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Show that every finite morphism of schemes is proper.

Here are first my thoughts:

Let $f:X \rightarrow Y$ be a finite morphism

Any finite morphism of schemes is of finite type, so we only need to show that $f$ is separated and universally closed.

Since properness is a local property, we can assume that $Y$ is affine and is equal to Spec $B$ for some ring $B$. But given that $f$ is finite, we also have that the pre-image of $Y$, i.e $X$, is affine and is equal to some Spec $A$. So we are reduced to a finite morphism of affine schemes, which is always separated.

Finally, since every finite morphism is closed, $f$ is closed. And given that finite morphisms are stable under base change, $f' : X \times _Y Y' \rightarrow Y'$ is also finite and is hence closed. So $f$ is unviersally closed.

My first question was just whether or not this works? Is there some subtlety I am missing? I ask because I don't fully understand finite and separated morphisms (hence, working on the exercises).

The second question I had was regarding the solution provided here

http://sv.20file.org/up1/1431_0.pdf

as well as this answer - Finite Morphism of Schemes is Proper ; They use the valuative criterion for properness. According to Hartshorne (Th. 3.7), to apply the valuative criterion on a morphism of finite type $f: X \rightarrow Y$, the scheme $X$ needs to be noetherian. Clearly we can assume they are affine, but I don't see why we can assume that they are noetherian.

I'm not really looking for an alternative solution. I'm just trying to understand if I missed something in my proof and what I am in fact missing in the suggested approaches in those solutions.

Any feedback\advice would be appreciated!

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Your proof looks good to me. The proof in the answers you refer to is also correct, once one uses a more general form of the Valuative Criterion for Properness, e.g. this MSE post.

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  • $\begingroup$ Thank you very much! $\endgroup$
    – User20354
    Commented Oct 30, 2019 at 0:20

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