Theorem 3

Let $A$ be an integrally closed domain and $A'$ a ring containing $A$ and integral over $A$. Suppose that $0$ is the only element of $A$ which is a divisor of $0$ in $A'$. Let $p, q$ be two prime ideals of $A$ such that $q \supset p$ and $q'$ a prime ideal of $A$ lying above $q$. Then there exists a prime ideal $p'$ of $A$ lying above $p$ and such that $q' \supset p'$.

Suppose first that $A'$ is an integral domain. Let $K, K'$ be the fields of fractions of $A$ and $A'$ respectively; let $K''$ be the algebraic closure of $K$ and $A$ the integral closure of $A$ in $K''$; then $A \subset A' \subset A''$. Let $p''$ be a prime ideal of $A''$ lying above $p$ (no. 1, Theorem 1), $q''$ a prime ideal of $A''$ lying above $q$ and such that $p'' \subset q''$ (no. 1, Corollary 2 to Theorem 1) and finally $q_1''$ a prime ideal of $A''$ lying above $q'$ (no. 1, Theorem 1). By no. 3, Proposition 6 (i), there exists a $K$-automorphism $\sigma$ of $K''$ such that $Q.q'' = q_1''$. Then $Q.p''$ is a prime ideal of $A''$ lying above $p$ such that $Q.p'' = q_1''$ and hence $p' = A' \cap \sigma.p''$ is a prime ideal of $A'$ lying above $p$ and contained in $A' \cap q_1'' = q'$.

We pass to the general case. As $A$ is an integral domain and $q'$ is prime, the subsets $A - \{0\}$ and $A' - q'$ of $A'$ are multiplicative; then their product $S = (A - \{0\})(A' - q')$ is a multiplicative subset of $A'$ which does not contain 0 since the non-zero elements of $A$ are not divisors of 0 in $A'$. Then there exists (Chapter II, §2, no. 5, Corollary 2 to Proposition 11) a prime ideal $m'$ of $A'$ disjoint from $S$, in other words such that $m' \subset q'$ and $m' \cap A = 0$. Let $h$ be the canonical homomorphism $A' \to A'/m'$. The restriction of $h$ to $A$ is injective and hence $h(A)$ is integrally closed. As $m' \subset q'$, $h(q')$ is a prime ideal of $A'/m'$ lying above $h(q)$; since $A'/m'$ is an integral domain, the first part of the proof proves that there exists a prime ideal $n'$ of $A'/m'$ such that $n' \cap h(A) = h(p)$ and $h(q') \supset n'$. The ideal $p' = h^{-1}(n')$ is a prime ideal of $A'$ and $q' \supset p'$, since $q'$ contains the kernel of $h$. As $n' \supset h(p)$, $p' \supset p$. Finally, for $x \in p' \cap A$, $h(x) \in n' \cap h(A) = h(p)$ and hence $x \in p$ since the restriction of $h$ to $A$ is injective; hence $p' \cap A = p$.