1. Terminology and Notation

Ring, module, and topological terminology and notation are standard. denotes a unital ring with identity , and denotes its countable product, that is, the set of all sequences of elements of . Of course . is both a left and right -module; it will be treated as a left -module unless otherwise stated. Results for left -modules carry over mutatis mutandis to right -modules.

The th coordinate of an element will be denoted . Let denote the submodule of , consisting of those elements of which have only finitely many nonzero coordinates. Let have coordinates when and otherwise; that is, , the Kronecker delta. is then the free submodule of , generated by the , which form a basis of .

Let denote the th canonical projection of onto ; that is, for , . Note that if and only if for all , and that .

is said to be slender if every -module homomorphism from to is on all but finitely many . As will be demonstrated, this condition determines the structure of the endomorphism ring of , .

As discussed in Section 2, the “original” slender ring was the integers . Familiar nonslender rings are the -adic integers for any prime . By using the method in [1, page 159, (d)] and known injective properties of the -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 and its countable product , 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 from the Baer-Specker group to satisfies for all but finitely many .

Proof. Suppose that for infinitely many ; it may be assumed that for all . There exists a prime, say , which does not divide . Define a sequence of elements of as follows, where .
Set and . Suppose that and the corresponding have been defined and primes have been chosen, with . Now choose a prime different from such that does not divide , . Then let be a multiple of such that is divisible by .
To see that the distinctiveness of the primes makes such a choice of possible, note that , so that the problem reduces to finding integers and that satisfy an integral equation of the form , in which and are relatively prime and . Being relatively prime, and generate all of , including , so that suitable k and indeed can be found. Set .
Let with . Now , so for all imply that is not divisible by . Thus . But for each , so that is divisible by because . The only integer divisible by an infinite number of primes is , so that , 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 from the Baer-Specker group to satisfies for all , then .

Proof. Clearly . Let and be distinct primes. Let be the subgroup of , consisting of all sequences of the form . The elements of are divisible by every power of . Thus necessitates , as otherwise would induce a nontrivial homomorphism from to , an impossibility in light of the divisibility of the elements of . Similarly, if consists of all sequences of the form , . Now each element can be written as with and , because for each , for some , so that . Thus and so .

Remark 2.3. Theorem 2.2 frequently is stated in terms of the homomorphism being on instead of just the . The important point is that the homomorphism is completely determined by its values on the (or on ).
The proofs of Theorems 2.1 and 2.2 use some number-theoretic properties of 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 [3, page 170].

Theorem 3.1. If is slender and is a homomorphism from to , such that for all , then .

Proof. Suppose that for some . Define a homomorphism from to by and for . Extend to an endomorphism of by defining . Now is a homomorphism from to such that for all , a contradiction to the slenderness of .

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

Theorem 3.2. If is slender, then every nonzero homomorphism from to has a unique expression of the form , with and the depending upon and each .

Proof. Because is slender, is except at some , in order if . Let . Now for all , so by Theorem 3.1, and thus has the form claimed. It is clear that cannot be expressed as any other linear combination of canonical projections with nonzero coefficients.

Note that if is slender, then every homomorphism from its product to is effectively a homomorphism from a finite subproduct to .

Theorem 3.3. If is a slender ring and is an endomorphism of , which satisfies for all , then .

Proof. If , then for some , so that for some . But would be a nonzero homomorphism from to , which is at each , thereby contradicting Theorem 3.1.

Theorem 3.3 simply says that, for a slender ring, each endomorphism of the product is determined by the endomorphism's values on the standard basis of , or on itself.

The results of this Section can be used to determine the structure of .

4. Infinite -Matrices and End

An infinite matrix over a ring is a -dimensional array of elements from , of the form . Such an infinite -matrix is said to be row-finite if, for each , for almost all ; that is, only finitely many for each .

If and are row-finite -matrices, their sum is defined to be the row-finite -matrix with entries . Their product is defined to be the row-finite -matrix with entries This order is required to encompass the case of noncommutative rings, when functions act from the left. The usual order of works when functions act from the right. Every such sum is finite because, for each , only finitely many are nonzero. Multiplication of row-finite -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 -matrices forms a ring with identity having entries .

For purposes of matrix-vector multiplication, the elements of are viewed as column vectors. If is a row-finite -matrix and , the product of and is denoted and is defined to be the sequence in with . It is easy to see that induces an endomorphism of via such matrix-vector multiplication.

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

What is most interesting is the converse for slender rings:

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

Proof. Let and define an -matrix by for . For each , is a homomorphism from to and so by slenderness is for almost all . But so that is row-finite.
By Theorem 4.1, induces an endomorphism of . According to Theorem 3.3, to prove that multiplication by is the same as mapping by , it suffices to demonstrate that for all ; that is, that . Now .

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

Proof. Note that the columns of the -matrix in the proof of Theorem 4.2 are the values of the endomorphism at the . is thus unique. There is then a - correspondence between the ring of row-finite infinite -matrices under addition and multiplication and under function addition and composition. It is easy to see that this correspondence preserves addition and multiplication.

5. Topologies for a Ring and Its Product

The foregoing algebraic properties of slender rings can be characterized topologically. Let a ring be equipped with the discrete topology, that is, the topology in which every subset is open. The singletons, , form a basis of this discrete topology, which is metrizable by defining for all and for distinct . Under the discrete topology of , addition and scalar multiplication are continuous.

Let be equipped with the product topology, that is, the topology in which the sets, , form a subbasis. In fact, the sets , form a subbasis. Under the product topology of , the canonical projections are continuous open maps and addition and scalar multiplication are continuous.

This product topology is metrizable by defining for all and, for distinct , if and first differ at . Thus, if , then for at least . In general, , an important fact used in the following discussions. Note also that for all and .

To see that induces the product topology on , let and consider the neighborhood of , . For all , , so that and so . Conversely, any satisfies , which means that so that .

A sequence of elements of ,, converges to if and only if the initial components of the become the initial components of as becomes large. It readily follows that converges to if and only if converges to . Clearly the sequence converges to .

For , if , is defined as if and for , then the sequence converges to . Thus is dense in . If , is defined as if and for , then the sequence converges to .

Observe that every Cauchy sequence in converges to an element of , so that is a complete metric space. To see this, simply observe that, at each coordinate , the become constant as becomes large, so that converges to with .

For any homomorphism of topological modules, continuity on the entire domain is determined by continuity at . Thus, if is a homomorphism from to , then is continuous on all of if and only if it is continuous at . This is easy to see because if the sequence converges to , then if and only if is continuous at . A similar result, of course, holds for any endomorphism of .

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 from its product is uniformly continuous on if and only if it is continuous at . Ditto for an endomorphism of .

Proof. Suppose that is a homomorphism from to , which is continuous at . Given , there exists such that provided . If converges to , so long as , then . This holds because and . Thus is uniformly continuous.
The proof for endomorphisms of is essentially the same.

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

Proof. Since is trivial, let be a nonzero homomorphism from to and let be a sequence in , which converges to . According to Theorem 3.2, is a linear combination of canonical projections, the last one of which is some . For sufficiently large, the first coordinates of the are and so . Thus is continuous at and hence uniformly continuous everywhere by Theorem 5.1.
Let be nonzero endomorphism of and let be the row-finite infinite -matrix which induces . Again, it suffices to check convergence at , so let be a sequence of elements of , which converges to . As gets large, the products have initial coordinates because is row-finite and the have initial coordinates . Specifically, given any , let be the largest such that , and let be such that coordinates - of are equal to for all . Then coordinates - of are for all so that . Again Theorem 5.1 completes the proof.

For slender , the continuity of an endomorphism of at is reflected by the row finiteness of the matrix inducing it. In particular, means that . Since is column of , the initial entries of the columns of (, , etc.) increasingly .

The ring properties previously described algebraically can be cast in topological terms.

Theorem 5.3. Let be a ring with . Then a homomorphism from to satisfies for all but finitely many if and only if it satisfies .

Proof. If for all , then for all , so that .
Conversely, if , there is a positive integer such that for all . But implies that because only satisfies in the metric on .

Corollary 5.4. A ring with is slender if and only if every homomorphism from to satisfies

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

Theorem 5.5. A ring with is slender if and only if every homomorphism from to is continuous at .

Proof. If is slender, then according to Theorem 5.2, every homomorphism from to is certainly continuous at .
Conversely, if every homomorphism from to is continuous at , then because . The slenderness of follows from Corollary 5.4.

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

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

Proof. If is an endomorphism of , which is induced by a row-finite matrix , and if is a sequence in , that converges to , then as get large, the initial coordinates of the become . Since is row-finite, the initial coordinates of the also become so that converges to , as in the proof of Theorem 5.2.
Conversely, suppose that every endomorphism of is continuous at , and let be a homomorphism from to . Define an endomorphism of by . By hypothesis, is continuous at . Since , ; that is, the initial coordinates of , which all , must be for sufficiently large. Thus , so by Corollary 5.4, 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 be a ring with , equipped with the discrete topology, and let its product be equipped with the product topology. Then conditions (I)–(VI) are equivalent. (I)Every homomorphism from to satisfies for all but finitely many .(II) is slender.(III)Every nonzero homomorphism from to has a unique expression of the form , with and the depending upon and each .(IV)Every endomorphism of is induced by multiplication of its vectors by a row-finite infinite matrix with entries from .(V)Every homomorphism from to is continuous at .(VI)Every endomorphism of is continuous at .

Proof. (I)(II): This equivalence is simply definitional.
(II)(III): This is Theorem 3.2.
(III)(II): If every homomorphism from to has the form stated, it is clear that for all but finitely many . 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 be an endomorphism of ; by Theorem 5.1, it suffices to demonstrate continuity at . Suppose that is a sequence in , which converges to . Each is a continuous homomorphism from to , which satisfies for all some . Given any , there exist such that for all , . If , then at least coordinates - of are for all ; that is, so that is continuous at .
(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 " 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)] using “continuous” instead of “at ”.

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 is slender, then every homomorphism from to satisfies for all but finitely many , for every infinite index set . Obviously, if every homomorphism from to satisfies for all but finitely many , for some infinite index set , then is slender. The problem for uncountable products, even when is slender, is that the analog of Theorem 3.1 may not hold; that is, there may exist a nonzero homomorphism from to , which annihilates all .

Whether uncountable products yield anything new depends upon the existence vel non of measurable cardinals. An uncountable cardinal number is said to be -measurable if there exists a set of cardinality and a measure on the subsets of , which (i) assumes only the values ; (ii) satisfies for all , and ; and (iii) is -additive in the sense that if is a collection of mutually disjoint subsets of with , then . 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 [1, page 161, Remark].

Proposition 7.1. Let be a ring with . If there exists a -measurable cardinal satisfying , and is an index set of cardinality and is a -additive measure on , then there exists a nonzero homomorphism from to , such that for all .

Proof. For every and , define . Then the are pairwise disjoint subsets of whose union is . Since , , so that exactly one of the sets, say , has measure . Set . Using the properties of , it will be shown that preserves addition and scalar multiplication and for all .
First, to see that for all , observe that , so that . Thus .
To confirm additivity, suppose that and ; then and . Now so that so . Since , ; that is, , so .
Finally, to check that for , note that so that or . Thus and is an -homomorphism.

Remark 7.2. In the Proof of Proposition 7.1, not only is it true that , but where is the submodule of consisting of all with . Thus has a large kernel.

Ulam has shown that if there exists an -measurable cardinal (the meaning of which should be clear), then there is a least one, call it , and it in fact is -measurable [5]. Thus often is referred to as the least measurable cardinal. Lady has shown that if is slender and is a homomorphism from to , satisfying for all , then , provided [4, Theorem (4)]. If there are no measurable cardinals, it then follows that, for slender , such a homomorphism is zero regardless of the cardinality of .

The state of uncountable products of rings can be summarized as follows.

Theorem 7.3. Let be a ring with , equipped with the discrete topology, let be an uncountable ordinal, and let the product be equipped with the product topology. If there are no measurable cardinals, then conditions (I)–(V) are equivalent. (I) is slender.(II)Every nonzero homomorphism from to has a unique expression of the form , with and the depending upon and each .(III)Every endomorphism of is induced by multiplication of its vectors by a row-finite infinite matrix with entries from and with rows and columns indexed by .(IV)Every homomorphism from to is continuous at .(V)Every endomorphism of is continuous at . If there exists a least measurable cardinal , then conditions (I)–(V) are equivalent for all ordinals of cardinality .

Proof. Assume that no measurable cardinal exists. Note that the proof of Theorem 3.2 requires only that a homomorphism which is on all itself be zero, a result which follows from [4, Theorem (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 is lost, the product topology has the property that an endomorphism of is continuous if and only if is continuous for all . Continuity at 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 be a continuous homomorphism from to . Suppose that , for all . As in previous Sections, let have canonical projections and let , be the basis of the free submodule . Inject into via defined for as , and otherwise. This injection is continuous because is the composition of continuous functions for all . Now is a continuous homomorphism from to and so, by Theorem 6.1, must be on all but finitely many . Since , must be except for those finitely many . Thus is slender and so (IV)(I).
Finally, suppose that (II) holds and let be a homomorphism from to and let ; it suffices to address subbasic open sets. For any , if and only if ; that is, if and only if . Now so that is open and 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 to annihilates all , and again [4, Theorem (4)] supplies the proof for .