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Assume we have a projective variety $X$ over some algebraically closed field $k$. How can we show that $O_{X}=k$? I tried to do it in simple examples but the proof is not clear to me.

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It would be helpful to know: a) WHICH proof you mean and b) WHICH steps in this proof are not clear to you. –  Nils Matthes Nov 20 '12 at 22:29
    
Is $X$ an integral variety ? –  user18119 Nov 20 '12 at 23:34
    
I asked a professor, who answered me "go to affine components, pass to the quotient, take the intersection", and I felt at lost. $X$ is a projective variety in $\mathbb{P}^{n}$. –  Bombyx mori Nov 21 '12 at 1:02
    
@user32240: In this case (if I am right about what a projective variety is in your context), a proof of this fact can be found in Hartshorne "Algebraic Geometry", Ch.1, Theorem 3.4. Does this help? –  Nils Matthes Nov 21 '12 at 11:23
    
This does help. The professor said I can follow Hartshorne, but I thought that must be over my head and did not really try that. –  Bombyx mori Nov 21 '12 at 12:25
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1 Answer

Maybe I'm being dumb, but I think you want $X$ to be irreducible, otherwise consider say $X = $2 points sitting around in $\mathbb{P}^1$, $\Gamma$ is $k \oplus k$.

I'm also working with $X$ a variety, i.e. $O_X$ nilpotent free, I hope that's cool.

For $X$ irreducible, a proof follows from knowing that $\mathbb{P}^n$ is proper over $k$, hence so is any closed subvariety, in particular $X$. With this, a global section is the same data as a regular map $X \rightarrow \mathbb{A}^1$, the image must be closed by properess, hence finitely many points, hence only one point by irreducibility.

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Your proof did not use $X$ being projective in any essential way. I have some doubt if it works. –  Bombyx mori Nov 21 '12 at 12:24
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The proof is 2 parts. The first (which I only cite) is that $\mathbb{P}^n$ is proper as a variety, and hence so is any closed subvariety namely $X$ (this is where we use $X$ projective). The second is that for any irreducible proper variety $Y$, $\Gamma(O_Y, Y) = k$. –  uncookedfalcon Nov 21 '12 at 22:54
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You can replace irreducible with connected (assuming $X$ is reduced). In the proof you should say why the image of $X\to \mathbb A^1$ is the whole space. –  user18119 Nov 22 '12 at 21:49
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