Are these two rings isomorphic?

Question

I have these two rings $k[x,y,1/y]/(x^2 +1 - y^2)$ and $k[u,v,1/v]/ (u^2 + v^2 -1)$, where $k$ is a field, and I was wondering if these two rings were isomorphic or not.

I would greatly appreciate any assistance with this! Thank you!

Answer 1

Let $A=k[x,y^{\pm1}]/(x^2+1-y^2)$ and $B=k[u,v^{\pm1}]/(u^2+v^2-1)$. Check that there is a map of $k$-algebras $f:A\to B$ such that $f(x)=u/v$ and $f(y)=1/v$. Construct, in a similar way, an inverse.

Answer 2

The answer depends on whether $k$ contains a square root of $-1$. This basically mirrors the fact that over $\mathbf C$ hyperbolas coincide with ellipses, in contrast to the case over $\mathbf R$. If you do have such an $i\in k,$ then consider the maps $y\mapsto v, x\mapsto iu$ and $v\mapsto y, u\mapsto -ix$.

Answer 3

Here is a reformulation of the proof sketched by Mariano.

Let $A$ be a commutative $k$-algebra. Then there is a natural bijection between solutions of $x^2+1=y^2$ in $A$ with $y \in A^\times$ and solutions of $u^2+v^2=1$ in $A$ with $v \in A^\times$. In fact, $u^2+v^2=1$ is equivalent to $(u/v)^2+1=(1/v)^2$, and conversely $x^2+1=y^2$ is equivalent to $(x/y)^2+(1/y)^2=1$. This shows that $(u,v) \mapsto (u/v,1/v)$ is a map with inverse map $(x,y) \mapsto (x/y,1/y)$.

This shows that the hom functors $\hom(k[X,Y^{\pm 1}]/(X^2+1=Y^2),-)$ and $\hom(k[U,V^{\pm 1}]/(U^2+V^2=1),-) : \mathsf{CAlg}(k) \to \mathsf{Set}$ are isomorphic. By the Yoneda Lemma, we get that $k[X,Y^{\pm 1}]/(X^2+1=Y^2) \cong k[U,V^{\pm 1}]/(U^2+V^2=1)$ (namely via mapping $X$ to $U/V$ and $Y$ to $1/V$.