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Let $k$ be algebraically closed. Show that $\mathbb{A}_k^1$ is not isomorphic to $C:= \{(x,y) \in \mathbb{A}_k^2 :x^2-y^2=1\}$.

I know that this is equivalent to showing that $k[t]$ and $k[x,y]/(x^2-y^2-1)$ are not isomorphic as $k$-algebras, but I don't know how to proceed.

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  • $\begingroup$ Hint: $y\cdot y=y^2=x^2-1=(x-1)(x+1)$ $\endgroup$ Commented Nov 5, 2018 at 16:22

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$k[x]$ is a UFD, but not is $k[x,y]/(x^2-y^2-1)$ since you have $y.y=(x+1)(x-1)$.

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  • $\begingroup$ If char $k=2$, the ring in question is not reduced. If char $k\neq 2$, the ring is a UFD, but has non-trivial units and thus can not be isomorphic to $k[x]$. Notice that $x^2-y^2-1=(x-y)(x+y)-1$. So changing variables $u=x-y,v=x+y$ gives $k[u,v]/(uv-1)=k[u,u^{-1}]$. $\endgroup$
    – Mohan
    Commented Nov 5, 2018 at 19:08

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