Lemma 15.9.8. Let $R \to S$ be a ring map. Let $I \subset R$ be an ideal of $R$ and let $J \subset S$ be an ideal of $S$. If the closure of the image of $V(J)$ in $\mathop{\mathrm{Spec}}(R)$ is disjoint from $V(I)$, then there exists an element $f \in R$ which maps to $1$ in $R/I$ and to an element of $J$ in $S$.

**Proof.**
Let $I' \subset R$ be an ideal such that $V(I')$ is the closure of the image of $V(J)$. Then $V(I) \cap V(I') = \emptyset $ by assumption and hence $I + I' = R$ by Algebra, Lemma 10.16.2. Write $1 = g + f$ with $g \in I$ and $f \in I'$. We have $V(f') \supset V(J)$ where $f'$ is the image of $f$ in $S$. Hence $(f')^ n \in J$ for some $n$, see Algebra, Lemma 10.16.2. Replacing $f$ by $f^ n$ we win.
$\square$

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