Stalk of Coherent Module Finite

168 Views Asked by At

I have a question about an argument used in a proof in following article:

https://amathew.wordpress.com/2010/12/23/the-baby-version-of-zariskis-main-theorem/#more-2181

here the relevant excerpt:

enter image description here

My question is: If we know that the push forward $f_*\mathcal{O}_X$ is coherent (therefore locally sits in exact sequence $\mathcal{O}_Y \vert_U \to f_*\mathcal{O}_X \vert _U \to 0$)

why is then the localisation $(f_*\mathcal{O}_X)_y$ at $y$ a finite $\mathcal{O}_y$-module?

1

There are 1 best solutions below

0
On BEST ANSWER

This is more or less part of the definition of a coherent sheaf. As $f_*\mathscr{O}_X$ is a coherent $\mathscr{O}_Y$-module, for all $y\in Y$ we can find a neighborhood $V$ of $y$ such that $f_*\mathcal{O}_X|V$ is finitely generated by sections, which is true if and only if the $\mathcal{O}_{Y,y}$-module $(f_*\mathcal{O}_X|V)_y$ is finitely generated.