Proposition 2

Let $h : A \to A'$ be a ring homomorphism such that $A'$ is integral over $A$, $\mathfrak{p}$ a prime ideal of $A$, $S = A - \mathfrak{p}$, $(p'_i)_{i \in I}$ the family of all the prime ideals $\mathfrak{p}$ of $A'$ lying above $\mathfrak{p}$ and $S' = \bigcap_{i \in I} (A' - p_i)$; then $S^{-1}A' = {S'}^{-1}A'$.

In fact, by definition $h(S) \subset S'$ and, as

it suffices to prove, by virtue of Chapter II, $§ 2$, no. 3, Proposition 8, that, if a prime ideal $q'$ of $A'$ does not meet $h(S)$, it does not meet $S'$ either. Now, suppose that $q' \cap h(S) = \varnothing$ and let $q = h^{-1}(q')$; then $q \cap S = \varnothing$, in other words $q \subset \mathfrak{p}$. As $q'$ lies above $q$ by definition, it follows from Corollary 2 to Proposition 1 that there is an index $c$ such that $q' \subset p'_c$ and hence $q' \cap S' = \varnothing$, which completes the proof.