Lemma 66.45.1. Let $S$ be a scheme. Let $f : X \to Y$ be a representable morphism of algebraic spaces over $S$. Then $f$ is integral, resp. finite (in the sense of Section 66.3), if and only if for all affine schemes $Z$ and morphisms $Z \to Y$ the scheme $X \times _ Y Z$ is affine and integral, resp. finite, over $Z$.

**Proof.**
This follows directly from the definition of an integral (resp. finite) morphism of schemes (Morphisms, Definition 29.44.1).
$\square$

## 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Your email address will not be published.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 (0)