Loading [MathJax]/extensions/tex2jax.js

The Stacks project

Lemma 59.73.13. Let $X$ be an integral normal scheme with function field $K$. Let $E$ be a set.

  1. Let $g : \mathop{\mathrm{Spec}}(K) \to X$ be the inclusion of the generic point. Then $g_*\underline{E} = \underline{E}$.

  2. Let $j : U \to X$ be the inclusion of a nonempty open. Then $j_*\underline{E} = \underline{E}$.

Proof. Proof of (1). Let $x \in X$ be a point. Let $\mathcal{O}^{sh}_{X, \overline{x}}$ be a strict henselization of $\mathcal{O}_{X, x}$. By More on Algebra, Lemma 15.45.6 we see that $\mathcal{O}^{sh}_{X, \overline{x}}$ is a normal domain. Hence $\mathop{\mathrm{Spec}}(K) \times _ X \mathop{\mathrm{Spec}}(\mathcal{O}^{sh}_{X, \overline{x}})$ is irreducible. It follows that the stalk $(g_*\underline{E})_{\overline{x}}$ is equal to $E$, see Theorem 59.53.1.

Proof of (2). Since $g$ factors through $j$ there is a map $j_*\underline{E} \to g_*\underline{E}$. This map is injective because for every scheme $V$ étale over $X$ the set $\mathop{\mathrm{Spec}}(K) \times _ X V$ is dense in $U \times _ X V$. On the other hand, we have a map $\underline{E} \to j_*\underline{E}$ and we conclude. $\square$


Comments (2)

Comment #9028 by DU Changjiang on

there is a typo in the proof: in the third line, a right parenthesis is missing. The right one should be


Your email address will not be published.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.