Lemma 67.17.5. Let $S$ be a scheme. Let $j : U \to X$ be an open immersion of algebraic spaces over $S$. Then $U$ is scheme theoretically dense in $X$ if and only if $\mathcal{O}_ X \to j_*\mathcal{O}_ U$ is injective.

**Proof.**
If $\mathcal{O}_ X \to j_*\mathcal{O}_ U$ is injective, then the same is true when restricted to any algebraic space $V$ étale over $X$. Hence the scheme theoretic closure of $U \times _ X V$ in $V$ is equal to $V$, see proof of Lemma 67.16.1. Conversely, assume the scheme theoretic closure of $U \times _ X V$ is equal to $V$ for all $V$ étale over $X$. Suppose that $\mathcal{O}_ X \to j_*\mathcal{O}_ U$ is not injective. Then we can find an affine, say $V = \mathop{\mathrm{Spec}}(A)$, étale over $X$ and a nonzero element $f \in A$ such that $f$ maps to zero in $\Gamma (V \times _ X U, \mathcal{O})$. In this case the scheme theoretic closure of $V \times _ X U$ in $V$ is clearly contained in $\mathop{\mathrm{Spec}}(A/(f))$ a 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Your email address will not be published.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)