Proper morphisms of schemes are defined to be separated, universally closed, and of finite type, for instance see The The Stacks Project - Tag 0CL4.

In section 10.3 of The Rising Sea: Foundations of Algebraic Geometry, starting with the notion of proper maps of topological spaces, Vakil motivates the requirement for being separated and emphasises the role of being universally closed.

What about finite type? Is there some important theorem about proper morphisms that does not work without the (locally of) finite type assumption, is it to prevent some counter-intuitive examples, or is it just because it is defined this way in EGA II?

  • $\begingroup$ I suppose this is the best translation of what we call proper map in topology (inverse images of compact sets are compact) into the language of algebraic geometry. $\endgroup$ Commented Sep 23, 2016 at 19:05
  • $\begingroup$ @HagenvonEitzen The word proper also occurs in Weil's foundations, is there any relation ? $\endgroup$ Commented Sep 23, 2016 at 19:12
  • 1
    $\begingroup$ @Hagen. A continuous map $X\to Y$ is called "proper" (according to Bourbaki) if it is universally closed. Translating this into algebraic geometry does not yield the notion of "proper" in algebraic geometry : this is precisely why the OP is puzzled. If $Y$ is locally compact, then toplogical properness is equivalent to the inverse image of compact subspaces being compact. The criterion "inverse image of compact subsets must be compact" is however completely unreasonable in algebraic geometry (almost all schemes, for example affine $n$-space $\mathbb A^n$ are quasi-compact) $\endgroup$ Commented Sep 23, 2016 at 19:56
  • $\begingroup$ @GeorgesElencwajg Ah, seems I had only a special case of (top.) proper in mind. Nevertheless, as "compact" is usually "as good as finite", we have a handwaving connection to morphisms of finite type :) $\endgroup$ Commented Sep 23, 2016 at 20:05
  • $\begingroup$ Dear @Hagen: actually the story is quite interesting and complicated. Among the early Bourbakistas were the best algebraic geometers on earth: Cartan, Weil, Dieudonné, Chevalley, Serre, Grothendieck, Cartier, .... Their redaction of Topologie Générale (and other Livres, like Algèbre Commutative) was influenced by this background of theirs, and this shows in their definitions of proper (=universally closed) or compact (=quasi-compact+Hausdorff), which must look rather unnatural to an analyst ! $\endgroup$ Commented Sep 23, 2016 at 20:27

1 Answer 1


1) Under the hypothesis that the scheme morphism $f:X\to S$ is separated and finite type we have the nice equivalence $$ f \operatorname {satisfies the valuation criterion} \iff f \operatorname {is proper }$$ To prove that the valuation criterion implies the universal closedness of $f$, the hypothesis that $f$ be of finite type is needed : Mumford-Oda, Chapter II, Proposition 6.8, page 78-80.

2) Given a proper map of schemes $f:X\to Y$ with $Y$ locally noetherian and a coherent sheaf $\mathcal F$ on $X$, the higher direct images $R^qf_*(\mathcal F) \;(q\geq0)$ are coherent sheaves on $Y$.
The proof requires $f$ to be of finite type: EGA III$_1$, Théorème 3.2.1, page 116.

  • 1
    $\begingroup$ You do not need finite type for the valuation criteria, it is enough to be quasi-compact and quasi-separated, see stacks.math.columbia.edu/tag/01KF and stacks.math.columbia.edu/tag/01KY. Both quasi-compact and quasi-separated are implied by universally closed and separated, and hence my question about the `extra assumption' of being (locally) of finite type. $\endgroup$
    – user24453
    Commented Sep 23, 2016 at 22:22
  • 3
    $\begingroup$ Well of course in every situation you can think of somewhat weaker hypotheses yielding the result but finite type seems to be a good general hypothesis: see 2). The fact that universally closed implies quasi-compact is a result proved by Poonen in order to answer a question of mine asked about six years ago on MathOverflow . See also Lemma 28.39.10 of the Stacks Project. So it is not clear that Mumford knew about it when he wrote the reference I gave. $\endgroup$ Commented Sep 23, 2016 at 22:44

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .