Lemma 29.44.9. Let $f : X \to Y$ be an integral morphism. Then $\dim (X) \leq \dim (Y)$. If $f$ is surjective then $\dim (X) = \dim (Y)$.
Proof. Since the dimension of $X$ and $Y$ is the supremum of the dimensions of the members of an affine open covering, we may assume $Y$ and $X$ are affine. The inequality follows from Algebra, Lemma 10.112.3. The equality then follows from Algebra, Lemmas 10.112.1 and 10.36.22. $\square$
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Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)