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Let $k$ be an algebraically closed field. Consider the polynomial ring $A=k[x,y]$. Consider the ideal $I=\langle x^2+y^2-1\rangle$ (which is just the vanishing set of the circle) and $J=\langle y^2-x^2-1\rangle$ (the vanishing set of the hyperbola). How to prove that $$A/I\cong A/J.$$

Consider $K=\langle y-x^2\rangle$. How to prove that $$A/K\ncong A/I.$$

random_guy
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    If $k=\mathbb C$, then you can do a linear change of coordinates sending $x^2+y^2$ to $y^2-x^2$. I think maybe $u=x+iy$ and $v=x-iy$ works. – hwong557 Jan 22 '17 at 15:52
  • I asked http://math.stackexchange.com/questions/1984063/circle-and-hyperbola-over-mathbfp2-bbb-r some time ago :-) – Alphonse Jan 22 '17 at 15:54
  • I have another idea (but its validity depends on proving that $I=\langle x^2+y^2-1\rangle$ is prime in $\mathbb{C}[x,y]$). Correct me if I am wrong. If $I$ is prime, then $R/I$ is integral domain. Then the ideal $L=\langle I+x, I+y\rangle$ is not principal in $R/I$. – random_guy Jan 22 '17 at 20:52
  • http://math.stackexchange.com/questions/244460/ring-of-trigonometric-functions-with-real-coefficients says that $\mathbb{C}[x,y]/\langle x^2+y^2-1\rangle$ is a PID. Then we cannot use our previous idea (that $A/I$ is not a UFD and $A/K$ is) to prove that $A/I\ncong A/K$. – random_guy Jan 22 '17 at 21:45
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    In $A/I$, $\langle x,y\rangle = \langle 1\rangle $. –  Jan 22 '17 at 21:48
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    The first comment solves the first question (which is pretty simple), and the second question is settled here: http://math.stackexchange.com/questions/1463572/show-that-bbb-cx-y-i-is-not-isomorphic-to-bbb-cz?noredirect=1&lq=1 – user26857 Jan 22 '17 at 22:15

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Geometrically, over an algebraically closed field, the line, circle, hyperbola, and parabola are all isomorphic projective varieties.

Up to isomorphism, the only difference between these rings, then, is which points at infinity they are missing. Your circle and hyperbola are each missing two points ($(1 : \pm i : 0)$ and $(1 : \pm 1 : 0)$ respectively), and the parabola is missing one point: $(0:1:0)$.

With one point removed, the coordinate ring of each of these curves is isomorphic to $k[t]$. Removing a second point corresponds to inverting the appropriate linear function. By a suitable change of variable, we can insist that the result is isomorphic to $k[t, t^{-1}]$.

For the parabola, the correspondence is $(x,y) = (t, t^2)$. For the hyperbola, one such correspondence comes from $t = x+y$ (and $t^{-1} = y-x$). For the circle, $t = x+iy$ (and $t^{-1} = x-iy$) works.

So, the remainder of the problem is to show that $k[t]$ and $k[t, t^{-1}]$ are not isomorphic as rings. There are probably lots of ways to do this (including some geometric argument using the count of missing points I described above); however, a simple method is to compute the unit group.

  • In the similar posts I've seen so far it's asked to prove that $k[t,t^{-1}]\not\simeq k[z]$ (I wouldn't write $k[t]$) as rings, not as $k$-algebras. Moreover, all the proofs assume (by contradiction) the isomorphism is of unitary rings. It would be interesting to find out a proof without this assumption (which means to find other properties of the two rings not involving units). – user26857 Jan 22 '17 at 23:27
  • To prove my point see here: http://math.stackexchange.com/questions/1050291/show-that-quotient-rings-are-not-isomorphic?noredirect=1&lq=1 – user26857 Jan 22 '17 at 23:30
  • @user26857: A ring that is also a $k$-algebra (i.e. a ring with a specified homomorphism to it from $k$, or a ring extension of $k$ or other equivalent formulation) is certainly what's intended by the question, but I did intend to edit my first draft to switch over to just plain rings, and apparently missed an occurrence; I've corrected. –  Jan 22 '17 at 23:33
  • I can understand this, but the question has been posted for few times and all the OPs have looked for a non-isomorphism as rings. Most likely the question is an exercise in a textbook (I think could be Vakil notes), and I suppose this was the requirement there. – user26857 Jan 22 '17 at 23:38