I'm studying algebraic geometry (it is my first course), by following Mumford's Red Book, and now I'm stacked in some (probably silly) questions about schemes, more precisely in the section 7 of Chapter 2, about proper morphisms and finite morphisms.

Following the definition of proper morphism (i.e., a finite type morphism $f:X\rightarrow Y$ such that for any morphism $g:K\rightarrow Y$, where $K$ is a prescheme, the projection $p:X\times_{Y}K\rightarrow K$ is closed), the author make some simple statements, that I can't figure out completely:

  1. In the definition of proper morphism, it is sufficient to look at affine $K$'s. Although it seems to be reasonable in face of many other properties I've seen so far, I don't know how to prove this statement without a deeper look at the topology of $X\times_{Y}K$.

  2. Closed immersions are proper: Actually I can prove that a closed immersion is of finite type, but why it is universally closed?

  3. The composition of proper morphisms is a proper morphism. Why?

For now I will try to figure out the statements of the rest of the section.

Thank you very much by your attention.


1 Answer 1

  1. This is just because given any $K$, you can cover by affine open subschemes. You can then use the fact that a subspace of a space is closed iff there is a cover of the space where the intersection if the subspace is closed in each element of the cover.
  2. It's closed because closed immersions are preserved under base change. You can verify closed embedding affine locally, in which case this comes down to the statement that tensoring preserves surjectivity: if $A\to B$ is surjective, then $A\otimes_A C\to B\otimes_A C$ is surjective, for $C$ some $A$-algebra.
  3. Composition of each properties is preserved. For example, let's suppose that if you have $f:X\to Y$ and $g:Y\to Z$. We want to show $g\circ f$ is universally closed. But, for any $Z$-scheme $T$ we have the following fibered diagram

$$\begin{matrix}T\times_Z X & \to & X\\ \downarrow & & \downarrow \\ T\times_Z Y & \to & Y\\ \downarrow & & \downarrow\\ T & \to & Z\end{matrix}$$

So, the maps $T\times_Z X\to T\times_Z Y$ and $T\times_Z Y\to T$ are closed, and so is there composition.


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