# Endomorphisms and Product Bases of the Baer-Specker Group

Authors: E. F. Cornelius
Journal: International Journal of Mathematics and Mathematical Sciences
 (2009)

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## Abstract

The endomorphism ring of the group of all sequences of integers, the Baer-Specker group, is isomorphic to the ring of row finite  infinite matrices over the integers. The product bases of that group are represented by the multiplicative group of invertible elements in that matrix ring. All products in the Baer-Specker group are characterized, and a lemma of László Fuchs regarding such products is revisited.


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## Body

## 1. Introduction

Abelian group notation and terminology are standard [1, 2]. ℕ denotes the nonnegative integers and ℕ+ the positive integers. ℤ denotes the ring of all integers. Let P=ℤℕ=∏ℕℤ, the group of all integer sequences, known as the Baer-Specker group [3, 4]. For purposes of matrix vector multiplication, an element x of P is viewed as a column vector, the  ith entry of which is denoted xi.


## 2. Infinite Integral Matrices and Their Operations

To prove the results claimed, some properties of infinite integral matrices are needed. These properties are stated without proof, as they are known or easily proved.The group of all infinite integral matrices is denoted byℳ, with addition being defined in the usual way. Matrix rows and columns are indexed by ℕ. Column operations (multiplying a column by -1, interchanging two columns, adding an integral multiple of one column to another) are carried out in the usual way.F∈ℳ is said to be row finite if, for each i∈ℕ,Fij=0 for all but finitely many j∈ℕ; that is, if for each i, there exists ki∈ℕ such that Fij=0 unless  j≤ki. A column finite matrix is defined analogously. The additive group of all row finite matrices in ℳ is denoted by ℱ. T∈ℳ is said to be lower triangular if Tij=0 unless i≥j. The additive group of all lower triangular matrices in ℳ is denoted by 𝒯. Obviously 𝒯⊂ℱ.Recall that multiplication of two infinite matrices may not be well defined and, even when defined, may not be associative. For examples of pathologies of infinite matrices, see Section5. However, for F∈ℱ and G∈ℳ, the product of F and G, denoted FG∈ℳ, is defined as(2.1)FGij=∑k∈ℕFikGkj,i,j∈ℕ.
Because F is row finite, the sum reduces to  ∑k≤niFikGkj  where ni∈ℕ. Under multiplication thus defined, ℱ and 𝒯  are rings with identity I satisfying Iij=δij, the Kronecker delta.F∈ℱ is said to be invertible if there exists G∈ℱ satisfying FG=I=GF. It clear that such a G is a unique two-sided inverse, so that F-1=G is well defined [5, pages 21–25]. The set of invertible matrices in ℱ is denoted 𝒢ℒ and forms a group under matrix multiplication.Proposition 2.1.
IfF′∈ℱ is obtained from F∈ℱ through a column operation, then F′=FE where E is obtained by performing the same operation on I. E is both row and column finite and invertible.This corollary is immediate.Corollary 2.2.
IfF′∈ℱ is obtained from F∈ℱ through a finite sequence of column operations, then F′=FE where E is obtained by performing the same sequence of operations on I. E is both row and column finite and invertible.LetU={uj:j∈ℕ} be a subset of P. If each expression ∑j∈ℕbjuj,bj∈ℤ, represents an element of P, then the set of all such elements forms a subgroup of P, denoted ∏j∈ℕ〈uj〉 and called a product in P [2, page 164]. Such a subset U forms a product if and only if the matrix with columns uj is row finite. If the representations are unique, then the uj are independent and U is said to be a basis of the product. One of the objectives of this paper is to characterize products and product bases in term of endomorphisms of P.ForF∈ℱ and x∈P, the matrix vector product of F and x, denoted simply Fx, is defined as Fx∈P, with Fxi=∑k∈ℕFikxk,i∈ℕ. As with matrix multiplication, because F is row finite, the sum reduces to ∑k≤niFikxk  where ni∈ℕ.  Each such F naturally induces an endomorphism of P, via matrix vector multiplication. Indeed, a product in P, ∏j∈ℕ〈uj〉, is simply the image of such an endomorphism, where the matrix F is constructed with columns uj,j∈ℕ.All endomorphisms ofP   are determined by their values on 𝚺, the free subgroup of P consisting of sequences which are 0 after awhile [2, Lemma  94.1]. The standard group basis of 𝚺 is {ej:(ej)i=δij,i,j∈ℕ}, and the ej also form the standard product basis of P. In particular, all endomorphisms of P are  determined by their values on  this standard basis. As a result, with each element f  in the endomorphism ring of P, End(P), is associated a unique matrix Mf∈ℳ, given by Mijf=f(ej)i,i,j∈ℕ.  The immediate goal is to prove that each such Mf is row finite, so that every endomorphism of P is effectively multiplication by a row finite matrix.


## 3. Endomorphisms and Product Bases

The main theorem  follows.Theorem 3.1.
Every endomorphism ofP is induced by the action of a row finite, infinite, integral matrix.Proof.
Letf∈End(P), and let Mf∈ℳ be the matrix associated with f, Mijf=f(ej)i,   i,j∈ℕ. Suppose that Mf is not row finite; that is, suppose there are i and distinct jk∈ℕ such that Mijkf≠0 for all  k∈ℕ. Let πi denote the canonical projection of the ith row of elements of P onto ℤ. Then πif  would be a homomorphism  of P into ℤ, which is nonzero at all ejk, k∈ℕ, an impossibility   [4, Satz III].   Hence it is not possible that an infinite number of the Mijf are nonzero; that is, Mf must be row finite.
To see thatMf acts on P to produce f, it suffices to check agreement on the standard basis of 𝚺. For i,j∈ℕ, (Mfej)i=∑k∈ℕMikf(ej)k=∑k∈ℕf(ek)iδjk=f(ej)i.Corollary 3.2.
(i)  The endomorphism ring ofP,  End(P), is isomorphic to the ring of row finite, infinite, integral matrices ℱ.  (ii)  The automorphism group of P, Aut(P), is isomorphic to the multiplicative group of invertible matrices in ℱ, 𝒢ℒ.Proof.
Forf∈End(P), f→Mf is a ring isomorphism with ℱ, which maps Aut(P) isomorphically onto 𝒢ℒ. The row finiteness of the elements of ℱ ensures that all sums are finite and that matrix multiplication is associative.Corollary 3.3.
There is a one-to-one correspondence between the matrices of𝒢ℒ and the product bases of P.Proof.
Let{uj:j∈ℕ} be a product basis for P, and let A∈Aut(P) be defined by A(x)i=ai for x∈P, when x=∑j∈ℕajuj,  aj∈ℤ. The uniqueness of expression of the elements of P in terms of the product basis guarantees that A is an automorphism. By Corollary 3.2(ii), A corresponds to a unique element of 𝒢ℒ. The converse is clear.


## 4. Products in P

Topological techniques are helpful in studying the endomorphisms ofP. Background material may be found in [6–8] and the references cited therein. For x≠y∈P, the distance between them, d(x,y), is defined to be 2-n, where n is the first place at which xn≠yn. Of course, d(x,x)=0. It is easy to check that d is a metric on P, which comports with the product topology on P, when ℤ is discrete.P is a separable metric space having 𝚺 as a countable dense subset. Finally, P is a complete metric space in which Cauchy sequences eventually become constant pointwise.The important aspect of this topology is that all endomorphisms ofP are continuous.Similarly,ℳ is a complete  separable metric space with the distance between distinct matrices F and G defined as 2-n, where n=min{i,j:Fij≠Gij}.Theorem 4.1.
Every product inP can be generated by a lower triangular matrix, the columns of which form a basis of the product.Proof.
LetF∈ℱ generate the product FP. The proof proceeds by transforming F via column operations into a lower triangular matrix T satisfying TP=FP. Since F=0 is trivial, assume F≠0. Let y∈FP, 0≠y=(y0,y1,…)t, where t denotes transpose, and suppose y=Fx for x=(x0,x1,…)t∈P.
If row0 of F consists entirely of 0's, simply set T1=F, E1=I, and x1=x and proceed to the next step. Otherwise, perform column operations on the nonzero entries of row 0 of F to obtain a new matrix T1 in which T0j1=0 for all j>0. This is tantamount to finding  the gcd of the entries in row 0, columns j≥0 of F. Perform the same column operations on I  to produce E1, so that T1=FE1 with E1 an invertible row and column finite matrix. Let G1 denote the inverse of E1. Since F=T1G1, y=Fx=T1(G1x). Set x1=G1x=(x01,x11,…)t so that y=T1x1.
IfT1j1=0 for all j≥1, simply set T2=T1, x2=x1, and E2=I  and proceed. Otherwise, perform column operations on the nonzero entries of row 1 in columns j≥1 of T1 to obtain a new matrix T2 in which T1j2=0 for all j>1. As before, this is tantamount to finding  the gcd of the entries in row 1, columns j≥1 of T1. Note that Ti02=Ti01 for all i; that is, column 0 of T2 is the same as column 0 of T1.
Perform the same column operations onI  to obtain E2, so that T2=T1E2 with E2 an invertible row and column finite matrix. Note that since no operation was performed on column 0 of T1, none was performed on column 0 of I. Moreover, rows 0 of T1 and I remained unchanged (because of 0 entries), so that E002=1 and the rest of the row 0 and column 0 entries of E2 are 0. Thus the same is true of G2, the inverse of E2. Since F=T2G2G1, y=Fx=T2G2(G1x)=T2(G2x1). Set x2=G2x1  so that y=T2x2. Because row 0 of G2 is 1,0,…, it follows that x02=x01.
Suppose that aftern>1 iterations, row finite matrices T1,…,Tn and invertible row and column finite matrices E1,…,En with inverses G1,…,Gn and x1,…,xn∈P have been obtained, such that(a)
Tijn=0 for all j>i, 0≤i≤n-1; that is, the matrices Tn are becoming increasingly lower triangular;(b)
columnj of Tn is identical to column j of Tn-1 for 0≤j<n-1; that is, after the nth  iteration, rows and columns <n do not change;(c)
Tn=Tn-1En=FE1⋯En;(d)
Eijn=δij=Gijn, 0≤i,j<n-1; that is, the En's and their inverses are becoming increasingly diagonal with 1's down the diagonal, tending toward the identity matrix;(e)
xn=Gnxn-1,  y=Tnxn; that is, y is always in the image of Tn;(f)
xin=xin-1,  0≤i<n-1; that is, after the nth iteration, the entries <n of the xn's do not change.
As before, ifTnjn=0 for all j≥n, simply set Tn+1=Tn, En+1=I, and xn+1=xn and continue. Otherwise, perform column operations on the nonzero entries of the nth row in columns j≥n of Tn to obtain a new matrix Tn+1 in which Tnjn+1=0 for all j>n. Because no operation is performed on columns 0 through (n-1), Tijn+1=Tijn,  i∈ℕ,  j=0,…,n-1; that is, columns 0 through (n-1) of Tn+1 are the same as columns 0 through (n-1) of Tn.
Perform the same column operations onI to produce En+1, so that Tn+1=TnEn+1 with En+1 an invertible row   and column finite  matrix having inverse Gn+1. Note that since no operation was performed on columns 0-(n-1) of Tn, none was performed on columns 0-(n-1) of I. Moreover, rows 0-(n-1) of Tn were not changed (because of 0's), and the same is true of I and Gn+1. (i)  As a result, Eijn+1=δij=Gijn+1, i,j<n. Since Tn+1=TnEn+1=(FE1⋯En)En+1, F=Tn+1Gn+1⋯G1, y=Fx=Tn+1Gn+1⋯G1x=Tn+1(Gn+1xn). (ii)  Set xn+1=Gn+1xn so that y=Fn+1xn+1. (i) and (ii) imply that  xin+1=xin,  i=0,…,n-1.
The Cauchy sequencexn converges to some x′∈P because P is a complete metric space. Since all endomorphisms of P are continuous, the sequence Tnxj converges to Tn(x′) for each n. Because of (b) and (f), the convergence is uniform. The Cauchy sequence of matrices Tn thus obtained converges to a lower triangular matrix T and Tx′=y so that TP⊇FP. From (b) it follows that  TP⊆FP and so   TP=FP.
It may, of course, happen that some of the diagonal terms ofT are 0 so that the first nonzero entry in a column may be below the diagonal. Should that occur, the first nonzero entry in the next column will be in a lower row. If  T has only finitely many nonzero columns, the result is elementary. If all columns of  T are nonzero, it is clear that an infinite linear combination of its columns can be 0only if all coefficients are 0, so that the columns of  T form a product basis.Remark 4.2.
From the results in Section3, a matrix ∈ℱ can be classified as epic or monic if the endomorphism of P which it induces is epic or monic, respectively. (1)  If F∈ℱ has a 0 row i, it cannot be epic because ei cannot be reached. If it has a 0 column j, it cannot be monic because ej is in the kernel. Here E={ei:i∈ℕ} denotes the usual basis of 𝚺⊂P.  (2)  Similarly, if a row (column) of F is a multiple of another row (column), then F cannot be epic (monic).  (3)  T∈𝒯  is invertible (2 sided) over the rationals if and only if each diagonal entry is nonzero [5, pages 19-20 ]. (4)  T  is integrally invertible if and only if each diagonal entry =±1.Lemma 4.3.
T∈𝒯  is epic if and only if each diagonal entry is ±1.Proof.
If each diagonal entry ofT  is ±1, then T∈𝒢ℒ and so is epic. Conversely let y=(y0,y1,…)t be a nonzero element of P. To solve Tx=T(x0,x1,…)t=y, induct on n. Let Tn denote the finite matrix (Tij:i,j=0,1,…,n). If n=0 and T00≠±1, then T00x0=y0 is not solvable in general. If T00=±1, then x0 is uniquely determined. Suppose that the diagonals Tkk=±1 for k=0,1,…,n, and suppose further that Tn(x0,…,xn)t=(y0,…,yn)t is uniquely solvable, (x0,…,xn)t=Sn(y0,…,yn)t, where Sn denotes  the inverse of Tn. Express
(4.1)Tn+1=(Tn0Tn+1,0⋯Tn+1,n+1)
so that
(4.2)Tn+1(x0,…,xn,xn+1)t=(Tn(x0,…,xn)t∑j=0j=n+1Tn+1,jxj),
with some abuse of notation for the sake of simplicity. Now by the induction hypothesis, (x0,…,xn)t=Sn(y0,…,yn)t uniquely. Thus the only solution to ∑j=0j=n+1Tn+1,jxj=yn+1 must come from Tn+1,n+1xn+1=yn+1-∑j=0j=nTn+1,jxj, which is impossible in general unless Tn+1,n+1=±1.Invertible lower triangular infinite integral matrices may therefore be characterized as follows.Theorem 4.4.
ForT∈𝒯  the following are equivalent.(1)
T  is epic.(2)
Tkk=±1 for all k∈ℕ.(3)
T∈𝒢ℒ.By now the following result is clear.Corollary 4.5.
T∈𝒯  is monic if and only if Tkk≠0 for all k∈ℕ.


## 5. Infinite Matrix Examples and Counterexamples

The next examples and counterexamples serve to justify some of the assumptions about matrix rings made in this paper. They illustrate some of the pitfalls and limitations of working with infinite matrices and further illustrate results from Section4. They are not new but continue to be discussed in various contexts: [9, section 2]; [10, page 437]; [7, pages 189–191].As before, the elements ofP are viewed as column vectors and E = {ej:j∈ℕ} denotes the usual basis of 𝚺⊂P; E is also the standard product basis of P. The same symbol E is used for the matrix with the ej as columns, that is, the identity matrix. The same liberty is taken with other bases and their matrices.Example 5.1.
E1={e0+e1+⋯+ei:i∈ℕ} also is a basis of 𝚺, having matrix
(5.1)E1=(1111⋯0111⋯0011⋯0001⋯⋮⋮⋮⋮⋱).
Because the matrix is not row finite, E1 cannot be a product basis of P.Example 5.2.
E2={e0,ei-ei-1:i∈ℕ+} also is a basis of 𝚺, having matrix
(5.2)E2=(1-100⋯01-10⋯001-1⋯0001⋯⋮⋮⋮⋮⋱).
Note that E2 is not lower triangular.Example 5.3.
Let1∈P  denote  the column vector having all entries =1 and let E3={e0+e2+⋯+ei-1-1:i∈ℕ+}. Note that E3 is not a basis of 𝚺; indeed, no element is even contained in 𝚺. The accompanying lower triangular matrix is
(5.3)E3=(0000⋯-1000⋯-1-100⋯-1-1-10⋯⋮⋮⋮⋮⋱).
The following relationships hold:(5.4)E1E2=E=E2E1,E2E3=E.
Observe that, although both E2 and E3 are row finite and even pure independent, E2 is not monic (1 is in the kernel), and E3 is not epic (e0 is not attained). In particular, (E1E2)1=E1=1, but E1(E21)=E1(0)=0, so that multiplication is not associative. Despite the fact that E2P=P, E2 nevertheless cannot serve as a product basis for P. The problem, of course, is that E2 is not invertible in ℱ. Restriction to multiplicatively associative matrix rings helps to eliminate much of this infinite matrix pathology [5, pages 19–22].
In the proof of Theorem4.1, although the columns of the resulting lower triangular matrix T obviously are independent, T nevertheless may be singular, as illustrated by E3 above. Moreover, it may not be the case  that the products E1⋯En in that proof will converge to a row finite matrix, as further illustrated by E2. Even though E2 itself is not monic, the resulting lower triangular matrix E, which results from column operations on E2, is monic.


## 6. Fuchs' Lemma  95.1 Redux

Products inP were introduced in [2] by beginning with a countable subset {xn:n∈ℕ}⊂P, with no condition other than that all sums of the form x=∑n=0∞snxn,sn∈ℤ, are well defined. As mentioned, this means that the infinite matrix with columns xn must be row finite.  Lemma  95.1 then was stated as follows in [2].LetX be a product in P. There are elements an∈P and integers kn(n∈ℕ) such that P=∏n=0∞〈an〉 and X=∏n=0∞〈knan〉 where (•) kn∣kn+i if kn≠0 for all n,i≥0.The goal was to establish a product analog of the well-known result for stacked bases of free groups of finite rank [1, Lemma  15.4]. However, a counterexample has been produced [11].Perhaps another approach is to utilize the weaker definition of stacked bases found in [12], which does not require divisibility as in (•). From Theorem 4.1, there would be no loss of generality in assuming that the xn∈P form a product basis and a lower triangular matrix.Lemma 6.1.
Let∏n=0∞〈xn〉 be a product  of infinite rank in P, let S=∑n=0∞〈xn〉, and let S* be the pure subgroup which the xn generate. Then there exist a product basis of P=∏n=0∞〈an〉 and positive integers kn(n∈ℕ) such that ∏n=0∞〈knan〉=∏n=0∞〈xn〉 if and only if there exist stacked bases of S and S* such that the matrix F* formed by the basis of S* is invertible and the matrix F formed by the basis of S satisfies FP=∏n=0∞〈xn〉.Proof.
If{an:n∈ℕ} is a product basis of P, then the an's are pure independent. If {kn:n∈ℕ)} are positive integers, then ∏n=0∞〈knan〉 is a product in P and if S=⊕n=0∞〈knan〉, then S*=⊕n=0∞〈an〉 so that {knan:n∈ℕ} and {an:n∈ℕ} are stacked bases for the respective subgroups of P. The matrix formed by the an's is in 𝒢ℒ by Corollary 3.3; the matrix F formed by the xn=knan certainly satisfies FP=∏n=0∞〈xn〉.
Conversely, let∏n=0∞〈xn〉 be a product in P, let S be the subgroup of P which the xn's generate, and let S* be the pure subgroup which they generate. Suppose there are stacked bases {knan:kn, n∈ℕ} of S and {an:n∈ℕ} of S*; because S*/S is torsion, all kn's must be positive. Suppose further that the matrix formed by the an's is invertible; then the an's are a product basis of P and ∏n=0∞〈knan〉 is a product in P. Finally, if the matrix F formed by the knan's satisfies FP = ∏n=0∞〈xn〉, the proof is concluded.Remark 6.2.
(1)  If∏n=0∞〈xn〉 is a product in P and {an:n∈ℕ} is a basis of the subgroup generated by the xn's, the matrix formed by the an's need not  be row finite. (2)  As previously noted, the lower triangular matrix T, obtained from the  column operations in Theorem 4.1, may be singular. However, Nunke   [8, page 199]  has shown that every endomorphic image of P is either free of finite rank or isomorphic to P itself. In the latter case, there then exists monic F∈ℱ satisfying FP = TP. Column operations will, of course, reveal matrix rank, finite or infinite.All products inP may therefore be characterized as follows.Theorem 6.3.
Every product inP is generated by a lower triangular, infinite, integral matrix, the columns of which form a product basis and the rank of which determines whether the product is free of finite rank or isomorphic to P.Generalizations
The author's initial draft of this paper was done for higher dimensions, along the lines of [13]. At the referee's suggestion, matrix dimension has been restricted to 2, to improve readability.



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