Is the ring $B=\mathbb{C}[x,y]/(xy-1)$ isomorphic with $C=\mathbb{C}[x,y]/(x^2+y^2-1)$?

I think they shouldn't but all my tryings fail to prove the fact. Are they in fact isomorphic so I may try to prove that. Thanks


Yes, they are isomorphic.

If $u = x+yi$, and $v = x-yi$ then $$\mathbb{C}[x,y]/(x^2+y^2-1)\simeq\mathbb C[u,v]/(uv-1).$$


This can be viewed as a question in algebraic geometry but also as a question in projective geometry. Both the hyperbola and the circle are conic sections, and are projectively equivalent. In homogeneous coordinates this follows from the fact that any pair of nondegenerate indefinite quadratic forms are equivalent.

  • $\begingroup$ Great conceptual note! $\endgroup$ – rschwieb Dec 8 '14 at 16:21
  • $\begingroup$ So all conics belong to an isomorphic set? $\endgroup$ – Narasimham Dec 8 '14 at 19:46
  • $\begingroup$ @Narasimham, Yes, from the point of view of projective geometry. Note that the ring of polynomials as given in the question has a quotient field called field of functions of the curve, which is an invariant of the projective curve. $\endgroup$ – Mikhail Katz Dec 9 '14 at 13:48
  • $\begingroup$ @Narasimham, also I should have mentioned that this applies to nondegenerate conics. Degenerate conics are of course different, for example a pair of lines. $\endgroup$ – Mikhail Katz Dec 9 '14 at 15:07

For fun:

Let $\ \phi : \mathbb C \rightarrow \mathbb C\ $ be an arbitrary bijection (doesn't have to be continuous or measurable or anything, just a set-theoretical bijection). Let $\ A\,\ B\subseteq \mathbb C^\mathbb C\ $ be such that $\ B=\{f\circ\phi: f\in A\}.\ $ Then the rings $\ \mathbb C[A]\,\ \mathbb C[B]\ $ are isomorphic (as subrings of $\ \mathbb C^\mathbb C).\ $

Thus, rings $\ \mathbb C[ch\,\ i\!\cdot\! sh]\ $ and $\ \mathbb C[cos\,\ sin]\ $ are isomorphic via bijection $\ z\ \mapsto\ i\cdot z;\ $ and these two are respectively isomorphic to $\ \mathbb C[ch\,\ sh]\ $ and $\ \mathbb C[cos\,\ sin],\ $ and these last two are respectively isomorphic to the rings from the OP's post.


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