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!
Best Answer
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.)