m } and use (4.1) to compute w : w = n X i =1 p l i − l g i + p l n +1 − l w n +1 . Application of ϕ gives w ϕ ≡ n − X i =1 p l i − l ( g i ϕ ) + p l n − l ( g n ϕ ) mod p l n +1 − l G ω +1 . If i < n , then g i ∈ G n and g i ϕ ∈ G n by the choice of ϕ . The last equality becomes w ϕ ≡ p l n − l ( g n ϕ ) mod p l n +1 − l G ω +1 + G n , hence q ( w ) ≡ p l n − l ( g n ϕ ) mod ( p l n +1 − l G ω +1 + G n ) . Finally we determine p l n − l ( g n ϕ ) in terms of B R M X n ∪{ g n } :From w ϕ = q ( w m ), n > m and the definition of w m in (4.2) we get w m = n − X i = m p l i − l m g i + p l n − l m g n + p l n +1 − l m w n +1 , thus q ( w m ) ≡ q ( n − X i = m p l i − l m g i + p l n − l m g n ) mod p l n +1 − l m G ω +1 , p l n − l ( g n ϕ ) ≡ q ( w m ) ≡ q ( n − X i = m p l i − l m g i + p l n − l m g n ) mod ( p l n +1 − l m G ω +1 + G n ) . Now we use (again) that g i ∈ G n for all i < n . The last equation reduces to p l n − l ( g n ϕ ) ∈ B R M X n ∪{ g n } + p l n +1 − l m G ω +1 , hence g n ϕ ∈ B ( G n , g n ) + p l n +1 − l m − l n G ω +1 .Note that l n +1 > l n by the choice of the l n s, hence l n +1 − l m − l n > l n , so we get aformula g n ϕ ∈ B R M X n ∪{ g n } + p l n G ω +1 that contradicts (5.1) and the Step Lemma 5.1 follows. Step Lemma 5.2 Let ϕ ∈ End R G be an endomorphism as in (4.4). Moreover sup-pose there are elements u n , g n ∈ X n +1 \ G n (for each n ∈ ω ) with u n ϕ = q n ( u n ) and g n ϕ = q n ( g n ) , where q n , q n are polynomials over B such that q n ( y ) − q n ( y ) / ∈ B i.e. y appears in the difference.If G ω +1 is defined with w n = w n ( g n + u n , l n , as in (4.1) for suitable elements l n ∈ ω ,then ϕ does not extend to an endomorphism in End R G ω +1 . Proof. Let c k := g k + u k . The set X ′ := ( X ω \ { u k | k < ω } ) ∪ { c k | k < ω } is nowa basis of B ω by Corollary 4.5, thus Proposition 4.10 applies and G ω +1 is well–defined.By definition of w n and (4.2) we have p l n +1 − l n w n +1 + c n = w n , and as in the proof ofStep Lemma 5.1 we get w ϕ = q ( w m ) = ⇒ X k ≥ p l k − l ( c k ϕ ) = q ( X k ≥ m p l k − l m c k )for n ∗ , m, q ( y ) as in Step Lemma 5.1. Furthermore, c k ϕ = ( g k + u k ) ϕ = g k ϕ + u k ϕ = q k ( g k ) + q k ( u k ) . Thus X k ≥ p l k − l ( q k ( g k ) + q k ( u k )) = q ( X k ≥ m p l k − l m ( g k + u k )) , where q k ( g k ) ∈ R B X ∪{ g k } and q k ( u k ) ∈ R B X ∪{ u k } , and arguments similar to Lemma4.9 apply: in every monomial of q ( y ) the variable y appears at most once as thereare no mixed monomials on the left–hand side, and the same holds for q k ( y ) , q k ( y ).Furthermore, the variable y does not appear in q k ( y ) − q k ( y ), which contradicts ourassumption on the q ik s. 23he next lemma is the only place where we will use that R is Σ S -incomplete inorder to find a sequence a n ∈ { , } , 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 < ℵ . Hence it will be sufficient if R + is free and S defines the usual p -adic topologyon R . Step Lemma 5.3 Let R be a Σ S -incomplete S -ring, let ϕ ∈ End R G be an endomor-phism as in (4.4) and let q = q ( y ) be a polynomial in y with coefficients in B such that gϕ − q ( g ) ∈ G for all g ∈ G . Moreover suppose that for all n ∈ ω there are elements g n ∈ X n +1 \ G n such that g n ϕ − q ( g n ) = 0 . If G ω +1 is defined with w n = w n ( g n ϕ − q ( g n ) , l n , as in (4.1) for suitable elements l n ∈ ω , then ϕ does not extend to an endomorphism in End R G ω +1 . Proof. Choose g n ∈ X n +1 as in the Lemma, and put h n = g n ϕ − q ( g n ) = 0. Byassumption on ϕ and q it follows that h n ∈ G . Let w n = w n ( g n , n, a n ) be defined as in(4.1) for a suitable sequence of elements a n ∈ { , } and l n = n for all n ∈ ω . We defineagain G ω +1 as in Proposition 4.10 using the new choice of elements w n . Note that G is a free R -module. By the assumption that R is Σ S -incomplete there is a sequence a n ∈ { , } with P k ∈ ω p k a k h k / ∈ G . However P k ∈ ω p k a k h k ∈ b G by the choice of h n , hence P k ∈ ω p k a k h k / ∈ G ω +1 by definition of G ω +1 . Recall w = P k ∈ ω p k a k g k andsuppose that w ϕ ∈ G ω +1 . We compute w ϕ = ( X k ∈ ω p k a k g k ) ϕ = X k ∈ ω p k a k ( g k ϕ )and X k ∈ ω p k a k h k = X k ∈ ω p k a k ( g k ϕ − q ( g k )) = X k ∈ ω p k a k g k − X k ∈ ω q ( g k ) p k a k = w − X k ∈ ω p k a k q ( g k ) . From P k ∈ ω p k a k h k / ∈ G ω +1 follows P k ∈ ω p k a k q ( g k ) / ∈ G ω +1 . However, by the definitionof bodies B α , the map taking g −→ q ( g ) for any g ∈ G ω +1 is an endomorphism of G ω +1 ,and also w = P k ∈ ω p k a k g k ∈ G ω +1 , hence P k ∈ ω p k a k q ( g k ) ∈ G ω +1 is a contradiction.We deduce w ϕ / ∈ G ω +1 and ϕ does not extend to an endomorphism of G ω +1 . E ( R )-Algebras Lemma 6.1 Let κ be a regular, uncountable cardinal and B = S α ∈ κ B α a κ -filtration ofbodies. Also let G α = R B α and G = R B . Then the following holds for any ϕ ∈ End R G . i) If there is g ∈ G such that gϕ / ∈ ( B α ) { g } , then there is also h ∈ G free over B α such that hϕ / ∈ ( B α ) { h } .(ii) If there are g ∈ G and a polynomial q ( y ) over B α such that gϕ − q ( g ) / ∈ ( B α ) { g } ,then there is also h ∈ G free over B α such that hϕ − q ( h ) / ∈ ( B α ) { h } . Proof. If g ∈ G satisfies the requirements in (i) or (ii), respectively, then chooseany element h ′ ∈ G which is free over B α . 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 α by Proposition 4.4. In this case h ′ ϕ ∈ ( B α ) { h ′ } or h ′ ϕ − q ( h ′ ) ∈ ( B α ) { h ′ } ,respectively. It follows hϕ / ∈ ( B α ) { h } or hϕ − q ( h ) / ∈ ( B α ) { 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 κ -filtration of B α s for application using ♦ κ E for some non-reflecting subset E ⊆ κ o . Recall that ♦ κ E holds for all regular,uncountable, not weakly compact cardinals κ and non-reflecting subsets E in V = L . Construction of a κ -filtration of free bodies: Let { ϕ ρ : ρ ∈ E } be the family of Jensenfunctions given by ♦ κ E . The body B and the R -module R B will be constructed as a κ –filtration B = S α ∈ κ B α of bodies. We choose | B α | = | α | + | R | = | B α +1 \ B α | and fix for each α ∈ E a strictly increasing sequence α n ∈ α \ E with sup n ∈ ω α n = α. This is possible, because E consists of limit ordinals cofinal to ω only and we can pick α n 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 α ν , ν ∈ cf( α ) with sup ν ∈ cf( α ) α ν = α and α ν ∈ α \ E if cf( α ) > ω . This is crucial, because the body B α of the (continuous) κ –filtration of B must be free in order to proceed by a transfinite construction. This case does notoccur for κ = ℵ .Using Lemma 5.1, Lemma 5.2 and Lemma 5.3 inductively, we define the bodystructure on B ν . We begin with B = 0, and by continuity of the ascending chain the25onstruction reduces to an inductive step passing from B ν to B ν +1 . We will carry onour induction hypothesis of the filtration at each step. In particular the following threeconditions must hold.(i) B ν is a free body.(ii) If ρ ∈ ν \ E , then B ν is a free body over B ρ .(iii) If ρ ∈ E then let ρ n ( n ∈ ω ) be the given sequence with sup n ∈ ω ρ n = ρ . Supposethat the hypothesis of one of the three step lemmas, Lemma 5.1 or Lemma 5.2or Lemma 5.3 holds for G n = R B ρ n ( n ∈ ω ). We identify B ρ +1 with B ω +1 fromthe Step Lemmas (so ϕ ρ does not extend to an endomorphism of B ρ +1 ).Following these rules we step to ν +1: If the hypotheses of condition (iii) are violated,for instance, if ν / ∈ E we choose B ν +1 := ( B α ) { v α } adding any new free variable v ν tothe body. However, next we must check that these conditions (i) to (iii) can be carriedover to ν + 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 ω +1 in Proposition 4.10. Condition (ii) needsthat B ν +1 is a free body over B ρ . However B ρ ⊆ B ν n for a large enough n ∈ ω .Hence (ii) follows from freeness of B ν n over B ρ (inductively) and B ν +1 over B ν n (byProposition 4.10 and Corollary 4.7).In the limit case γ we have two possibilities: If cf( γ ) = ω then sup n ∈ ω γ n = γ ,hence B γ = S n ∈ ω B γ n and B γ is a free body with the help of (i) and (ii) by induction,see Corollary 4.7. If cf( γ ) > ω , then by our set theoretic assumption ( E is non–reflecting) we have a limit sup α ∈ cf( γ ) γ α = γ of ordinals not in E . The union of thechain B γ = S α ∈ cf( γ ) B γ α by (i) and (ii) is again a free body, see Corollary 4.7. Thus weproceed and obtain B = S ν ∈ κ B ν which is a κ –filtration of free bodies. It remains toshow the Main Lemma 6.2 Assume ♦ κ E . Let κ be a regular, uncountable cardinal and B = S α ∈ κ B α be the κ -filtration of bodies just constructed. Also let G α = R B α and G = R B .Suppose that ϕ ∈ End R G does not satisfy the following conditions (i) or (ii) for any α ∈ κ and any polynomial q ( y ) over B α :(i) There is g ∈ G such that gϕ / ∈ R ( B β ) { g } .(ii) There is g ∈ G such that gϕ − q ( g ) / ∈ R ( B β ) { g } Then ϕ is represented in B . roof. Suppose for contradiction that ϕ is not represented in B . Let E ⊆ κ o begiven from ♦ κ E , let { ϕ δ : δ ∈ E } be the family of Jensen functions and define astationary subset E ′ ϕ = { δ ∈ E : ϕ ↾ G δ = ϕ δ } . Note that C = { δ ∈ κ : G δ ϕ ⊆ G δ } is acub, thus E ϕ := E ′ ϕ ∩ C is also stationary.As a consequence we see that there is δ ∈ E ϕ satisfying one of the following condi-tions.(i) For every α < δ ∈ E ϕ there is g ∈ G δ such that gϕ / ∈ R ( B α ) { g } .(ii) There is α < δ ∈ E ϕ such that gϕ ∈ R ( B α ) { g } for all g ∈ G δ (not case (i)) but forevery α < δ and every polynomial q ( y ) over B α represented by an endomorphismof G there is g ∈ G δ with gϕ − q ( g ) / ∈ G α .(iii) There is α < δ ∈ E ϕ such that gϕ ∈ R ( B α ) { g } and there is a polynomial q ( y ) over B α with gϕ − q ( g ) ∈ G α for all g ∈ G δ (neither (i) nor (ii) holds), but ϕ is notrepresented by B . Thus there are a sequence δ n < δ ( n ∈ ω ) with sup n ∈ ω δ n = δ and g n ∈ G δ n such that g n ϕ − q ( g n ) = 0 for all n ∈ ω .By Lemma 6.1 we may assume that the elements g existing by (i) and (ii), respec-tively, are free over B α . Moreover, if g ∈ G , then by cf κ > ω we can choose δ ∈ E ϕ such that g ∈ G δ .If (i) holds, then we can choose a proper ascending sequence δ n ∈ E ϕ with sup n ∈ ω δ n = δ and elements g n ∈ G δ n +1 such that g n is free over B δ n and g n ϕ / ∈ R ( B δ n ) { g n } for all n ∈ ω. We identify G δ n with G n in Step Lemma 5.1 and note that (by δ n ∈ E ϕ ) ϕ ↾ G n is an endomorphism with G n ϕ ⊆ G n which is predicted as a Jensen function. Byconstruction of G δ +1 (as a copy of G ω +1 from Lemma 5.1), the endomorphism ϕ ↾ G δ does not extend to End R G δ +1 . However ϕ ∈ End G , thus G δ +1 ϕ ⊆ G α for some α < κ .Finally note that G δ +1 is the S -adic closure of G δ in G because G δ is S -dense in G δ +1 and G δ +1 is a summand of G α hence S -closed in G α . We derive the contradiction thatindeed ϕ ↾ G δ +1 ∈ End R G δ +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 δ n ∈ E ϕ with sup n ∈ ω δ n = δ as above andthere are free elements g n , u n ∈ G δ n +1 (also free over B δ n and polynomials q n , q n over B δ such that u n ϕ = q n ( u n ) = g n ϕ = q n ( g n ). Moreover, the polynomials q n ( y ) − q n ( y )are not constant over B δ . 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 α < κ and q ( y ) a polynomial over B α such that gϕ − q ( g ) ∈ G α for all g ∈ G .27he polynomial q ( y ) is represented by an endomorphism of G α . Moreover (from (iii))we find g n ∈ G δ n +1 free over G δ n for a suitable sequence δ n with sup n ∈ ω δ n = δ suchthat g n ϕ − q ( g n ) = 0. We now apply Step Lemma 5.3; the argument from case (i) givesa final contradiction. Thus the Main Lemma holds. 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