The Stacks project

Lemma 29.28.2. Let $f : X \to Y$ and $g : Y \to S$ be morphisms of schemes. Let $x \in X$ and set $y = f(x)$, $s = g(y)$. Assume $f$ and $g$ locally of finite type. Then

\[ \dim _ x(X_ s) \leq \dim _ x(X_ y) + \dim _ y(Y_ s). \]

Moreover, equality holds if $\mathcal{O}_{X_ s, x}$ is flat over $\mathcal{O}_{Y_ s, y}$, which holds for example if $\mathcal{O}_{X, x}$ is flat over $\mathcal{O}_{Y, y}$.

Proof. Note that $\text{trdeg}_{\kappa (s)}(\kappa (x)) = \text{trdeg}_{\kappa (y)}(\kappa (x)) + \text{trdeg}_{\kappa (s)}(\kappa (y))$. Thus by Lemma 29.28.1 the statement is equivalent to

\[ \dim (\mathcal{O}_{X_ s, x}) \leq \dim (\mathcal{O}_{X_ y, x}) + \dim (\mathcal{O}_{Y_ s, y}). \]

For this see Algebra, Lemma 10.112.6. For the flat case see Algebra, Lemma 10.112.7. $\square$

Comments (0)

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Your email address will not be published.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 02JS. Beware of the difference between the letter 'O' and the digit '0'.