Exercise 5, § 3

Let A be an integral domain and $B \supseteq A$ a subring of the field of fractions K of A; suppose that A is integrally closed in B and that B contains an element $t$ such that B is a finitely generated $A[t]$-module.

(a) Let F be a polynomial $\neq 0$ of $A[X]$ such that $F(t)$ belongs to the conductor f of B with respect to $A[t]$; show that, if c is the dominant coefficient of F, c belongs to the radical off. (By considering the rings $A[c^{-1}]$ and $B[c^{-1}]$, reduce it to showing that, if $c = 1$, necessarily $B = A[t]$. On the other hand we may assume that $F(t) \neq 0$, otherwise $t \in A$. For all $x \in B$, by hypothesis

where Q and R are polynomials in $A[X]$ and $\deg(R) < \deg(F)$; attention may be confined to the case where $R(t) \neq 0$; if $y = R(t)(F(t))^{-1} = x - Q(t)$, show that $t$ is integral over $A[y^{-1}]$ and deduce that y is integral over A.)

(b) Suppose further that A is Noetherian; let q be a prime ideal of B belonging to $\operatorname{Ass}(B/f)$. Show that, if $\omega$ is the canonical homomorphism from B to the field of fractions of $B/q$, $\omega(t)$ is transcendental over the field of fractions of $A/(q \cap A)$. (If $p = q \cap A[t]$, there is an element $y \in B$ such that q is the set of $z \in B$ for which $zy \in f$; show that p is the conductor of $B = A[t] + By$ with respect to $A[t]$. Then argue by reductio ad absurdurn using (a).)

(c) Suppose that A is Noetherian. Let n be a maximal ideal of B and $p = A \cap n$. Show that, if $A_p \neq B_n$, there exists a prime ideal of B contained in n, distinct from n and containing p. (Let f be the conductor of B with respect to $A[t]$. Suppose first that $n \supset f$; consider a minimal element q in the set of prime ideals of B contained in n and containing f; consider the canonical image of $A$ in $B/q$ and use (b) and Exercise 4. If $n \not\subset f$, the local ring $B_n$ is equal to the local ring $A[t]_{n_1}$, where $n_1 = n \cap A[t]$ (§ 1, Exercise 16). If $t \in A_p$, show that $A_p = B_{n'}$. If on the contrary $A_p \neq B_{n'}$, deduce from § 2, Exercise 14 (c) that $pA_p[t]$ is a non-maximal prime ideal of $A_p[t]$ and conclude that $pB_n$ is a prime ideal of $B_n$ distinct from $nB_n$.