Properties of Slender Rings

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.

1. Terminology and Notation

Ring, module, and topological terminology and notation are standard. ℛ denotes a unital ring with identity 1, and ℛ^𝜔 denotes its countable product, that is, the set of all sequences of elements of ℛ. Of course 𝜔={0,1,…}. ℛ^𝜔 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.

The 𝑖th coordinate of an element 𝑥∈ℛ^𝜔 will be denoted 𝑥(𝑖). Let ℱ denote the submodule of ℛ^𝜔, consisting of those elements of ℛ^𝜔 which have only finitely many nonzero coordinates. Let 𝑒^𝑛∈ℱ have coordinates 𝑒^𝑛(𝑖)=1 when 𝑖=𝑛 and 0 otherwise; that is, 𝑒^𝑛(𝑖)=𝛿_𝑛𝑖, the Kronecker delta. ℱ is then the free submodule of ℛ^𝜔, generated by the 𝑒^𝑛, which form a basis of ℱ.

Let 𝜋_𝑛 denote the 𝑛th canonical projection of ℛ^𝜔 onto ℛ; that is, for 𝑥=(𝑥(𝑖)∶𝑖∈𝜔)∈ℛ^𝜔, 𝜋_𝑛(𝑥)=𝑥(𝑛). Note that 𝑥=0∈ℛ^𝜔 if and only if 𝜋_𝑛(𝑥)=0 for all 𝑛, and that 𝜋_𝑛(𝑒^𝑖)=𝛿_𝑛𝑖.

ℛ is said to be slender if every ℛ-module homomorphism from ℛ^𝜔 to ℛ is 0 on all but finitely many 𝑒^𝑛. As will be demonstrated, this condition determines the structure of the endomorphism ring of ℛ^𝜔, End_ℛ(ℛ^𝜔).

As discussed in Section 2, 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 ℤ^𝜔, 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 homomorphism ℎ from the Baer-Specker group to ℤ satisfies ℎ(𝑒^𝑛)=0 for all but finitely many 𝑛.

Proof. Suppose that ℎ(𝑒^𝑛)≠0 for infinitely many 𝑒^𝑛; it may be assumed that ℎ(𝑒^𝑛)≠0 for all 𝑛. There exists a prime, say 𝑝₀, which does not divide ℎ(𝑒⁰). Define a sequence 𝑏^𝑛=𝑎_𝑛𝑒^𝑛 of elements of ℤ^𝜔 as follows, where 𝑎_𝑛∈ℤ.
Set 𝑏⁰=𝑒⁰ and 𝑎₀=1. Suppose that 𝑏⁰,…,𝑏^𝑛 and the corresponding 𝑎₀,…,𝑎_𝑛 have been defined and primes 𝑝₀,…,𝑝_𝑛 have been chosen, with 𝑝_𝑖∣ℎ(𝑏⁰+⋯+𝑏^𝑖),𝑖=0,…,𝑛≠0. Now choose a prime 𝑝_𝑛+1 different from 𝑝₀,…,𝑝_𝑛 such that 𝑝_𝑛+1 does not divide ℎ(𝑒^𝑛+1), 𝑝_𝑛+10𝑥0008𝑐∤ℎ(𝑒^𝑛+1). Then let 𝑏^𝑛+1 be a multiple 𝑘 of 𝑝₀⋯𝑝_𝑛𝑒^𝑛+1 such that ℎ(𝑏⁰+⋯+𝑏^𝑛+𝑏^𝑛+1) is divisible by 𝑝_𝑛+1.
To see that the distinctiveness of the primes 𝑝_𝑖 makes such a choice of 𝑏^𝑛+1possible, note that ℎ(𝑏⁰+⋯+𝑏^𝑛+𝑏^𝑛+1)=ℎ(𝑏⁰+⋯+𝑏^𝑛)+𝑘𝑝₀⋯𝑝_𝑛ℎ(𝑒^𝑛+1), so that the problem reduces to finding integers 𝑘 and 𝑙 that satisfy an integral equation of the form 𝑘𝑝_𝑛+1=𝑗+𝑙𝑚, in which 𝑝_𝑛+1 and 𝑚=𝑝₀⋯𝑝_𝑛ℎ(𝑒^𝑛+1) are relatively prime and 𝑗=ℎ(𝑏⁰+⋯+𝑏^𝑛). Being relatively prime, 𝑚 and 𝑝_𝑛+1 generate all of ℤ, including 𝑗=ℎ(𝑏⁰+⋯+𝑏^𝑛), so that suitable k and 𝑙 indeed can be found. Set 𝑎_𝑛+1=𝑘𝑝₀⋯𝑝_𝑛.
Let 𝑎∈ℤ^𝜔 with 𝑎(𝑖)=𝑎_𝑖. Now 𝑝₀∤0𝑥0008𝑐ℎ(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+⋯+𝑏^𝑖)+𝑝_𝑖ℎ(0,…,0,𝑎(𝑖+1)/𝑝_𝑖,𝑎(𝑖+2)/𝑝_𝑖,…) so that ℎ(𝑎) is divisible by 𝑝_𝑖 because 𝑝_𝑖∣ℎ(𝑏⁰+⋯+𝑏^𝑖). 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 Theorem 3.1.

Theorem 2.2. If a homomorphism ℎ from the Baer-Specker group to ℤ satisfies ℎ(𝑒^𝑛)=0 for all 𝑛, then ℎ=0.

Proof. Clearly ℎ(ℱ)=0. Let 𝑝 and 𝑞 be distinct primes. Let 𝐴 be the subgroup of ℤ^𝜔, consisting of all sequences of the form (𝑎_𝑛𝑝^𝑛+1∶𝑎_𝑛∈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 (𝑐_𝑛𝑞^𝑛+1), ℎ(𝐶)=0. Now each element 𝑏∈ℤ^𝜔 can be written as 𝑏=𝑎+𝑐 with 𝑎∈𝐴 and 𝑐∈𝐶, because for each 𝑛, 1=𝑎_𝑛𝑝^𝑛+1+𝑐_𝑛𝑞^𝑛+1 for some 𝑎_𝑛,𝑐_𝑛∈ℤ, so that 𝑏(𝑛)=𝑏(𝑛)𝑎_𝑛𝑝^𝑛+1+𝑏(𝑛)𝑐_𝑛𝑞^𝑛+1. Thus ℤ^𝜔=𝐴+𝐶 and so ℎ(ℤ^𝜔)=ℎ(𝐴)+ℎ(𝐶)=0+0=0.

Remark 2.3. Theorem 2.2 frequently is stated in terms of the homomorphism being 0 on ℱ instead of just the 𝑒^𝑛. The important point is that the homomorphism is completely determined by its values on the 𝑒^𝑛 (or on ℱ).
The proofs of Theorems 2.1 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 for ℤ [3, page 170].

Theorem 3.1. If ℛ is slender and ℎ is a homomorphism from ℛ^𝜔 to ℛ, such that ℎ(𝑒^𝑛)=0 for all 𝑛, then ℎ=0.

Proof. Suppose that ℎ≠0 for some 𝑝∈ℛ^𝜔. Define a homomorphism ℎ from ℱ to ℛ^𝜔 by ℎ(𝑒⁰)=𝑝 and ℎ(𝑒^𝑛)=𝑝−𝑝(0)𝑒⁰−⋯−𝑝(𝑛−1)𝑒^𝑛−1for 𝑛>0. Extend ℎ to an endomorphism of ℛ^𝜔 by defining [ℎ(𝑥)](𝑛)=∑^𝑛_𝑖=0𝑥(𝑖)[ℎ(𝑒^𝑖)](𝑖). Now ℎℎ is a homomorphism from ℛ^𝜔 to ℛ such that ℎℎ(𝑒^𝑛)=ℎ[𝑝−𝑝(0)𝑒⁰−⋯−𝑝(𝑛−1)𝑒^𝑛−1]=ℎ(𝑝)−𝑝(0)ℎ(𝑒⁰)−⋯−𝑝(𝑛−1)ℎ(𝑒^𝑛−1)=ℎ(𝑝)≠0 for all 𝑛, a contradiction to the slenderness of ℛ.

Specker proved a less precise formulation of the next theorem for ℤ and ℤ^𝜔 [2, Satz III]. The proof given here seems shorter and simpler.

Theorem 3.2. If ℛ is slender, then every nonzero homomorphism ℎ from ℛ^𝜔 to ℛ has a unique expression of the form ℎ=𝜋_𝑛0ℎ(𝑒^𝑛0)+⋯+𝜋_𝑛𝑘ℎ(𝑒^𝑛𝑘), with 𝑘 and the 𝑛_𝑖 depending upon ℎ and each ℎ(𝑒^𝑛𝑖)≠0.

Proof. Because ℛ is slender, ℎ is 0 except at some 𝑒^𝑛0,…,𝑒^𝑛𝑘, in order 𝑛₀<⋯<𝑛_𝑘 if 𝑘>0. Let 𝑔=ℎ−𝜋_𝑛0ℎ(𝑒^𝑛0)−⋯−𝜋_𝑛𝑘ℎ(𝑒^𝑛𝑘). Now 𝑔(𝑒^𝑖)=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 if ℛ is slender, then every homomorphism from its product ℛ^𝜔 to ℛ is effectively a homomorphism from a finite subproduct to ℛ.

Theorem 3.3. If ℛ is a slender ring and 𝑔 is an endomorphism of ℛ^𝜔, which satisfies 𝑔(𝑒^𝑛)=0 for all 𝑛, then 𝑔=0.

Proof. If 𝑔≠0, then for some 𝑝∈ℛ^𝜔, 𝑔(𝑝)≠0 so that 𝜋_𝑘(𝑔(𝑝))≠0 for some 𝑘. But 𝜋_𝑘𝑔 would be a nonzero homomorphism from ℛ^𝜔 to ℛ, which is 0 at each 𝑒^𝑛, thereby contradicting Theorem 3.1.

Theorem 3.3 simply says that, for a slender ring, each endomorphism of the product ℛ^𝜔 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 of End_ℛ(ℛ^𝜔).

4. Infinite ℛ-Matrices and End_ℛ(ℛ^𝜔)

An infinite matrix over a ring ℛ 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.

If 𝑀 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 entries𝑀𝑁(𝑖,𝑗)=_𝑘∈𝜔𝑁(𝑘,𝑗)𝑀(𝑖,𝑘). (4.1) 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)=𝛿_𝑖𝑗.

For purposes of matrix-vector multiplication, the elements of ℛ^𝜔 are viewed as column vectors. If 𝑀 is a row-finite ℛ-matrix and 𝑥∈ℛ^𝜔, the product of 𝑀 and 𝑥 is denoted 𝑀𝑥 and is defined to be the sequence in ℛ^𝜔 with 𝑀𝑥(𝑖)=∑_𝑘∈𝜔𝑥(𝑘)𝑀(𝑖,𝑘). It is easy to see that 𝑀 induces an endomorphism of ℛ^𝜔 via such matrix-vector multiplication.

Theorem 4.1. Every row-finite infinite ℛ-matrix induces an endomorphism of ℛ^𝜔 via matrix-vector multiplication.

What is most interesting is the converse for slender rings:

Theorem 4.2. If ℛ is slender, then every endomorphism of ℛ^𝜔 is induced by multiplication of its vectors by a row-finite infinite ℛ-matrix.

Proof. Let 𝑔∈End_ℛ(ℛ^𝜔) and define an ℛ-matrix 𝑀 by 𝑀(𝑖,𝑗)=[𝑔(𝑒^𝑗)](𝑖) for 𝑖,𝑗∈𝜔. For each 𝑖, 𝜋_𝑖𝑔 is a homomorphism from ℛ^𝜔 to ℛ and so by slenderness is 0 for almost all 𝑒^𝑗,𝑗∈𝜔. But 𝜋_𝑖𝑔(𝑒^𝑗)=[𝑔(𝑒^𝑗)](𝑖)=𝑀(𝑖,𝑗) so that 𝑀 is row-finite.
By Theorem 4.1, 𝑀 induces an endomorphism of ℛ^𝜔. According to Theorem 3.3, to prove that multiplication by 𝑀 is the same as mapping by 𝑔, it suffices to demonstrate that 𝑀𝑒^𝑗=𝑔(𝑒^j) for all 𝑗; that is, that (𝑀𝑒^𝑗)(𝑖)=[𝑔(𝑒^𝑗)](𝑖). Now (𝑀𝑒^𝑗)(𝑖)=∑_𝑘∈𝜔𝑒^𝑗(𝑘)𝑀(𝑖,𝑘)=∑_𝑘∈𝜔𝛿_𝑗𝑘𝑀(𝑖,𝑘)=𝛿_𝑗𝑗𝑀(𝑖,𝑗)=𝑀(𝑖,𝑗)=[𝑔(𝑒^𝑗)](𝑖).

Corollary 4.3. If ℛ is a slender ring, then the endomorphism ring 𝐸𝑛𝑑_ℛ(ℛ^𝜔) under function addition and composition is isomorphic to the ring of row-finite infinite ℛ-matrices.

Proof. Note that the columns of the ℛ-matrix 𝑀 in the proof of Theorem 4.2 are the values of the endomorphism 𝑔 at the 𝑒^𝑛. 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 End_ℛ(ℛ^𝜔) under function addition and composition. It is easy to see that this correspondence preserves addition and multiplication.

5. Topologies for a Ring ℛ and Its Product ℛ^𝜔

The foregoing algebraic properties of slender rings can be characterized topologically. Let a ring ℛ 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.

Let ℛ^𝜔 be equipped with the product topology, that is, the topology in which the sets, 𝜋_𝑛⁻¹(𝑈),𝑛∈𝜔,𝑈⊆ℛ, form a subbasis. In fact, the sets 𝜋_𝑛⁻¹({𝑟}),𝑛∈𝜔,𝑟∈ℛ, form a subbasis. Under the product topology of ℛ^𝜔, the canonical projections are continuous open maps and addition and scalar multiplication are continuous.

This product topology is metrizable by defining 𝑑(𝑥,𝑥)=0 for all 𝑥∈ℛ^𝜔 and, for distinct 𝑥,𝑦∈ℛ^𝜔, 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∈ℛ^𝜔.

To see that 𝑑 induces the product topology on ℛ^𝜔, let 𝑥∈ℛ^𝜔 and consider the neighborhood of 𝑥, 𝑁={𝑦∈ℛ^𝜔∶𝑑(𝑥,𝑦)<2^−𝑛}. For all 𝑦∈𝑁, 𝑥(𝑖)=𝑦(𝑖),𝑖=0,…,𝑛, so that 𝑦∈𝑂=𝜋₀⁻¹({𝑥(0)})∩⋯∩𝜋_𝑛⁻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 of ℛ^𝜔,(𝑥^𝑛∶𝑛∈𝜔), converges to 𝑥∈ℛ^𝜔 if and only if the initial components of the 𝑥^𝑛 become the initial components of 𝑥 as 𝑛 becomes large. It readily follows that 𝑥^𝑛 converges to 𝑥 if and only if (𝑥−𝑥^𝑛∶𝑛∈𝜔) converges to 0∈ℛ^𝜔. Clearly the sequence (𝑒^𝑛) converges to 0.

For 𝑥=(𝑥(𝑖)∶𝑖∈𝜔)∈ℛ^𝜔, if 𝑥^𝑛∈ℱ,𝑛∈𝜔, is defined as 𝑥^𝑛(𝑖)=𝑥(𝑖) if 𝑖≤𝑛 and 𝑥^𝑛(𝑖)=0 for 𝑖>𝑛, then the sequence (𝑥^𝑛) converges to 𝑥. Thus ℱ is dense in ℛ^𝜔. If ^⟵𝑥^𝑛∈ℛ^𝜔,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 (𝑥^𝑛∶𝑛∈𝜔) in ℛ^𝜔 converges to an element of ℛ^𝜔, so that ℛ^𝜔 is a complete metric space. To see this, simply observe that, at each coordinate 𝑖, the 𝑥^𝑛(𝑖) become constant as 𝑛 becomes large, so that 𝑥^𝑛 converges to 𝑥 with 𝑥(𝑖)=lim_𝑛→∞𝑥^𝑛(𝑖).

For any homomorphism of topological modules, continuity on the entire domain is determined by continuity at 0. Thus, if ℎ is a homomorphism from ℛ^𝜔 to ℛ, then ℎ is continuous on all of ℛ^𝜔 if and only if it is continuous at 0. This is easy to see because if the sequence 𝑥^𝑛 converges to 𝑥∈ℛ^𝜔, then lim_𝑛→∞[ℎ(𝑥^𝑛)−ℎ(𝑥)]=lim_𝑛→∞ℎ(𝑥^𝑛−𝑥)=0 if and only if ℎ is continuous at 0. A similar result, of course, holds for any endomorphism of ℛ^𝜔.

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 ring ℛ from its product ℛ^𝜔 is uniformly continuous on ℛ^𝜔 if and only if it is continuous at 0. Ditto for an endomorphism of ℛ^𝜔.

Proof. Suppose that ℎ is a homomorphism from ℛ^𝜔 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 (𝑦^𝑛∶𝑛∈𝜔) converges to 𝑦∈ℛ^𝜔, so long as 𝑑(𝑦,𝑦^𝑛)<𝛿, then 𝑑(ℎ(y),ℎ(𝑦^𝑛))<𝜖. This holds because 𝑑(ℎ(𝑦),ℎ(𝑦^𝑛))=𝑑(ℎ(𝑦)−ℎ(𝑦^𝑛),0)=𝑑(ℎ(𝑦−𝑦^𝑛),0) and 𝑑(𝑦,𝑦^𝑛)=𝑑(𝑦−𝑦^𝑛,0). Thus ℎ is uniformly continuous.
The proof for endomorphisms of ℛ^𝜔 is essentially the same.

Theorem 5.2. If a ring ℛ is slender, then every homomorphism from its product ℛ^𝜔 to ℛ is uniformly continuous, as is every endomorphism of ℛ^𝜔.

Proof. Since 0 is trivial, let ℎ be a nonzero homomorphism from ℛ^𝜔 to ℛ and let (𝑥^𝑛) be a sequence in ℛ^𝜔, which converges to 0. According to Theorem 3.2, ℎ is a linear combination of canonical projections, the last one of which is some 𝑛_𝑘. For 𝑛 sufficiently large, the first 𝑛_𝑘 coordinates of the 𝑥^𝑛 are 0 and so ℎ(𝑥^𝑛)=0. Thus ℎ is continuous at 0 and hence uniformly continuous everywhere by Theorem 5.1.
Let 𝑔 be nonzero endomorphism of ℛ^𝜔 and let 𝑀 be the row-finite infinite ℛ-matrix which induces 𝑔. Again, it suffices to check convergence at 0, so let (x^𝑛) be a sequence of elements of ℛ^𝜔, which converges to 0. As 𝑛 gets large, the products 𝑀𝑥^𝑛 have initial coordinates =0 because 𝑀 is row-finite and the 𝑥^𝑛 have initial coordinates =0. Specifically, given any 𝑘∈𝜔, let 𝑛_𝑘 be the largest 𝑗 such that 𝑀(𝑖,𝑗)≠0,𝑖≤𝑘, and let 𝑚_𝑘∈𝜔 be such that coordinates 0-𝑛_𝑘 of 𝑥^𝑛 are equal to 0 for all 𝑛>𝑚_𝑘. Then coordinates 0-𝑘 of 𝑀𝑥^𝑛 are 0 for all 𝑛>𝑚_𝑘 so that lim_𝑛→∞𝑀𝑥^𝑛=0. Again Theorem 5.1 completes the proof.

For slender ℛ, the continuity of an endomorphism 𝑔 of ℛ^𝜔 at 0 is reflected by the row finiteness of the matrix 𝑀 inducing it. In particular, lim_𝑛→∞𝑒^𝑛=0 means that lim_𝑛→∞𝑀𝑒^𝑛=0. Since 𝑀𝑒^𝑛 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. Let ℛ be a ring with 1. Then a homomorphism ℎ from ℛ^𝜔 to ℛ satisfies ℎ(𝑒^𝑛)=0 for all but finitely many 𝑛 if and only if it satisfies lim_𝑛→∞ℎ(𝑒^𝑛)=0.

Proof. If ℎ(𝑒^𝑛)=0 for all 𝑛>𝑘, then 𝑑(ℎ(𝑒^𝑛),0)=0 for all 𝑛>𝑘, so that lim_𝑛→∞ℎ(𝑒^𝑛)=0.
Conversely, if lim_𝑛→∞ℎ(𝑒^𝑛)=0, there is a positive integer 𝑘 such that 𝑑(ℎ(𝑒^𝑛),0)<1 for all 𝑛>𝑘. But 𝑑(ℎ(𝑒^𝑛),0)<1 implies that ℎ(𝑒^𝑛)=0 because only 𝑥=0 satisfies 𝑑(𝑥,0)<1 in the metric on ℛ.

Corollary 5.4. A ring ℛ with 1 is slender if and only if every homomorphism ℎ from ℛ^𝜔 to ℛ satisfies lim_𝑛→∞ℎ(𝑒^𝑛)=0.

The next theorem is a casting of slenderness in topological terms.

Theorem 5.5. A ring ℛ with 1 is slender if and only if every homomorphism from ℛ^𝜔 to ℛ is continuous at 0.

Proof. If ℛ is slender, then according to Theorem 5.2, every homomorphism from ℛ^𝜔 to ℛ is certainly continuous at 0.
Conversely, if every homomorphism ℎ from ℛ^𝜔 to ℛ is continuous at 0, then lim_𝑛→∞ℎ(𝑒^𝑛)=0 because lim_𝑛→∞𝑒^𝑛=0. The slenderness of ℛ follows from Corollary 5.4.

The next theorem is the topological analog of Theorems 4.1 and 4.2.

Theorem 5.6. Let ℛ be a ring with 1. Then every endomorphism of its product ℛ^𝜔 is induced by a row-finite matrix if and only if every endomorphism of ℛ^𝜔 is continuous at 0.

Proof. If 𝑔 is an endomorphism of ℛ^𝜔, which is induced by a row-finite matrix 𝑀, and if (𝑥^𝑛) is a sequence in ℛ^𝜔, that converges to 0, then as 𝑛 get large, the initial coordinates of the 𝑥^𝑛 become 0. Since 𝑀 is row-finite, the initial coordinates of the 𝑀𝑥^𝑛 also become 0 so that (𝑀𝑥^𝑛)=(𝑔(𝑥^𝑛)) converges to 0, as in the proof of Theorem 5.2.
Conversely, suppose that every endomorphism of ℛ^𝜔 is continuous at 0, and let ℎ be a homomorphism from ℛ^𝜔 to ℛ. Define an endomorphism 𝑔 of ℛ^𝜔 by 𝑔(𝑥)=(ℎ(𝑥),ℎ(𝑥),…). By hypothesis, 𝑔 is continuous at 0. Since lim_𝑛→∞𝑒^𝑛=0, lim_𝑛→∞𝑔(𝑒^𝑛)=0; that is, the initial coordinates of 𝑔(𝑒^𝑛), which all =ℎ(𝑒^𝑛), must be 0 for 𝑛 sufficiently large. Thus lim_𝑛→∞ℎ(𝑒^𝑛)=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. Let ℛ be a ring with 1, equipped with the discrete topology, and let its product ℛ^𝜔 be equipped with the product topology. Then conditions (I)–(VI) are equivalent. (I)Every homomorphism ℎ from ℛ^𝜔 to ℛ satisfies ℎ(𝑒^𝑛)=0 for all but finitely many 𝑛.(II)ℛ is slender.(III)Every nonzero homomorphism ℎ from ℛ^𝜔 to ℛ has a unique expression of the form ℎ=𝜋_𝑛0ℎ(𝑒^𝑛0)+⋯+𝜋_𝑛𝑘ℎ(𝑒^𝑛𝑘), with 𝑘 and the 𝑛_𝑖 depending upon ℎ and each ℎ(𝑒^𝑛𝑖)≠0.(IV)Every endomorphism of ℛ^𝜔 is induced by multiplication of its vectors by a row-finite infinite matrix with entries from ℛ.(V)Every homomorphism from ℛ^𝜔 to ℛ is continuous at 0.(VI)Every endomorphism of ℛ^𝜔 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 ℛ^𝜔 to ℛ has the form stated, it is clear that ℎ(𝑒^𝑛)=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 ℛ^𝜔; by Theorem 5.1, it suffices to demonstrate continuity at 0. Suppose that (𝑥^𝑖) is a sequence in ℛ^𝜔, which converges to 0. Each 𝜋_𝑛𝑔 is a continuous homomorphism from ℛ^𝜔 to ℛ, which satisfies 𝜋_𝑛𝑔(𝑥^𝑖)=0 for all 𝑖> some 𝑘_𝑛∈𝜔. Given any 𝑙∈𝜔, there exist 𝑘_𝑛∈𝜔 such that 𝜋_𝑛𝑔(𝑥^𝑖)=0 for all 𝑖>𝑘_𝑛, 𝑛=0,…,𝑙. If 𝑚=max{𝑘₀,…,𝑘_𝑙}, then at least coordinates 0-𝑙 of 𝑔(𝑥^𝑖) are 0 for all 𝑖>𝑚; that is, lim_𝑖→∞𝑔(𝑥^𝑖)=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 Theorem 6.1 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 if ℛ is slender, then every homomorphism ℎ from ℛ^𝐼 to ℛ satisfies ℎ(𝑒^𝑖)=0 for all but finitely many 𝑖, for every infinite index set 𝐼. Obviously, if every homomorphism ℎ from ℛ^𝐼 to ℛ satisfies ℎ(𝑒^𝑖)=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 ℛ^𝐼 to ℛ, which annihilates all 𝑒^𝑖.

Whether uncountable products yield anything new depends upon the existence vel 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 {𝐽_𝑘∶𝑘∈𝐾} is a collection of mutually disjoint subsets of 𝐼 with |𝐾|<𝜅, then 𝜇(⋃_𝑘∈𝐾𝐽_𝑘)=∑_𝑘∈𝐾𝜇(𝐽_𝑘). 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. Let ℛ 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 ℛ^𝐼 to ℛ, such that ℎ(𝑒^𝑖)=0 for all 𝑖∈𝐼.

Proof. For every 𝑎=(𝑎(𝑖)∶𝑖∈𝐼)∈ℛ^𝐼 and 𝑟∈ℛ, define 𝑋_𝑟(𝑎)={𝑖∈𝐼∶𝑎(𝑖)=𝑟}. Then the 𝑋_𝑟(𝑎) are pairwise disjoint subsets of 𝐼 whose union is 𝐼. Since |ℛ|<𝜅, 1=𝜇(𝐼)=𝜇(⋃_𝑟∈ℛ𝑋_𝑟(𝑎))=∑_𝑟∈𝑅𝜇(𝑋_𝑟(𝑎)), so that exactly one of the sets, say 𝑋_𝑠(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(𝑒^𝑖)=0 for all 𝑖∈𝐼.
First, to see that 𝜂(𝑒^𝑖)=0 for all 𝑖∈𝐼, observe that 𝐼=𝑋₀(𝑒^𝑖)∪𝑋₁(𝑒^𝑖)={𝐼⧵{𝑖}}∪{𝑖}, so that 𝜇(𝑋₀(𝑒^𝑖))=1. Thus 𝜂(𝑒^𝑖)=0.
To confirm additivity, suppose that 𝜂(𝑎)=𝑠 and 𝜂(𝑏)=𝑡; then 𝜇({𝑖∈𝐼∶𝑎(𝑖)≠𝑠})=0 and 𝜇({𝑖∈𝐼∶𝑏(𝑖)≠𝑡})=0. Now 𝐼⊆(𝑋_𝑠(𝑎)∩𝑋_𝑡(𝑏))∪{𝑖∈𝐼∶𝑎(𝑖)≠𝑠}∪{𝑖∈𝐼∶𝑏(𝑖)≠𝑡} so that 1=𝜇(𝐼)≤𝜇(𝑋_𝑠(𝑎)∩𝑋_𝑡(𝑏))+𝜇{𝑖∈𝐼∶𝑎(𝑖)≠𝑠}+𝜇{𝑖∈𝐼∶𝑏(𝑖)≠𝑡}=𝜇(𝑋_𝑠(𝑎)∩𝑋_𝑡(𝑏))≤1, (7.1) so 𝜇(𝑋_𝑠(𝑎)∩𝑋_𝑡(𝑏))=1. Since 𝑋_𝑠(𝑎)∩𝑋_𝑡(𝑏)⊆𝑋_𝑠+𝑡(𝑎+𝑏), 1=𝜇(𝑋_𝑠(𝑎)∩𝑋_𝑡(𝑏))≤𝜇(𝑋_𝑠+𝑡(𝑎+𝑏))≤1; that is, 𝜇(𝑋_𝑠+𝑡(𝑎+𝑏))=1, so 𝜂(𝑎+𝑏)=𝑠+𝑡.
Finally, to check that 𝜂(𝑟𝑎)=𝑟𝑠=𝑟𝜂(𝑎) for 𝑟∈ℛ, note that 𝑋_𝑠(𝑎)⊆𝑋_𝑟𝑠(𝑟𝑎) so that 1=𝜇(𝑋_𝑠(𝑎))≤𝜇(𝑋_𝑟𝑠(𝑟𝑎))≤1 or 𝜇(𝑋_𝑟𝑠(𝑟𝑎))=1. Thus 𝜂(𝑟𝑎)=𝑟𝑠=𝑟𝜂(𝑎) and 𝜂 is an ℛ-homomorphism.

Remark 7.2. In the Proof of Proposition 7.1, not only is it true that 𝜂(ℱ)=0, but 𝜂(ℱ_𝜅)=0 where ℱ_𝜅 is the submodule of ℛ^𝐼 consisting of all 𝑥∈ℛ^𝐼 with 𝜇({𝑖∈𝐼∶𝑥(𝑖)≠0})=0. Thus 𝜂 has a large kernel.

Ulam has shown that if there exists an ℵ₁-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 ℛ^𝐼 to ℛ, satisfying ℎ(𝑒^𝑖)=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. Let ℛ be a ring with 1, equipped with the discrete topology, let 𝜎 be an uncountable ordinal, and let the product ℛ^𝜎 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 homomorphism ℎ from ℛ^𝜎 to ℛ has a unique expression of the form ℎ=𝜋_𝜄0ℎ(𝑒^𝜄0)+⋯+𝜋_𝜄𝑘ℎ(𝑒^𝜄𝑘), with 𝑘 and the 𝜄_𝑗 depending upon ℎ and each ℎ(𝑒^𝜄𝑗)≠0.(III)Every endomorphism of ℛ^𝜎 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 from ℛ^𝜎 to ℛ is continuous at 0.(V)Every endomorphism of ℛ^𝜎 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 Theorem 3.2 requires only that a homomorphism which is 0 on all 𝑒^𝑖 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 on ℛ^𝜎 is lost, the product topology has the property that an endomorphism 𝑔 of ℛ^𝜎 is continuous if and only if 𝜋_𝑖𝑔 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 let ℎ be a continuous homomorphism from ℛ^𝜎 to ℛ. Suppose that ℎ(𝑒^𝜄𝑛)≠0,𝜄_𝑛<𝜎, for all 𝑛∈𝜔. As in previous Sections, let ℛ^𝜔 have canonical projections 𝑝_𝑛 and let 𝑒^𝑛,𝑛∈𝜔, be the basis of the free submodule ℱ⊂ℛ^𝜔. Inject ℛ^𝜔 into ℛ^𝜎 via 𝑔 defined for 𝑥∈ℛ^𝜔 as [𝑔(𝑥)](𝜄_𝑛)=𝑥(𝑛),𝑛∈𝜔, and [𝑔(𝑥)](𝜄)=0 otherwise. This injection is continuous because 𝑝_𝑛=𝜋_𝜄𝑛𝑔 is the composition of continuous functions for all 𝑛∈𝜔. Now ℎ𝑔 is a continuous homomorphism from ℛ^𝜔 to ℛ and so, by Theorem 6.1, must be 0 on all but finitely many 𝑒^𝑛. Since 𝑔(𝑒^𝑛)=𝑒^𝜄𝑛, ℎ must be 0 except for those finitely many 𝑒^𝜄𝑛. Thus ℛ is slender and so (IV)⇒(I).
Finally, suppose that (II) holds and let ℎ=𝜋_𝜄0ℎ(𝑒^𝜄0)+⋯+𝜋_𝜄𝑘ℎ(𝑒^𝜄𝑘) be a homomorphism from ℛ^𝜎 to ℛ and let 𝑟∈ℛ; it suffices to address subbasic open sets. For any 𝑥∈ℛ^𝜎, 𝑥∈ℎ⁻¹({𝑟}) if and only if ℎ(𝑥)=𝑟; that is, if and only if 𝑥(𝜄₀)ℎ(𝑒^𝜄0)+⋯+𝑥(𝜄_𝑘)ℎ(𝑒^𝜄𝑘)=𝑟. Now 𝑥∈𝜋_𝜄⁻¹₀({𝑥(𝜄₀)})∩⋯∩𝜋_𝜄⁻¹_𝑘({𝑥(𝜄_𝑘)})⊆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 from ℛ^𝜎 to ℛ annihilates all 𝑒^𝜄, 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 Theorem 7.3 to use an ordinal for indexing, as any uncountable index set would suffice.

Acknowlegment

The author thanks Professor Phill Schultz of the University of Western Australia for discussions via email.