Lemma 81.7.4. In Situation 81.2.1 let $X, Y/B$ be good and let $f : X \to Y$ be a morphism over $B$. Assume $X$, $Y$ integral and $\dim _\delta (X) = \dim _\delta (Y)$. Then either $f$ factors through a proper closed subspace of $Y$, or $f$ is dominant and the extension of function fields $R(X) / R(Y)$ is finite.

**Proof.**
By Lemma 81.7.1 there is a unique integral closed subspace $Z \subset Y$ such that $f$ factors through a dominant morphism $X \to Z$. Then $Z = Y$ if and only if $\dim _\delta (Z) = \dim _\delta (Y)$. On the other hand, by our construction of dimension functions (see Situation 81.2.1) we have $\dim _\delta (X) = \dim _\delta (Z) + r$ where $r$ the transcendence degree of the extension $R(X)/R(Z)$. Combining this with Spaces over Fields, Lemma 71.5.1 the lemma follows.
$\square$

## 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Your email address will not be published.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)