Lemma 1

Let $h : \mathbf{A} \to \mathbf{A}'$ be a ring homomorphism, $S$ a multiplicative subset of $\mathbf{A}$, $i = i_A^S : \mathbf{A} \to S^{-1}\mathbf{A}$, $i' = i_{A'}^{h(S)} : \mathbf{A}' \to S^{-1}\mathbf{A}' = (h(S))^{-1}\mathbf{A}'$ the canonical homomorphisms and $h_1 = S^{-1}h : S^{-1}\mathbf{A} \to S^{-1}\mathbf{A}'$, so that there is a commutative diagram

Let $p$ be a prime ideal of $\mathbf{A}$ such that $p \cap S = \varnothing$. Then $a' \mapsto S^{-1}a'$ is a surjective mapping of the set $\mathscr{F}$ of ideals of $\mathbf{A}'$ lying above $p$ onto the set $\mathscr{F}_1$ of ideals of $S^{-1}\mathbf{A}'$ lying above $S^{-1}p$ and the mapping $a'_1 \mapsto i'(a'_1)$ is a bijection of $\mathscr{F}_1$ onto the set of ideals belonging to $\mathscr{F}$ and is saturated with respect to $h(S)$; in particular $p' \mapsto S^{-1}p'$ is a bijection of the set of prime ideals of $\mathbf{A}'$ lying above $p$ onto the set of prime ideals of $S^{-1}\mathbf{A}'$ lying above $S^{-1}p$.

We know that $S^{-1}p$ is a prime ideal of $S^{-1}\mathbf{A}$ and that $i_1(S^{-1}p) = p$ (Chapter 11, § 2, no. 5, Proposition 11); if there exists an ideal $b'$ of $S^{-1}\mathbf{A}'$ lying above $S^{-1}p$, then $h(i_1(b')) = i(h_1(b')) = p$; as $S^{-1}i'(b') = b'$ (loc. cit.), this already shows that the image of $\mathscr{F}$ under the mapping $a' \mapsto S^{-1}a'$ contains $\mathscr{F}_1$. On the other hand, if $a' \in \mathscr{F}$, $a \in A$ and $s \in S$, there are the following equivalences

$\iff$ there exists $t \in S$ such that $h(t)h(a) \in a'$ $\iff$ there exists $t \in S$ such that $ta \in \mathfrak{p}$ $\iff a/z \in S^{-1}\mathfrak{p}$.

Hence $h_1^{-1}(S^{-1}a') = S^{-1}\mathfrak{p}$, which completes the proof that the image of $\mathscr{F}$ under the mapping $a' \mapsto S^{-1}a'$ is equal to $\mathscr{F}_1$; the other assertions follow from Chapter 11, § 2, no. 5, Proposition 11.