## Tag `01RE`

Chapter 28: Morphisms of Schemes > Section 28.7: Scheme theoretic closure and density

Lemma 28.7.5. 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 28.6.1. 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. $\square$

The code snippet corresponding to this tag is a part of the file `morphisms.tex` and is located in lines 1065–1070 (see updates for more information).

```
\begin{lemma}
\label{lemma-characterize-scheme-theoretically-dense}
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.
\end{lemma}
\begin{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 \ref{lemma-scheme-theoretic-image}.
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 $\Spec(A) = V \subset X$
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 $\Spec(A/(f))$
a contradiction.
\end{proof}
```

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