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Let $f:X\rightarrow Y$ be a finite morphism of scheme. For $x\in X$, $y=f(x)\in Y$. Is it true that the map $$f^\sharp_x:\mathcal{O}_{Y,y}\rightarrow \mathcal{O}_{X,x}$$ is finite? (i.e $\mathcal{O}_{X,x}$ is a finite $\mathcal{O}_{Y,y}$-module).

Equivalently, if $\varphi:A\rightarrow B$ is a finite morphism of rings and $P\in \text{Spec}(B)$ then $A_{\varphi^{-1}(P)}\rightarrow B_{P}$ is a finite morphism as well.

If you change finite by finite type this is false. Take for example $A=k$ a field, $B=k[x]$ and $P=(x)$.

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    $\begingroup$ No, it is not finite. For example, take $\phi:k[x]\to k[x]$, $x\mapsto x^2$ and take $P=(x-1)$. $\endgroup$
    – Mohan
    Mar 5, 2019 at 13:50

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