# Coordinate ring of zero locus of unit circle

How to find the coordinate ring of $Z(x^{2}+y^{2}-1)$? I'm not sure how to proceed. Can you please help? Here $k$ is an algebraically closed field with char k $\neq 2$. I'm trying to find out if we can prove/disprove if the coordinate ring is isomorphic to $k[x,x^{-1}]$.

-
You have to be more specific what you are looking for. Otherwise, Andrew's answer (with Dylan's caveat) is the explicit description of the coordinate ring. –  Michael Joyce Feb 22 '12 at 22:07
@Michael Joyce: Sorry, just updated it. –  user6495 Feb 24 '12 at 1:54
If $\operatorname{char}k = 2$, then $x^2 + y^2 - 1 = (x + y)^2 - 1$. –  Dylan Moreland Feb 24 '12 at 2:21
@Dylan Moreland: sorry, typo, char $k \neq 2$. –  user6495 Feb 24 '12 at 2:27

For the second part of your question, it is true that $k[x,y]/(x^2+y^2-1)$ is isomorphic to $k[t,t^{-1}]$. For instance, let $i$ be one of the square roots of $-1$ in the algebraically closed field $k$, and send $t$ to $y-ix$, $t^{-1}$ to $y+ix$. The inverse isomorphism sends $x$ to $i(t-t^{-1})/2$ and $y$ to $(t+t^{-1})/2$.
By definition, the coordinate ring is $k[x,y]/(x^2+y^2-1)$.
Maybe we should be a little more careful. We have some algebraic set $Y$ and we want to find $k[x, y]/I(Y)$, where $I(Y)$ is the ideal of $Y$. Certainly $(x^2 + y^2 - 1) \subset I(Y)$, but equality here doesn't seem to be a matter of definition. [And maybe the OP should specify what $k$ is.] –  Dylan Moreland Feb 22 '12 at 16:09