Lemma 29.29.6. Let $f : X \to Y$ be a morphism of locally Noetherian schemes which is flat, locally of finite type and of relative dimension $d$. For every point $x$ in $X$ with image $y$ in $Y$ we have $\dim _ x(X) = \dim _ y(Y) + d$.

Proof. After shrinking $X$ and $Y$ to open neighborhoods of $x$ and $y$, we can assume that $\dim (X) = \dim _ x(X)$ and $\dim (Y) = \dim _ y(Y)$, by definition of the dimension of a scheme at a point (Properties, Definition 28.10.1). The morphism $f$ is open by Lemmas 29.21.9 and 29.25.10. Hence we can shrink $Y$ to arrange that $f$ is surjective. It remains to show that $\dim (X) = \dim (Y) + d$.

Let $a$ be a point in $X$ with image $b$ in $Y$. By Algebra, Lemma 10.112.7,

$\dim (\mathcal{O}_{X,a}) = \dim (\mathcal{O}_{Y,b}) + \dim (\mathcal{O}_{X_ b, a}).$

Taking the supremum over all points $a$ in $X$, it follows that $\dim (X) = \dim (Y) + d$, as we want, see Properties, Lemma 28.10.2. $\square$

## Comments (2)

Comment #2305 by Daniel on

Typo in the 3rd line from the bottom: The last local ring in the equation.

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Your email address will not be published.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