Lemma 10.36.17. Suppose that $R \to S$ is an integral ring extension with $R \subset S$. Then $\varphi : \mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ is surjective.

Proof. Let $\mathfrak p \subset R$ be a prime ideal. We have to show $\mathfrak pS_{\mathfrak p} \not= S_{\mathfrak p}$, see Lemma 10.17.9. The localization $R_{\mathfrak p} \to S_{\mathfrak p}$ is injective (as localization is exact) and integral by Lemma 10.36.11 or 10.36.13. Hence we may replace $R$, $S$ by $R_{\mathfrak p}$, $S_{\mathfrak p}$ and we may assume $R$ is local with maximal ideal $\mathfrak m$ and it suffices to show that $\mathfrak mS \not= S$. Suppose $1 = \sum f_ i s_ i$ with $f_ i \in \mathfrak m$ and $s_ i \in S$ in order to get a contradiction. Let $R \subset S' \subset S$ be such that $R \to S'$ is finite and $s_ i \in S'$, see Lemma 10.36.4. The equation $1 = \sum f_ i s_ i$ implies that the finite $R$-module $S'$ satisfies $S' = \mathfrak m S'$. Hence by Nakayama's Lemma 10.20.1 we see $S' = 0$. Contradiction. $\square$

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