# A curve is proper iff the space of global sections is finite-dimensional

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?

• The open embedding $\mathbf{A}^2 - \{(0,0)\} \hookrightarrow \mathbf{A}^2$ already gives you a counterexample over arbitrary base. – HYL May 2 at 18:43
• Doesn't projective plane minus a point give a counter example to the second question? – Asvin May 2 at 23:20
• @HYL I don't understand. Counterexample to which statement? – user74900 May 3 at 1:53

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.
• 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. – RP_ May 2 at 14:08
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.