Lemma 29.14.6. Let $P$ be a property of ring maps. Assume $P$ is local and stable under base change. The base change of a morphism locally of type $P$ is locally of type $P$.

Proof. Let $f : X \to S$ be a morphism locally of type $P$. Let $S' \to S$ be any morphism. Denote $f' : X_{S'} = S' \times _ S X \to S'$ the base change of $f$. For every $s' \in S'$ there exists an open affine neighbourhood $s' \in V' \subset S'$ which maps into some open affine $V \subset S$. By Lemma 29.14.4 the open $f^{-1}(V)$ is a union of affines $U_ i$ such that the ring maps $\mathcal{O}_ S(V) \to \mathcal{O}_ X(U_ i)$ all satisfy $P$. By the material in Schemes, Section 26.17 we see that $f^{-1}(U)_{V'} = V' \times _ V f^{-1}(V)$ is the union of the affine opens $V' \times _ V U_ i$. Since $\mathcal{O}_{X_{S'}}(V' \times _ V U_ i) = \mathcal{O}_{S'}(V') \otimes _{\mathcal{O}_ S(V)} \mathcal{O}_ X(U_ i)$ we see that the ring maps $\mathcal{O}_{S'}(V') \to \mathcal{O}_{X_{S'}}(V' \times _ V U_ i)$ satisfy $P$ as $P$ is assumed stable under base change. $\square$

## 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Your email address will not be published.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.

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 01SW. Beware of the difference between the letter 'O' and the digit '0'.