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!
Best Answer
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.