# Hartshorne II.4.1 proof verification and question

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.