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Let $X$ be complete variety over an algebraically closed field $k$. It is an immediate consequence of the definition that $\mathcal{O}_X=k$. Is the converse true as well ? I suspect this to be not true, but so far I have not been able to find any reference. I appreciate any hints/nudges.

Hmm.
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    If you remove a point from a projective variety of large dimension, you do not change the ring of regular functions. – Mariano Suárez-Álvarez Oct 15 '16 at 17:41
  • I suspect I'm missing something. If you delete a point, corresponding to the homogeneous maximal ideal say $m$, the ring of regular functions becomes $k[x_0,\dotso,x_n]_{(m)}$, which I don't see being isomorphic to $k$. – Hmm. Oct 15 '16 at 17:51

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Let me expand upon Mariano's comment.

I claim that if $X$ is a variety of dimension $n \geq$ 2, then $\Gamma(\mathcal O_X,X)$ and $\Gamma(\mathcal O_X,X \backslash pt)$ are isomorphic.

(since removing a point makes a variety incomplete, this would disprove the assertion that checking on global sections is enough)

We have a natural map $\varphi:\mathcal O_X \to \mathcal O_{X\backslash pt}$ by restriction. It is injective, since if a function is zero on an open set, it is zero everywhere.

So the question becomes: can every function defined on $X \backslash pt$ be extended to a function on $X$?

This is answered in the affirmative for normal varieties in this question answer.

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Fredrik Meyer
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  • So deleting a point from a normal complete variety would do the trick. Perfect, thanks. – Hmm. Oct 15 '16 at 18:22
  • @Hmm. Minor nitpick: the point must be of codimension at least two. So removing a point from a curve won't do the trick. – Ariyan Javanpeykar Oct 16 '16 at 07:50
  • @Ariyan Yes, that is why I said that $X$ should be of dimension $\geq 2$ (and I was implicitly only talking about closed points). – Fredrik Meyer Oct 16 '16 at 11:27
  • @FredrikMeyer Yes, your answer is absolutely clear in that matter. I was only commenting Hmm.'s remark that removing a point from a normal complete variety would "do the trick". – Ariyan Javanpeykar Oct 17 '16 at 10:58