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Let $k$ be a field, $X\rightarrow \mathrm{Spec}\,k$ be a separated morphism of finite type of relative dimension$\leq 1$ (as defined here). Is it true that $f$ is proper iff $f_* \mathcal{O}_X$ is coherent? One direction is clear because proper morphisms preserve coherence.

Is this true over more general bases?

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  • $\begingroup$ The open embedding $\mathbf{A}^2 - \{(0,0)\} \hookrightarrow \mathbf{A}^2$ already gives you a counterexample over arbitrary base. $\endgroup$ – HYL May 2 at 18:43
  • $\begingroup$ Doesn't projective plane minus a point give a counter example to the second question? $\endgroup$ – Asvin May 2 at 23:20
  • $\begingroup$ @HYL I don't understand. Counterexample to which statement? $\endgroup$ – user74900 May 3 at 1:53
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The answer to the first question is certainly yes, because if a curve is non-proper, it must be affine, and hence its ring of global sections is not finite as a $k$-module. See this link to a M.SE topic. The answer to your second question I do not know.

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  • $\begingroup$ technically, the link talks about "geometrically connected" curves, which is not necessarily the case for this question. I am not claiming that this is a problem but this answer is strictly speaking not complete. $\endgroup$ – user138661 May 2 at 13:59
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    $\begingroup$ The $k$-dimension of the ring of global sections is stable under extension of the ground field, so we may assume the curve is a union of geometrically connected curves. If the union is not proper, then one of the components must be non-proper, and its ring of global sections injects into the ring of global sections of the original curve. $\endgroup$ – RP_ May 2 at 14:08
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As pointed out in the comments, this is false for general bases. Let $k$ be a field, $S = \mathrm{Spec}(k[t])$, let $\overline{X} = \mathbb{P}^1 \times_{\mathrm{Spec} k} S$ with projection $\overline{f} \colon \overline{X} \rightarrow S$. and $X = \overline{X} - \{(0,0)\}$ and $f = \overline{f}|_X$. Then $f_* \mathscr{O}_X$ is the sheaf on $S$ associated to the $k[t]$-module $\Gamma(X, \mathscr{O}_X)$. Now by Hartog's theorem (or a calculation with a standard open cover of $X$), we see that $\Gamma(X, \mathscr{O}_X) = \Gamma(\overline{X}, \mathscr{O}_{\overline{X}}) = k[t]$, which is certainly a finite $k[t]$-module.

Here's a positive result: let $f \colon X \rightarrow S$ be a "relative curve", i.e. a finite type separated flat morphism with geometrically connected fibers of dimension $1$. Furthermore, suppose $S$ is locally noetherian (one could probably eliminate this hypothesis by a limit argument). Then if $\mathrm{R}^i f_* \mathscr{O}_X$ is coherent for all $i$, $f$ is proper. (It seems likely that it suffices to assume this just for $i = 0, 1$, but I don't see the argument right now).

Indeed, EGA IV_3 Corollaire 15.7.10 says that since $f$ is flat with geometrically connected fibers, properness of $f$ may be checked on the fibers. This lets us reduce to the case that $S$ is a field, once we ensure that $(f_s)_* \mathscr{O}_{X_s} = \Gamma(X_s, \mathscr{O}_{X_s})$ is a finite-dimensional $k(s)$-vector space for any $s \in S$.

Let $i_s \colon \mathrm{Spec }\ k(s) \rightarrow S$ be the inclusion. By a fancy version of cohomology and base change (see Tag 08IB in the Stacks Project: the "Tor-independence" hypothesis is satisfied because $f$ is flat), we get an isomorphism in the derived category $\mathrm{L} i_s^* \mathrm{R} f_* \mathscr{O}_X \rightarrow \mathrm{R} (f_s)_* \mathscr{O}_{X_s}$ (flatness of $f$ ensures that $\mathscr{O}_{X_s}$ is the derived pullback of $\mathscr{O}_X$ to $X_s$). Passing to cohomology shows that the $k(s)$-vector space $(f_s)_* \mathscr{O}_{X_s}$ has a filtration by $\mathrm{Tor}_{i, \mathscr{O}_{S, s}}(k(s), (\mathrm{R}^i f_* \mathscr{O}_X)_s)$. By assumption, the $(\mathrm{R}^i f_* \mathscr{O}_X)_s$ are finite type $\mathscr{O}_{S,s}$-modules, so these Tor-groups are finite-dimensional $k(s)$-vector spaces and thus $(f_s)_* \mathscr{O}_{X_s}$ is as well. The same argument shows that the higher cohomology groups $\mathrm{H}^i(X_s, \mathscr{O}_{X_s})$ are finite-dimensional over $k(s)$ as well.

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