Lemma 10.118.6. Let $R \to S$ be a ring map. Let $M$ be an $S$-module. Let $U \subset \mathop{\mathrm{Spec}}(R)$ be a dense open. Assume there is a covering $U = \bigcup _{i \in I} D(f_ i)$ of opens such that $U(R_{f_ i} \to S_{f_ i}, M_{f_ i})$ is dense in $D(f_ i)$ for each $i \in I$. Then $U(R \to S, M)$ is dense in $\mathop{\mathrm{Spec}}(R)$.

**Proof.**
In view of Lemma 10.118.5 this is a purely topological statement. Namely, by that lemma we see that $U(R \to S, M) \cap D(f_ i)$ is dense in $D(f_ i)$ for each $i \in I$. By Topology, Lemma 5.21.4 we see that $U(R \to S, M) \cap U$ is dense in $U$. Since $U$ is dense in $\mathop{\mathrm{Spec}}(R)$ we conclude that $U(R \to S, M)$ is dense in $\mathop{\mathrm{Spec}}(R)$.
$\square$

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