Lemma 37.43.4. Let $\pi : X \to Y$ be a finite morphism. Let $x \in X$ with $y = \pi (x)$ such that $\pi ^{-1}(\{ y\} ) = \{ x\}$. Then

1. For every neighbourhood $U \subset X$ of $x$ in $X$, there exists a neighbourhood $V \subset Y$ of $y$ such that $\pi ^{-1}(V) \subset U$.

2. The ring map $\mathcal{O}_{Y, y} \to \mathcal{O}_{X, x}$ is finite.

3. If $\pi$ is of finite presentation, then $\mathcal{O}_{Y, y} \to \mathcal{O}_{X, x}$ is of finite presentation.

4. For any quasi-coherent $\mathcal{O}_ X$-module $\mathcal{F}$ we have $\mathcal{F}_ x = \pi _*\mathcal{F}_ y$ as $\mathcal{O}_{Y, y}$-modules.

Proof. The first assertion is purely topological; use that $\pi$ is a continuous and closed map such that $\pi ^{-1}(\{ y\} ) = \{ x\}$. To prove the second and third parts we may assume $X = \mathop{\mathrm{Spec}}(B)$ and $Y = \mathop{\mathrm{Spec}}(A)$. Then $A \to B$ is a finite ring map and $y$ corresponds to a prime $\mathfrak p$ of $A$ such that there exists a unique prime $\mathfrak q$ of $B$ lying over $\mathfrak p$. Then $B_{\mathfrak q} = B_{\mathfrak p}$, see Algebra, Lemma 10.41.11. In other words, the map $A_{\mathfrak p} \to B_{\mathfrak q}$ is equal to the map $A_{\mathfrak p} \to B_{\mathfrak p}$ you get from localizing $A \to B$ at $\mathfrak p$. Thus (2) and (3) follow from simple properties of localization (some details omitted). For the final statement, suppose that $\mathcal{F} = \widetilde M$ for some $B$-module $M$. Then $\mathcal{F} = M_{\mathfrak q}$ and $\pi _*\mathcal{F}_ y = M_{\mathfrak p}$. By the above these localizations agree. Alternatively you can use part (1) and the definition of stalks to see that $\mathcal{F}_ x = \pi _*\mathcal{F}_ y$ directly. $\square$

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