Polynomial ring over integrally closed subring is integrally closed in polynomial ring

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Statement

Suppose R is an integrally closed subring of a commutative unital ring S. Then, the polynomial ring R[x] is an integrally closed subring of S[x].

Proof

Given: A ring S, an integrally closed subring R.

To prove: R[x] is an integrally closed subring of S[x].

Proof: Suppose f \in S[x] satisfies a monic polynomial over R[x]:

f^n + g_{n-1}f^{n-1} + g_{n-2}f^{n-2} + \dots + g_0 = 0.

Here, g_i \in R[x] for all math>i</math>. Write f = h + x^r, for r greater than the maximum of the degrees of the g_i. Note that f \in R[x] if and only if h \in R[x], so it suffices to show that h \in R[x].

h satisfies the monic polynomial:

(h + x^r)^n + g_{n-1}(h + x^r)^{n-1} + g_{n-2}(h + x^r)^{n-2} + \dots + g_0 = 0.

The constant term of this, viewed as a polynomial in h, is:

(x^r)^n + g_{n-1}(x^r)^{n-1} + \dots + g_0.

Since all the g_i are in R[x], this whole constant term is an element of R[x].

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