# Integral Closure of Rings of Integer-Valued Polynomials on Algebras

Chapter

## Abstract

Let D be an integrally closed domain with quotient field K. Let A be a torsion-free D-algebra that is finitely generated as a D-module. For every a in A we consider its minimal polynomial μ a (X) ∈ D[X], i.e. the monic polynomial of least degree such that μ a (a) = 0. The ring Int K (A) consists of polynomials in K[X] that send elements of A back to A under evaluation. If D has finite residue rings, we show that the integral closure of Int K (A) is the ring of polynomials in K[X] which map the roots in an algebraic closure of K of all the μ a (X), aA, into elements that are integral over D. The result is obtained by identifying A with a D-subalgebra of the matrix algebra M n (K) for some n and then considering polynomials which map a matrix to a matrix integral over D. We also obtain information about polynomially dense subsets of these rings of polynomials.

## Keywords

Integer-valued polynomial Matrix Triangular matrix Integral closure Pullback Polynomially dense

## MSC(2010) classification:

[]13B25 13B22 11C20

## 1 Introduction

Let D be a (commutative) integral domain with quotient field K. The ring Int(D) of integer-valued polynomials on D consists of polynomials in K[X] that map elements of D back to D. More generally, if E ⊆ K, then one may define the ring Int(E, D) of polynomials that map elements of E into D.

One focus of recent research ([4, 5, 6, 9, 10, 11]) has been to generalize the notion of integer-valued polynomial to D-algebras. When A ⊇ D is a torsion-free module finite D-algebra we define $$\mathrm{Int}_{K}(A):=\{ f \in K[X]\mid f(A) \subseteq A\}$$. The set Int K (A) forms a commutative ring. If we assume that $$K \cap A = D$$, then Int K (A) is contained in Int(D) (these two facts are indeed equivalent), and often Int K (A) shares properties similar to those of Int(D) (see the references above, especially ).

When A = M n (D), the ring of n × n matrices with entries in D, Int K (A) has proven to be particularly amenable to investigation. For instance, [9, Thm. 4.6] shows that the integral closure of $$\mathrm{Int}_{\mathbb{Q}}(M_{n}(\mathbb{Z}))$$ is $$\mathrm{Int}_{\mathbb{Q}}(\mathcal{A}_{n})$$, where $$\mathcal{A}_{n}$$ is the set of algebraic integers of degree at most $$n$$, and $$\mathrm{Int}_{\mathbb{Q}}(\mathcal{A}_{n}) =\{ f \in \mathbb{Q}[X]\mid f(\mathcal{A}_{n}) \subseteq \mathcal{A}_{n}\}$$. In this paper, we will generalize this theorem and describe the integral closure of Int K (A) when D is an integrally closed domain with finite residue rings. Our description (Theorem 13) may be considered an extension of both [9, Thm. 4.6] and [2, Thm. IV.4.7] (the latter originally proved in [7, Prop. 2.2]), which states that if D is Noetherian and D′ is its integral closure in K, then the integral closure of Int(D) equals $$\mathrm{Int}(D,D^{\prime}) =\{ f \in K[X]\mid f(D) \subset D^{\prime}\}$$.

Key to our work will be rings of polynomials that we dub integral-valued polynomials and which act on certain subsets of M n (K). Let $$\overline{K}$$ be an algebraic closure of K. We will establish a close connection between integral-valued polynomials and polynomials that act on elements of $$\overline{K}$$ that are integral over D. We will also investigate polynomially dense subsets of rings of integral-valued polynomials.

In Sect. 2, we define what we mean by integral-valued polynomials, discuss when sets of such polynomials form a ring, and connect them to the integral elements of $$\overline{K}$$. In Sect. 3, we apply the results of Sect. 2 to Int K (A) and prove the aforementioned theorem about its integral closure. Section 4 covers polynomially dense subsets of rings of integral-valued polynomials. We close by posing several problems for further research.

## 2 Integral-Valued Polynomials

Throughout, we assume that D is an integrally closed domain with quotient field K. We denote by $$\overline{K}$$ a fixed algebraic closure of K. When working in M n (K), we associate K with the scalar matrices, so that we may consider K (and D) to be subsets of M n (K).

For each matrix M ∈ M n (K), we let μ M (X) ∈ K[X] denote the minimal polynomial of M, which is the monic generator of $$N_{K[X]}(M) =\{ f \in K[X]\mid f(M) = 0\}$$, called the null ideal of M. We define Ω M to be the set of eigenvalues of M considered as a matrix in $$M_{n}(\overline{K})$$, which are the roots of μ M in $$\overline{K}$$. For a subset $$S \subseteq M_{n}(K)$$, we define $$\varOmega _{S}:=\bigcup _{M\in S}\varOmega _{M}$$. Note that a matrix in M n (K) may have minimal polynomial in D[X] even though the matrix itself is not in M n (D). A simple example is given by $$\big(\begin{matrix}0&q \\ 0&0\end{matrix}\big) \in M_{2}(K)$$, where q ∈ K∖ D.

### Definition 1.

We say that M ∈ M n (K) is integral over D (or just integral, or is an integral matrix) if M solves a monic polynomial in D[X]. A subset S of M n (K) is said to be integral if each M ∈ S is integral over D.

Our first lemma gives equivalent definitions for a matrix to be integral.

### Lemma 2.

Let M ∈ M n (K). The following are equivalent:
1. (i)

M is integral over D

2. (ii)

μ M ∈ D[X]

3. (iii)

Each α ∈Ω M is integral over D

### Proof.

(i) ⇒ (iii)

Suppose M solves a monic polynomial f(X) with coefficients in D. As μ M (X) divides f(X), its roots are then also roots of f(X). Hence, the elements of Ω M are integral over D.

(iii) ⇒ (ii)

The coefficients of μ M  ∈ K[X] are the elementary symmetric functions of its roots. Assuming (iii) holds, these roots are integral over D; hence, the coefficients of μ M are integral over D. Since D is integrally closed, we must have μ M  ∈ D[X].

(ii) ⇒ (i)

Obvious. □

For the rest of this section, we will study polynomials in K[X] that take values on sets of integral matrices. These are the integral-valued polynomials mentioned in the introduction.

### Definition 3.

Let S ⊆ M n (K). Let K[S] denote the K-subalgebra of M n (K) generated by K and the elements of S. Define S′: = { M ∈ K[S]∣M is integral} and Int K (S, S′): = { f ∈ K[X]∣f(S) ⊆ S′}. We call Int K (S, S′) a set of integral-valued polynomials.

### Remark 4.

In the next lemma, we will prove that forming the set S′ is a closure operation in the sense that (S′)′ = S′. We point out that this construction differs from the usual notion of integral closure in several ways. First, if S itself is not integral, then S ⊈ S′. Second, S′ need not have a ring structure. Indeed, if $$D = \mathbb{Z}$$ and $$S = M_{2}(\mathbb{Z})$$, then both $$\big(\begin{matrix}1&0 \\ 0&0\end{matrix}\big)$$ and $$\big(\begin{matrix}1/2\;&1/2 \\ 1/2\;&1/2\end{matrix}\big)$$ are in S′, but neither their sum nor their product is integral. Lastly, if S is a commutative ring, then S′ need not be the same as the integral closure of S in K[S], because we insist that the elements of S satisfy a monic polynomial in D[X] rather than S[X].

However, if S is a commutative D-algebra and it is an integral subset of M n (K), then S′ is equal to the integral closure of S in K[S] (see Corollary 1 to Proposition 2 and Proposition 6 of [1, Chapt. V]).

### Lemma 5.

Let S ⊆ M n (K). Then, (S′)′ = S′.

### Proof.

We just need to show that K[S′] = K[S]. By definition, S′ ⊆ K[S], so K[S′] ⊆ K[S]. For the other containment, let M ∈ K[S]. Let d ∈ D be a common multiple for all the denominators of the entries in M. Then, dM ∈ S′ ⊆ K[S′]. Since 1∕d ∈ K, we get M ∈ K[S′]. □

An integral subset of M n (K) need not be closed under addition or multiplication, so at first glance it may not be clear that Int K (S, S′) is closed under these operations. As we now show, Int K (S, S′) is in fact a ring.

### Proposition 6.

Let S ⊆ M n (K). Then, IntK (S,S′) is a ring, and if D ⊆ S, then $$\mathrm{Int}_{K}(S,S^{\prime}) \subseteq \mathrm{ Int}(D)$$ .

### Proof.

Let M ∈ S and f, g ∈ Int K (S, S′). Then, f(M), g(M) are integral over D. By Corollary 2 after Proposition 4 of [1, Chap. V], the D-algebra generated by f(M) and g(M) is integral over D. So, f(M) + g(M) and f(M)g(M) are both integral over D and are both in K[S]. Thus, f(M) + g(M), f(M)g(M) ∈ S′ and f + g, fg ∈ Int K (S, S′). Assuming D ⊆ S, let f ∈ Int K (S, S′) and d ∈ D. Then, f(d) is an integral element of K. Since D is integrally closed, f(d) ∈ D. Thus, Int K (S, S′) ⊆ Int(D). □

We now begin to connect our rings of integral-valued polynomials to rings of polynomials that act on elements of $$\overline{K}$$ that are integral over D. For each n > 0, let
$$\displaystyle{\varLambda _{n}:=\{\alpha \in \overline{K}\mid \alpha \text{ solves a monic polynomial in }D[X]\text{ of degree }n\}.}$$
In the special case $$D = \mathbb{Z}$$, we let $$\mathcal{A}_{n}:=\varLambda _{n} =\{ \text{ algebraic integers of degree}\}\{\text{at most}$$ n$$\}\subset \overline{\mathbb{Q}}$$.
For any subset $$\mathcal{E}$$ of Λ n , define
$$\displaystyle{\mathrm{Int}_{K}(\mathcal{E},\varLambda _{n}):=\{ f \in K[X]\mid f(\mathcal{E}) \subseteq \varLambda _{n}\}.}$$
to be the set of polynomials in K[X] mapping elements of $$\mathcal{E}$$ into Λ n . If $$\mathcal{E} =\varLambda _{n}$$, then we write simply Int K (Λ n ). As with Int K (S, S′), $$\mathrm{Int}_{K}(\mathcal{E},\varLambda _{n})$$ is a ring despite the fact that Λ n is not.

### Proposition 7.

For any $$\mathcal{E}\subseteq \varLambda _{n}$$, $$\mathrm{Int}_{K}(\mathcal{E},\varLambda _{n})$$ is a ring and is integrally closed.

### Proof.

Let Λ be the integral closure of D in $$\overline{K}$$. We set $$\mathrm{Int}_{\overline{K}}(\mathcal{E},\varLambda _{\infty }) =\{ f \in \overline{K}[X]\mid f(\mathcal{E}) \subseteq \varLambda _{\infty }\}$$. Then, $$\mathrm{Int}_{\overline{K}}(\mathcal{E},\varLambda _{\infty })$$ is a ring, and by [2, Prop. IV.4.1] it is integrally closed.

Let $$\mathrm{Int}_{K}(\mathcal{E},\varLambda _{\infty }) =\{ f \in K[X]\mid f(\mathcal{E}) \subseteq \varLambda _{\infty }\}$$. Clearly, $$\mathrm{Int}_{K}(\mathcal{E},\varLambda _{n}) \subseteq \mathrm{ Int}_{K}(\mathcal{E},\varLambda _{\infty })$$. However, if $$\alpha \in \mathcal{E}$$ and f ∈ K[X], then [K(f(α)): K] ≤ [K(α): K] ≤ n, so in fact $$\mathrm{Int}_{K}(\mathcal{E},\varLambda _{n}) =\mathrm{ Int}_{K}(\mathcal{E},\varLambda _{\infty })$$. Finally, since $$\mathrm{Int}_{K}(\mathcal{E},\varLambda _{\infty }) =\mathrm{ Int}_{\overline{K}}(\mathcal{E},\varLambda _{\infty }) \cap K[X]$$ is the contraction of $$\mathrm{Int}_{\overline{K}}(\mathcal{E},\varLambda _{\infty })$$ to K[X], it is an integrally closed ring, proving the proposition. □

Theorem 4.6 in  shows that the integral closure of $$\mathrm{Int}_{\mathbb{Q}}(M_{n}(\mathbb{Z}))$$ equals the ring $$\mathrm{Int}_{\mathbb{Q}}(\mathcal{A}_{n})$$. As we shall see (Theorem 9), this is evidence of a broader connection between the rings of integral-valued polynomials Int K (S, S′) and rings of polynomials that act on elements of Λ n . The key to this connection is the observation contained in Lemma 2 that the eigenvalues of an integral matrix in M n (K) lie in Λ n and also the well-known fact that if M ∈ M n (K) and f ∈ K[X], then the eigenvalues of f(M) are exactly f(α), where α is an eigenvalue of M. More precisely, if $$\chi _{M}(X) =\prod _{i=1,\ldots,n}(X -\alpha _{i})$$ is the characteristic polynomial of M (the roots α i are in $$\overline{K}$$ and there may be repetitions), then the characteristic polynomial of f(M) is $$\chi _{f(M)}(X) =\prod _{i=1,\ldots,n}(X - f(\alpha _{i}))$$. Phrased in terms of our Ω-notation, we have
$$\displaystyle{ \mbox{ if}M \in M_{n}(K)\mbox{ and}f \in K[X],\mbox{ then}\varOmega _{f(M)} = f(\varOmega _{M}) =\{ f(\alpha )\mid \alpha \in \varOmega _{M}\}. }$$
(8)

Using this fact and our previous work, we can equate Int K (S, S′) with a ring of the form $$\mathrm{Int}_{K}(\mathcal{E},\varLambda _{n})$$.

### Theorem 9.

Let S ⊆ M n (K). Then, $$\mathrm{Int}_{K}(S,S^{\prime}) =\mathrm{ Int}_{K}(\varOmega _{S},\varLambda _{n})$$ , and in particular IntK (S,S′) is integrally closed.

### Proof.

We first prove this for S = { M}. Using Lemma 2 and (8), for each f ∈ K[X] we have
$$\displaystyle{f(M) \in S^{\prime}\;\Longleftrightarrow\;f(M)\text{ is integral }\;\Longleftrightarrow\;\varOmega _{f(M)} \subseteq \varLambda _{n}\;\Longleftrightarrow\;f(\varOmega _{M}) \subseteq \varLambda _{n}.}$$
This proves that $$\mathrm{Int}_{K}(\{M\},\{M\}^{\prime}) =\mathrm{ Int}_{K}(\varOmega _{M},\varLambda _{n})$$. For a general subset S of M n (K), we have
$$\displaystyle{\mathrm{Int}_{K}(S,S^{\prime}) =\bigcap _{M\in S}\mathrm{Int}_{K}(\{M\},S^{\prime}) =\bigcap _{M\in S}\mathrm{Int}_{K}(\varOmega _{M},\varLambda _{n}) =\mathrm{ Int}_{K}(\varOmega _{S},\varLambda _{n}).}$$
□
The above proof shows that if a polynomial is integral-valued on a matrix, then it is also integral-valued on any other matrix with the same set of eigenvalues. Note that for a single integral matrix M we have these inclusions:
$$\displaystyle{D \subset D[M] \subseteq \{ M\}^{\prime} \subset K[M].}$$
Moreover, {M}′ is equal to the integral closure of D[M] in K[M] (because D[M] is a commutative algebra).

## 3 The Case of a D-Algebra

We now use the results from Sect. 2 to gain information about Int K (A), where A is a D-algebra. In Theorem 13 below, we shall obtain a description of the integral closure of Int K (A).

As mentioned in the introduction, we assume that A is a torsion-free D-algebra that is finitely generated as a D-module. Let B = A D K be the extension of A to a K-algebra. Since A is a faithful D-module, B contains copies of D, A, and K. Furthermore, K is contained in the center of B, so we can evaluate polynomials in K[X] at elements of B and define
$$\displaystyle{\mathrm{Int}_{K}(A):=\{ f \in K[X]\mid f(A) \subseteq A\}.}$$

Letting n be the vector space dimension of B over K, we also have an embedding $$B\hookrightarrow M_{n}(K)$$, $$b\mapsto M_{b}$$. More precisely, we may embed B into the ring of K-linear endomorphisms of B (which is isomorphic to M n (K)) via the map $$B\hookrightarrow \text{End}_{K}(B)$$ sending b ∈ B to the endomorphism $$x\mapsto b \cdot x$$. Consequently, starting with just D and A, we obtain a representation of A as a D-subalgebra of M n (K). Note that n may be less than the minimum number of generators of A as a D-module.

In light of the aforementioned matrix representation of B, several of the definitions and notations we defined in Sect. 2 will carry over to B. Since the concepts of minimal polynomial and eigenvalue are independent of the representation $$B\hookrightarrow M_{n}(K)$$, the following are well defined:
• For all b ∈ B, μ b (X) ∈ K[X] is the minimal polynomial of b. So, μ b (X) is the monic polynomial of minimal degree in K[X] that kills b. Equivalently, μ b is the monic generator of the null ideal N K[X](b) of b. This is the same as the minimal polynomial of M b  ∈ M n (K), since for all f ∈ K[X] we have $$f(M_{b}) = M_{f(b)}$$. To ease the notation, from now on we will identify b with M b .

• By the Cayley-Hamilton Theorem, $$\deg (\mu _{b}) \leq n$$, for all b ∈ B.

• For all b ∈ B, $$\varOmega _{b} =\{ \text{roots of }\mu _{b}\text{ in }\overline{K}\}$$. The elements of Ω b are nothing else than the eigenvalues of b under any matrix representation $$B\hookrightarrow M_{n}(K)$$. If S ⊆ B, then $$\varOmega _{S} =\bigcup _{b\in S}\varOmega _{b}$$.

• b ∈ B is integral over D (or just integral) if b solves a monic polynomial in D[X].

• B = K[A], since B is formed by extension of scalars from D to K.

• A′ = { b ∈ Bb is integral}. By [1, Theorem 1, Chapt. V] A ⊆ A′. In particular, this implies $$A \cap K = D$$ (because D is integrally closed), so that $$\mathrm{Int}_{K}(A) \subseteq \mathrm{ Int}(D)$$ (see the remarks in the introduction).

• $$\mathrm{Int}_{K}(A,A^{\prime}) =\{ f \in K[X]\mid f(A) \subseteq A^{\prime}\}$$.

Working exactly as in Proposition 6, we find that Int K (A, A′) is another ring of integral-valued polynomials. Additionally, Lemma 2 and (8) hold for elements of B. Consequently, we have

### Theorem 10.

IntK (A,A′) is an integrally closed ring and is equal to $$\mathrm{Int}_{K}(\varOmega _{A},\varLambda _{n})$$ .

By generalizing results from , we will show that if D has finite residue rings, then Int K (A, A′) is the integral closure of Int K (A). This establishes the analogue of [2, Thm. IV.4.7] (originally proved in [7, Prop. 2.2]) mentioned in the introduction.

We will actually prove a slightly stronger statement and give a description of Int K (A, A′) as the integral closure of an intersection of pullbacks. Notice that
$$\displaystyle{\bigcap _{a\in A}(D[X] +\mu _{a}(X) \cdot K[X]) \subseteq \mathrm{ Int}_{K}(A)}$$
because if $$f \in D[X] +\mu _{a}(X) \cdot K[X]$$, then f(a) ∈ D[a] ⊆ A. We thus have a chain of inclusions:
$$\displaystyle{ \bigcap _{a\in A}(D[X] +\mu _{a}(X) \cdot K[X]) \subseteq \mathrm{ Int}_{K}(A) \subseteq \mathrm{ Int}_{K}(A,A^{\prime}) }$$
(11)
and our work below will show that this is actually a chain of integral ring extensions.

### Lemma 12.

Let f ∈ IntK (A,A′), and write $$f(X) = g(X)/d$$ for some g ∈ D[X] and some nonzero d ∈ D. Then, for each h ∈ D[X], $$d^{n-1}h(f(X)) \in \bigcap _{a\in A}(D[X] +\mu _{a}(X) \cdot K[X])$$ .

### Proof.

Let a ∈ A. Since f ∈ Int K (A, A′), m: = μ f(a) ∈ D[X], and deg(m) ≤ n.

Now, m is monic, so we can divide h by m to get $$h(X) = q(X)m(X) + r(X)$$, where q, r ∈ D[X], and either r = 0 or deg(r) < n. Then,
$$\displaystyle{d^{n-1}h(f(X)) = d^{n-1}q(f(X))m(f(X)) + d^{n-1}r(f(X))}$$
The polynomial $$d^{n-1}q(f(X))m(f(X)) \in K[X]$$ is divisible by μ a (X) because m(f(a)) = 0. Since deg(r) < n, d n−1 r(f(X)) ∈ D[X]. Thus, $$d^{n-1}h(f(X)) \in D[X] +\mu _{a}(X) \cdot K[X]$$, and since a was arbitrary the lemma is true. □

For the next result, we need an additional assumption. Recall that a ring D has finite residue rings if for all proper nonzero ideal I ⊂ D, the residue ring DI is finite. Clearly, this condition is equivalent to asking that for all nonzero d ∈ D, the residue ring DdD is finite.

### Theorem 13.

Assume that D has finite residue rings. Then, $$\mathrm{Int}_{K}(A,A^{\prime}) =\mathrm{ Int}_{K}(\varOmega _{A},\varLambda _{n})$$ is the integral closure of both $$\bigcap _{a\in A}(D[X] +\mu _{a}(X) \cdot K[X])$$ and IntK (A).

### Proof.

Let $$R =\bigcap _{a\in A}(D[X] +\mu _{a}(X) \cdot K[X])$$. By (11), it suffices to prove that Int K (A, A′) is the integral closure of R. Let $$f(X) = g(X)/d \in \mathrm{ Int}_{K}(A,A^{\prime})$$. By Theorem 10, Int K (A, A′) is integrally closed, so it is enough to find a monic polynomial ϕ ∈ D[X] such that ϕ(f(X)) ∈ R.

Let $$\mathcal{P}\subseteq D[X]$$ be a set of monic residue representatives for {μ f(a)(X)} a ∈ A modulo (d n−1)2. Since D has finite residue rings, $$\mathcal{P}$$ is finite. Let ϕ(X) be the product of all the polynomials in $$\mathcal{P}$$. Then, ϕ is monic and is in D[X].

Fix a ∈ A and let m = μ f(a). There exists $$p(X) \in \mathcal{P}$$ such that p(X) is equivalent to m mod (d n−1)2, so $$p(X) = m(X) + (d^{n-1})^{2}r(X)$$ for some r ∈ D[X]. Furthermore, p(X) divides ϕ(X), so there exists q(X) ∈ D[X] such that ϕ(X) = p(X)q(X). Thus,
$$\displaystyle\begin{array}{rcl} \phi (f(X))& =& p(f(X))q(f(X)) {}\\ & =& m(f(X))q(f(X)) + (d^{n-1})^{2}r(f(X))q(f(X)) {}\\ & =& m(f(X))q(f(X)) + d^{n-1}r(f(X)) \cdot d^{n-1}q(f(X)) {}\\ \end{array}$$

As in Lemma 12, $$m(f(X))q(f(X)) \in \mu _{a}(X) \cdot K[X]$$ because m(f(a)) = 0. By Lemma 12, d n−1 r(f(X)) and d n−1 q(f(X)) are in $$D[X] +\mu _{a}(X) \cdot K[X]$$. Hence, $$\phi (f(X)) \in D[X] +\mu _{a}(X) \cdot K[X]$$, and since a was arbitrary, ϕ(f(X)) ∈ R. □

Theorem 13 says that the integral closure of Int K (A) is equal to the ring of polynomials in K[X] which map the eigenvalues of all the elements a ∈ A to integral elements over D.

### Remark 14.

By following essentially the same steps as in Lemma 12 and Theorem 13, one may prove that Int K (A, A′) is the integral closure of Int K (A) without the use of the pullbacks D[X] +μ a (X) ⋅ K[X]. However, employing the pullbacks gives a slightly stronger theorem without any additional difficulty.

In the case A = M n (D), Int K (M n (D)) is equal to the intersection of the pullbacks $$D[X] +\mu _{M}(X) \cdot K[X]$$, for M ∈ M n (D). Indeed, let f ∈ Int K (M n (D)) and M ∈ M n (D). By [11, Remark 2.1 & (3)], Int K (M n (D)) is equal to the intersection of the pullbacks D[X] +χ M (X)K[X], for M ∈ M n (D), where χ M (X) is the characteristic polynomial of M. By the Cayley-Hamilton Theorem, μ M (X) divides χ M (X) so that $$f \in D[X] +\chi _{M}(X)K[X] \subseteq D[X] +\mu _{M}(X)K[X]$$ and we are done.

### Remark 15.

The assumption that D has finite residue rings implies that D is Noetherian (and in fact that D is a Dedekind domain, because D is integrally closed). Given that [2, Thm. IV.4.7] (or [7, Prop. 2.2]) requires only the assumption that D is Noetherian, it is fair to ask if Theorem 13 holds under the weaker condition that D is Noetherian.

Note that $$\varOmega _{M_{n}(D)} =\varLambda _{n}$$ (and in particular, $$\varOmega _{M_{n}(\mathbb{Z})} = \mathcal{A}_{n}$$). Hence, we obtain the following (which generalizes [9, Thm. 4.6]):

### Corollary 16.

If D has finite residue rings, then the integral closure of IntK (M n (D)) is IntK n ).

The algebra of upper triangular matrices yields another interesting example.

### Corollary 17.

Assume that D has finite residue rings. For each n > 0, let T n (D) be the ring of n × n upper triangular matrices with entries in D. Then, the integral closure of IntK (T n (D)) equals Int (D).

### Proof.

For each a ∈ T n (D), μ a splits completely and has roots in D, so $$\varOmega _{T_{n}(D)} = D$$. Hence, the integral closure of Int K (T n (D)) is $$\mathrm{Int}_{K}(\varOmega _{T_{n}(D)},\varLambda _{n}) =\mathrm{ Int}_{K}(D,\varLambda _{n})$$. But, polynomials in K[X] that move D into Λ n actually move D into $$\varLambda _{n} \cap K = D$$. Thus, $$\mathrm{Int}_{K}(D,\varLambda _{n}) =\mathrm{ Int}(D)$$. □

Since $$\mathrm{Int}_{K}(T_{n}(D)) \subseteq \mathrm{ Int}_{K}(T_{n-1}(D))$$ for all n > 0, the previous proposition proves that
$$\displaystyle{\cdots \subseteq \mathrm{ Int}_{K}(T_{n}(D)) \subseteq \mathrm{ Int}_{K}(T_{n-1}(D)) \subseteq \cdots \subseteq \mathrm{ Int}(D)}$$
is a chain of integral ring extensions.

## 4 Matrix Rings and Polynomially Dense Subsets

For any D-algebra A, we have (A′)′ = A′, so Int K (A′) is integrally closed by Theorem 10. Furthermore, Int K (A′) is always contained in Int K (A, A′). One may then ask: when does Int K (A′) equal Int K (A, A′)? In this section, we investigate this question and attempt to identify polynomially dense subsets of rings of integral-valued polynomials. The theory presented here is far from complete, so we raise several related questions worthy of future research.

### Definition 18.

Let $$S \subseteq T \subseteq M_{n}(K)$$. Define $$\mathrm{Int}_{K}(S,T):=\{ f \in K[X]$$f(S) ⊆ T} and Int K (T): = Int K (T, T). To say that S is polynomially dense in T means that Int K (S, T) = Int K (T).

Thus, the question posed at the start of this section can be phrased as: is A polynomially dense in A′?

In general, it is not clear how to produce polynomially dense subsets of A′, but we can describe some polynomially dense subsets of M n (D)′.

### Proposition 19.

For each Ω ⊂Λ n of cardinality at most n, choose M ∈ M n (D)′ such that Ω M = Ω. Let S be the set formed by such matrices. Then, S is polynomially dense in M n (D)′. In particular, the set of companion matrices in M n (D) is polynomially dense in M n (D)′.

### Proof.

We know that $$\mathrm{Int}_{K}(M_{n}(D)^{\prime}) \subseteq \mathrm{ Int}_{K}(S,M_{n}(D)^{\prime})$$, so we must show that the other containment holds. Let f ∈ Int K (S, M n (D)′) and N ∈ M n (D)′. Let M ∈ S such that Ω M  = Ω N . Then, f(M) is integral, so by Lemma 2 and (8), f(N) is also integral.

The proposition holds for the set of companion matrices because for any Ω ⊂ Λ n , we can find a companion matrix in M n (D) whose eigenvalues are the elements of Ω. □

By the proposition, any subset of M n (D) containing the set of companion matrices is polynomially dense in M n (D)′. In particular, M n (D) is polynomially dense in M n (D)′.

When $$D = \mathbb{Z}$$, we can say more. In  it is shown that $$\mathrm{Int}_{\mathbb{Q}}(\mathcal{A}_{n}) =\mathrm{ Int}_{\mathbb{Q}}(A_{n},\mathcal{A}_{n})$$, where A n is the set of algebraic integers of degree equal to n. Letting $$\mathcal{I}$$ be the set of companion matrices in $$M_{n}(\mathbb{Z})$$ of irreducible polynomials, we have $$\varOmega _{\mathcal{I}} = A_{n}$$. Hence, by Corollary 16 and Theorem 9, $$\mathcal{I}$$ is polynomially dense in $$M_{n}(\mathbb{Z})^{\prime}$$.

Returning to the case of a general D-algebra A, the following diagram summarizes the relationships among the various polynomial rings we have considered:
$$\displaystyle{ \begin{array}{cccc} \mathrm{Int}_{K}(A) \subseteq & \mathrm{Int}_{K}(A,A^{\prime}) & =&\mathrm{Int}_{K}(\varOmega _{A},\varLambda _{n}) \\ & \subseteq & & \subseteq \\ & \mathrm{Int}_{K}(A^{\prime}) & =&\mathrm{Int}_{K}(\varOmega _{A^{\prime}},\varLambda _{n}) \\ & \subseteq & & \subseteq \\ &\mathrm{Int}_{K}(M_{n}(D)^{\prime})& =& \mathrm{Int}_{K}(\varLambda _{n}) \end{array} }$$
(20)
From this diagram, we deduce that A is polynomially dense in A′ if and only if Ω A is polynomially dense in Ω A.

It is fair to ask what other relationships hold among these rings. We present several examples and a proposition concerning possible equalities in the diagram. Again, we point out that such equalities can be phrased in terms of polynomially dense subsets. First, we show that Int K (A) need not equal Int K (A, A′) (i.e., A need not be polynomially dense in A′).

### Example 21.

Take $$D = \mathbb{Z}$$ and $$A = \mathbb{Z}[\sqrt{-3}]$$. Then, $$A^{\prime} = \mathbb{Z}[\theta ]$$, where $$\theta = \frac{1+\sqrt{-3}} {2}$$. The ring $$\mathrm{Int}_{K}(A,A^{\prime})$$ contains both Int K (A) and Int K (A′).

If Int K (A, A′) equaled Int K (A′), then we would have Int K (A) ⊆ Int K (A′). However, this is not the case. Indeed, working mod 2, we see that for all $$\alpha = a + b\sqrt{-3} \in A$$, $$\alpha ^{2} \equiv a^{2} - 3b^{2} \equiv a^{2} + b^{2}$$. So, $$\alpha ^{2}(\alpha ^{2} + 1)$$ is always divisible by 2, and hence $$\frac{x^{2}(x^{2}+1)} {2} \in \mathrm{ Int}_{K}(A)$$. On the other hand, $$\frac{\theta ^{2}(\theta ^{2}+1)} {2} = -\frac{1} {2}$$, so $$\frac{x^{2}(x^{2}+1)} {2} \notin \mathrm{Int}_{K}(A^{\prime})$$. Thus, we conclude that $$\mathrm{Int}_{K}(A^{\prime}) \subsetneq \mathrm{ Int}_{K}(A,A^{\prime})$$.

The work in the previous example suggests the following proposition.

### Proposition 22.

Assume that D has finite residue rings. Then, A is polynomially dense in A′ if and only if IntK (A) ⊆ IntK (A′).

### Proof.

This is similar to [2, Thm. IV.4.9]. If A is polynomially dense in A′, then Int K (A′) = Int K (A, A′), and we are done because Int K (A, A′) always contains Int K (A). Conversely, assume that Int K (A) ⊆ Int K (A′). Then, Int K (A) ⊆ Int K (A′) ⊆ Int K (A, A′). Since Int K (A′) is integrally closed by Theorem 10 and Int K (A, A′) is integral over Int K (A) by Theorem 13, we must have Int K (A′) = Int K (A, A′). □

By Proposition 19, $$\mathrm{Int}_{K}(M_{n}(D)^{\prime}) =\mathrm{ Int}_{K}(M_{n}(D),M_{n}(D)^{\prime})$$. There exist algebras other than matrix rings for which Int K (A′) = Int K (A, A′). We now present two such examples.

### Example 23.

Let A = T n (D), the ring of n × n upper triangular matrices with entries in D. Define T n (K) similarly. Then, A′ consists of the integral matrices in T n (K), and since D is integrally closed, such matrices must have diagonal entries in D. Thus, $$\varOmega _{A^{\prime}} = D =\varOmega _{A}$$. It follows that $$\mathrm{Int}_{K}(T_{n}(D),T_{n}(D)^{\prime}) =\mathrm{ Int}_{K}(T_{n}(D)^{\prime})$$.

### Example 24.

Let i, j, and k be the standard quaternion units satisfying $$\mathbf{i}^{2} = \mathbf{j}^{2} = \mathbf{k}^{2} = -1$$ and $$\mathbf{i}\mathbf{j} = \mathbf{k} = -\mathbf{j}\mathbf{i}$$ (see, e.g., [8, Ex. 1.1, 1.13] or  for basic material on quaternions).

Let A be the $$\mathbb{Z}$$-algebra consisting of Hurwitz quaternions:
$$\displaystyle{A =\{ a_{0} + a_{1}\mathbf{i} + a_{2}\mathbf{j} + a_{3}\mathbf{k}\mid a_{\ell} \in \mathbb{Z}\mbox{ for all}\ell\mbox{ or }a_{\ell} \in \mathbb{Z} + \tfrac{1} {2}\mbox{ for all}\ell\}}$$
Then, for B we have
$$\displaystyle{B =\{ q_{0} + q_{1}\mathbf{i} + q_{2}\mathbf{j} + q_{3}\mathbf{k}\mid q_{\ell} \in \mathbb{Q}\}}$$
It is well known that the minimal polynomial of the element $$q = q_{0} + q_{1}\mathbf{i} + q_{2}\mathbf{j} + q_{3}\mathbf{k} \in B\setminus \mathbb{Q}$$ is $$\mu _{q}(X) = X^{2} - 2q_{0}X + (q_{0}^{2} + q_{1}^{2} + q_{2}^{2} + q_{3}^{2})$$, so A′ is the set
$$\displaystyle{A^{\prime} =\{ q_{0} + q_{1}\mathbf{i} + q_{2}\mathbf{j} + q_{3}\mathbf{k} \in B\mid 2q_{0},\;q_{0}^{2} + q_{ 1}^{2} + q_{ 2}^{2} + q_{ 3}^{2} \in \mathbb{Z}\}}$$

As with the previous example, by (20), it is enough to prove that Ω A = Ω A .

Let $$q = q_{0} + q_{1}\mathbf{i} + q_{2}\mathbf{j} + q_{3}\mathbf{k} \in A^{\prime}$$ and $$N = q_{0}^{2} + q_{1}^{2} + q_{2}^{2} + q_{3}^{2} \in \mathbb{Z}$$. Then, $$2q_{0} \in \mathbb{Z}$$, so q 0 is either an integer or a half-integer. If $$q_{0} \in \mathbb{Z}$$, then $$q_{1}^{2} + q_{2}^{2} + q_{3}^{2} = N - q_{0}^{2} \in \mathbb{Z}$$. It is known (see for instance [12, Lem. B p. 46]) that an integer which is the sum of three rational squares is a sum of three integer squares. Thus, there exist $$a_{1},a_{2},a_{3} \in \mathbb{Z}$$ such that $$a_{1}^{2} + a_{2}^{2} + a_{3}^{2} = N - q_{0}^{2}$$. Then, $$q^{\prime} = q_{0} + a_{1}\mathbf{i} + a_{2}\mathbf{j} + a_{3}\mathbf{k}$$ is an element of A such that Ω q = Ω q .

If q 0 is a half-integer, then $$q_{0} = \tfrac{t} {2}$$ for some odd $$t \in \mathbb{Z}$$. In this case, $$q_{1}^{2} + q_{2}^{2} + q_{3}^{2} = \frac{4N-t^{2}} {4} = \frac{u} {4}$$, where u ≡ 3 mod 4. Clearing denominators, we get $$(2q_{1})^{2} + (2q_{2})^{2} + (2q_{3})^{2} = u$$. As before, there exist integers a 1, a 2, and a 3 such that $$a_{1}^{2} + a_{2}^{2} + a_{3}^{2} = u$$. But since u ≡ 3 mod 4, each of the a must be odd. So, $$q^{\prime} = (t + a_{1}\mathbf{i} + a_{2}\mathbf{j} + a_{3}\mathbf{k})/2 \in A$$ is such that Ω q = Ω q .

It follows that Ω A = Ω A and thus that Int K (A, A′) = Int K (A′).

### Example 25.

In contrast to the last example, the Lipschitz quaternions $$A_{1} = \mathbb{Z} \oplus \mathbb{Z}\mathbf{i} \oplus \mathbb{Z}\mathbf{j} \oplus \mathbb{Z}\mathbf{k}$$ (where we only allow $$a_{\ell} \in \mathbb{Z}$$) are not polynomially dense in A 1′. With A as in Example 24, we have A 1 ⊂ A, and both rings have the same B, so A 1′ = A′. Our proof is identical to Example 21. Working mod 2, the only possible minimal polynomials for elements of $$A_{1}\setminus \mathbb{Z}$$ are X 2 and X 2 + 1. It follows that $$f(X) = \frac{x^{2}(x^{2}+1)} {2} \in \mathrm{ Int}_{K}(A_{1})$$. Let $$\alpha = \frac{1+\mathbf{i}+\mathbf{j}+\mathbf{k}} {2} \in A^{\prime}$$. Then, the minimal polynomial of α is $$X^{2} - X + 1$$ (note that this minimal polynomial is shared by $$\theta = \frac{1+\sqrt{-3}} {2}$$ in Example 21). Just as in Example 21, $$f(\alpha ) = -\frac{1} {2}$$, which is not in A′. Thus, $$\mathrm{Int}_{K}(A_{1})\not\subseteq \mathrm{Int}(A^{\prime})$$, so A 1 is not polynomially dense in A 1′ = A′ by Proposition 22.

## 5 Further Questions

Here, we list more questions for further investigation.

### Question 26.

Under what conditions do we have equalities in (20)? In particular, what are necessary and sufficient conditions on A for A to be polynomially dense in A′? In Examples 23 and 24, we exploited the fact that if Ω A  = Ω A, then Int K (A, A′) = Int K (A′). It is natural to ask whether the converse holds. If $$\mathrm{Int}_{K}(A,A^{\prime}) =\mathrm{ Int}_{K}(A^{\prime})$$, does it follow that Ω A  = Ω A? In other words, if A is polynomially dense in A′, then is Ω A equal to Ω A?

### Question 27.

By [2, Proposition IV.4.1] it follows that Int(D) is integrally closed if and only if D is integrally closed. By Theorem 10 we know that if A = A′, then Int K (A) is integrally closed. Do we have a converse? Namely, if Int K (A) is integrally closed, can we deduce that A = A′?

### Question 28.

In our proof (Theorem 13) that Int K (A, A′) is the integral closure of Int K (A), we needed the assumption that D has finite residue rings. Is the theorem true without this assumption? In particular, is it true whenever D is Noetherian?

### Question 29.

When is $$\mathrm{Int}_{K}(A,A^{\prime}) =\mathrm{ Int}_{K}(\varOmega _{A},\varLambda _{n})$$ a Prüfer domain? When $$D = \mathbb{Z}$$, $$\mathrm{Int}_{\mathbb{Q}}(A,A^{\prime})$$ is always Prüfer by [9, Cor. 4.7]. On the other hand, even when A = D is a Prüfer domain, Int(D) need not be Prüfer (see [2, Sec. IV.4]).

### Question 30.

In Remark 14, we proved that $$\mathrm{Int}_{K}(M_{n}(D))$$ equals an intersection of pullbacks:
$$\displaystyle{\bigcap _{M\in M_{n}(D)}(D[X] +\mu _{M}(X) \cdot K[X]) =\mathrm{ Int}_{K}(M_{n}(D))}$$
Does such an equality hold for other algebras?

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