The Stacks project

Lemma 29.28.7. Let $f : X \to S$ be a morphism of schemes. Let $x \leadsto x'$ be a nontrivial specialization of points in $X$ lying over the same point $s \in S$. Assume $f$ is locally of finite type. Then

  1. $\dim _ x(X_ s) \leq \dim _{x'}(X_ s)$,

  2. $\dim (\mathcal{O}_{X_ s, x}) < \dim (\mathcal{O}_{X_ s, x'})$, and

  3. $\text{trdeg}_{\kappa (s)}(\kappa (x)) > \text{trdeg}_{\kappa (s)}(\kappa (x'))$.

Proof. Part (1) follows from the fact that any open of $X_ s$ containing $x'$ also contains $x$. Part (2) follows since $\mathcal{O}_{X_ s, x}$ is a localization of $\mathcal{O}_{X_ s, x'}$ at a prime ideal, hence any chain of prime ideals in $\mathcal{O}_{X_ s, x}$ is part of a strictly longer chain of primes in $\mathcal{O}_{X_ s, x'}$. The last inequality follows from Algebra, Lemma 10.116.2. $\square$


Comments (1)

Comment #734 by Keenan Kidwell on

The second of the second line of the proof should be . Also, "at a prime ideal" instead of "in a prime ideal?"


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Your email address will not be published.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 06RU. Beware of the difference between the letter 'O' and the digit '0'.