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. $\square$

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).