isomorphism $\mathbb{C} [X,Y]/(X^2 + Y^2 - 1)$

Question

I have to show $\mathbb{C} [X,Y]/(X^2 + Y^2 - 1)$ is isomorphic to $\mathbb{C} [U,V]/(UV - 1)$.

First, I would apply the first isomorphism theorem by considering $ \mathbb{C} [X,Y] \rightarrow \mathbb{C} [U,V]/(UV - 1) $. But I'm not sure it will need somewhere.

Second, I tried to show $\mathbb{C} [X,Y]/(X^2 + Y^2 - 1)$ and $\mathbb{C} [X,Y]/(XY - 1)$ are both isomorphic to $\mathbb{C}[T]$. (then $\mathbb{C} [X,Y]/(X^2 + Y^2 - 1) \simeq \mathbb{C}[T] \simeq \mathbb{C} [X,Y]/(XY - 1) $) but I didn't find a morphism to do it.

If somebody have a hint, I would be grateful to him. Thank you

It is proved here that it is isomorphic to $\Bbb C[U,U^{-1}]$. So we are done. — Dietrich Burde, Oct 25 '21 at 14:21
Does this answer your question? $A=\frac{\mathbb{C}[X,Y]}{(X^2+Y^2-1)}$ is a PID. — Dietrich Burde, Oct 25 '21 at 14:23
From the above duplicate: "so the ring in question is $\Bbb C[u,v]/(uv-1)$". — Dietrich Burde, Oct 25 '21 at 14:23
Define $\phi : \Bbb C[X,Y] \to \Bbb C[T,\frac{1}{T}]$ by, $\phi(X+iY)=T, \phi(X-iY)=\frac{1}{T}$ and extend as it as $\Bbb C$-algebra homomorphism. Prove that the morphism induced by the Universal Property of Quotients define an isomorphism between $\frac{\Bbb C[X,Y]}{(X^2+Y^2 -1)} \cong \Bbb C[T,\frac{1}{T}]$ . Conclude by observing that $\Bbb C[T,\frac{1}{T}] \cong \frac{\Bbb C[U,V]}{(UV-1)}$ — Brozovic, Oct 25 '21 at 14:26
It's more clear now, I did not think about the changes of variables... Thank you so much both, I appreciate it. — Tohiea, Oct 25 '21 at 14:43

isomorphism $\mathbb{C} [X,Y]/(X^2 + Y^2 - 1)$

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