Lemma 29.7.5. Let $j : U \to X$ be an open immersion of schemes. 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 open $V$ of $X$. Hence the scheme theoretic closure of $U \cap V$ in $V$ is equal to $V$, see proof of Lemma 29.6.1. Conversely, suppose that the scheme theoretic closure of $U \cap V$ is equal to $V$ for all opens $V$. Suppose that $\mathcal{O}_ X \to j_*\mathcal{O}_ U$ is not injective. Then we can find an affine open, say $\mathop{\mathrm{Spec}}(A) = V \subset X$ and a nonzero element $f \in A$ such that $f$ maps to zero in $\Gamma (V \cap U, \mathcal{O}_ X)$. In this case the scheme theoretic closure of $V \cap U$ 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Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)