Exercise 16, § 4

Let A be a noetherian integral domain, $\overline{A}$ its integral closure in a finite extension of its field of fractions, B a ring containing A and contained in $\overline{A}$. Put Y = Spec(B) and X = Spec(A). We propose to prove that Y is a noetherian space. Let $(f_n)_{n \in \mathbf{N}}$ be a sequence of elements of B. For $n \in \mathbf{N}$, let $U_n = \operatorname{Spec}(B_{f_n})$ be the open subset of Y defined by $f_n$. Put $U = \bigcup_{n \in \mathbf{N}} U_n$, $F_n = U - \bigcup_{0 \leq m \leq n} U_m$. We denote by $\overline{F}_n$ the closure of $F_n$ in Y, by $G_n$ the image of $F_n$ in X, by $\overline{G}_n$ the closure of $G_n$ in X, by $B_n$ the reduced quotient ring of B such that $\operatorname{Spec}(B_n) = \overline{F}_n$, by $\overline{f}_n$ the image of $f_n$ in $B_n$. a) We have $F_n = V(\overline{f}_n B_n) \cap U$ for every $n \in \mathbf{N}$. b) Suppose that for an integer $n \in \mathbf{N}$, $\overline{F}_n$ has only a finite number of irreducible components. Then $V(\overline{f}_n B_n)$ has only a finite number of irreducible components. (Apply the result of Exercise 15, using also Exercise 11, a).) c) For every $n \in \mathbf{N}$, $F_n$ has only a finite number of irreducible components. d) If $A_n$ is the reduced quotient ring of A such that $\operatorname{Spec}(A_n) = \overline{G}_n$, then the minimal prime ideals of $A_n$ belong to $G_n$. e) Let $n \in \mathbf{N}$ and x be such a minimal prime ideal of $A_n$. Then there exists an integer $n' > n$ such that $F_{n'}$ contains no point of Y lying over x, and x does not belong to $\overline{G}_{n'}$. (Use Exercise 11, a).) f) For n sufficiently large, $F_n$ is empty.