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The quotient $k[x,y]/(x-y^2)$ is isomorphic to $k[y]$ as a ring. Suppose, $g$ is a polynomial in $y^2$. Is there a "nice" ring that is isomorphic to $k[x,y^2]/(x^2-gy^2)$ assuming $g$ is not a unit?

Edit: Sorry I meant to write a different second ring.

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The quotient of a polynomial ring by a single polynomial is already a pretty nice ring. I'm not sure what kind of answer you're expecting. What do you actually want to know? – Qiaochu Yuan May 10 '11 at 5:11

I assume you mean $k[x,y]/(x^2-gy^2)$ . I think you won't be able to find a "nice ring" isomorphic to your ring because geometrically you have an affine curve of positive genus (except in some degenerate cases e.g. $g=y$) .These curves are not rational and so not isomorphic to an open subset of $\mathbb A^1_k$. In particular the ring of these curves can't be something as simple as, say, $k[x]$ or $k[x,x^{-1}]$.[We have to be careful with the notion of genus since the curve is singular and not complete, but this is technical and not really crucial ]

Conclusion I'm not sure how happy you'll be with this answer, since "niceness" is in the eye of the beholder. Let me sum up by saying that geometers have come up with an invariant for an algebraic curve, its geometric genus, which is an integer that gives a rough indication of the complexity of that curve, from an algebraic, geometric and arithmetic point of view.

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Thanks for the response, but my question was about $k[x,y^2]/(x^2-gy^2)$. – Florian J May 10 '11 at 15:49
But Florian, you write that $g$ is a polynomial in $y$.This is not possible if you really mean $k[x, y^2]$. For example if $g=y$, what does $k[x,y^2]/(x^2-y.y^2)$ mean? – Georges Elencwajg May 10 '11 at 16:44
Sorry, $g$ is a polynomial in $y^2$. – Florian J May 10 '11 at 17:09
@Florian: If there are no occurrences of $y$ by itself, then it seems to me we might as well take $$k[x,y^2]\cong k[x,z]$$ by sending $y^2$ to $z$ and then consider $$k[x,y^2]/(x^2-gy^2)\cong k[x,z]/(x^2-hz)$$ where $h(z)=g(y^2)$. – Zev Chonoles May 10 '11 at 17:19
@Zev: I was trying to relate it to a subring of $k[x,y]$, basically – Florian J May 10 '11 at 17:23

The only "nice" way I can think of to write that would be $k[y,y\sqrt{g}]$. I don't think it can be simplified any further unless $g$ is a square in $k[y]$.

Edit: My answer applies to a previous version of the question.

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I am not sure I understand this. If I relax the assumption that $g$ be a non-unit and let's say $g=1$, then, by what you wrote, $k[x,y]/(x^2-y^2)\cong k[y] \cong k[x,y]/(x-y)$. Am I missing something here? – Florian J May 10 '11 at 5:22
@ Florian : No, what's meant is $k[x,y]/(x-y^2) \simeq k[y] \simeq k[x,y]/(x-y)$. No square on the $x$. – Joel Cohen May 10 '11 at 5:27
@Joel: Thanks, but I had $(x^2-gy^2)$ as the ideal with respect to which the quotient was taken. Anyway, I meant to ask about a different ring as edited above. Also, I don't see why $\sqrt{g}$ will appear in the above answer if there is no square on the $x$. – Florian J May 10 '11 at 5:30
# Florian : Well I think $k[x,y]/(x^2-y^2)$ is pretty simple in itself, it's just a ring where $x^2 = y^2$ (you can also write it $(x-y)(x+y) = 0$). But this relation does not uniquely determine $x$ as "a function of" $y$ nor $y$ as a function of $x$ if that's your question. So I really think you need both variables to define the ring. – Joel Cohen May 10 '11 at 5:45
@Zev: I still don't see how your answer applies to the previous version of the question. My first comment is related to the previous version. – Florian J May 10 '11 at 17:24

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