Let A be an nbyn matrix with entries in a commutative integral domain R with quotient field F. We are interested in the question of when the (monic) minimum polynomial of A over F has all its coefficients in R. This is true if R is integrally closed because the minimum polynomial is a factor of the characteristic polynomial. It is also true in the following two simple cases:
These are the only cases for n = 2, so every 2by2 matrix has this property. For a 3by3 matrix there is one other case: The characteristic polynomial of A is (Xa)^{2}(Xb) and the minimum polynomial is (Xa)(Xb) with a,b Î F integral over R (possibly a = b). In [1] William C. Brown considered the matrix

We saw that the question for 3by3 matrices reduces to the case where the minimum polynomial is quadratic. Let us say that R is 3closed if the minimum polynomial of every 3by3 matrix over R has its coefficients in R, and that R is quadratically closed if that is true for quadratic minimum polynomials of any square matrix over R. So if R is quadratically closed, then R is 3closed.
Let D be a domain and M a submonoid of t^{N} = {t^{n}:n Î N}, where t is an indeterminate. Consider the monoid algebra R = D[M]. We can represent M by an infinite binary string, like 10111¼ for the monoid generated by t^{2} and t^{3}. Brown's example of the preceding section can be elaborated upon.
Theorem 1 If the binary string of the monoid M contains the pattern 11011, then the monoid algebra R = D[M] is not 3closed.
Proof. Suppose the pattern 11011 starts at position k (for that to happen, k must be at least 3). Consider the matrix


Similarly, replacing t by t^{n} in B, we see that any string containing k, k+n, k+3n, k+4n and missing k+2n, is not 3closed. So the pattern

For the purpose of deciding whether or not D[M] is 3closed, we might as well assume that the string for M ends in all 1's. Otherwise, M = (M^{¢})^{n}, where n > 1, and D[M] @ D[M^{¢}].
If we look at Brown's example from another angle, we can construct a number ring that is not 3closed. Instead of thinking of k[t^{3},t^{4}], we can think of k[x,y]/(x^{4}y^{3}). In the latter ring we can identify t as y/x = t^{4}/t^{3}. So we look for a number ring with elements x and y so that x^{4} = y^{3}.
Let a = ^{3}Ö{16}, so a^{3} = 2^{4}, whence (a/2)^{3} = 2. Let t = a/2, so t^{3} = 2 and t^{4} = a are in Z[a], and t^{5} = a^{2}/2 is not in Z[a]. Our matrix is

This summarizes what we know of the negative side of the problem. We now turn to the positive side.
Let R Ì R^{¢} be integral domains. We say that an element x Î R^{¢} is strongly integral over R if there is a nonzero finitely generated ideal J of R such that xJ Ì J^{2}. In particular, if x is strongly integral over R, then x is in the quotient field of R, and x is integral over R. We say that R is weakly integrally closed in R^{¢} if each element of R^{¢} that is strongly integral over R is in R. By the preceding observation, if R is an integrally closed domain, then R is weakly integrally closed in R^{¢}.
To show (2), let J be a nonzero finitely generated ideal of R such that xJ Ì J^{2} for x Î R^{¢¢}. Then J^{¢} = R^{¢}J is a nonzero finitely generated ideal of R^{¢} such that xJ^{¢} Ì J^{¢2}, so x Î R^{¢}, so x Î R.
Let J be a nonzero finitely generated ideal of R and xJ Ì J^{2} for some x Î R^{¢}. Then xJ Ì J, so x is integral over R, whence x Î R. This proves (3).
To show (4a), suppose R is weakly integrally closed in R^{¢}. Let x Î R_{S}^{¢} and J be a nonzero finitely generated ideal of R_{S}. There is s Î S such that sJ is a finitely generated ideal of R, and sx Î R^{¢}. If xJ Ì J^{2}, then sxsJ Ì (sJ)^{2}, so sx Î R whence x Î R_{S}. Therefore R_{S} is weakly integrally closed in R_{S}^{¢}.
For (4b), suppose R_{M} is weakly integrally closed in R_{M}^{¢} for every maximal ideal M of R. Let x Î R^{¢} and J be a nonzero finitely generated ideal of R such that xJ Ì J^{2}. Let C = {r Î R:rx Î R}. If M is any maximal ideal of R, then x Î R_{M}, so rx Î R for some r Î R\M. Therefore C is not contained in M for any maximal ideal M, so C = R, whence x Î R.
Because of (1) we can talk about the weak integral closure of R in R^{¢}. If R is weakly integrally closed in its quotient field, then we say simply that R is weakly integrally closed. We shall see later that the weak integral closure need not coincide with the set of strongly integral elements (Example 1).
Theorem 3 If R is weakly integrally closed, then R is quadratically closed. That is, if A is a nonzero square matrix with entries in R that satisfies a monic quadratic polynomial with coefficients in the quotient field of R, then those coefficients are in R.
Proof. Suppose


So D[t^{3},t^{4}] is not weakly integrally closed, because it is not 3closed, being a monoid algebra with string 100110111¼ .
What are some examples of rings that are weakly integrally closed but not integrally closed? For monoid algebras, the following is our best positive result. First a lemma.
Lemma 4 If D is an integrally closed domain with quotient field K, and M is a submonoid of t^{N}, then D[M] is weakly integrally closed if and only if K[M] is.
Proof. We have D[M] = K[M]ÇD[t], and D[t] is integrally closed. Thus if K[M] is weakly integrally closed, so is D[M]. Conversely, K[M] is a localization of D[M].
Theorem 5 Let D be an integrally closed domain and t an indeterminate. Then the domain

Proof. From Lemma 4, we may assume that D is a field.
The theorem is clearly true for i = 1 or m = 1. Fix i > 1, write R_{m} for R_{i,m}, and proceed by induction on m. Note that

For any element e Î D[t] let e = e_{0}+e_{1}t+e_{2}t^{2}+¼ and define the value v_{t}(e) to be the least j such that e_{j} ¹ 0. Let n be the minimum of v_{t}(e), as e ranges over J. As v_{t}(xe) ³ 2n for each e Î J, it follows that v_{t}(x) ³ n.
We pause for an observation. Let a Î R_{m1} and b Î R_{m}. Suppose v_{t}(a) ³ n and v_{t}(b) ³ n, where n < m, and i divides n. Then


Now suppose n < m, so n is a multiple of i. Recall that v_{t}(x) ³ n. If e and e^{¢} are in J, then (ee^{¢})_{m1+n} = e_{m1}e_{n}^{¢} = 0, by the observation above. So if f is any element of J^{2}, then f_{m1+n} = 0. Therefore, if e Î J, then (xe)_{m1+n} = 0.
If e Î J has value n, then 0 = (xe)_{m1+n} = x_{m1}e_{n} from the observation again. So x_{m1} = 0, whence x Î R_{m}.
We can get a simple characterization of when D[t^{m},t^{n}] is weakly integrally closed (or 3closed) when D is integrally closed.
Theorem 6 Let R = D[t^{m},t^{n}] with D a domain and 1 £ m < n. If m divides 2n, and D is integrally closed, then R is weakly integrally closed. If m does not divide 2n, then R is not 3closed.
Proof. As the hypothesis and the conclusion are invariant under dividing m and n by a common factor, we may assume that m and n are relatively prime. If m divides 2n, then m divides 2, so m = 1 or m = 2. If m = 1, then R = D[t] is integrally closed, hence weakly integrally closed. If m = 2, then R = D[t^{2},t^{n}] where n is odd. Then Theorem 5 shows that R is weakly integrally closed.
So suppose that m ³ 3 is relatively prime to n > m. We will show that R is not 3closed by finding the pattern 11011 in the binary string of the monoid M generated by m and n (Theorem 1). Consider the mbyn array

If n = m+1, then the first column of the smaller array, and no other, consists entirely of numbers that are not divisible by m. As the first entry of the last column is equal to m, this means that (m2)n+1 is not in M, while the four integers closest to it are in M (this uses m > 2): the pattern 11011.
If n > m+1, then there is a column of the smaller array, other than the first, which consists of numbers that are not divisible m. Pick the rightmost such column. Its last entry is the 0 of 11011.
The weak integral closure does not always coincide with the set of strongly integral elements, which need not be closed under addition (but is closed under multiplication).
Example 1 Let D be an integrally closed domain. The weak integral closure of R = D[t^{3},t^{4}] is R^{¢} = D[t^{3},t^{4},t^{5}]. The set of elements in R^{¢} that are strongly integral over R is RÈM^{¢}, where M^{¢} = {f Î R^{¢}:f(0) = 0}.
Proof. We see from Theorem 5 that R^{¢} is weakly integrally closed. Let M = M^{¢}ÇR. If x Î M^{¢}, then xM Ì M^{2}, while if x Î R, then xR Ì R^{2}, so the elements of RÈM^{¢} are strongly integral over R, and R^{¢} = DÅM^{¢} is the weak integral closure of R.
Now suppose x Î R^{¢} but not in RÈM^{¢}, and J is a nonzero ideal of R. If J is not contained in M, then xJ is not contained in R. If J Ì M, let the smallest tvalue of an element of J be k ³ 3. The smallest tvalue of an element of J^{2} is then 2k > k, so xJ cannot be contained in J^{2}.
In particular, although 1 and t^{5} are strongly integral over D[t^{3},t^{4}], the sum 1+t^{5} is not.
Recall that the integral closure of an ideal I in a ring R consists of those elements x Î R that satisfy an equation of the form

The conductor of R in R^{¢} is {x Î R:xR^{¢} Ì R}. It is the largest R^{¢}ideal contained in R. Here is a general sufficient condition for R to be weakly integrally closed in R^{¢}.
Theorem 7 Let R Ì R^{¢} be integral domains, and C the conductor of R in R^{¢}. Suppose that
Then R is weakly integrally closed in R^{¢}.
Proof. Let J be a finitely generated nonzero ideal of R, and x Î R^{¢} such that xJ Ì J^{2}. Suppose x Ï R. As C = {r Î R:rx Î R}, we have J Ì C. As xJ Ì J^{2}, the element x is integral over C, hence in C Ì R.
Note that the first condition in Theorem 7 holds if C is a maximal ideal in R.
Corollary 8 If the conductor of R in R^{¢} is a finite intersection of maximal ideals of R, and is integrally closed in R^{¢}, then R is weakly integrally closed in R^{¢}.
Proof. By Theorem 2 it suffices to prove that R_{M} is weakly integrally closed in R_{M}^{¢} for each maximal ideal M of R. Let C be the conductor of R in R^{¢}. If M does not contain C, then R_{M}^{¢} = R_{M} because rR^{¢} Ì R for any r Î C\M. If M contains C, then CR_{M} = MR_{M} because C is a finite intersection of maximal ideals. Moreover, the maximal ideal CR_{M} is the conductor of R_{M} in R_{M}^{¢}. Finally, CR_{M} is integrally closed as an ideal of R_{M}^{¢} because C is integrally closed as an ideal of R^{¢}. The result now follows from Theorem 7.
Corollary 9 If R^{¢} is a Dedekind domain, and the conductor of R in R^{¢} is a finite intersection of maximal ideals of R, then R is weakly integrally closed in R^{¢}.
Corollary 9 covers the rings K[t^{m},t^{m+1},t^{m+2},¼], for K a field (and so for K integrally closed), and also the rings Z[Öd] where d is squarefree and congruent to 1 modulo 4, as we will see in the next section. We can use Theorem 7 to extend Theorem 5 via the next theorem.
Theorem 10 Let M be a submonoid of t^{N} and t^{m} the smallest nonunit power of t in M. Let D be an integrally closed domain, R = D[M\{t^{m}}] and R^{¢} = D[M]. If R^{¢} is weakly integrally closed, then so is R.
Proof. From Lemma 4 we may assume that D is a field. The conductor C of R in R^{¢} is the linear span of M\{1,t^{m}}, a maximal ideal in R. Moreover, the obvious discrete valuation v on R^{¢} has the property that C = {r Î R^{¢}:vr > m}, so C is integrally closed in R^{¢}. The result now follows from Corollary 8.
Iterating this theorem gives yet another proof that D[t^{m},t^{m+1},t^{m+2},¼] is weakly integrally closed, if D is integrally closed, by deleting t,t^{2},¼,t^{m1} successively from D[t]. We can also apply it to the other rings in Theorem 5 to show, for example, that D[t^{4},t^{6},t^{9},t^{11}] is weakly integrally closed (from D[t^{2},t^{9}] by deleting t^{2}).
Let 1,q be an integral basis for the integers in a quadratic field Q(q). Any subring of the integers of Q(q) has the form R = Z[nq] = ZÅnZq. In particular, when d is squarefree and congruent to 1 modulo 4, then Z[Öd] = Z[2q] where q = (1+Öd)/2. The conductor of R in its integral closure Z[q] is C = nZ[q], and R/C is isomorphic to Z/nZ. So the conductor is maximal in R if and only if n is a prime, in which case R is weakly integrally closed by Corollary 9.
What about Z[4i]? It suffices to show that Z[4i] is weakly integrally closed in Z[2i]. The conductor is the unique maximal ideal M = 2ZÅ4Zi = 2Z[2i] of Z[4i] that contains 2. Is M integrally closed in Z[2i]? No because (2i)^{2} = 2^{2} Î M^{2} shows that 2i is integral over M. So this is not covered by Theorem 7. Nevertheless, Z[4i] is weakly integrally closed, as we shall see.
If D is a subring of an integral domain R, we say that R is a separable quadratic extension of D if R is a subring of a separable quadratic extension field of the quotient field K of D. By the primitive element theorem, it suffices to require that each element of R satisfies a separable quadratic polynomial over K.
Theorem 11 Any separable quadratic integral extension of a discrete valuation domain is weakly integrally closed.
Proof. Let D be a discrete valuation domain with prime p, and R a separable quadratic integral extension of D. Let 1,q be an integral basis for the integral closure D[q] of R (see [2]). Then either R = D[q], which is integrally closed, or R = D[p^{k+1}q] for some nonnegative integer k. By induction and transitivity, it suffices to show that R = D[p^{k+1}q] is weakly integrally closed in R^{¢} = D[p^{k}q].
Note that D[q] is a principal ideal domain with at most two primes. Note also that R is a local ring, and its unique maximal ideal, pR^{¢}, is the conductor of R in R^{¢}.
Write q^{2} = s+tq where s,t Î D. Suppose J is a proper finitely generated ideal of R. For x Î D, let vx denote the value of x. We can generate J as a Dmodule by a+bq and c, where va ³ 1 and vb ³ k+1. We may assume that J is not principal, so c ¹ 0. If vc £ va, then we may, and do, take a = 0. Note that J^{2} is generated by a^{2}+sb^{2}+(2ab+tb^{2})q, and c^{2} and ca+cbq.
Suppose (r+p^{k}q)J Ì J^{2}. Then a+bq Î J and


So we may assume that va > k and vc > k. Note that p^{k+1} does not divide r+p^{k}q. Let p be the prime in D[q] that divides r+p^{k}q the least. Let n be the pvalue of p. Then the pvalue of r+p^{k}q is less than n(k+1). As va > k and vc > k, we have J Ì p^{m}D[q]\p^{m+1}D[q] for some m ³ n(k+1). So J^{2} Ì p^{2m}D[q] but (r+p^{k}q)J is not contained in p^{2m}D[q].
We can eliminate the requirement that the extension be integral in Theorem 11.
Theorem 12 Any separable quadratic extension of a discrete valuation domain is either integral, a field, or a principal ideal domain.
Proof. Suppose D is a discrete valuation domain with prime p and quotient field K, and R is a separable quadratic extension of D. Suppose D[q] = DÅDq is integrally closed, where q Î KR = K[q], the quotient field of R, and q^{2} = sq+t with s,t Î D. Let l be an element of R Ì K[q]. We will show that if l is not in D[q], then D[l] contains either K or q. Thus either R Ì D[q], in which case R is integral over D; or R contains K, in which case R = KR is a field; or R contains q, in which case R is between D[q], a principal ideal domain, and its quotient field K[q], so is a principal ideal domain.
Multiplying l by an appropriate power of p we may assume that



Having done the discrete valuation domain case, we proceed to the global case.
Theorem 13 Let R^{¢} be a separable quadratic extension of a domain R. Let F be a family of prime ideals of R such that R_{P} is a discrete valuation domain for each P Î F. If R^{¢} = Ç_{P Î F}R_{P}^{¢}, then R^{¢} is weakly integrally closed.
Proof. It suffices to show that R_{P}^{¢} is weakly integrally closed for each P in F. But R_{P}^{¢} is a separable quadratic extension of the discrete valuation domain R_{P}.
Corollary 14 Let R^{¢} be a separable quadratic extension of a Krull domain R. If R is a Dedekind domain, or if R^{¢} is a free Rmodule, then R^{¢} is weakly integrally closed.
Proof. Let F be the set of heightone prime ideals of R. In both cases, localization at an ideal in F gives a discrete valuation domain. When R is Dedekind, these ideals are maximal, so R^{¢} is equal to the indicated intersection. When R^{¢} is a free Rmodule, this equality follows from the fact that R = Ç_{P Î F}R_{P}.
Corollary 15 If R is a Dedekind domain with quotient field K, and L is a separable quadratic extension of K, then any ring between R and L is weakly integrally closed. In particular, any ring of quadratic algebraic numbers is weakly integrally closed.
Recall from Section 2.1 that the ring Z[^{3}Ö{16}], which is a free cubic extension of Z, is not weakly integrally closed.
The conductor, M = pR^{¢}, of R in R^{¢} need not be integrally closed for rings of quadratic integers. For Z[pi] Ì Z[i], the conductor is maximal and integrally closed. For Z[p^{2}i] Ì Z[pi], the conductor is maximal but not integrally closed as (pi)^{2} = p^{2} Î M^{2}. So Theorem 7 does not apply. For Z[p^{2}i] Ì Z[i] the conductor p^{2}Z[i] is integrally closed but not maximal.
As a nonintegral example of Corollary 15, consider the ring Z[(3+4i)/5]. If we localize at a prime other than 2 or 5 we get Z_{(p)}[i], which is integrally closed. Localizing at 2 gives Z_{(2)}[4i], which is weakly integrally closed. If we localize at 5 we get Z[i] localized at the prime ideal (2+i), a discrete valuation domain, so integrally closed.
We address the question as to whether R[X] is weakly integrally closed whenever R is. First we observe that all of our constructions of weakly integrally closed domains R have the property that R[X] is also weakly integrally closed.
If R is a separable quadratic integral extension of a discrete valuation domain D, then R is free over D, so R[X] is a free separable quadratic extension of the Krull domain D[X]. So R[X] is weakly integrally closed. This also works for other separable quadratic extensions R of discrete valuation domains as these are either fields or principal ideal domains. If D is a Dedekind domain, then R_{P}[X] is weakly integrally closed for each maximal ideal P of D, so R[X] is weakly integrally closed.
As for the monoid rings, if R = D[M] where D is integrally closed, then R[X] = D[X][M] and D[X] is integrally closed.
Here are some general facts about being strongly integral over R[X] which fall short of showing that R[X] is weakly integrally closed whenever R is.
Lemma 16 Let R Ì R^{¢} be integral domains. If f in R^{¢}[X] is strongly integral over R[X], then f(r) is strongly integral over R for each element r in R.
Proof. Let J be a nonzero finitely generated ideal of R[X] and f Î R^{¢}[X] such that fJ Ì J^{2}. For r in R, let J_{r} be the image of J in R under evaluation at r. Then J_{r} is a finitely generated ideal of R. Write J = (Xr)^{m}I where I is finitely generated, m ³ 0, and I_{r} ¹ 0. Then f·(Xr)^{m}I Ì (Xr)^{2m}I^{2}, so fI Ì I^{2}. As f(r)I_{r} Ì I_{r}^{2}, it follows that f(r) is strongly integral over R.
Theorem 17 Let R be a domain that contains an infinite set U, all of whose nonzero differences are units. If R is weakly integrally closed in R^{¢}, then R[X] is weakly integrally closed in R^{¢}[X].
Proof. If f Î R^{¢}[X] is strongly integral over R[X], then it follows from the lemma that f(u) Î R for each u in U. Using 1+degf of these elements, a Vandermonde determinant argument shows that f Î R[X].
Corollary 18 If R is a weakly integrally closed domain that contains an infinite field, then R[X] is weakly integrally closed.
The ring R need not contain an infinite field to satisfy the condition of Theorem 17. Consider the subring R of the field Q(X) consisting of quotients a/b of polynomials in Z[X] where b is primitive. We can let U be the set {1,1+X,1+X+X^{2},¼}. Of course R is integrally closed, being a unique factorization domain, but the ring R[2i] is not, and R[2i] is weakly integrally closed because it is a free quadratic extension of the Krull domain R.
Theorem 19 If R is a Noetherian domain, and f is strongly integral over R[X], then the leading coefficient of f is strongly integral over R.
Proof. Suppose fJ Ì J^{2} for J a nonzero finitely generated ideal of R[X]. Let I be the ideal in R of leading coefficients of J. As R is Noetherian, I is finitely generated. The ideal of leading coefficients of J^{2} is I^{2}. So if a is the leading coefficient of f, then aI Ì I^{2}, whence a is strongly integral over R.
Corollary 20 If R is a Noetherian weakly integrally closed domain, and f is strongly integral over R[X] with degf £ 2, then f Î R[X].
Proof. Let f = a_{0}+a_{1}X+a_{2}X^{2}. Then a_{2} Î R as is f(0) = a_{0} and f(1) = a_{0}+a_{1}+a_{2}.
