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I would like to show that

  1. The circle $X : x^2+y^2=1$ and the hyperbola $Y : u^2-v^2=1$ are not isomorphic over $\Bbb R$, i.e. there is no isomorphism $f:X(\Bbb C) \to Y(\Bbb C)$ having real coefficients.
  2. Their projective closure $X' : x^2+y^2=z^2$ and $Y': u^2-v^2=w^2$ are isomorphic over $\Bbb R$.

For part 1., I know that $f(x,y)=(x,iy)$ is an isomorphism, but it has non-real coefficients. I wasn't sure how to prove that any isomorphism must have non-real coefficients.

For part 2., I tried to prove that the rings $\Bbb R[x,y,z]/(x^2+y^2-z^2)$ and $\Bbb R[u,v,w]/(u^2-v^2-w^2)$ are isomorphic. I don't know whether this is true, but at least I was told that the projective closures are isomorphic over $\Bbb R$, as this seems to be confirmed by this answer. But I'm not sure if there is an isomorphism with homogeneous real polynomials.

Thank you!

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  • $\begingroup$ @hwong557 : Thank you for comment. But if we assume that there exists an isomorphism $f:X(\Bbb C) \to Y(\Bbb C)$ having real coefficients, how do we know that the restriction $f\vert_{X(\Bbb R)} : X(\Bbb R) \to Y(\Bbb R)$ is surjective (and then, as it is injective, it would be an isomorphism, so an homeomorphism)? $\endgroup$ – Alphonse Oct 25 '16 at 17:58
  • $\begingroup$ Inverses are unique, so since the complex conjugate of the inverse to $f$ is also an inverse, we see that the inverse also has real coeffs. Thus $f$ does induce an isomorphism on real points. $\endgroup$ – tracing Oct 26 '16 at 3:50
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    $\begingroup$ $\mathbb R[x,y]/(x^2+y^2-1)$ is not a UFD while $\mathbb R[x,y]/(x^2-y^2-1)\simeq\mathbb R[u,v]/(uv-1)\simeq\mathbb R[u,u^{-1}]$ is a UFD. $\endgroup$ – user26857 Jan 22 '17 at 19:53
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For part 2 note that $\mathbb{R}[x,y,z]\cong\mathbb{R}[v,w,u]$, and this implies an isomorphism between your desired coordinate rings. Just write out the relations.

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    $\begingroup$ I would have added more explanation to your answer. For instance, the ideal $(u^2-v^2-w^2)$ is the same as $(v^2+w^2-u^2)$. That's why we can think to $x \mapsto v, y \mapsto w, z \mapsto u$. $\endgroup$ – Alphonse Oct 25 '16 at 18:05
  • $\begingroup$ I only accept your answer since my first question is answered in the comments : "$\mathbb R[x,y]/(x^2+y^2-1)$ is not a UFD while $\mathbb R[x,y]/(x^2-y^2-1)\simeq\mathbb R[u,v]/(uv-1)\simeq\mathbb R[u,u^{-1}]$ is a UFD." $\endgroup$ – Alphonse Oct 23 '18 at 12:05

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