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A well-known result by Serre is that properness of a noetherian scheme $(X,\mathscr O_X)$ over $k$ implies finite dimensionality of $H^i(X,\mathscr O_X)$ for all $i \geq 0$. For a projective variety it is easy to prove that $H^0(X,\mathscr O_X) = \mathscr O_X(X) = k$. I was wondering if there's an elementary proof for finite-dimensionality of higher cohomology groups. Maybe, for the easiest case of $X$ being an smooth curve and finite-dimensionality of $H^1(X,\mathscr O_X)$ one could reduce it to computing it for $\mathbb{P}^1$?

  • Could you expand on what you feel is elementary? For example, if you just mean "doesn't use fancy scheme theory" you could use the Hodge decomposition and finite dimensionality of de Rham cohomology for compact complex manifolds. But of course this is really non-elementary in a different way and has more restrictive hypotheses. – Samir Canning Oct 30 '18 at 05:02
  • what they call classical alg. geonetry, e.g. Hartshorne Chapter 1. Obviously, one needs to know about cohomology and a bit of sheaf theory. My point is that it should be easier for the structure sheaf and not just “every coherent sheaf”. –  Oct 30 '18 at 10:59

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Elementary can mean many things, but here is a way to reduce to projective space case. Given a projective variety $X$ of dimension $n$, you can find a finite map $f:X\to\mathbb{P}^n$. Since finite maps are affine, one checks that for any quasi-coherent sheaf $F$ on $X$, $H^i(X,F)=H^i(\mathbb{P}^n, f_*F)$. Finally, using finiteness, for any coherent sheaf $F$ on $X$, one checks $f_*F$ is coherent and thus you are reduced to the projective space case.

Mohan
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  • Hi @Mohan ,may I ask a question why we can find such finite map $X\to \Bbb{P}^n$, I know there is some closed immersion from $X\to \Bbb{P}^N$ which is also finite, how to construct a map with same dimension? – yi li Mar 16 '23 at 03:17
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    @yili This is standard. If $X\subset \mathbb{P}^m$ is a closed embedding and dimension of $X$ is $n$, a general linear subspace of dimension $m-n-1$ in $\mathbb{P}^m$ does not intersect $X$ and project from this linear space to get such a finite map. – Mohan Mar 16 '23 at 03:26
  • Thank you Mohan, I also found some post here : https://math.stackexchange.com/q/679768/360262 – yi li Mar 16 '23 at 04:13