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Let $X=\mathbb{P}^2$ and $U=X\backslash\mathbb{V}(x_0^2+x_1^2+x_2^2)$, could anyone show me how to find $\mathcal{O}_X(U)$?

I see examples in affine case, but have no idea how to calculate the ring in the projective case.

hxhxhx88
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1 Answers1

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The ring $\mathcal{O}_X(U)$ of the affine open subset $U\subset \mathbb P^2$ consists of all fractions of the form $$\frac {P(x_0,x_1,x_2)}{(x_0^2+x_1^2+x_2^2)^r}$$ with $r\geq 0$ an arbitrary integer and $P(x_0,x_1,x_2)$ a homogeneous polynomial of degree $2r$.

Georges Elencwajg
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  • Yeah I can guess this, but I am wondering is there any "close" form? As in the affine case, the ring of regular function can be expressed as a localization. – hxhxhx88 Nov 23 '13 at 10:39
  • Dear hxhxhx88, the ring $\mathcal O_X(U)$ is a subring of the fraction ring $k[x_0,x_1,x_2]f$, where $f=x_0^2+x_1^2+x_2^2$ . One writes $\mathcal O_X(U)=k[x_0,x_1,x_2]{(f)}\subset k[x_0,x_1,x_2]_f $ – Georges Elencwajg Nov 23 '13 at 11:53