Journal of Algebra and Its Applications c World Scientific Publishing Company

POLYNOMIALS THAT KILL EACH ELEMENT OF A FINITE RING

NICHOLAS J. WERNER Department of Mathematics, University of Evansville, 1800 Lincoln Avenue Evansville, IN 47722, USA [email protected] Received (Day Month Year) Revised (Day Month Year) Accepted (Day Month Year) Communicated by (xxxxxxxxx) Given a finite (associative, unital) ring R, let K(R) denote the set of polynomials in R[x] that send each element of R to 0 under evaluation. We study K(R) and its elements. We conjecture that K(R) is a two-sided ideal of R[x] for any finite ring R, and prove the conjecture for several classes of finite rings (including commutative rings, semisimple rings, local rings, and all finite rings of odd order). We also examine a connection to sets of integer-valued polynomials. Keywords: Finite ring; integer-valued polynomial 2000 Mathematics Subject Classification: 16P10, 13F20, 11C99

Notation and Conventions All rings under consideration are associative and contain unity, but need not be commutative. Given a ring A, A[x] denotes the ring of polynomials in one indeterminate with coefficients in A. We assume that the indeterminate x commutes with all elements of A and that polynomials are evaluated with x on the right. If f (x), g(x) ∈ A[x], then (f g)(x) denotes the product of f (x) and g(x) in A[x]. So, (f g)(x) always equals f (x)g(x); note, however, that if A is noncommutative P and a ∈ A, then (f g)(a) need not equal f (a)g(a). If f (x) = i ai xi , then we may P P express (f g)(x) as (f g)(x) = i ai g(x)xi , and we have (f g)(a) = i ai g(a)ai for any a ∈ A. In particular, if g is central, then (f g)(a) = f (a)g(a). Unless otherwise noted, ideals are two-sided. 1. Introduction Let D be a commutative integral domain with field of fractions k. Then, the ring of integer-valued polynomials over D, denoted by Int(D), is defined to be Int(D) = {f ∈ k[x] | f (D) ⊆ D}. The ring Int(D) has been extensively studied; the book 1

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[3] is a good reference. Recently, analogous constructions have been studied for noncommutative rings ([8], [9], [4]). We begin by considering such a construction for a torsion-free Z-algebra that is finitely generated as a Z-module. Definition 1.1. Let A be a torsion-free Z-algebra that is finitely generated as a Z-module, and let B = Q ⊗Z A be the extension of A to a Q-algebra. By identifying A and Q with their isomorphic copies in B, we can evaluate polynomials in B[x] at elements of A. We define Int(A) to be Int(A) = {f ∈ B[x] | f (A) ⊆ A}, called the set of integer-valued polynomials over A. Remark 1.2. The reason we require A to be torsion-free is so that A contains a copy of Z and B contains a copy of Q. The fact that A is finitely generated guarantees that the residue ring A/nA is finite for any integer n > 1; this will come into play shortly when we investigate a connection between Int(A) and finite rings. We will always assume that A is a Z-algebra, but much of our work will carry over to the more general case where A is an algebra over a commutative integral domain with all residue rings finite. It is easy to check that the set Int(A) always has the structure of a left A[x]module. However, when B is noncommutative, Int(A) consists of polynomials with non-commuting coefficients, and it is nontrivial to determine whether Int(A) is closed under multiplication (because, in general, (f g)(a) will not equal f (a)g(a)). We are interested in situations where Int(A) is closed under multiplication, and thus has a ring structure. The following theorem gives a sufficient condition for this to occur. Theorem 1.3 ([8] Thm. 1.2). Let A and B be as in Definition 1.1. Assume that P each a ∈ A may be written as a finite sum a = i ci ui for some ci , ui ∈ A such that each ui is a unit in A and each ci is central in B. Then, Int(A) is closed under multiplication, and hence is a ring. Theorem 1.3 applies to several common types of rings, such as the matrix rings Mn (Z) and the group algebras ZG, where G is a finite group. In this paper, we will use a connection between Int(A) and polynomials over finite rings to give a different characterization of when Int(A) is a ring. 2. A Connection to Finite Rings Given A and B as in Definition 1.1, any polynomial f (x) ∈ B[x] may be written as f (x) = g(x) n , where g(x) ∈ A[x] and n is a positive integer. Then, f (x) ∈ Int(A) ⇐⇒ g(a) ∈ nA for all a ∈ A ⇐⇒ g(a) = 0 in A/nA for each residue a ∈ A/nA. This leads us to the following definition. Definition 2.1. Let R be a nonzero finite ring. We define K(R) = {f (x) ∈ R[x] | f (a) = 0 for all a ∈ R}, the set of polynomials in R[x] that kill each element of R.

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We formalize the connection between Int(A) and K(R) in the following correspondence, which we will use freely. Correspondence. Let A be a torsion-free Z-algebra that is finitely generated as a Z-module and let B = Q ⊗Z A. Let f (x) ∈ B[x] and write f (x) = g(x) n for some g ∈ A[x] and some integer n > 0. Then, f ∈ Int(A) if and only if g ∈ K(A/nA). While we have defined the sets Int(A) and K(R), we have not mentioned whether they contain any nontrivial elements. An easy computation involving K(R) answers this question in the affirmative. Proposition 2.2. (1) Let R be a finite ring. Then, K(R) 6= {0}; in fact, K(R) contains a monic polynomial that is central in R[x] (2) Let A be a torsion-free Z-algebra that is finitely generated as a Z-module. Then, A[x] $ Int(A). Proof. For (1), consider a ∈ R. Since R is finite, the elements produced by taking powers of a will eventually repeat. That is, there exist positive integers n1 > n2 such that an1 − an2 = 0. It follows that xn1 − xn2 is a monic, central polynomial that kills a. Taking the product of all such polynomials as a runs across R will produce the desired element of K(R). For (2), we always have A[x] ⊆ Int(A). Since A is finitely generated, A/nA is a finite ring for each integer n > 1. Taking g ∈ K(A/nA) \ {0}, we have g(x) n ∈ Int(A) \ A[x]. As with Int(A) always being a left A[x]-module, K(R) is always a left R[x]module. Further parallels are noted in the following results. Lemma 2.3. (1) Int(A) is a ring if and only if for all f ∈ Int(A) and all a ∈ A, we have f a ∈ Int(A) (2) K(R) is an ideal of R[x] if and only if for all f ∈ K(R) and all a ∈ R, we have f a ∈ K(R) Proof. For (1), the forward implication is clear, so assume that f a ∈ Int(A) whenever f ∈ Int(A) and a ∈ A. It suffices to show that Int(A) is closed under multipliP P cation. Let f, g ∈ Int(A), and write f (x) = i ai xi . Then, (f g)(x) = i ai g(x)xi . P P Let b ∈ A, and let a = g(b) ∈ A. Then, (f g)(b) = i ai g(b)bi = i ai abi = (f a)(b), which is in A because f a ∈ Int(A). Since K(R) is a left R[x]-module, the proof of (2) is the same once we take f ∈ K(R) and g ∈ R[x]. Theorem 2.4. Let A be a torsion-free Z-algebra that is finitely generated as a Z-module. Then, the following are equivalent.

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(1) Int(A) is a ring (2) K(A/nA) is an ideal of A/nA[x] for each integer n > 1 (3) K(A/pλ A) is an ideal of A/pλ A[x] for each prime p and each λ > 0 Proof. (2) ⇒ (1) Assume that K(A/nA) is an ideal for each n > 1. Let f ∈ Int(A), and write f (x) = g(x) n , where g ∈ A[x] and n > 1. Let a ∈ A. By passing to A/nA, we get ga ∈ K(A/nA). Thus, (f a)(x) = (ga)(x) ∈ Int(A), as required. n (1) ⇒ (3) Assume that Int(A) is a ring. Fix pλ , let g ∈ K(A/pλ A), and let ∈ Int(A) and (ga)(x) ∈ a ∈ A/pλ A. Pulling g and a back to A[x], we see that g(x) pλ pλ λ λ Int(A) because Int(A) is a ring. Hence, ga ∈ K(A/p A) and K(A/p A) is an ideal by Lemma 2.3. (3) ⇒ (2) Assume that K(A/pλ A) is an ideal for each prime power pλ . Let n > 1 with prime factorization n = pλ1 1 pλ2 2 · · · pλt t . Let R = A/nA, g ∈ K(R), and a ∈ R. Note that for each i, A/pλi i A ∼ = R/pλi i R. So, passing from R to R/pλi i R, we see that ga ∈ K(R/pλi i R). Thus, for each b ∈ R, we have (ga)(b) ∈ pλi i R. Since this holds for each i, we conclude that (ga)(b) ∈

t \

pλi i R = (0).

i=1

Thus, ga ∈ K(R), as required. In light of the above theorem, when determining whether Int(A) is a ring we may work instead with K(R). This will be our focus for the rest of this paper. 3. K-rings We begin to study K(R) in its own right. In fact, we consider the following more general sets of polynomials. Definition 3.1. Let R be a nonzero finite ring and I an ideal of R. We define K(R, I) = {f (x) ∈ R[x] | f (a) ∈ I for all a ∈ R}; then, K(R) = K(R, (0)). If K(R) is an ideal of R[x], we say that R is a K-ring. If K(R, I) is an ideal of R[x] for each ideal I of R, then we say that R is a strong K-ring. There are several connections among K(R), K(R, I), and K(R/I). Proposition 3.2. Let R be a finite ring and I a nonzero ideal of R. (1) K(R, I) is an ideal of R[x] if and only if K(R/I) is an ideal of R/I[x]. (2) Let L be an ideal of R such that I ⊆ L. Then, K(R, L) is an ideal of R[x] if and only if K(R/I, L/I) is an ideal of R/I[x]. (3) If R is a strong K-ring, then R/I is a strong K-ring. (4) If every finite ring is a K-ring, then every finite ring is a strong K-ring.

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Proof. Parts (1) and (2) are straightforward, and (3) follows from (2). Finally, if every finite ring is a K-ring, then R/I is a K-ring for any ideal I of R, and hence K(R, I) is always an ideal of R[x]. Our main conjecture is the following. Conjecture 3.3. Every finite ring is a strong K-ring (and hence a K-ring). At the present time, we do not have a complete proof of this conjecture, although we will show that is holds for several classes finite rings (see Theorem 3.7). If the converse of part (3) of the Proposition 3.2 holds, then we can use induction to prove that every finite ring is a strong K-ring (and hence a K-ring). The problem we run into is that even if R/I is a strong K-ring for each nonzero ideal I, it does not immediately follow that K(R) is an ideal of R[x]. We will examine this situation further in Proposition 4.4. The first observation we make regarding Conjecture 3.3 is that we may assume that R is indecomposable. Lemma 3.4. Let R1 and R2 be finite rings. If R1 and R2 are strong K-rings, then so is R1 × R2 . Proof. Let I be an ideal of R1 × R2 , and decompose I as I = I1 × I2 , where I1 and I2 are ideals of R1 and R2 , respectively. By exploiting the isomorphism (R1 × R2 )[x] ∼ = R1 [x] × R2 [x], we get K(R1 × R2 , I) ∼ = K(R1 , I1 ) × K(R2 , I2 ). By Lemma 2.3, to show that R is a K-ring, it suffices to show that K(R) is closed under multiplication on the right by elements of R. The analogous result holds for K(R, I). It turns out that we may always multiply on the right by units and remain in K(R, I). Lemma 3.5. Let I be an ideal of R, f ∈ K(R, I), and u ∈ R× . Then, f u ∈ K(R, I). ai xi , and let a ∈ R. Then, X X (f u)(a) = ai uai = ai (uau−1 )i u = f (uau−1 )u ∈ I.

Proof. Write f (x) =

P

i

i

i

When f ∈ R[x] and a, b ∈ R, we have (f (a + b))(x) = (f a)(x) + (f b)(x). So, by Lemma 3.5, K(R, I) will be an ideal whenever each element of R is a sum of units. The following characterization of such rings is well known and is an easy consequence of [6, Thms. 3, 11]. Proposition 3.6. Let R be a finite ring with Jacobson radical J(R). Then, each element of R is a sum of units if and only if R/J(R) contains no factor isomorphic to F2 × F2 (here, F2 is the field with 2 elements).

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For a finite ring R, the Jacobson radical J(R) of R is equal to the intersection of the maximal ideals of R. It is well known that a finite ring R is simple if and only if R is isomorphic to a ring Mn (F) of n × n matrices over a finite field F. We will consider a finite ring R to be semisimple if J(R) = (0); equivalently, R is semisimple if R is isomorphic to a direct product of matrix rings over finite fields [7, Thm. VIII.4]. Combining the previous results, we see that several types of finite rings are strong K-rings. Theorem 3.7. The finite ring R is a strong K-ring in any of the following cases. (1) (2) (3) (4) (5) (6) (7) (8)

R is commutative R∼ = R1 × R2 , where R1 and R2 are strong K-rings each element of R is a sum of units of R R is semisimple R = Mn (S) is a matrix ring over a commutative ring S R is local R has odd order R has order 2n, where n is odd

Proof. Case 1 is trivial, Case 2 is Lemma 3.4, and Case 3 follows from Lemma 3.5. Lemma 3.6 applies to simple rings, so Case 4 follows by Cases 2 and 3. For Case 5, observe that R = S if n = 1, and if n > 1, then R/J(R) will be isomorphic to a direct product of n × n matrix rings over finite fields. The remaining cases follow from Lemma 3.6 and Case 3. Using Theorem 3.7, we can prove that there exist rings A that do not satisfy the conditions of Theorem 1.3, but for which Int(A) is a ring. Example 3.8. We construct a ring A in which the following two properties hold: (1) for each prime power pλ , every element of A/pλ A is a sum of units (2) there exist elements of A that are not sums of products of central elements and units Let s and t be odd integers less than −1. Let i, j, and k satisfy i2 = s, j 2 = t, ij = k, and ji = −k. Take A to be the generalized integral quaternion algebra A = Z ⊕ Zi ⊕ Zj ⊕ Zk. For α = a + bi + cj + dk ∈ A, we let α = a − bi − cj − dk and N (α) = αα = a2 − sb2 − tc2 + std2 . When α is a unit, α−1 is equal to N α(α) . Now, since s and t are odd, i, j, and k are units in A/2λ A for each λ > 0. Furthermore, when p is an odd prime, the order of A/pλ A equals p4λ , which is odd. Thus, we conclude that for any prime power pλ , each element of A/pλ A is a sum of units. It follows from Theorems 3.7 and 2.4 that Int(A) is a ring. However, we claim that A× = {±1}. Indeed, when α = a + bi + cj + dk ∈ A, we have N (α) ≥ max{|a|, |b|, |c|, |d|}; and when α ∈ / {0, ±1}, we have N (α) >

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max{|a|, |b|, |c|, |d|}. So, if α ∈ / {0, ±1}, then N (α) cannot divide any of a, b, c, or a−bi−cj−dk d; thus, ∈ / A and α ∈ / A× . It follows that A× = {±1}. N (α) Finally, the center of A is Z, so the only elements of A that are sums of products of central elements and units are the integers. Hence, Theorem 1.3 does not apply to A. 4. Sufficient Conditions for a Ring to be a K-ring In this section, we outline a plan of attack to prove Conjecture 3.3, and derive more sufficient conditions under which a ring R is a strong K-ring. First, in light of the Chinese Remainder Theorem and Lemma 3.4, it suffices to prove that R is a strong K-ring when |R| = pλ , where p is prime and λ > 0. The cases where p is odd are handled by Theorem 3.7, so we may assume that |R| is a power of 2 and R has characteristic 2λ for some positive integer λ. Second, we know that R is a strong K-ring when each element of R is a sum of units. Lemma 3.6 tells us exactly when this occurs. In fact, the lemma tells us exactly which elements of R are sums of units. It is well known that both idempotents and units can be lifted from R/J(R) to R. Assume that R/J(R) has t factors isomorphic to F2 . Then, we may decompose Qt R/J(R) as ( i=1 F2 ) × R0 , where R0 contains no factors isomorphic to F2 . Under this decomposition, when a ∈ R we may write a mod J(R) as (a1 , a2 , . . . , at , a0 ), where each ai ∈ {0, 1}. Note that in F2 × F2 , the elements (0, 0) and (1, 1) are sums of units, while (1, 0) and (0, 1) are not. It follows that a ∈ R is a sum of units ⇐⇒ (a1 , . . . , at ) = (0, . . . , 0) or (1, . . . , 1). For each i, let ei be the idempotent lifted from the element of R/J(R) with 1 in the ith entry and 0 elsewhere. That is, e1 mod J(R) = (1, 0, 0, . . . , 0), e2 mod J(R) = (0, 1, 0, . . . , 0), etc. Let E(R) = {e1 , e2 , . . . , et }. Our first sufficient condition is the following. Proposition 4.1. If f e ∈ K(R, I) for each ideal I of R, each f ∈ K(R, I) and each e ∈ E(R), then R is a strong K-ring. Proof. Note that each element of R is a sum of units and elements of E(R). The result follows from Lemma 2.3. As shown in the next lemma, when f ∈ K(R, I) and e is an idempotent, there are always some elements of R that are sent into I by f e, and sometimes this is enough to prove that R is a strong K-ring. Given an ideal I of R and an element a ∈ R, we let (I : a)r = {b ∈ R | ab ∈ I}. When I = (0), we let r.ann(a) = (I : a)r be the right annihilator of a in R. Lemma 4.2. Let I be an ideal of R, let f ∈ K(R, I), and let e be an idempotent of R. Then, f e sends eR and Re + (I : e)r into I.

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Proof. By working in R/I, we may assume WLOG that I = (0) and (I : e)r = r.ann(e). When a ∈ R, for each n > 0 we have e(ea)n = (ea)n and e(ae)n = (eae)n . Since f (0) = 0, the constant term of f is 0. So, (f e)(ea) = f (ea) = 0 and (f e)(ae) = f (eae) = 0; thus, f e kills eR and Re. Furthermore, let y + b ∈ Re + r.ann(e). Then, yb = 0, so for each n > 0 we can find y 0 ∈ R such that (y + b)n = y n + by 0 . Then, e(y + b)n = ey n , and (f e)(y + b) = (f e)(y) = 0. Thus, f e kills Re + r.ann(e) as well. Example 4.3. Let S be a finite commutative local ring with residue field F2 . Let
T be an ideal of S. Let R = S0 TS be a subring of upper triangular matrices in ∼ M2 (S). We show that R is a strong K-ring. We have R/J(R) = F2 × F2 and we can

1 0 0 0 take E(R) = {e1 , e2 } where e1 = 0 0 and e2 = 0 1 . Let I be an ideal of R and f ∈ K(R, I). Since e1 =
1 − e2 ,
it suffices to prove that f e2 ∈ K(R, I). We have Re2 + r.ann(e2 ) = 00 TS + S0 T0 = R. It follows from Lemma 4.2 and Proposition 4.1 that R is a strong K-ring. Note that using the same steps with e1 in place of e2 will prove the analogous result for a subring of lower triangular matrices in M2 (S). Also, we obtain the following as a corollary: let A be the algebra of 2 × 2 upper (or lower) triangular matrices with entries in Z. Then, Int(A) is a ring, even though A does not meet the conditions of Theorem 1.3. Unfortunately, the strategy employed in the previous example does not generalize to Mn (S) with n > 2. For instance, if R consists of 3 × 3 matrices, then Re3 + r.ann(e3 ) = R, but Rei + r.ann(ei ) 6= R for i = 1 or 2. As mentioned after Conjecture 3.3, knowing that K(R/I) is an ideal does not necessarily help us determine that K(R) is an ideal. However, we can prove that K(R) is an ideal if K(R) is the intersection of ideals of the form K(R, I). This situation can occur when R has more than one minimal ideal. Proposition 4.4. If every finite ring with a unique minimal ideal is a K-ring, then every finite ring is a strong K-ring. Proof. By Proposition 3.2 part (4), it suffices to prove that if every finite ring with a unique minimal ideal is a K-ring, then every finite ring is a K-ring. Assume that a finite ring is a K-ring if it has a unique minimal ideal. Let R be a finite ring with t proper ideals. We will use induction on t to prove that R is a K-ring. If t = 1, then R is simple and we are done by Theorem 3.7. Assume that t > 1 and that any finite ring with less than t proper ideals is a K-ring (in particular, this assumption holds for R/I whenever I is a nonzero ideal of R). If R has a unique minimal ideal, then we are done by assumption. If not, then there exist two distinct minimal ideals I1 and I2 of R. By our inductive hypothesis, both R/I1 and R/I2 are K-rings, so K(R/I1 ) and K(R/I2 ) are ideals in their respective polynomials rings. By Proposition 3.2, both K(R, I1 ) and K(R, I2 ) are ideals of R[x]. Thus, K(R) = K(R, I1 ) ∩ K(R, I2 ) is also an ideal of R[x].

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We can get one more reduction by using part (3) of Proposition 3.2. By [7, Thm. XIX.3], any finite ring R of characteristic pλ is a residue ring of a subring of a matrix ring Mn (S), where S is a finite commutative local ring with maximal ideal pS. We have handled the cases where p 6= 2, so we have: Proposition 4.5. If, whenever S is a finite commutative local ring of characteristic 2λ with maximal ideal 2S, every subring of Mn (S) is a strong K-ring, then every finite ring is a strong K-ring. Thus, to prove Conjecture 3.3, it suffices to prove it for every ring of order 2 having a unique minimal ideal, or to prove it for subrings of the matrix rings described above. λ

5. Generators in the Case of Matrix Rings Throughout this section, S denotes a finite commutative ring and R the matrix ring Mn (S). We associate S with the scalar matrices in R. By Theorem 3.7, K(R) is an ideal of R[x]. This section examines elements of K(R). In particular, if S is a local ring with principal maximal ideal, then we will exhibit a generating set for K(R). Definition 5.1. By Proposition 2.2, K(R) contains monic polynomials in S[x]. We define φR to be a monic polynomial in K(R) ∩ S[x] that is of minimal degree among all monic polynomials in K(R) ∩ S[x]. In general, φR need not be unique. For example, one may check that both (x2 −x)2 and (x2 −x)2 +2(x2 −x) have minimal degree among all monic polynomials in K(Z/4Z). However, for our purposes uniqueness is not needed, so we will assume from this point on that φR is fixed for each R. Also, note that when I is an ideal of R, we may lift φR/I to R[x] and consider it to be an element of K(R, I). In the case of matrix rings over a commutative ring, previous authors have investigated what we are calling K(R) and K(R) ∩ S[x]. Proposition 5.2. Let Pn be the set of monic polynomials in S[x] of degree n. Then, (1) (2) (3) (4)

f ∈ K(R) ∩ S[x] if and only if f is divisible by every polynomial in Pn K(R) is generated by polynomials in S[x] φR is of minimal degree among all the monic least common multiples for Pn n n−1 If S = Fq , then φR = (xq − x)(xq − x) · · · (xq − x) and K(R) is generated by φR

Proof. (1) is [5, Lems. 3.3, 3.4] and (3) follows from (1). (2) is essentially the same as [8, Thm. 1.6]. The expression for φR in (4) is [2, Thm. 3], and K(R) = (φR )R[x] because of (2) and the fact that S is a field.

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By the content of a polynomial f ∈ R[x], we mean the ideal of R generated by the coefficients of f . We denote the content of f by c(f ). If c(f ) = (1), we say that f has content 1. When I is a nilpotent ideal, the nilpotency of I refers to the smallest positive integer N such that I N = (0). If S is local with maximal ideal m, then R = Mn (S) is also local, and has maximal ideal M = Mn [m]. It is well known that the Jacobson radical of a finite ring is nilpotent, so both m and M are nilpotent and have the same nilpotency. Lemma 5.3. Let S be a finite commutative local ring with maximal ideal m, let R = Mn (S), and let M = Mn [m] be the maximal ideal of R. If f ∈ K(R) ∩ S[x] has content 1, then deg(f ) ≥ deg(φR ). Proof. Since φR has content 1, we may assume WLOG that f ∈ K(R) ∩ S[x] is of minimal degree among all content 1 polynomials in K(R) ∩ S[x]. We will show that deg(f ) = deg(φR ). Let N be the nilpotency of M. We will proceed by induction on N. If N = 1, then S is a field and K(R) = (φR )R[x] by Proposition 5.2. Hence, φR divides f and the lemma is true when N = 1. Assume that N > 1 and that the result holds for nilpotencies less than N . Then, in particular, the result holds for R/Mi ∼ = Mn [S/mi ] whenever 1 ≤ i ≤ N − 1. By way of contradiction, suppose that deg(f ) < deg(φR ), and let a be the leading coefficient of f . If a ∈ / m, then a−1 f is a monic element of K(R) ∩ S[x] with degree less than deg(φR ), which is a contradiction. So, assume that a ∈ m, and let i ∈ {1, 2, . . . , N − 1} be such that a ∈ mi \ mi+1 . Let ψ = φR/MN −i ; then, aψ ∈ K(R) ∩ S[x]. Now, f mod MN −i is a content 1 polynomial in K(R/MN −i ) ∩ S/mN −i [x], so by induction deg(f ) ≥ deg(f mod MN −i ) ≥ deg(ψ). Let d = deg(f ) − deg(ψ) and g = f − axd ψ ∈ K(R) ∩ S[x]. Then, deg(g) < deg(f ), so c(g) ⊆ m. Hence, f = g + axd ψ and every coefficient of f is in m. This contradicts the fact that f has content 1. Having reached a contradiction in all cases, we conclude that deg(f ) ≥ deg(φR ). When S is a finite (hence Artinian) commutative local ring with principal maximal ideal πS, [1, Prop. 8.8] shows that S is a principal ideal ring, and furthermore that each ideal of S is generated by a power of π. These conditions imply that R and all of the residue rings of R and S are also principal ideal rings whose ideals are generated by powers of π. Theorem 5.4. Let S be a finite commutative local ring with principal maximal ideal πS, and let N be the nilpotency of πS. Let R = Mn (S), and for each 1 ≤ i ≤ N − 1, let Ri = R/π i R. Then, K(R) is generated by {φR , πφRN −1 , . . . , π N −1 φR1 }. Proof. We use induction on N . When N = 1, we are done by Proposition 5.2, so assume that N > 1 and that the result holds for all nilpotencies less than N . Let

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I be the ideal of R[x] generated by {φR , πφRN −1 , . . . , π N −1 φR1 }; then, I ⊆ K(R). By Proposition 5.2 part (2), K(R) is generated by K(R) ∩ S[x], so it suffices to show that K(R) ∩ S[x] ⊆ I. Let f ∈ K(R) ∩ S[x]. Since φR is monic, we may divide f by φR over S[x] to get f = gφR + h for some g, h ∈ S[x] such that either h = 0 or deg(h) < deg(φR ). Since gφR ∈ I, we just need to show that h ∈ I. If h = 0, then we are done, so assume that h 6= 0. By Lemma 5.3, c(h) ⊆ πR, so h = πh0 for some h0 ∈ S[x]. Since h ∈ K(R), we have h0 ∈ K(R, π N −1 R). By our inductive hypothesis and the observations in the paragraph preceding this theorem, the result is true mod π N −1 R, so h0 mod π N −1 R is in the ideal of RN −1 [x] generated by {φRN −1 , πφRN −2 , . . . , π N −1 φR1 }. Lifting h0 mod π N −1 R back to R, we obtain h = πh0 ∈ I, which completes the proof. We close this paper with one more interesting result. In what follows, let Zpλ = Z/pλ Z, where p is a prime and λ > 0. We will establish a condition under which K(Mn (S)) can be generated not just by polynomials in S[x], but by polynomials in Zpλ [x]. By a Galois extension of Zpλ , we mean a ring S of the form S = Zpλ [α], where α is a root of a monic irreducible polynomial f ∈ Zpλ [x]. Basic results about Galois extensions include the following (proofs can be found in [7, §XV, XVI]): S ∼ = Fpd , where = Fp (α) ∼ = Zpλ [x]/(f ); S is local, with maximal ideal pS; S/pS ∼ d = deg(f ); the group G = Gal(S/Zpλ ) of S-automorphisms that fix Zpλ is cyclic of order d; f has exactly d roots in S, and the elements of G transitively permute these roots; and Zpλ is the subring of S fixed by G. Let S be a Galois extension of Zpλ , let Fq be the residue field of S, and let ι ∈ Fq [x] be a monic irreducible polynomial. We say that f ∈ S[x] is ι-primary if f is monic and the residue of f in Fq [x] is positive power of ι. Let R = Mn (S), and let Pn be the set of monic polynomials in S[x] of degree n. By Proposition 5.2, φR is a monic least common multiple (lcm) for Pn . The following proposition (a restatement of [10, Thm. 5.1]) provides a more precise expression for φR . Proposition 5.5. Let X be the set of all the monic irreducible polynomials in Fq [x] of degree less than or equal to n. For each ι ∈ X, let   n  Pι = ι-primary polynomials in S[x] of degree deg(ι) deg(ι) , Q and let Lι ∈ S[x] be a monic lcm for Pι . Then, ι∈X Lι is a monic lcm for Pn . Q In light of this proposition, we can take φR = ι∈X Lι . Moreover, by employing the Galois theory of S over Zpλ , we can pick φR ∈ Zpλ [x]. We have G = Gal(S/Zpλ ) ∼ = Gal(Fq /Fp ), so G acts on the set X: for each σ ∈ G and each ι ∈ X, σ(ι) is a monic irreducible polynomial in Fq [x]. Furthermore, there is a corresponding action on the set of lcms: if Lι is a monic lcm for Pι , then σ(Lι ) is a monic lcm for Pσ(ι) .

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Break X into G-orbits X1 , X2 , . . . , Xt , and index the elements of X so that ιij ∈ Xi for 1 ≤ j ≤ |Xi |. Choose lcms Lij that are compatible with the action of Q G: Lik = σ(Lij ) if and only if ιik = σ(ιij ). For each 1 ≤ i ≤ t, let fi = j Lij . Then, Q each fi is stable under the action of G, and hence is in Zpλ [x]. Finally, φR = i fi is in Zpλ [x] as well. We have shown: Theorem 5.6. Let S be a Galois extension of Zpλ and let R = Mn (S). Then, we may assume that φR ∈ Zpλ [x]. Thus, by Theorem 5.4, K(R) is generated by polynomials in Zpλ [x]. References [1] M. Atiyah and I. MacDonald, Introduction to Commutative Algebra (AddisonWesley, Reading, MA, 1969). [2] J. V. Brawley and L. Carlitz, Scalar polynomial functions on the n × n matrices over a finite field, Linear Algebra and Appl. 10 (1975) 199–217. [3] P.-J. Cahen and J.-L. Chabert, Integer-valued polynomials (Amer. Math. Soc. Surveys and Monographs, Providence, RI, 1997). [4] S. Frisch, Integer-valued polynomials on algebras, J. Algebra 373 (2013) 414–425. [5] S. Frisch, Polynomial separation of points in algebras, in Arithmetical Properties of Commutative Rings and Monoids, Lect. Notes Pure Appl. Math. 241 (Chapman & Hall, Boca Raton, FL, 2005). [6] M. Henriksen, Two classes of rings generated by their units, J. Algebra 31 (1974) 182-193. [7] B. R. McDonald, Finite rings with identity, Pure and Applied Mathematics Vol. 28 (Marcel Dekker, New York, 1974). [8] N. J. Werner, Integer-valued polynomials over matrix rings, Comm. Algebra 40 (2012) no. 12, 4717–4726. [9] N. J. Werner, Integer-valued polynomials over quaternion rings, J. Algebra 324 (2010) no. 7 1754–1769. [10] N. J. Werner, On least common multiples of polynomials in Z/nZ[x], Comm. Algebra 40 (2012) no. 6 2066–2080.