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In Hartshorne (and pretty much everywhere else I know) a proper morphism is defined to be

A morphism of schemes $f:X\to Y$ is said to be proper if it is sperated, of finite type and universally closed.

But then I started reading "Introduction to Moduli Problems and Orbit Spaces" by Newstead. He defines a proper morphism between two separated varieties (which he simply calls variety, calling non-seperated varieties prevarieties) quite differently:

Let $X,Y$ be (separated) varieties. A morphism $f:X\to Y$ is proper if for any other (separated) variety $Z$, the map $f\times 1: X\times Z\to Y\times Z$ is closed.

I can't figure out why these two definitions are equivalent to each other when the Harthsorne definition is restricted separated varieties over $k$. The finite type in the definition of Hartshorne is clearly redundant when dealing with varieties.


Edit: After a few comments, I think I should say what parts of this equivalence is problematic for me to prove (although the comments already helped me resolve parts of the problem/confusion).

  • If $f:X\to Y$ is universally closed, i.e. for all separated $k$-varieties $Z$, the projection map $Z\times_Y X\to Z$ is closed (after base change), then $f\times 1: X\times Z\to Y\times Z$ is closed:

Consider the base change of $f:X\to Y$ under the projection map $\pi_Y: Z\times Y\to Y$ with $Z$ arbitrary. One easily check that $X\times Z$ with $\pi_X: X\times Z\to X$ and $f\times 1: X\times Z\to Y\times Z$ satisfies the UMP of fibered product. Then universally closed yields $f\times 1$ is closed.

So what remains to prove is

  • If $f\times 1: X\times Z\to Y\times Z$ is closed for all separated $k$-varieties $Z$, then $Z\times_Y X\to Z$ is closed for all $Z$ too.

  • Also I can really get some help on why any morphism of separated $k$-varieties is separated. I think it should be simple and I'm just thinking over-complicated.

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    $\begingroup$ As you said, finite type is redundant, and every morphism of $k$-varieties is separated. Thus, what Newstead gives as the definition of proper is just the definition of universally closed. $\endgroup$ – Tabes Bridges Jul 27 '16 at 15:55
  • $\begingroup$ That's exactly the two things I couldn't prove. Every morphism of (separated?) varieties is separated? I suspected that to be true... Moreover universally closed means the projection $X\times_Y Z\to Z$ is closed for any $g:Z\to Y$, not $f\times 1$ is closed. What I'm asking is how closeness of one of them results the other. $\endgroup$ – Hamed Jul 27 '16 at 16:51
  • $\begingroup$ Going (1) $\Rightarrow$ (2) seems okay, right? $\endgroup$ – Hoot Jul 27 '16 at 19:02
  • $\begingroup$ @Hoot yeah, I added the proof of $1\to 2$ in my question. $\endgroup$ – Hamed Jul 27 '16 at 19:56
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    $\begingroup$ Re "Every morphism of separated varieties is separated": given $f : X \to Y$ and $g : Y \to Z$, if $g \circ f$ is separated, then so is $f$ (Hartshorne Corollary II.4.6(e)). Taking $Z = \operatorname{Spec}(\mathbb{Z})$ then gives $X$ separated $\implies f$ separated $\endgroup$ – Jay Jul 28 '16 at 0:16
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I think the two definitions are equivalent if one works in the category of (separated) varieties. Hartshorne's definition says that, for any morphism $g: Z \to Y$, the morphism from the fiber product $X \times_Y Z \to Z$ is closed. In fact, $g: Z \to Y$ can be factored as $Z \to Y \times Z \to Y$, the first part being given by the graph of $g$, the second by projection to $Y$. Now, $Z \to Y \times Z$ is a closed embedding, since we are dealing with separated spaces, and so is $X \times_Y Z \to X \times Z$. So $X \times Z \to Y \times Z$ being closed implies the same for $X \times_Y Z \to Z$.

After a few comments, I think I should say what parts of this equivalence is problematic for me to prove (although the comments already helped me resolve parts of the problem/confusion).

  • If $f:X\to Y$ is universally closed, i.e. for all separated $k$-varieties $Z$, the projection map $Z\times_Y X\to Z$ is closed (after base change), then $f\times 1: X\times Z\to Y\times Z$ is closed:

Consider the base change of $f:X\to Y$ under the projection map $\pi_Y: Z\times Y\to Y$ with $Z$ arbitrary. One easily check that $X\times Z$ with $\pi_X: X\times Z\to X$ and $f\times 1: X\times Z\to Y\times Z$ satisfies the UMP of fibered product. Then universally closed yields $f\times 1$ is closed.

So what remains to prove is

  1. If $f\times 1: X\times Z\to Y\times Z$ is closed for all separated $k$-varieties $Z$, then $Z\times_Y X\to Z$ is closed for all $Z$ too.

  2. Also I can really get some help on why any morphism of separated $k$-varieties is separated. I think it should be simple and I'm just thinking over-complicated.

I can at least answer your question 2. This is covered by Corollary II.4.6(e) of Hartshorne by taking $Z = \text{Spec}\,k$. In fact, if $X$ is any Noetherian separated $k$-scheme and $f: X \to Y$ is any morphism, then $f$ is separated.

I think 1 is more complicated. The Grothendieck definition of properness can not be used in the category of varieties because fiber products of varieties are not necessarily varieties. That is why Newstead used the simpler definition in his book, which is the classical definition and I think it is okay for what he wants. Actually, classical algebraic geometry did not use properness a great deal, if at all, using only the theorem that says that the image of a projective variety under a regular map (i.e. morphism) is projective. Of course one needs an algebraically closed field for this. However, there may still be an argument to prove something like Grothendieck's definition from that of Newstead. Suppose $f: X \to Y$ is a morphism of separated varieties and $g: Z \to Y$ is a morphism with $Z$ a separated variety. Then the graph $Z \to Z \times Y$ is a closed immersion (Exercise II.4.4 of Hartshorne). If one can prove that $Z \times_Y X \to Z \times X$ is also a closed immersion, that should be enough. I am not sure about this and I am certainly not sure what happens when $Z$ is not separated.

So I think basically I am telling you not to worry. Both definitions are okay in context, but they may not be equivalent.

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