Question 1.217: Suppose R ⊂ S is an integral ring extension. Let D be a mult......
Suppose R \subset S is an integral ring extension. Let D be a multiplicative submonoid of R. Prove that
(a) D^{-1} S is integral over D^{-1} R;
(b) If P is a prime ideal of S, then S / P is integral over R / P \cap R.
(a) First of all, if D \subset R is multiplicatively closed, then it is multiplicatively closed in S as well, therefore, it makes sense to localize S at D.
Let \alpha / d, \alpha \in S, d \in D be an element of D^{-1} S. Since \alpha is integral over R, there exists a monic polynomial f(x)=x^{m}+a_{m-1} x^{m-1}+\cdots+a_{0} \in R[x] such that f(\alpha)=0.
Dividing both sides with d^{m} we obtain
\frac{f(\alpha)}{d^{m}}=\left(\frac{\alpha}{d}\right)^{m}+\frac{a_{m-1}}{d}\left(\frac{\alpha}{d}\right)^{m-1}+\cdots+\frac{a_{0}}{d^{m}}=0
Since the coefficients of the left hand side is from D^{-1} R and this is a monic expression, we see that \alpha / d is an integral element of D^{-1} S over D^{-1} R.
(b) If \alpha \in S and f(x) \in R[x] is a monoic polynomial such that f(\alpha)=0, then the image of \alpha in S / P satisfies the polynomial \tilde{f}(x) \in(R / R \cap P)[x] that is obtained from f(x) by reducing its coefficients modulo P \cap R. Since f(x) has a monic leading term, \tilde{f}(x) is not identically zero. Therefore, the image of \alpha in S / P is integral over R / R \cap P.