Lemma 38.17.3. Let $f : X \to S$ be a finite type, flat morphism of schemes with geometrically integral fibres. Then $X$ is universally pure over $S$.

Proof. Let $\xi \in X$ with $s' = f(\xi )$ and $s' \leadsto s$ a specialization of $S$. If $\xi$ is an associated point of $X_{s'}$, then $\xi$ is the unique generic point because $X_{s'}$ is an integral scheme. Let $\xi _0$ be the unique generic point of $X_ s$. As $X \to S$ is flat we can lift $s' \leadsto s$ to a specialization $\xi ' \leadsto \xi _0$ in $X$, see Morphisms, Lemma 29.25.9. The $\xi \leadsto \xi '$ because $\xi$ is the generic point of $X_{s'}$ hence $\xi \leadsto \xi _0$. This means that $(\text{id}_ S, s' \to s, \xi )$ is not an impurity of $\mathcal{O}_ X$ above $s$. Since the assumption that $f$ is finite type, flat with geometrically integral fibres is preserved under base change, we see that there doesn't exist an impurity after any base change. In this way we see that $X$ is universally $S$-pure. $\square$

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.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 05K5. Beware of the difference between the letter 'O' and the digit '0'.