By this notes, a morphism of varieties $f: X \to Y$ is proper if for every morphism $g: Z \to Y$, the induced morphism $X\times_Y Z\to Z$ is closed, in other words, it's universally closed.

But a morphism of schemes is proper if it is separated, of finite type, and universally closed.

I wonder why there two definitions are coherent? Why do we need to add separated and of finite type for the schemes?


1 Answer 1


Mustata is assuming his varieties are separated schemes of finite type over a field. This implies that every morphism of these varieties is of finite type and separated: if $f:X\to Y$ and $g:Y\to Z$ are morphisms with $g\circ f$ separated (resp. finite type), then $f$ is separated (resp. finite type). This applies to our situation by letting $Z$ be the spectrum of the field we're working over. So Mustata's definition is the same as the usual definition.

As to your question about why we want to "add" separated and of finite type for the scheme side, this is somewhat backwards - this is the typical definition one takes, Mustata just doesn't need to say either of finite type or separated, because the restriction on the types of varieties he's working with already imply that all morphisms of varieties he considers have these properties.

  • $\begingroup$ But I think Mustata doesn't consider his varieties as schemes but just affine or projective varieties. If under this setting, do we have the same thing? $\endgroup$
    – 6666
    Commented Apr 18, 2020 at 23:13
  • $\begingroup$ Your claim that "Mustata doesn't consider his varieties as schemes but just affine or projective varieties" is not correct, see this previous entry in his lecture notes for the same class. $\endgroup$
    – KReiser
    Commented Apr 18, 2020 at 23:20
  • $\begingroup$ I see, that marks sense $\endgroup$
    – 6666
    Commented Apr 18, 2020 at 23:29
  • 2
    $\begingroup$ First of all, there are finiteness conditions in the topological case: the target ought to be locally compact Hausdorff for proper to be the same as universally closed. Next, what do you mean, "need"? One can certainly consider separated, universally closed morphisms, but they do not have some of the nice properties that proper morphisms have. For instance, in a proper morphism, the direct image of a coherent sheaf is coherent, but this is not necessarily true for universally closed + separated, as the example of $\coprod^\infty X\to X$ shows. $\endgroup$
    – KReiser
    Commented Apr 19, 2020 at 0:39
  • 1
    $\begingroup$ @KReiser I agree with everything you said, but just a nitpick : I don't think $\coprod^\infty X\to X$ is closed. $\endgroup$
    – Roland
    Commented Apr 19, 2020 at 9:53

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