Lemma 29.25.16. Let $f : X \to Y$ be a flat morphism of schemes. Let $g : V \to Y$ be a quasi-compact morphism of schemes. Let $Z \subset Y$ be the scheme theoretic image of $g$ and let $Z' \subset X$ be the scheme theoretic image of the base change $V \times _ Y X \to X$. Then $Z' = f^{-1}Z$.

** Taking scheme theoretic images commutes with flat base change in the quasi-compact case **

**Proof.**
Recall that $Z$ is cut out by $\mathcal{I} = \mathop{\mathrm{Ker}}(\mathcal{O}_ Y \to g_*\mathcal{O}_ V)$ and $Z'$ is cut out by $\mathcal{I}' = \mathop{\mathrm{Ker}}(\mathcal{O}_ X \to (V \times _ Y X \to X)_*\mathcal{O}_{V \times _ Y X})$, see Lemma 29.6.3. Hence the question is local on $X$ and $Y$ and we may assume $X$ and $Y$ affine. Note that we may replace $V$ by $\coprod V_ i$ where $V = V_1 \cup \ldots \cup V_ n$ is a finite affine open covering. Hence we may assume $g$ is affine. In this case $(V \times _ Y X \to X)_*\mathcal{O}_{V \times _ Y X}$ is the pullback of $g_*\mathcal{O}_ V$ by $f$. Since $f$ is flat we conclude that $f^*\mathcal{I} = \mathcal{I}'$ and the lemma holds.
$\square$

## 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Your email address will not be published.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 (1)

Comment #3833 by slogan_bot on

There are also: