# global sections of structure sheaf of a projective scheme X over a field k?

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!

• 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. ;) – Pece Mar 21 '14 at 6:34
• Thanks so much for your advice and edition! – user48537 Mar 21 '14 at 13:34
• See Corollary 3.21 in Qing Liu's book Algebraic geometry and arithmetic curves, Oxford University Press (2006). – Watson Nov 16 '18 at 21:16

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.)