Lemma 10.118.5. Let $R \to S$ be a ring map. Let $M$ be an $S$-module. Let $f \in R$. Using the identification $\mathop{\mathrm{Spec}}(R_ f) = D(f)$ we have $U(R_ f \to S_ f, M_ f) = D(f) \cap U(R \to S, M)$.
Proof. Suppose that $u \in U(R_ f \to S_ f, M_ f)$. Then there exists an element $g \in R_ f$ such that $u \in D(g)$ and such that (10.118.3.1) holds for the pair $((R_ f)_ g \to (S_ f)_ g, (M_ f)_ g)$. Write $g = a/f^ n$ for some $a \in R$. Set $h = af$. Then $R_ h = (R_ f)_ g$, $S_ h = (S_ f)_ g$, and $M_ h = (M_ f)_ g$. Moreover $u \in D(h)$. Hence $u \in U(R \to S, M)$. Conversely, suppose that $u \in D(f) \cap U(R \to S, M)$. Then there exists an element $g \in R$ such that $u \in D(g)$ and such that (10.118.3.1) holds for the pair $(R_ g \to S_ g, M_ g)$. Then it is clear that (10.118.3.1) also holds for the pair $(R_{fg} \to S_{fg}, M_{fg}) = ((R_ f)_ g \to (S_ f)_ g, (M_ f)_ g)$. Hence $u \in U(R_ f \to S_ f, M_ f)$ and we win. $\square$
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