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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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    Thanks for the response, but my question was about $k[x,y^2]/(x^2-gy^2)$.2011-05-10
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    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?2011-05-10
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    Sorry, $g$ is a polynomial in $y^2$.2011-05-10
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    @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 $h(z)=g(y^2)$.2011-05-10
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    @Zev: I was trying to relate it to a subring of $k[x,y]$, basically2011-05-10
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    @Florian: But we are only talking about rings up to isomorphism here - that is how we related $$k[x,y]/(x-y^2)\cong k[y^2,y]=k[y]$$ at the beginning of your question. Also, note that most rings of the form $k[x,y]/I$ for some ideal $I$ cannot be found as subrings of $k[x,y]$. In fact I would guess it is only possible when $I=(0),(x),(y),$ or $(x,y)$.2011-05-10
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    @Zev: Yes, I understand that we are dealing with rings up to isomorphism. In my previous version, the ring I had wrote was $k[x,y]/(x^2-gy^2)$, but from your comment I now gather, you were talking about the ring $k[x,y]/(x-gy^2)$. In this case, I don't see why there should be a $\sqrt{g}$ in your answer. Since, $k[x,y]/(x-gy^2)\cong k[y]$. I also understand that most quotients of polynomial rings are not isomorphic to subrings of polynomial rings. I am asking about a very specific situation and I am looking at a quotient of $k[x,y^2]$ and looking to relate it to a subring of $k[x,y]$.2011-05-10
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    @Zev: Also, regarding your last line. If I take $I=(x-gy)$ where $g$ is a polynomial in $y$, then $k[x,y]/I\cong k[y] \subseteq k[x,y]$. That's the kind of thing I mean, when I say "I want to relate it to a subring of a polynomial ring".2011-05-10
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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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    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?2011-05-10
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    @ 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 $x$.2011-05-10
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    @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 $x$.2011-05-10
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    # 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.2011-05-10
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    @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.2011-05-10
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    @Florian: You included the assumption that $g$ was not a unit in the previous version of the question. Because $g$ was assumed not to be a unit, then I'm pretty sure $x^2-gy^2$ is irreducible in $k[x,y]$ unless $g$ is a square in $k[y]$. If $x^2-gy^2$ is irreducible, then $(x^2-gy^2)$ is prime, hence $k[x,y]/(x^2-gy^2)$ would be a $k$-algebra that is an integral domain, generated over $k$ by the symbol $y$ and a symbol $x$ with the property that $x^2=gy^2$. Isn't it valid to write $k[y,y\sqrt{g}]$ for this ring?2011-05-10
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    @Zev: Sorry, I was not challenging your answer. I just did not see how you arrived it. I didn't think the assumption of $g$ being a non-unit was used. Now, it makes sense to me. Thanks for persisting with your comments.2011-05-10
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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