For any commutative ring B and subring A, the following are equivalent:

(1) A is integrally closed in B

(2) If F in A[X] factorizes as F = G.H in B[X] with G and H monic, then G and H are in A[X]

These conditions imply:

(3) If F in A[X] is monic and irreducible, then F remains irreducible in B[X]

When B is an integral domain, all three conditions are equivalent.

Proof: (2) ==> (1) is trivial: if F(b) = 0 for some b in B and F in A[X] monic, then F = (X-b).G in B[X] for some polynomial G (remainder theorem). Then G is monic, so by (2) X-b and G are in A[X]. In particular, b is in A.

(1) ==> (2) is folklore. Take a commutative ring S that contains B, over which G and H can be written as products of linear factors: G = (X-x1)...(X-xn), H = (X-y1)...(X-ym). Then in S the xi and yj are zeroes of the monic polynomial F, so they are integral over A. But the coefficients of G and H are (elementary symmetric) polynomials in the xi and yj, respectively, hence they are again integral over A. As these coefficients are in B, by (1) they must lie in A. (Basically, one can construct such a ring S in the same way as a splitting field for a polynomial over a field is found: first consider S1 := B[X]/(G); then G = (X-x1).G1 in S1[X], where we write x1 := X mod (G) in S1. Then "adjoin another root x2 of G" by passing to S2 := S1[X]/(G1), etc, until we arrive at a ring Sn over which G completely splits into linear factors. Then proceed to adjoin roots for H in the same manner. Note that B remains a subring throughout, i.e. no non-zero element of B will map to 0 in S.)

Condition (3) immediately follows from (2), for if F = G.H in B[X], the leading coefficients of G and H are inverse units of B because F is monic. So we can rewrite this as F = G1.H1 with G1 and H1 monic in B[X].

When A and B are domains, (3) implies (1): if F(b) = 0 with b in B and F in A[X] monic, factor F as F1...Fr with the Fi monic and irreducible in A[X]. (Factoring F as a product of monic polynomials reduces the degree, so eventually we wind up with factors that are irreducible.) Since B is a domain, it follows that Fi(b) = 0 for some i. By (3), Fi is still irreducible in B[X]. But it is divisible by X-b there, and so Fi = X-b. Hence X-b is in A[X] and b belongs to A.

Q.e.d.

(Matthe van der Lee, Amsterdam.)