# The Stacks Project

## Tag 081H

Lemma 28.24.14. Let $f : X \to Y$ be a flat morphism of schemes. Let $V \subset Y$ be a retrocompact open which is scheme theoretically dense. Then $f^{-1}V$ is scheme theoretically dense in $X$.

Proof. We will use the characterization of Lemma 28.7.5. We have to show that for any open $U \subset X$ the map $\mathcal{O}_X(U) \to \mathcal{O}_X(U \cap f^{-1}V)$ is injective. It suffices to prove this when $U$ is an affine open which maps into an affine open $W \subset Y$. Say $W = \mathop{\rm Spec}(A)$ and $U = \mathop{\rm Spec}(B)$. Then $V \cap W = D(f_1) \cup \ldots \cup D(f_n)$ for some $f_i \in A$, see Algebra, Lemma 10.28.1. Thus we have to show that $B \to B_{f_1} \times \ldots \times B_{f_n}$ is injective. We are given that $A \to A_{f_1} \times \ldots \times A_{f_n}$ is injective and that $A \to B$ is flat. Since $B_{f_i} = A_{f_i} \otimes_A B$ we win. $\square$

The code snippet corresponding to this tag is a part of the file morphisms.tex and is located in lines 4474–4479 (see updates for more information).

\begin{lemma}
\label{lemma-flat-morphism-scheme-theoretically-dense-open}
Let $f : X \to Y$ be a flat morphism of schemes. Let $V \subset Y$ be
a retrocompact open which is scheme theoretically dense. Then $f^{-1}V$
is scheme theoretically dense in $X$.
\end{lemma}

\begin{proof}
We will use the characterization of
Lemma \ref{lemma-characterize-scheme-theoretically-dense}.
We have to show that for any open $U \subset X$ the map
$\mathcal{O}_X(U) \to \mathcal{O}_X(U \cap f^{-1}V)$ is injective.
It suffices to prove this when $U$ is an affine open which maps into
an affine open $W \subset Y$. Say $W = \Spec(A)$ and $U = \Spec(B)$.
Then $V \cap W = D(f_1) \cup \ldots \cup D(f_n)$ for some
$f_i \in A$, see
Algebra, Lemma \ref{algebra-lemma-qc-open}.
Thus we have to show that
$B \to B_{f_1} \times \ldots \times B_{f_n}$ is injective.
We are given that $A \to A_{f_1} \times \ldots \times A_{f_n}$ is injective
and that $A \to B$ is flat. Since $B_{f_i} = A_{f_i} \otimes_A B$ we win.
\end{proof}

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