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Some may have already asked this question. What are the global sections of structure sheaf of a projective scheme $X$ over a field $k$?

By Hartshorne page 18, Chapter 1, Theorem 3.4, global sections will be $k$ when $k$ is algebraically closed and $X$ in some $\mathbb P^n$ is a projective variety.

Could any one give a counter example for the case $k$ is not algebraically closed? (What will happen if $k=\mathbb R$, the real numbers, and $X=\mathrm{Proj}(\mathbb R[x,y] \mathop{/} {(x^2+y^2)})$ ?)

If $X$ is an integral projective scheme of finite dimension over $k$ (algebraically closed), then $X$ is a projective variety by Hartshorne page 104, Chapter 2, Proposition 4.10. And its global sections should be $k$. Do I need to fix some very ample sheaf to give an embedding to define its structure sheaf?

This is my first time to ask a question. Welcome any advice.

Thanks!

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  • $\begingroup$ I edited your post directly. For any mathematical part, you must use $\LaTeX$ : it is much more readable. I emphasized the references. And the guy is Hartshorne, not Harsthorne. ;) $\endgroup$
    – Pece
    Commented Mar 21, 2014 at 6:34
  • $\begingroup$ Thanks so much for your advice and edition! $\endgroup$
    – user48537
    Commented Mar 21, 2014 at 13:34
  • $\begingroup$ See Corollary 3.21 in Qing Liu's book Algebraic geometry and arithmetic curves, Oxford University Press (2006). $\endgroup$
    – Watson
    Commented Nov 16, 2018 at 21:16

1 Answer 1

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If $X/k$ is proper, integral, then $\Gamma(X, \mathcal O_X)$ will be a finite extension of $k$.

In the example you ask for, namely $X=\text{Proj}(\mathbf{R}[x,y]/(x^2+y^2))$ (graded with $x,y$ in degree $1$), the global sections of $\mathcal O_X$ will be (up to the choice of $x/y$ or $y/x$ as $i$) isomorphic to $\mathbf C$. This scheme is actually affine, isomorphic as an $\mathbf R$-scheme to $\text{Spec } \mathbf C$.

To answer your second question: if $X/k$ is a scheme, projective or not, then it has a structure sheaf by definition of a scheme. You do not need to fix an embedding into projective space. (Anyways, if $X/k$ is a projective scheme, then by definition, it comes with an embedding into projective space, so you don't need to fix one yourself.)

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  • $\begingroup$ Thanks! I am wondering in the case that k is not algebraically closed, is it still true that negative degree line bundles do not have global section? $\endgroup$
    – user48537
    Commented Mar 21, 2014 at 4:43
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    $\begingroup$ I think that finite type is part of the definition of proper :) Also, if the OP is curious, your claim about global sections of proper things being finite can be found in Liu--corollary 3.19. Of course, it also easier to see if one uses more sophisticated machinery $\endgroup$ Commented Mar 21, 2014 at 4:43
  • $\begingroup$ @AlexYoucis You are right, proper already implies finite type. Thanks for the heads up $\endgroup$ Commented Mar 21, 2014 at 5:07
  • $\begingroup$ @user48537 You are welcome. If you have another question, you should post it separately. $\endgroup$ Commented Mar 21, 2014 at 5:13
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    $\begingroup$ "then by definition, it comes with an embedding into projective space" is wrong! it just admits some, but a projective scheme is not a couple! $\endgroup$
    – Juan Fran
    Commented Aug 3, 2016 at 13:52

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