# Is the unit circle always an infinite set if the field is infinite?

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 $$t = y/(x+1)$$.