# The Stacks Project

## Tag: 0832

This tag has label spaces-morphisms-lemma-scheme-theoretically-dense-representable and it points to

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Lemma 45.17.1. Let $S$ be a scheme. Let $W \subset S$ be a scheme theoretically dense open subscheme (Morphisms, Definition 25.7.1). Let $f : X \to S$ be a morphism of schemes which is flat, locally of finite presentation, and locally quasi-finite. Then $f^{-1}(W)$ is scheme theoretically dense in $X$.

Proof. We will use the characterization of Morphisms, Lemma 25.7.5. Assume $V \subset X$ is an open and $g \in \Gamma(V, \mathcal{O}_V)$ is a function which restricts to zero on $f^{-1}(W) \cap V$. We have to show that $g = 0$. Assume $g \not = 0$ to get a contradiction. By More on Morphisms, Lemma 33.31.11 we may shrink $V$, find an open $U \subset S$ fitting into a commutative diagram $$\xymatrix{ V \ar[r] \ar[d]_\pi & X \ar[d]^f \\ U \ar[r] & S, }$$ a quasi-coherent subsheaf $\mathcal{F} \subset \mathcal{O}_U$, an integer $r > 0$, and an injective $\mathcal{O}_U$-module map $\mathcal{F}^{\oplus r} \to \pi_*\mathcal{O}_V$ whose image contains $g|_V$. Say $(g_1, \ldots, g_r) \in \Gamma(U, \mathcal{F}^{\oplus r})$ maps to $g$. Then we see that $g_i|_{W \cap U} = 0$ because $g|_{f^{-1}W \cap V} = 0$. Hence $g_i = 0$ because $\mathcal{F} \subset \mathcal{O}_U$ and $W$ is scheme theoretically dense in $S$. This implies $g = 0$ which is the desired contradiction. $\square$

\begin{lemma}
\label{lemma-scheme-theoretically-dense-representable}
Let $S$ be a scheme. Let $W \subset S$ be a scheme theoretically
dense open subscheme
(Morphisms, Definition \ref{morphisms-definition-scheme-theoretically-dense}).
Let $f : X \to S$ be a morphism of schemes which is flat, locally of
finite presentation, and locally quasi-finite.
Then $f^{-1}(W)$ is scheme theoretically dense in $X$.
\end{lemma}

\begin{proof}
We will use the characterization of Morphisms, Lemma
\ref{morphisms-lemma-characterize-scheme-theoretically-dense}.
Assume $V \subset X$ is an open and $g \in \Gamma(V, \mathcal{O}_V)$
is a function which restricts to zero on $f^{-1}(W) \cap V$.
We have to show that $g = 0$. Assume $g \not = 0$ to get a
More on Morphisms, Lemma \ref{more-morphisms-lemma-go-down-with-annihilators}
we may shrink $V$, find an open $U \subset S$ fitting into a
commutative diagram
$$\xymatrix{ V \ar[r] \ar[d]_\pi & X \ar[d]^f \\ U \ar[r] & S, }$$
a quasi-coherent subsheaf $\mathcal{F} \subset \mathcal{O}_U$, an integer
$r > 0$, and an injective $\mathcal{O}_U$-module map
$\mathcal{F}^{\oplus r} \to \pi_*\mathcal{O}_V$
whose image contains $g|_V$. Say
$(g_1, \ldots, g_r) \in \Gamma(U, \mathcal{F}^{\oplus r})$ maps to $g$.
Then we see that $g_i|_{W \cap U} = 0$ because $g|_{f^{-1}W \cap V} = 0$.
Hence $g_i = 0$ because $\mathcal{F} \subset \mathcal{O}_U$ and
$W$ is scheme theoretically dense in $S$.
This implies $g = 0$ which is the desired contradiction.
\end{proof}


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