Exercise 17, § 4

Let A be a noetherian integral domain and $a \neq 0$ an element of the radical of A. Assume the following: (i) $\operatorname{Ass}_A(A/aA)$ contains a single minimal element $p$; (ii) $aA_p = pA_p$; (iii) $A/p$ is integrally closed; (iv) the integral closure $\overline{A}$ of A is a finite A-algebra; (v) if q is a prime ideal of height 1 of $\overline{A}$, then $q \cap A$ is a prime ideal of height 1 of A. Prove that one has $\overline{A} = A$ (A is integrally closed) and $aA = p$ (hence $A/aA$ is integrally closed). To do this, establish successively the following assertions: a) $A_p$ and $\overline{A}_p$ are isomorphic and $\overline{p} = p\overline{A}$ is the unique prime ideal of $\overline{A}$ lying over $p$. b) Moreover, $p$ is the unique minimal element of $\operatorname{Ass}_A(\overline{A}/a\overline{A})$.

c) In fact $\bar{p}$ is the only element of $\mathrm{Ass}_A(\overline{A}/a\overline{A})$. d) $\overline{A}/a\overline{A}$ is an integral domain and $a\overline{A} = \bar{p}$. e) $A/p$ and $\overline{A}/\bar{p}$ are isomorphic and the canonical homomorphism of $A/aA$ into $\overline{A}/a\overline{A}$ is surjective.