The other direction is well known. I think it is true and I was told by several other guys doing algebraic geometry that it is indeed true but they did not know how to prove. I am also wondering whether there is a general nonsense style proof. It looks like the statement can be proved by playing adjunctions in several categories. Note that a scheme $X$ is quasi compact iff the structure sheaf $O_X$ is a compact object in category of quasi coherent sheaves $Qcoh(X)$. We also know that if $X$ is noetherian scheme, then coherent sheaves are exactly the compact objects in $Qcoh(X)$, so it looks like this statement is equivalent to say if $f_*:Qcoh(X)\rightarrow Qcoh(Y)$ preseves compact objects, then $f$ itself is proper morphism of scheme. But this somehow means that $f^{-1}$ preserves compactness in the topological sense. (I think this can also be phrased into general nonsense language,say the compact objects in category of topological space.) Then one might be able play the commutativity of $Hom(M,?)$ functor and filtered colimits game to get proof.

All the comments are welcome Thanks

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    $\begingroup$ i guess that the first thing to do is prove that if you have a non proper variety over a field k, then it has a coherent sheaf whose global sections are infinite dimensional $\endgroup$ Sep 14, 2014 at 20:41
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    $\begingroup$ A good example of such would be the structure sheaf of an incomplete (so affine) closed curve on this non-proper variety. You get such curves thanks to the valuative criterion of properness. $\endgroup$ Sep 14, 2014 at 20:58
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    $\begingroup$ Or maybe even just do this. Take a compactification $X'$ of $X$, choose a curve going through a point of $X' \setminus X$. That should do the job... $\endgroup$ Sep 14, 2014 at 22:23
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    $\begingroup$ why should it even be separated? the map from the affine line with the doubled origin to the affine line seems like it ought to preserve coherent sheaves $\endgroup$ Sep 15, 2014 at 7:16
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    $\begingroup$ Vivek is right, although I think if you also demand that the higher direct images are coherent you can deduce separatedness of f. $\endgroup$
    – Tom Graber
    Sep 16, 2014 at 19:38

2 Answers 2


Well, as already explained in the comment by Vivek Shende, the answer is no: there are non-proper morphisms with push-forward that preserves coherence. I've thought a bit on this the last couple of years and on the "positive" side we have:

Theorem 1. If $f\colon X\to Y$ is a universally closed morphism of finite type between noetherian schemes, then $f_*$ preserves coherence.

This is quite easy to show with Raynaud–Gruson's refined Chow Lemma (also when $X$ and $Y$ are algebraic spaces) and standard dévissage arguments. With a completely different proof I have also extended this to algebraic stacks.

The morphism $\mathbb{P}^2\setminus 0\to *$ does not preserve coherence as there exist non-proper curves. If $f_*$ preserves coherence after arbitrary noetherian base change, then $f$ is universally closed: the valuative criterion is easily seen to hold (as indicated by Piotr Achinger) since the fraction field of a DVR is not coherent. However, it turns out that it is not necessary to involve base change and it is also possible to characterize properness in terms of $R^1f_*$ (cf comment by Tom Graber).

Theorem 2. Let $f\colon X\to Y$ be a morphism of finite type between noetherian schemes. Then
(i) $f$ is universally closed if and only if $f_*$ preserves coherence, and
(ii) $f$ is proper if and only if $f_*$ and $R^1f_*$ preserves coherence.

The necessity of the conditions is Theorem 1 in (i) and well-known in (ii). To see that they are sufficient we need some lemmas that say that the valuative criteria can be checked using curves in $X$. A curve is a $1$-dimensional noetherian integral scheme. First we show that it is enough with curves of finite type over $Y$.

Lemma 1. Let $f\colon X\to Y$ be a morphism of finite type between noetherian schemes. If $f$ is not universally closed (resp. not separated) then there exists
(i) a separated curve C together with a morphism $C\to Y$ of finite type; and
(ii) a closed subscheme $C'$ of $X\times_Y C$ such that $C'$ is a curve and $C'\to C$ is birational and not surjective (resp. not separated).

Proof. By the valuative criterion, there exists a DVR $D$, a morphism $Y_0:=\operatorname{Spec} D\to Y$ and an integral closed subscheme $Z_0\subseteq X_0:=X\times_Y Y_0$ such that $Z_0\to Y_0$ is an open immersion: the inclusion of the generic point (resp. not separated). In the latter case, we can assume that there are two different sections of $Z_0\to Y_0$ that only agree over the generic point $U_0=\operatorname{Spec} K(D)$.

We now approximate the map $Y_0\to Y$ and the data $U_0\subseteq Y_0$, $Z_0\to Y_0$: there exists

  • $Y'\to Y$ of finite type with $Y'$ integral,
  • an (affine) open immersion $U'\subseteq Y'$, and
  • a closed integral subscheme $Z'\subseteq X'=X\times_Y Y'$ such that $Z'\to Y'$ is an isomorphism over $U'$.

    Moreover, we can arrange so that $Z'\to Y'$ is equal to $U'\to Y'$ (resp. there are $2$ sections of $Z'\to Y'$ which only coincide over $U'$).

    Pick a closed point $y'$ in $Y'\setminus U'$ and a generization $u'\in U'$ such that $C:=\overline{u'}$ has dimension $1$. Let $C'=Z\cap X'\times_{Y'} C$. QED

    Now we refine Lemma 1 and show that it is enough to consider either "vertical curves" (contained in a fiber) or "horizontal curves" (birational to a curve in the base).

    Lemma 2. Let $f\colon X\to Y$ be a morphism of finite type between noetherian schemes. If $f$ is not universally closed (resp. not separated) then there exists a birational morphism of curves $C'\to C$ which is not universally closed (resp. not separated) such that either:
    (i) there is a finite morphism $C'\to X_y$ for a point $y\in Y$ (of finite type) and $C$ is a projective curve over $y$; or
    (ii) there is a finite morphism $C\to \operatorname{Spec} \mathcal{O}_{Y,y}$ and $C'$ is a closed subscheme of $X\times_Y C$.

    Proof. Let $C\to Y$ be a curve as in Lemma 1. The image is either a point or the morphism is quasi-finite.

    Case (i): the image of $C$ is a point $y\in Y$. We may replace $C$ by a compactification and assume that $C\to y$ is projective. If the image of $C'\to X_y$ is a point, then it is a closed point and it follows that $C'\to C$ is finite which gives a contradiction since $C'\to C$ is not proper. Thus the image of $C'\to X_y$ is $1$-dimensional and $C'\to X_y$ is quasi-finite and proper, hence finite.

    Case (ii): $C\to Y$ is quasi-finite. We may replace $Y$ with the schematic image of $C$ and localize at the image of a suitable closed point of $C$. Then $Y$ is integral and $1$-dimensional. Using Zariski's main theorem, we have $C\to \overline{C}\to Y$ where $C\to \overline{C}$ is an open immersion and $\overline{C}\to Y$ is finite. We may replace $C$ with $\overline{C}$ and $C'$ with its closure in $X\times_Y \overline{C}$. QED

    Remark. The localization in (ii) is only necessary when $Y$ is not Jacobson.

    The third lemma shows that curves as in Lemma 2 give rise to non-coherent cohomology.

    Lemma 3. Let $f\colon C'\to C$ be a birational morphism of finite type of curves. If $f$ is not universally closed (resp. not separated), then $f_*\mathcal{O}_{C'}$ is not coherent (resp. $R^1f_*\mathcal{O}_{C'}$ is not coherent). If $C$ is projective over $k$, then in addition $\Gamma(C',\mathcal{O}_{C'})$ (resp. $H^1(C',\mathcal{O}_{C'})$) is infinite-dimensional.

    Proof. For the first statement, it is enough to show that the sheaf is not coherent after passing to the henselization of a suitable point $c$ in $C$. We may thus assume that $C$ is local and henselian (although not necessarily irreducible). If $f$ is not universally closed, then (after replacing $C'$ with a connected component if it is reducible) $C'\to C$ is an open immersion. Then $f_*\mathcal{O}_{C'}$ is not coherent.

    If $f$ is not separated, then (after replacing $C'$ with a connected component if it is reducible) $C'\to C$ is not separated and there is an open covering of $C'$ consisting of the local rings at the points above $c$. A Cech calculation gives that $H^0(C',\mathcal{O}_{C'})$ is coherent whereas $H^1(C',\mathcal{O}_{C'})$ is not coherent.

    For the second statement, we note that there is a dense open subscheme $U\subseteq C$ such that $f_*\mathcal{O}_{C'}$ is coherent over $U$ and $R^1f_*\mathcal{O}_{C'}$ is zero over $U$. If $f_*\mathcal{O}_{C'}$ is not coherent, then choose a coherent subsheaf $\mathcal{F}\subseteq f_*\mathcal{O}_X$ that is an isomorphism over $U$. Then the cokernel $\mathcal{Q}$ is a non-coherent sheaf that is zero over $U$. It follows that $H^0(C,\mathcal{Q})$ is infinite, hence that $H^0(C,f_*\mathcal{O}_{C'})$ is infinite since $H^1(C,\mathcal{F})$ is finite. If $R^1f_*\mathcal{O}_{C'}$ is not coherent, then $H^0(C,R^1f_*\mathcal{O}_{C'})$ is infinite. QED.

    Theorem 2 is an easy consequence of the lemmas above:

    Proof of Theorem 2. We have seen that the conditions are necessary. Conversely, assume that $f$ is not universally closed (resp. not separated). We have to show that $f_*$ does not preserve coherence (resp. $R^1f_*$ does not preserve coherence). Let $C'\to C$ be as in Lemma 2 and choose a coherent sheaf $\mathcal{F}\in \mathrm{Coh}(X)$ restricting to the push-forward of $\mathcal{O}_{C'}$ in $\mathrm{Coh}(X\times \operatorname{Spec} \mathcal{O}_{Y,y})$. By Lemma 3, we have that $f_*\mathcal{F}$ (resp. $R^1f_*\mathcal{F}$) is not coherent at $y$. QED.

    Remark 1. With small modifications, the proof of Theorem 2 also holds for algebraic spaces and Deligne–Mumford stacks: first one takes a finite cover of $Y$ to reduce to $Y$ a scheme, then uses the valuative criterion for stacks. An alternative proof for schemes, algebraic spaces and Deligne–Mumford stacks would be to use Raynaud–Gruson's Chow lemma to reduce the question to morphisms of the form $f\colon X\to \mathbb{P}^n_Y \to Y$ where $f\colon X\to \mathbb{P}^n_Y$ is étale and birational and $Y$ is a scheme.

    Remark 2. The situation for algebraic stacks with infinite-dimensional stabilizers is more subtle. Theorem 1 holds as mentioned above and Theorem 2 (i) is probably ok. Theorem 2 (ii) however has to be modified. If $X$ has a good moduli space then it follows from Theorem 2 that $X$ has coherent cohomology if and only if $X_\mathrm{gms}$ is proper. For example, note that $BGL_n$ has coherent cohomology but is not separated. This is all well-known and one may argue that such stacks are "almost proper" (cf. paper by Halpern-Leistner–Preygel, arXiv:1402.3204).

  • $\endgroup$

    If $f:X\to Y$ is separated of finite type between noetherian schemes and $f_*$ preserves coherence, then $f$ is proper. Here is a proof that follows the geometric idea (given in the comments of Piotr Achinger and Karl Schwede) to use a curve in $X$ that is not proper over $Y$ :

    Since it is possible to extend a coherent sheaf on an open of $X$ to the whole of $X$ [Stacks project, Tag 01PD], we see that the statement is local on $Y$, and we are reduced to the case where $Y$ is the spectrum of a local ring.

    By Nagata's compactification theorem, we may find a compactification of $f$: it is the composition of an open immersion with dense image $i:X\to \overline{X}$ and a proper morphism $\overline{f}:\overline{X}\to Y$. We suppose for contradiction that $Z=\overline{X}\setminus X$ is not empty.

    Let us reduce to the case where $Z$ is finite and for every $z\in Z$, $\mathcal{O}_{\overline{X},z}$ is of dimension $1$. We do it by applying iteratively the following procedure. Choose a closed point $z\in Z$ at which $\mathcal{O}_{\overline{X},z}$ is not of dimension $1$. Using prime avoidance, choose a function $\phi\in \mathcal{O}_{\overline{X},z}$ that vanishes on $z$, but does not vanish identically on any irreducible component of $\overline{X}$ through $z$ nor on any positive-dimensional irreducible component of $Z$ through $z$. Replace $\overline{X}$, $X$ and $Z$ by their intersection with the closure of $\{\phi=0\}$ in $\overline{X}$. The hypothesis of preserving coherence still holds because closed immersions preserve coherence. The Hauptidealsatz ensures that $X$ is still dense in $\overline{X}$.

    Using noetherian induction on $Z$, and induction on the dimensions of the local rings $(\mathcal{O}_{\overline{X},z})_{z\in Z}$, we see that the procedure terminates. Hence, from now on, we suppose that $Z$ is finite and for every $z\in Z$, $\mathcal{O}_{\overline{X},z}$ is of dimension $1$. Replacing moreover $\overline{X}$ by one of its components through $z$, we may assume that it is integral of generic point $\eta$.

    Now we distinguish two cases. If $f(\eta)$ is the closed point of $Y$, then $X$ is an affine curve over the residue field of $Y$, and $f_*\mathcal{O}_X$ is not coherent: a contradiction.

    Otherwise, consider the Stein factorization $\overline{X}\to S\to Y$ of $\overline{f}$. Up to replacing $Y$ by $S$ (we do not lose the hypothesis of preserving coherence, because $S\to Y$ is finite), we may suppose that $\overline{f}$ satisfies $\overline{f}_*\mathcal{O}=\mathcal{O}$. In particular, it has connected fibers. Localizing $Y$ at a point in the image of $Z$ as in the first step, we may still assume that $Y$ is local.

    The points of $Z$ are closed points, hence sent to the closed point $y$ of $Y$ by properness of $\overline{f}$. Since their only generization in $\overline{X}$ is $\eta$, and since the fibers of $\overline{f}$ are connected, $\overline{f}^{-1}(y)$ consists of a single point $z=Z$. It then follows that $\overline{X}$ has only two points $z$ and $\eta$, and that $Y$ has also only two points: their images. Consequently, $\overline{f}$ is quasi-finite and proper, hence finite, and in fact an isomorphism because $\overline{f}_*\mathcal{O}=\mathcal{O}$.

    At this point, the hypothesis that $f_*\mathcal{O}$ is coherent means exactly that $\textrm{Frac}(\mathcal{O}_{Y,y})$ is of finite type over $\mathcal{O}_{Y,y}$, which is absurd.

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      $\begingroup$ Nice proof. It easily extends to the case where $X$ is a separated algebraic stack by taking a proper covering by a scheme. Making the base a general stack is more difficult though (exactly as in my proof). Nitpick: why do you write "qcqs" when all schemes are noetherian? $\endgroup$
      – David Rydh
      Oct 8, 2014 at 19:38
    • $\begingroup$ @DavidRydh Thanks ! Of course, your proof is easier in the sense that Zariski-Riemann spaces are somehow hidden in the use of Nagata compactification. (And my qs hypothesis was a consequence of my lack of knowledge of non-separated phenomena ;-) ). $\endgroup$ Oct 8, 2014 at 20:11
    • $\begingroup$ @Olivier: nice proof, but I'm missing something on your reduction step towards finiteness of $Z$: do you ask something to the function $f\in\mathcal{O}_{\overline{X},z}$ (which you might call $g$, you already had an $f$ around, right? ) or simply not to vanish? What if I take $g=1$ and have $\overline{X},Z,X$ empty? Am I producing nonsense? $\endgroup$ Oct 10, 2014 at 7:28
    • $\begingroup$ @FilippoAlbertoEdoardo : hi ! You're of course right : I want $f$ to vanish on the point $z$ I have chosen. This ensures that $Z$ is still not empty. To prevent other degenerate situations (like $X$ being empty, or $X$ not being dense anymore in $\overline{X}$), I also require that $f$ does not vanish on various subschemes : but this I did not forget to write... $\endgroup$ Oct 10, 2014 at 8:07
    • $\begingroup$ @OlivierBenoist: Just another small question: do you really need that a coherent sheaf can be extended (which I agree is true, of course)? After all, in your situation, what you only need is that it can be restricted (preserving coherence), no? If $f: X\to Y$ is as above and you want to prove it proper (local condition), pick a point $y\in Y$, restrict $f$ to $X_0=f^{-1}(\mathrm{Spec}(\text{local containing }y))$ and get that $f\vert X_0 :X_0\to \mathrm{Spec}(\text{this local})$ is proper, hence $f$ is proper. No? $\endgroup$ Oct 10, 2014 at 9:35

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