TITLE: If the direct image of f preserves coherent sheaves on noetherian schemes, how to show f is proper?
QUESTION [14 upvotes]: 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

REPLY [10 votes]: 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.