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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?2011-05-10

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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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    @Florian: No,$I$was not talking about $k[x,y]/(x-gy^2)$, I was talking about $k[x,y]/(x^2-gy^2)$. The indeterminate $x$ now has the property that $x^2=gy^2$, hence my writing $x=y\sqrt{g}$. Also your second comment is correct, we can set $x$ to any polynomial in $y$ - I was too restrictive in my previous comment.2011-05-10
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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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    @Florian: It is no problem :) Also, I am sorry, I should have been clearer in my answer that "$k[y,y\sqrt{g}]$" would not be a correct expression if $g$ were a square, my answer could be read incorrectly.2011-05-10