Lemma 10.35.17. 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 10.16.9. The localization $R_{\mathfrak p} \to S_{\mathfrak p}$ is injective (as localization is exact) and integral by Lemma 10.35.11 or 10.35.13. 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 10.35.4. 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 10.19.1 we see $S' = 0$. Contradiction.
$\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)