Commutative Algebra pp 293305  Cite as
Integral Closure of Rings of IntegerValued Polynomials on Algebras
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Abstract
Let D be an integrally closed domain with quotient field K. Let A be a torsionfree Dalgebra that is finitely generated as a Dmodule. 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), a ∈ A, into elements that are integral over D. The result is obtained by identifying A with a Dsubalgebra 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
Integervalued polynomial Matrix Triangular matrix Integral closure Pullback Polynomially denseMSC(2010) classification:
[]13B25 13B22 11C201 Introduction
Let D be a (commutative) integral domain with quotient field K. The ring Int(D) of integervalued 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 integervalued polynomial to Dalgebras. When A ⊇ D is a torsionfree module finite Dalgebra 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 [6]).
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 integralvalued 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 integralvalued 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 integralvalued polynomials.
In Sect. 2, we define what we mean by integralvalued 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 integralvalued polynomials. We close by posing several problems for further research.
2 IntegralValued 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 \(\big(\begin{matrix}\scriptstyle 0&\scriptstyle q \\ \scriptstyle 0&\scriptstyle 0\end{matrix}\big) \in M_{2}(K)\), 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.
 (i)
M is integral over D
 (ii)
μ _{M} ∈ D[X]
 (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 integralvalued polynomials mentioned in the introduction.
Definition 3.
Let S ⊆ M _{ n }(K). Let K[S] denote the Ksubalgebra 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 integralvalued 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 \(\big(\begin{matrix}\scriptstyle 1&\scriptstyle 0 \\ \scriptstyle 0&\scriptstyle 0\end{matrix}\big)\) and \(\big(\begin{matrix}\scriptstyle 1/2\;&\scriptstyle 1/2 \\ \scriptstyle 1/2\;&\scriptstyle 1/2\end{matrix}\big)\) 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 Dalgebra 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, Int_{K} (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 Dalgebra 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). □
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. □
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 Int_{K} (S,S′) is integrally closed.
Proof.
3 The Case of a DAlgebra
We now use the results from Sect. 2 to gain information about Int_{ K }(A), where A is a Dalgebra. In Theorem 13 below, we shall obtain a description of the integral closure of Int_{ K }(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 Klinear 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 Dsubalgebra of M _{ n }(K). Note that n may be less than the minimum number of generators of A as a Dmodule.

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 CayleyHamilton 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 ∈ B∣b 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 integralvalued polynomials. Additionally, Lemma 2 and (8) hold for elements of B. Consequently, we have
Theorem 10.
Int_{K} (A,A′) is an integrally closed ring and is equal to \(\mathrm{Int}_{K}(\varOmega _{A},\varLambda _{n})\) .
By generalizing results from [9], 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.
Lemma 12.
Let f ∈ Int_{K} (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], \(d^{n1}h(f(X)) \in \bigcap _{a\in A}(D[X] +\mu _{a}(X) \cdot K[X])\) .
Proof.
Let a ∈ A. Since f ∈ Int_{ K }(A, A′), m: = μ _{ f(a)} ∈ D[X], and deg(m) ≤ n.
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 D∕I is finite. Clearly, this condition is equivalent to asking that for all nonzero d ∈ D, the residue ring D∕dD 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 Int_{K} (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].
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 CayleyHamilton 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 Int_{K} (M _{n} (D)) is Int_{K} (Λ _{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 Int_{K} (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, \(\mathrm{Int}_{K}(D,\varLambda _{n}) =\mathrm{ Int}(D)\). □
4 Matrix Rings and Polynomially Dense Subsets
For any Dalgebra 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 integralvalued 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 [10] 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 \(M_{n}(\mathbb{Z})^{\prime}\).
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 Int_{K} (A) ⊆ Int_{K} (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 [3] for basic material on quaternions).
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 halfinteger. 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 halfinteger, then \(q_{0} = \tfrac{t} {2}\) for some odd \(t \in \mathbb{Z}\). In this case, \(q_{1}^{2} + q_{2}^{2} + q_{3}^{2} = \frac{4Nt^{2}} {4} = \frac{u} {4}\), 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 _{ ℓ } 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.
Notes
Acknowledgements
The authors wish to thank the referee for his/her suggestions. The first author wishes to thank Daniel Smertnig for useful discussions during the preparation of this paper about integrality in noncommutative settings. The same author was supported by the Austrian Science Foundation (FWF), Project Number P23245N18.
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