# Properties of Slender Rings

Authors: E. F.  Cornelius
Journal: International Journal of Mathematics and Mathematical Sciences
 (2010)

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## Abstract

Slender rings and their products are characterized in terms of algebraic and topological properties. Possible limitations on uncountable products of such rings are discussed.


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## Body

## 1. Terminology and Notation

Ring, module, and topological terminology and notation are standard.ℛ denotes a unital ring with identity 1, and ℛω denotes its countable product, that is, the set of all sequences of elements of ℛ. Of course ω={0,1,…}. ℛω is both a left and right ℛ-module; it will be treated as a left ℛ-module unless otherwise stated. Results for left ℛ-modules carry over mutatis mutandis to right ℛ-modules.Theith coordinate of an element x∈ℛω will be denoted x(i). Let ℱ denote the submodule of ℛω, consisting of those elements of ℛω which have only finitely many nonzero coordinates. Let en∈ℱ have coordinates en(i)=1 when i=n and 0 otherwise; that is, en(i)=δni, the Kronecker delta. ℱ is then the free submodule of ℛω, generated by the en, which form a basis of ℱ.Letπn denote the nth canonical projection of ℛω onto ℛ; that is, for x=(x(i):i∈ω)∈ℛω, πn(x)=x(n). Note that x=0∈ℛω if and only if πn(x)=0 for all n, and that πn(ei)=δni.ℛ is said to be slender if every ℛ-module homomorphism from ℛω to ℛ is 0 on all but finitely many en. As will be demonstrated, this condition determines the structure of the endomorphism ring of ℛω, Endℛ(ℛω).As discussed in Section2, the “original” slender ring was the integers ℤ. Familiar nonslender rings are the p-adic integers for any prime p∈ℤ. By using the method in [1, page 159, (d)] and known injective properties of the p-adics, it is easy to see that they are not slender.


## 2. The Origin of Slender Rings

Slender rings had their origin in the seminal paper by Specker [2], who proved the results in this Section for the ring of integers ℤ and its countable product ℤω, which has become known as the Baer-Specker group.The proof of the first theorem is an elaboration on the proof of a somewhat different theorem [3, page 172].Theorem 2.1.
Every homomorphismh from the Baer-Specker group to  ℤ satisfies  h(en)=0 for all but finitely many  n.Proof.
Suppose thath(en)≠0 for infinitely many en; it may be assumed that h(en)≠0 for all n. There exists a prime, say p0, which does not divide h(e0). Define a sequence bn=anen of elements of ℤω as follows, where an∈ℤ.
Setb0=e0 and a0=1. Suppose that b0,…,bn and the corresponding a0,…,an have been defined and primes p0,…,pn have been chosen, with pi∣h(b0+⋯+bi),i=0,…,n≠0. Now choose a prime pn+1 different from p0,…,pn such that pn+1 does not divide h(en+1), pn+1∤h(en+1). Then let bn+1 be a multiple k of p0⋯pnen+1 such that h(b0+⋯+bn+bn+1) is divisible by pn+1.
To see that the distinctiveness of the primespi makes such a choice of bn+1possible, note that h(b0+⋯+bn+bn+1)=h(b0+⋯+bn)+kp0⋯pnh(en+1), so that the problem reduces to finding integers k and l that satisfy an integral equation of the form kpn+1=j+lm, in which pn+1 and m=p0⋯pnh(en+1) are relatively prime and j=h(b0+⋯+bn). Being relatively prime, m and pn+1 generate all of ℤ, including j=h(b0+⋯+bn), so that suitable k and l indeed can be found. Set an+1=kp0⋯pn.
Leta∈ℤω with a(i)=ai. Now p0∤h(e0), so p0∣a(i) for all i>0 imply that h(a)=h(b0)+h(0,a(1),a(2),…)=h(e0)+p0h(0,a(1)/p0,a(2)/p0,…) is not divisible by p0. Thus h(a)≠0. But for each i>0, h(a)=h(b0+⋯+bi)+pih(0,…,0,a(i+1)/pi,a(i+2)/pi,…) so that h(a) is divisible by pi because pi∣h(b0+⋯+bi). The only integer divisible by an infinite number of primes is 0, so that h(a)=0, a contradiction.The following proof has been utilized in a number of contexts; for a more general proof, see Theorem3.1.Theorem 2.2.
If a homomorphismh from the Baer-Specker group to  ℤ satisfies  h(en)=0 for all  n,  then  h=0.Proof.
Clearlyh(ℱ)=0. Let p and q be distinct primes. Let A be the subgroup of ℤω, consisting of all sequences of the form (anpn+1:an∈ℤ,n∈ω). The elements of A/ℱ are divisible by every power of p. Thus h(ℱ)=0 necessitates h(A)=0, as otherwise h  would induce a nontrivial homomorphism from A/ℱ to ℤ, an impossibility in light of the divisibility of the elements of A/ℱ. Similarly, if C consists of all sequences of the form (cnqn+1), h(C)=0. Now each element b∈ℤω can be written as b=a+c with a∈A and c∈C, because for each n, 1=anpn+1+cnqn+1 for some an,cn∈ℤ, so that b(n)=b(n)anpn+1+b(n)cnqn+1. Thus ℤω=A+C and so h(ℤω)=h(A)+h(C)=0+0=0.Remark 2.3.
Theorem2.2 frequently is stated in terms of the homomorphism being 0 on ℱ instead of just the en. The important point is that the homomorphism is completely determined by its values on the en (or on ℱ).
The proofs of Theorems2.1 and 2.2 use some number-theoretic properties of ℤ and so can be extended to only slender rings having the requisite primes, such as rings of polynomials over fields. Nevertheless, the results below hold for any ring with identity.


## 3. Homomorphisms of Countable Products of Slender Rings

The proof of the following theorem is adapted from a proof forℤ [3, page 170].Theorem 3.1.
Ifℛ is slender and  h is a homomorphism from  ℛω to  ℛ, such that  h(en)=0 for all  n, then  h=0.Proof.
Suppose thath≠0 for some p∈ℛω. Define a homomorphism h̃ from ℱ to ℛω by h̃(e0)=p and h̃(en)=p-p(0)e0-⋯-p(n-1)en-1for n>0. Extend h̃ to an endomorphism of ℛω by defining [h̃(x)](n)=∑i=0nx(i)[h̃(ei)](i). Now hh̃ is a homomorphism from ℛω to  ℛ such that hh̃(en)=h[p-p(0)e0-⋯-p(n-1)en-1]=h(p)-p(0)h(e0)-⋯-p(n-1)h(en-1)=h(p)≠0 for all n, a contradiction to the slenderness of ℛ.Specker proved a less precise formulation of the next theorem forℤ and ℤω [2, Satz III]. The proof given here seems shorter and simpler.Theorem 3.2.
Ifℛ is slender, then every nonzero homomorphism  h from  ℛω to  ℛ has a unique expression of the form  h=πn0h(en0)+⋯+πnkh(enk), with  k and the  ni depending upon  h and each  h(eni)≠0.Proof.
Becauseℛ is slender, h is 0 except at some en0,…,enk, in order n0<⋯<nk if k>0. Let g=h-πn0h(en0)-⋯-πnkh(enk). Now g(ei)=0 for all i, so g=0 by Theorem 3.1, and thus h has the form claimed. It is clear that h cannot be expressed as any other linear combination of canonical projections with nonzero coefficients.Note that ifℛ is slender, then every homomorphism from its product ℛω to ℛ is effectively a homomorphism from a finite subproduct to ℛ.Theorem 3.3.
Ifℛ is a slender ring and  g is an endomorphism of  ℛω, which satisfies  g(en)=0 for all  n, then  g=0.Proof.
Ifg≠0, then for some p∈ℛω, g(p)≠0 so that πk(g(p))≠0 for some k. But πkg would be a nonzero homomorphism from ℛω to ℛ, which is 0 at each en, thereby contradicting Theorem 3.1.Theorem3.3 simply says that, for a slender ring, each endomorphism of the product ℛω is determined by the endomorphism's values on the standard basis of ℱ, or on ℱ itself.The results of this Section can be used to determine the structure ofEndℛ(ℛω).


## 4. Infiniteℛ-Matrices and Endℛ(ℛω)

An infinite matrix over a ringℛ is a 2-dimensional array of elements from ℛ, of the form M=(M(i,j):M(i,j)∈ℛ;i,j∈ω). Such an infinite ℛ-matrix is said to be row-finite if, for each i, M(i,j)=0 for almost all j; that is, only finitely many M(i,j)≠0 for each i.IfM and N are row-finite ℛ-matrices, their sum M+N is defined to be the row-finite ℛ-matrix with entries (M+N)(i,j)=M(i,j)+N(i,j). Their product MN is defined to be the row-finite ℛ-matrix with entries(4.1)MN(i,j)=∑k∈ωN(k,j)M(i,k).
This order is required to encompass the case of noncommutative rings, when functions act from the left. The usual order of M(i,k)N(k,j) works when functions act from the right. Every such sum is finite because, for each i, only finitely many M(i,k) are nonzero. Multiplication of row-finite ℛ-matrices is associative, a property not always found in the multiplication of infinite matrices. With these definitions of addition and multiplication, the set of row-finite infinite ℛ-matrices forms a ring with identity I having entries I(i,j)=δij.For purposes of matrix-vector multiplication, the elements ofℛω are viewed as column vectors. If M is a row-finite ℛ-matrix and x∈ℛω, the product of M and x is denoted Mx and is defined to be the sequence in ℛω with Mx(i)=∑k∈ωx(k)M(i,k). It is easy to see that M induces an endomorphism of ℛω via such matrix-vector multiplication.Theorem 4.1.
Every row-finite infiniteℛ-matrix induces an endomorphism of  ℛω via matrix-vector multiplication.What is most interesting is the converse for slender rings:Theorem 4.2.
Ifℛ is slender, then every endomorphism of  ℛω is induced by multiplication of its vectors by a row-finite infinite  ℛ-matrix.Proof.
Letg∈Endℛ(ℛω) and define an ℛ-matrix M by M(i,j)=[g(ej)](i) for i,j∈ω. For each i, πig is a homomorphism from ℛω to ℛ and so by slenderness is 0 for almost all ej,j∈ω. But πig(ej)=[g(ej)](i)=M(i,j) so that M is row-finite.
By Theorem4.1, M induces an endomorphism of ℛω. According to Theorem 3.3, to prove that multiplication by M is the same as mapping by g, it suffices to demonstrate that Mej=g(ej) for all j; that is, that (Mej)(i)=[g(ej)](i). Now (Mej)(i)=∑k∈ωej(k)M(i,k)=∑k∈ωδjkM(i,k)=δjjM(i,j)=M(i,j)=[g(ej)](i).Corollary 4.3.
Ifℛ is a  slender ring, then the endomorphism ring  Endℛ(ℛω) under function addition and composition is isomorphic to the ring of row-finite infinite  ℛ-matrices.Proof.
Note that the columns of theℛ-matrix M in the proof of Theorem 4.2 are the values of the endomorphism g at the en. M is thus unique. There is then a 1-1 correspondence between the ring of row-finite infinite ℛ-matrices under addition and multiplication and Endℛ(ℛω) under function addition and composition. It is easy to see that this correspondence preserves addition and multiplication.


## 5. Topologies for a Ringℛ and Its Product ℛω

The foregoing algebraic properties of slender rings can be characterized topologically. Let a ringℛ be equipped with the discrete topology, that is, the topology in which every subset is open. The singletons, {r},r∈ℛ, form a basis of this discrete topology, which is metrizable by defining d(r,r)=0 for all r∈ℛ and d(r,s)=1 for distinct r,s∈ℛ. Under the discrete topology of ℛ, addition and scalar multiplication are continuous.Letℛω be equipped with the product topology, that is, the topology in which the sets, πn-1(U),n∈ω,U⊆ℛ, form a subbasis. In fact, the sets πn-1({r}),n∈ω,r∈ℛ, form a subbasis. Under the product topology of ℛω, the canonical projections are continuous open maps and addition and scalar multiplication are continuous.This product topology is metrizable by definingd(x,x)=0 for all x∈ℛω and, for distinct x,y∈ℛω, d(x,y)=2-n if x(i) and y(i) first differ at n∈ω. Thus, if d(x,y)<2-n, then x(i)=y(i) for at least i=0,…,n. In general, d(x,y)=d(x-y,0), an important fact used in the following discussions. Note also that d(rx,0)≤d(x,0) for all r∈ℛ and x∈ℛω.To see thatd induces the product topology on ℛω, let x∈ℛω and consider the neighborhood of x, N={y∈ℛω:d(x,y)<2-n}. For all y∈N, x(i)=y(i),i=0,…,n, so that y∈O=π0-1({x(0)})∩⋯∩πn-1({x(n)}) and so N⊆O. Conversely, any y∈O satisfies y(i)=x(i),i=0,…,n, which means that d(x,y)<2-n so that O⊆N.A sequence of elements ofℛω,(xn:n∈ω), converges to x∈ℛω if and only if the initial components of the xn become the initial components of x as n becomes large. It readily follows that xn converges to x if and only if (x-xn:n∈ω) converges to 0∈ℛω. Clearly the sequence (en) converges to 0.Forx=(x(i):i∈ω)∈ℛω, if x⃗n∈ℱ,n∈ω, is defined as x⃗n(i)=x(i) if i≤n and x⃗n(i)=0 for i>n, then the sequence (x⃗n) converges to x. Thus ℱ  is dense in ℛω. If x⟵n∈ℛω,n∈ω, is defined as x⟵n(i)=0 if i≤n and x⟵n(i)=x(i) for i>n, then the sequence (x⟵n) converges to 0.Observe that every Cauchy sequence(xn:n∈ω) in ℛω converges to an element of ℛω, so that ℛω is a complete metric space. To see this, simply observe that, at each coordinate i, the xn(i) become constant as n becomes large, so that xn converges to x with x(i)=limn→∞xn(i).For any homomorphism of topological modules, continuity on the entire domain is determined by continuity at0. Thus, if h is a homomorphism from ℛω to ℛ, then h is continuous on all of ℛω if and only if it is continuous at 0. This is easy to see because if the sequence xn converges to x∈ℛω, then limn→∞[h(xn)-h(x)]=limn→∞h(xn-x)=0 if and only if h is continuous at 0. A similar result, of course, holds for any endomorphism of ℛω.Convention re Topologies
Unless stated otherwise, the topology on a ring is always the discrete topology, and the topology on its product is always the product topology.Theorem 5.1.
A homomorphism to a ringℛ  from its product  ℛω is uniformly continuous on  ℛω if and only if it is continuous at  0. Ditto for an endomorphism of  ℛω.Proof.
Suppose thath is a homomorphism from ℛω to ℛ, which is continuous at 0. Given ϵ>0, there exists δ>0 such that d(h(x),0)<ϵ provided d(x,0)<δ. If (yn:n∈ω) converges to y∈ℛω, so long as d(y,yn)<δ, then d(h(y),h(yn))<ϵ. This holds because d(h(y),h(yn))=d(h(y)-h(yn),0)=d(h(y-yn),0) and d(y,yn)=d(y-yn,0). Thus h  is uniformly continuous.
The proof for endomorphisms ofℛω is essentially the same.Theorem 5.2.
If a ringℛ is slender, then every homomorphism from its product  ℛω to   ℛ is uniformly continuous, as is every endomorphism of  ℛω.Proof.
Since0 is trivial, let h be a nonzero homomorphism from ℛω to ℛ and let (xn) be a sequence in ℛω, which converges to 0. According to Theorem 3.2, h is a linear combination of canonical projections, the last one of which is some nk. For n sufficiently large, the first nk coordinates of the xn are 0 and so h(xn)=0. Thus h is continuous at 0 and hence uniformly continuous everywhere by Theorem 5.1.
Letg be nonzero endomorphism of ℛω and let M be the row-finite infinite ℛ-matrix which induces g. Again, it suffices to check convergence at 0, so let (xn) be a sequence of elements of ℛω, which converges to 0. As n gets large, the products Mxn have initial coordinates =0 because M is row-finite and the xn have initial coordinates =0. Specifically, given any k∈ω, let nk be the largest j such that M(i,j)≠0,i≤k, and let mk∈ω be such that coordinates 0-nk of xn are equal to 0 for all n>mk. Then coordinates 0-k of Mxn are 0 for all n>mk so that limn→∞Mxn=0. Again Theorem 5.1 completes the proof.For slenderℛ, the continuity of an endomorphism g of ℛω at 0 is reflected by the row finiteness of the matrix M inducing it. In particular, limn→∞en=0 means that limn→∞Men=0. Since Men is column n of M, the initial entries of the columns of M (M(0,n), M(1,n), etc.) increasingly =0.The ring properties previously described algebraically can be cast in topological terms.Theorem 5.3.
Letℛ be a ring with  1. Then a homomorphism  h from  ℛω to  ℛ satisfies  h(en)=0 for all but finitely many  n if and only if it  satisfies  limn→∞h(en)=0.Proof.
Ifh(en)=0 for all n>k, then d(h(en),0)=0 for all n>k, so that limn→∞h(en)=0.
Conversely, iflimn→∞h(en)=0, there is a positive integer k such that d(h(en),0)<1 for all n>k. But d(h(en),0)<1 implies that h(en)=0 because only x=0  satisfies d(x,0)<1 in the metric on ℛ.Corollary 5.4.
A ringℛ with  1 is slender if and only if every homomorphism  h from  ℛω to  ℛ satisfies  limn→∞h(en)=0.The next theorem is a casting of slenderness in topological terms.Theorem 5.5.
A ringℛ with   1 is slender if and only if every homomorphism from  ℛω to  ℛ is continuous at   0.Proof.
Ifℛ is slender, then according to Theorem 5.2, every homomorphism from ℛω to ℛ  is certainly continuous at 0.
Conversely, if every homomorphismh from ℛω  to ℛ is continuous at 0, then limn→∞h(en)=0 because limn→∞en=0. The slenderness of ℛ follows from Corollary 5.4.The next theorem is the topological analog of Theorems4.1 and 4.2.Theorem 5.6.
Letℛ be a ring with  1. Then every endomorphism of its product  ℛω is induced by a row-finite matrix if and only if every endomorphism of  ℛω is continuous at  0.Proof.
Ifg is an endomorphism of ℛω, which is induced by a row-finite matrix M, and if (xn) is a sequence in ℛω, that converges to 0, then as n get large, the initial coordinates of the xn become 0. Since M is row-finite, the initial coordinates of the Mxn also become 0 so that (Mxn)=(g(xn)) converges to 0, as in the proof of Theorem 5.2.
Conversely, suppose that every endomorphism ofℛω is continuous at 0, and let h be a homomorphism from ℛω to ℛ. Define an endomorphism g of ℛω by g(x)=(h(x),h(x),…). By hypothesis, g is continuous at 0. Since limn→∞en=0, limn→∞g(en)=0; that is, the initial coordinates of g(en), which all =h(en), must be 0 for n sufficiently large. Thus limn→∞h(en)=0, so by Corollary 5.4, ℛ is slender. Theorem 4.2 completes the proof.


## 6. Equivalent Conditions for Slenderness

It is convenient to draw together some of the more useful conditions for a ring to be slender. There are, of course, other equivalences.Theorem 6.1.
Letℛ be a ring with  1, equipped with the discrete topology, and let its product  ℛω be equipped with the product topology. Then conditions (I)–(VI) are equivalent. (I)
Every homomorphismh from  ℛω to  ℛ satisfies  h(en)=0 for all but finitely many  n.(II)
ℛ is slender.(III)
Every nonzero homomorphismh from  ℛω to  ℛ has a unique expression of the form  h=πn0h(en0)+⋯+πnkh(enk), with  k and the  ni depending upon  h and each  h(eni)≠0.(IV)
Every endomorphism ofℛω is induced by multiplication of its vectors by a row-finite infinite matrix with entries from  ℛ.(V)
Every homomorphism fromℛω to  ℛ is continuous at  0.(VI)
Every endomorphism ofℛω is continuous at  0.Proof.
(I)⇔(II):  This equivalence is simply definitional.
(II)⇒(III):  This is Theorem 3.2.
(III)⇒(II):  If every homomorphism h from ℛω to ℛ has the form stated, it is clear that h(en)=0 for all but finitely many n. Thus (II)⇔(III).
(II)⇔(V):  This is Theorem 5.5. Thus (I), (II), (III), and (V) are equivalent.
(IV)⇔(VI):  This is Theorem 5.6. It now suffices to demonstrate the equivalence of (V) and (VI).
(V)⇒(VI):  Let g be an endomorphism of ℛω; by Theorem 5.1, it suffices to demonstrate continuity at 0. Suppose that (xi) is a sequence in ℛω, which converges to 0. Each πng is a continuous homomorphism from ℛω to ℛ, which satisfies πng(xi)=0 for all i> some kn∈ω. Given any l∈ω, there exist kn∈ω such that πng(xi)=0 for all i>kn, n=0,…,l. If m=max{k0,…,kl}, then at least coordinates 0-l of g(xi) are 0 for all i>m; that is, limi→∞g(xi)=0 so that g is continuous at 0.
(VI)⇒(V): This was shown in the proof of Theorem 5.6.Remark 6.2.
Conditions (V) and (VI) of Theorem6.1 could, of course, be stated by deleting "at 0" inasmuch as Theorem 5.1 covers uniform continuity everywhere. Similarly, “continuous” could be strengthened to “uniformly continuous” without weakening the theorem. The equivalence of (I) and (V) is shown in [4, Theorem 3(3)] using “continuous” instead of “at 0”.


## 7. Extensions to Uncountable Products of Slender Rings

Although there is nothing sacrosanct about countable products of slender rings, it turns out that little may be gained from considering uncountable products. Notation used in the countable case will be carried over to the uncountable without further comment.It is clear that ifℛ is slender, then every homomorphism h from ℛI to ℛ satisfies h(ei)=0 for all but finitely many i, for every infinite index set I. Obviously, if every homomorphism h from ℛI to ℛ satisfies h(ei)=0 for all but finitely many i, for some infinite index set I, then ℛ is slender. The problem for uncountable products, even when ℛ is slender, is that the analog of Theorem 3.1 may not hold; that is, there may exist a nonzero homomorphism from ℛI to ℛ, which annihilates all ei.Whether uncountable products yield anything new depends upon the existencevel non of measurable cardinals. An uncountable cardinal number κ is said to be κ-measurable if there exists a set I of cardinality κ and a measure μ on the subsets of I, which (i) assumes only the values 0,1; (ii) satisfies μ(∅)=0=μ({i}) for all i∈I, and μ(I)=1; and (iii) is κ-additive in the sense that if {Jk:k∈K} is a collection of mutually disjoint subsets of I with |K|<κ, then μ(⋃k∈KJk)=∑k∈Kμ(Jk). With this definition of measurable cardinals, uncountable products of rings may be subject to the limitation discussed in the proposition below, the proof of which is modeled after one for ℤ [1, page 161, Remark].Proposition 7.1.
Letℛ be a ring with  1. If there exists a  κ-measurable cardinal satisfying  κ>|ℛ|, and  I  is an index set of cardinality  κ and  μ is a  κ-additive measure on  I,  then there exists a nonzero homomorphism  h from  ℛI to  ℛ, such that  h(ei)=0 for all  i∈I.Proof.
For everya=(a(i):i∈I)∈ℛI and r∈ℛ, define Xr(a)={i∈I:a(i)=r}. Then the Xr(a) are pairwise disjoint subsets of I whose union is I. Since |ℛ|<κ, 1=μ(I)=μ(⋃r∈ℛXr(a))=∑r∈Rμ(Xr(a)), so that exactly one of the sets, say Xs(a), has measure 1. Set η(a)=s. Using the properties of μ, it will be shown that η preserves addition and scalar multiplication and η(ei)=0 for all i∈I.
First, to see thatη(ei)=0 for all i∈I, observe that I=X0(ei)∪X1(ei)={I∖{i}}∪{i}, so that μ(X0(ei))=1. Thus η(ei)=0.
To confirm additivity, suppose thatη(a)=s and η(b)=t; then μ({i∈I:a(i)≠s})=0 and μ({i∈I:b(i)≠t})=0. Now I⊆(Xs(a)∩Xt(b))∪{i∈I:a(i)≠s}∪{i∈I:b(i)≠t} so that
(7.1)1=μ(I)≤μ(Xs(a)∩Xt(b))+μ{i∈I:a(i)≠s}+μ{i∈I:b(i)≠t}=μ(Xs(a)∩Xt(b))≤1,
so μ(Xs(a)∩Xt(b))=1. Since Xs(a)∩Xt(b)⊆Xs+t(a+b), 1=μ(Xs(a)∩Xt(b))≤μ(Xs+t(a+b))≤1; that is, μ(Xs+t(a+b))=1, so η(a+b)=s+t.
Finally, to check thatη(ra)=rs=rη(a) for r∈ℛ, note that Xs(a)⊆Xrs(ra) so that 1=μ(Xs(a))≤μ(Xrs(ra))≤1 or μ(Xrs(ra))=1. Thus η(ra)=rs=rη(a) and η is an ℛ-homomorphism.Remark 7.2.
In the Proof of Proposition7.1, not only is it true that η(ℱ)=0, but η(ℱκ)=0 where ℱκ is the submodule of ℛI consisting of all x∈ℛI with μ({i∈I:x(i)≠0})=0. Thus η has a large kernel.Ulam has shown that if there exists anℵ1-measurable cardinal (the meaning of which should be clear), then there is a least one, call it 𝔪, and it in fact is 𝔪-measurable [5]. Thus 𝔪 often is referred to as the least measurable cardinal. Lady has shown that if ℛ is slender and h is a homomorphism from ℛI to ℛ, satisfying h(ei)=0 for all i∈I, then h=0, provided |I|<m [4, Theorem 3(4)]. If there are no measurable cardinals, it then follows that, for slender ℛ, such a homomorphism h is zero regardless of the cardinality of I.The state of uncountable products of rings can be summarized as follows.Theorem 7.3.
Letℛ be a ring with  1, equipped with the discrete topology, let  σ  be an uncountable ordinal, and let the product  ℛσ be equipped with the product topology. If there are no measurable cardinals, then conditions (I)–(V) are equivalent. (I)
ℛ is slender.(II)
Every nonzero homomorphismh from  ℛσ to  ℛ has a unique expression of the form  h=πι0h(eι0)+⋯+πιkh(eιk), with  k  and the  ιj depending upon  h and each  h(eιj)≠0.(III)
Every endomorphism ofℛσ is induced by multiplication of its vectors by a row-finite infinite matrix with entries from  ℛ and with rows and columns indexed by  σ.(IV)
Every homomorphism fromℛσ to  ℛ is continuous at  0.(V)
Every endomorphism ofℛσ is continuous at  0. If there exists a least measurable cardinal  𝔪, then conditions (I)–(V) are equivalent for all ordinals  σ of cardinality  <m.Proof.
Assume that no measurable cardinal exists. Note that the proof of Theorem3.2 requires only that a homomorphism which is 0 on all ei itself be zero, a result which follows from [4, Theorem 3(4)] in the absence of measurable cardinals. Further note that Theorems 4.1 and 4.2 do not depend upon a countable product of rings. From earlier Sections, the equivalence of conditions (I)–(III) is apparent.
Although metrizability of the product topology onℛσ is lost, the product topology has the property that an endomorphism g of ℛσ is continuous if and only if πig is continuous for all i. Continuity at 0 continues to be the litmus test for overall continuity (as with any topological module). Thus conditions (IV) and (V) are readily seen to be equivalent.
All that remains is to connect (I)–(III) with (IV)-(V). Suppose that (IV) holds and leth be a continuous homomorphism from ℛσ to ℛ. Suppose that h(eιn)≠0,ιn<σ, for all n∈ω. As in previous Sections, let ℛω have canonical projections pn and let en,n∈ω, be the basis of the free submodule ℱ⊂ℛω. Inject ℛω into ℛσ via g defined for x∈ℛω as [g(x)](ιn)=x(n),n∈ω, and [g(x)](ι)=0 otherwise. This injection is continuous because pn=πιng is the composition of continuous functions for all n∈ω. Now hg is a continuous homomorphism from ℛω to ℛ and so, by Theorem 6.1, must be 0 on all but finitely many en. Since g(en)=eιn, h must be 0 except for those finitely many eιn. Thus ℛ is slender and so (IV)⇒(I).
Finally, suppose that (II) holds and leth=πι0h(eι0)+⋯+πιkh(eιk) be a homomorphism from ℛσ to ℛ and let r∈ℛ; it suffices to address subbasic open sets. For any x∈ℛσ, x∈h-1({r}) if and only if h(x)=r; that is, if and only if x(ι0)h(eι0)+⋯+x(ιk)h(eιk)=r. Now x∈πι0-1({x(ι0)})∩⋯∩πιk-1({x(ιk)})⊆h-1({r}) so that h-1({r}) is open and h is continuous. Thus (II)⇒(IV), making conditions (I)–(V) equivalent, under the assumption that there are no measurable cardinals.
If measurable cardinals exist, the key to the equivalence of (I)–(V) remains demonstrating that only the zero homomorphism fromℛσ to ℛ annihilates all eι, and again [4, Theorem 3(4)] supplies the proof for |σ|<𝔪.Remark 7.4.
The equivalence of conditions (I)–(V) for all uncountable ordinals is equivalent to the nonexistence of measurable cardinals. It is, of course, unnecessary in Theorem7.3 to use an ordinal for indexing, as any uncountable index set would suffice.



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