Integral closure of rings of integer-valued polynomials on algebras

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1 Integral closure of rings of integer-valued polynomials on algebras Giulio Peruginelli Nicholas J. Werner July 4, 013 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), a A, 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 Primary: 13F0 Secondary: 13B5, 13B 11C0. 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], [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 Int K (A) := {f K[X] f(a) A}. The set Int K (A) forms a Institut für Analysis und Comput. Number Theory, Technische Univ., Steyrergasse 30, A-8010 Graz, Austria. peruginelli@math.tugraz.at. University of Evansville, 1800 Lincoln Avenue, Evansville, IN, USA. nw89@evansville.edu 1

2 commutative ring contained in Int(D), and often shares properties similar to those of Int(D) (see the references above, especially [5]). 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 Int Q (M n (Z)) is Int Q (A n ), where A n is the set of algebraic integers of degree at most n, and Int Q (A n ) = {f Q[X] f(a n ) 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 3.3) may be considered an extension of both [9, Thm. 4.6] and [, Thm. IV.4.7] (the latter originally proved in [7, Prop..]), which states that if D is Noetherian and D is its integral closure in K, then the integral closure of Int(D) equals Int(D, D ) = {f K[X] f(d) D }. 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 K be an algebraic closure of K. We will establish a close connection between integral-valued polynomials and polynomials that act on elements of K that are integral over D. We will also investigate polynomially dense subsets of rings of integral-valued polynomials. In Section, 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 K. In Section 3, we apply the results of Section 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. Integral-valued polynomials Throughout, we assume that D is an integrally closed domain with quotient field K. We denote by 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 K[X] 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 (K), which are the roots of µ M in K. For a subset S M n (K), we define Ω S := M S Ω 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 ( ) 0 q 0 0 M (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.. Let M M n (K). The following are equivalent:

3 (i) M is integral over D (ii) µ M D[X] (iii) each α Ω M is integral over D Proof. (i) (ii) By assumption, there exists a monic p D[X] which is contained in N K[X] (M). So, µ M is a monic polynomial in K[X] that divides p. Since D is integrally closed, by the Corollary after Proposition 11 of [1, Chap. V], µ M D[X]. (ii) (iii) This is clear, since the elements of Ω M are the roots of µ M. (iii) (i) 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]. 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 = Z and S = M (Z), then both ( ) and ( 1/ 1/ 1/ 1/ ) are in S, but neither their sum nor their product is integral. Lastly, even 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 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. 3

4 Proposition.6. Let S M n (K). Int K (S, S ) Int(D). Then, Int K (S, S ) is a ring, and if D S, then Proof. Let M S and f, g Int K (S, S ). Then, f(m), g(m) are integral over D. By Corollary 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, the containment Int K (S, S ) Int(D) holds because D is integrally closed. We now begin to connect our rings of integral-valued polynomials to rings of polynomials that act on elements of K that are integral over D. For each n > 0, let Λ n := {α K α solves a monic polynomial in D[X] of degree n} For any subset E of Λ n, define Int K (E, Λ n ) := {f K[X] f(e) Λ n } to be the set of polynomials in K[X] mapping elements of E into Λ n. If E = Λ n, then we write simply Int K (Λ n ). As with Int K (S, S ), Int K (E, Λ n ) is a ring despite the fact that Λ n is not. Proposition.7. For any E Λ n, Int K (E, Λ n ) is a ring, and is integrally closed. Proof. Let Λ be the integral closure of D in K. We set Int K (E, Λ ) = {f K[X] f(e) Λ }. Then, Int K (E, Λ ) is a ring, and by [, Prop. IV.4.1] it is integrally closed. Let Int K (E, Λ ) = {f K[X] f(e) Λ }. Clearly, Int K (E, Λ n ) Int K (E, Λ ). However, if α E and f K[X], then [K(f(α)) : K] [K(α) : K] n, so in fact Int K (E, Λ n ) = Int K (E, Λ ). Finally, since Int K (E, Λ ) = Int K (E, Λ ) K[X] is the contraction of Int K (E, Λ ) to K[X], it is an integrally closed ring, proving the proposition. Theorem 4.6 in [9] shows that the integral closure of Int Q (M n (Z)) equals the ring Int Q (A n ), where A n is the set of algebraic integers of degree at most n. As we shall see (Theorem.9), this is evidence of a broader connection between the rings of integralvalued 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. 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 χ M (X) = i=1,...,n (X α i) is the characteristic polynomial of M (the roots α i are in K and there may be repetitions), then the characteristic polynomial of f(m) is χ f(m) (X) = i=1,...,n (X f(α i)). Phrased in terms of our Ω-notation, we have: if M M n (K) and f K[X], then Ω f(m) = f(ω M ) = {f(α) α Ω M }. (.8) 4

5 Using this fact and our previous work, we can equate Int K (S, S ) with a ring of the form Int K (E, Λ n ). Theorem.9. Let S M n (K). Then, Int K (S, S ) = Int K (Ω S, Λ n ), and in particular Int K (S, S ) is integrally closed. Proof. We first prove this for S = {M}. Using Lemma. and (.8), for each f K[X] we have: f(m) S f(m) is integral Ω f(m) Λ n f(ω M ) Λ n. This proves that Int K ({M}, {M} ) = Int K (Ω M, Λ n ). For a general subset S of M n (K), we have Int K (S, S ) = Int K ({M}, S ) = Int K (Ω M, Λ n ) = Int K (Ω S, Λ n ). M S M S 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 matrix M we have these inclusions: D D[M] {M} K[M] so if M is integral, {M} is equal to the integral closure of D[M] in K[M]. 3 The case of a D-algebra We now use the results from Section to gain information about Int K (A), where A is a D-algebra. In Theorem 3.3 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 Int K (A) := {f K[X] f(a) A}. Letting n be the vector space dimension of B over K, we also have an embedding B M n (K), b 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 End K (B) sending b B to the endomorphism x b 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. 5

6 In light of the aforementioned matrix representation of B, several of the definitions and notations we defined in Section will carry over to B. Since the concepts of minimal polynomial and eigenvalue are independent of the representation B 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(µ b ) n, for all b B. for all b B, Ω b = {roots of µ b in K}. The elements of Ω b are nothing else than the eigenvalues of b under any matrix representation B M n (K). If S B, then Ω S = b S Ω 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} Int K (A, A ) = {f K[X] f(a) A } Working exactly as in Proposition.6, we find that Int K (A, A ) is another ring of integral-valued polynomials. Additionally, Lemma. and (.8) hold for elements of B. Consequently, Int K (A, A ) is an integrally closed ring and is equal to Int K (Ω A, Λ n ), which is the ring of integral-valued polynomials on the set of eigenvalues of all the elements of A. 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 [, Thm. IV.4.7] (originally proved in [7, Prop..]) 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 (D[X] + µ a (X) K[X]) Int K (A) a A because if f D[X] + µ a (X) K[X], then f(a) D[a] A. We thus have a chain of inclusions (D[X] + µ a (X) K[X]) Int K (A) Int K (A, A ) (3.1) a A and our work below will show that this is actually a chain of integral ring extensions. 6

7 Lemma 3.. 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 n 1 h(f(x)) a A (D[X] + µ a(x) 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, 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)) 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)) D[X] + µ a (X) K[X], and since a was arbitrary the lemma is true. Theorem 3.3. Assume that D has finite residue rings. Then, Int K (A, A ) = Int K (Ω A, Λ n ) is the integral closure of both a A (D[X] + µ a(x) K[X]) and Int K (A). Proof. Let R = a A (D[X] + µ a(x) K[X]). By (3.1), it suffices to prove that Int K (A, A ) is the integral closure of R. Let f(x) = g(x)/d Int K (A, A ). By Theorem.9, 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 P = {µ f(a) (X) a A} and let P D[X] be a set of monic representatives for the residues in P modulo (d n 1 ). Since D has finite residue rings, P is finite. Let φ(x) be the product of all the polynomials in P. Then, φ is monic and is in D[X]. Fix a A and let m = µ f(a). There exists p(x) P such that p(x) is equivalent to m mod (d n 1 ), so p(x) = m(x) + (d n 1 ) 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, φ(f(x)) = p(f(x))q(f(x)) = m(f(x))q(f(x)) + (d n 1 ) r(f(x))q(f(x)) = m(f(x))q(f(x)) + d n 1 r(f(x)) d n 1 q(f(x)) As in Lemma 3., m(f(x))q(f(x)) µ a (X) K[X] because m(f(a)) = 0. By Lemma 3., d n 1 r(f(x)) d n 1 q(f(x)) is in D[X]. Hence, φ(f(x)) D[X] + µ a (X) K[X], and since a was arbitrary, φ(f(x)) R. Theorem 3.3 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 3.4. By following essentially the same steps as in Lemma 3. and Theorem 3.3, 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. 7

8 In the case A = M n (D), Int K (M n (D)) is equal to the intersection of the pullbacks D[X] + µ M (X) K[X], for M M n (D). Indeed, let f Int K (M n (D)) and M M n (D). By [11, Remark.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) χ M (X) so that f D[X] + χ M (X)K[X] D[X] + µ M (X)K[X] and we are done. Remark 3.5. The assumption that D has finite residue rings implies that D is Noetherian. Given that [, Thm. IV.4.7] (or [7, Prop..]) requires only the assumption that D is Noetherian, it is fair to ask if Theorem 3.3 holds under the weaker condition that D is Noetherian. Since Ω Mn(D) = Λ n, we obtain [9, Thm. 4.6] as a corollary: Corollary 3.6. Let A n be the set of algebraic integers of degree at most n. The integral closure of Int Q (M n (Z)) is Int Q (A n ). The algebra of upper triangular matrices yields another interesting example. Proposition 3.7. 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. Let A = T n (D). For each a A, µ a splits completely and has roots in D, so Ω A = D. Hence, the integral closure of Int K (T n (D)) is Int K (Ω A, Λ n ) = Int K (D, Λ n ). But, polynomials in K[X] that move D into Λ n actually move D into Λ n K = D. Thus, Int K (D, Λ n ) = Int(D). Since Int K (T n (D)) Int K (T n 1 (D)) for all n > 0, the previous proposition proves that is a chain of integral ring extensions. Int K (T n (D)) Int K (T n 1 (D)) Int(D) 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.9. 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. 8

9 Definition 4.1. Let S T M n (K). Define Int K (S, T ) := {f 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 4.. 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 Int K (M n (D) ) Int K (S, M n (D) ), 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. 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). Thus we have Corollary 4.3. When D has finite residue rings, Int K (M n (D) ) = Int K (M n (D), M n (D) ) = Int K (Λ n ) is the integral closure Int K (M n (D)). When D = Z, we can say more. In [10] it is shown that Int Q (A n ) = Int Q (A n, A n ), where A n is the set of algebraic integers of degree equal to n. Letting I be the set of companion matrices in M n (Z) of irreducible polynomials, we have Ω I = A n. Hence, by Corollary 3.6 and Theorem.9, I is polynomially dense in M n (Z). Returning to the case of a general D-algebra A, the following diagram summarizes the relationships among the various polynomial rings we have considered: Int K (A) Int K (A, A ) = Int K (Ω A, Λ n ) Int K (A ) = Int K (Ω A, Λ n ) (4.4) Int K (M n (D) ) = Int K (Λ n ) 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 ) (that is, A need not be polynomially dense in A ). 9

10 Example 4.5. Take D = Z and A = Z[ 3]. Then, A = Z[θ], where θ = The ring Int K (A, A ) 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, we see that for all α = a + b 3 A, α a 3b a +b. So, α (α +1) is always divisible by, and hence x (x +1) Int K (A). θ On the other hand, (θ +1) = 1, so x (x +1) / Int K (A ). Thus, we conclude that Int K (A ) Int K (A, A ). The work in the previous example suggests the following proposition. Proposition 4.6. 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 [, 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.9 and Int K (A, A ) is integral over Int K (A) by Theorem 3.3, we must have Int K (A ) = Int K (A, A ). By Proposition 4., Int K (M n (D) ) = Int K (M n (D), M n (D) ). There exist algebras other than matrix rings for which Int K (A ) = Int K (A, A ). We now present two such examples. Example 4.7. 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, Ω A = D = Ω A. It follows that Int K (T n (D), T n (D) ) = Int K (T n (D) ). Example 4.8. Let i, j, and k be the standard quaternion units satisfying i = j = k = 1 and ij = k = ji (see e.g. [8, Ex. 1.1, 1.13] or [3] for basic material on quaternions). Let A be the Z-algebra consisting of Hurwitz quaternions: A = {a 0 + a 1 i + a j + a 3 k a l Z for all l or a l Z + 1 for all l} Then, for B we have B = {q 0 + q 1 i + q j + q 3 k q l Q} It is well known that the minimal polynomial of the element q = q 0 +q 1 i+q j+q 3 k B \Q is µ q (X) = X q 0 X + (q 0 + q 1 + q + q 3 ), so A is the set A = {q 0 + q 1 i + q j + q 3 k B q 0, q 0 + q 1 + q + q 3 Z} As with the previous example, we will prove that Ω A = Ω A. Let q = q 0 +q 1 i+q j+q 3 k A and N = q0 + q 1 + q + q 3 Z. Then, q 0 Z, so q 0 is either an integer or a halfinteger. If q 0 Z, then q1 + q + q 3 = N q 0 Z. It is known (see for instance [1, 10

11 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, a 3 Z such that a 1 + a + a 3 = N q 0. Then, q = q 0 + a 1 i + a j + a 3 k is an element of A such that Ω q = Ω q. If q 0 is a half-integer, then q 0 = t for some odd t Z. In this case, q 1 + q + q 3 = 4N t 4 = u 4, where u 3 mod 4. Clearing denominators, we get (q 1) +(q ) +(q 3 ) = u. As before, there exist integers a 1, a, and a 3 such that a 1 + a + a 3 = u. But since u 3 mod 4, each of the a l must be odd. So, q = (t + a 1 i + a j + a 3 k)/ 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 4.9. In contrast to the last example, the Lipschitz quaternions A 1 = Z Zi Zj Zk (where we only allow a l Z) are not polynomially dense in A 1. With A as in Example 4.8, we have A 1 A, and both rings have the same B, so A 1 = A. Our proof is identical to Example 4.5. Working mod, the only possible minimal polynomials for elements of A 1 \ Z are X and X + 1. It follows that f(x) = x (x +1) Int K (A 1 ). Let α = 1+i+j+k A. Then, the minimal polynomial of α is X X + 1 (note that this minimal polynomial is shared by θ = 1+ 3 in Example 4.5). Just as in Example 4.5, f(α) = 1, which is not in A. Thus, Int K (A 1 ) Int(A ), so A 1 is not polynomially dense in A 1 = A by Proposition Further questions Here, we list more questions for further investigation. Question 5.1. Under what conditions do we have equalities in (4.4)? In particular, what are necessary and sufficient conditions on A for A to be polynomially dense in A? In Examples 4.7 and 4.8, 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 Int K (A, A ) = Int K (A ), does it follow that Ω A = Ω A? In other words, if A is polynomially dense in A, then is Ω A polynomially dense in Ω A (with respect to mappings into Λ n )? Question 5.. By [, Proposition IV.4.1] it follows that Int(D) is integrally closed if and only if D is integrally closed. By Theorem.9 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 5.3. In our proof (Theorem 3.3) 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 5.4. When is Int K (A, A ) = Int K (Ω A, Λ n ) a Prüfer domain? When D = Z, Int Q (A, A ) is always Prüfer by [9, Cor. 4.7]. However, the case where D is a general integrally closed domain with finite residue rings is unresolved. 11

12 Question 5.5. In Remark 3.4, we proved that Int K (M n (D)) equals an intersection of pullbacks: (D[X] + µ M (X) K[X]) = Int K (M n (D)) M M n(d) Does such an equality hold for other algebras? References [1] N. Bourbaki. Commutative Algebra. Hermann, Paris; Addison-Wesley Publishing Co., Reading, Mass., 197. [] J.-P. Cahen, J.-L. Chabert. Integer-valued polynomials. Amer. Math. Soc. Surveys and Monographs, 48, Providence, [3] L. E. Dickson. Algebras and their arithmetics. Dover Publications, New York, [4] S. Evrard, Y. Fares, K. Johnson. Integer valued polynomials on lower triangular integer matrices. Monats. für Math. 170 (013), no., [5] S. Frisch. Integer-valued polynomials on algebras. J. of Algebra 373 (013), [6] S. Frisch. Corrigendum to Integer-valued polynomials on algebras. J. of Algebra, in press (013). [7] R. Gilmer, W. Heinzer, D. Lantz. The Noetherian property in rings of integer-valued polynomials. Trans. Amer. Math. Soc. 338 (1993), no. 1, [8] T. Y. Lam. A First Course in Noncommutative Rings. Springer, New York, [9] K. Alan Loper, Nicholas J. Werner. Generalized Rings of Integer-valued Polynomials. J. Number Theory 13 (01), no. 11, [10] G. Peruginelli. Integral-valued polynomials over sets of algebraic integers of bounded degree. submitted, [11] G. Peruginelli. Integer-valued polynomials over matrices and divided differences. published in Monatshefte für Mathematik, [1] J. P. Serre. A Course in Arithmetic. Springer, New York,