Let $k$ be a field. Consider the unit circle $X=V(x^2+y^2-1):=\{(x,y) \in k^2 \mid x^2+y^2-1=0\}$.
Question: Show that $V(x^2+y^2-1) \cong k$ as affine varieties if and only if char $(k)=2$.
My Attempt: I already know how to prove $V(x^2+y^2-1) \cong k$ if char $(k)=2$. Now, suppose char $(k) \neq 2$. Then $k$ has characteristic zero or characteristic $p>2$.
If char $(k)=0$, then $X=V(x^2+y^2-1)$ is infinite. So the coordinate ring of $X$ is just $k[x,y]/\langle x^2+y^2-1 \rangle$ which is not isomorphic to $k[t]$ (the coordinate ring of $k$ if $k$ is infinite)
If char $(k)=p>2$, we consider two cases. If $k$ is infinite, then the coordinate ring of $k$ is still $k[t]$, but is the coordinate ring of $X$ still $k[x,y]/\langle x^2+y^2-1 \rangle$? That is, is $X$ still an infinite set? And I don't know how to deal with the case that char $(k) \neq 2$ and $k$ is finite.
Can anyone give me some hints?
It might help to note that the variety can be parametrized as $x = (1-t^2)/(1+t^2), y = 2t/(1+t^2)$ (for $t^2 \ne -1$) with $t = y/(x+1)$.