# What is the ring of global sections of this variety?

Let $C$ be the affine cone over the projective conic $xy-z^2$. Give an affine plane the coordinates $x',z'$. Construct a variety $X=C\cup \mathbb{A}^2$ by gluing the portion of $C$ with $x\neq 0$ to the portion of $\mathbb{A}^2$ with $x'\neq 0$ using the identification $z'=z, x'=x^{-1}$.

I want to find $\mathcal{O}_X(X)$. I believe it can be seen as $k[y,z]$ (using the coordinates on $C$). Please let me know if my thinking about this is correct. Here it is:

$\mathcal{O}_X(X)\subset k[C]=k[x,y,z]/(xy-z^2)$. For a function in $k[x,y,z]/(xy-z^2)$ to be in $\mathcal{O}_X(X)$, it has to also be regular when it is translated into a function on $\mathbb{A}^2$ by changing coordinates on $C\cap \mathbb{A}^2$. A polynomial $f(x,y,z)$ becomes $f(x'^{-1},x'z'^2,z')$ with respect to the coordinates on $\mathbb{A}^2$, since $x=x'^{-1}$ and $z=z'$ and on $C\setminus\{x=0\}$ we have $xy=z^2$ so that $x'^{-1}y=z'^2$ and $y=x'z'^2$. Thus for $f$ to remain regular on $\mathbb{A}^2$ we need $\deg_x M\leq \deg_y M$ for every monomial $M$ of $f(x,y,z)$. Then every occurrence of $x$ in $f$ is accompanied by $y$; we can replace $xy$ with $z^2$ and we will have gotten rid of all $x$'s. Thus $f$ can be represented by a polynomial in $y,z$ alone. In the other direction, any polynomial in $y,z$ alone becomes $f(x'z'^2,z')$, so it remains regular on $\mathbb{A}^2$. This shows $\mathcal{O}_X(X)=k[y,z]$. Yes?

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