The Stacks project

Lemma 106.3.2. Let $U \to X$ be a smooth morphism of locally Noetherian schemes. Let $T'$ is an irreducible component of $U$. Let $T$ be the irreducible component of $X$ obtained as the closure of the image of $T'$. Then $m_{T', U} = m_{T, X}$.

Proof. Write $\xi '$ for the generic point of $T'$, and $\xi $ for the generic point of $T$. Let $A = \mathcal{O}_{X, \xi }$ and $B = \mathcal{O}_{U, \xi '}$. We need to show that $\text{length}_ A A = \text{length}_ B B$. Since $A \to B$ is a flat local homomorphism of rings (since smooth morphisms are flat), we have

\[ \text{length}_ A(A) \text{length}_ B(B/\mathfrak m_ A B) = \text{length}_ B(B) \]

by Algebra, Lemma 10.52.13. Thus it suffices to show $\mathfrak m_ A B = \mathfrak m_ B$, or equivalently, that $B/\mathfrak m_ A B$ is reduced. Since $U \to X$ is smooth, so is its base change $U_{\xi } \to \mathop{\mathrm{Spec}}\kappa (\xi )$. As $U_{\xi }$ is a smooth scheme over a field, it is reduced, and thus so its local ring at any point (Varieties, Lemma 33.25.4). In particular,

\[ B/\mathfrak m_ A B = \mathcal{O}_{U, \xi '}/\mathfrak m_{X, \xi }\mathcal{O}_{U, \xi '} = \mathcal{O}_{U_\xi , \xi '} \]

is reduced, as required. $\square$

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Your email address will not be published.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 0DR6. Beware of the difference between the letter 'O' and the digit '0'.