$k$ is a field, char$(k)\neq 2$.

Can $k[x,y]/(y-x^2)$ be isomorphic to $k[x,y]/(x^2+y^2-1)$ by some ring homomorphism?

Edit: Since jspecter gives a nice and quick proof, I will ask another question

Can $k[x,y]/(y^2-x^2(x+1))$ be isomorphic to $k[x,y]/(x^2+y^2-1)$?

MY motivation is that two picture defined $y-x^2$ and $x^2+y^2-1$ are not "homeomorphic" as we view in $\mathbb{R}^2$, maybe it will induce that the coordinate rings are not isomorphic?


They are not isomorphic as $k$-algebras, at least. (From a geometric perspective, this is the more natural question....)

The first ring -- say $R$ -- is isomorphic to $k[t]$, so is the ring of regular functions on the affine line $\mathbb{A}^1$ over $k$. In other words, it corresponds to the projective line minus a single point at infinity.

The second ring -- say $S$ -- is the ring of polynomial functions on the unit circle. Geometrically, it is the projective line minus two points at infinity. If $\sqrt{-1} \in k$, these points are both rational and $S \cong k[t,t^{-1}]$. Thus the unit group of $S$ is larger than just $k^{\times}$ and so they cannot be isomorphic as $k$-algebras.

In case $\sqrt{-1}$ does not lie in $k$, if they were isomorphic as $k$-algebras then by tensoring with $k(\sqrt{-1})$ they would also be isomorphic as $k(\sqrt{-1})$-algebras and one can reduce to the previous case. But actually in this case one can do better: the Picard group of $S$ has order $2$, whereas the Picard group of $R$ is trivial, so $R$ and $S$ are not isomorphic even as rings. (See e.g. page 6 of this paper for a treatment of the Picard group of $S$.)

This answer leaves open the possibility that $R \cong S$ as rings when $\sqrt{-1} \in k$. Probably someone else will address this...

Added: The third ring $T = k[x,y]/(y^2-x^2(x+1))$ is the coordinate ring of an affine curve which is singular at the origin, so is not even a Dedekind domain, unlike $R$ and $S$.

  • $\begingroup$ In posts (but not comments, for some reason), – and — work roughly like -- and --- do in $\TeX$. $\endgroup$ – cardinal Jun 12 '11 at 13:45
  • $\begingroup$ Yup. $T$ also has the rational field as its field of quotients $k(t), t=y/x$. However, $T$ is not integrally closed, as $t$ is integral over $T$: $t^2=x+1$, but $t\notin T$. This is, of course, just a manifestation of the singularity of the alpha-curve in the origin. $\endgroup$ – Jyrki Lahtonen Jun 12 '11 at 14:31

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