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Thread: Quotient of polynomial ring is a localization of a polynomial ring

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    Quotient of polynomial ring is a localization of a polynomial ring

    Let $\displaystyle p(x,y,z)=z^3-y(y^2-x^2)(x-1)\in k[x,y,z]$, where $\displaystyle k$ is an algebraically closed field. The goal is to show that some ring obtained by inverting certain elements of $\displaystyle T=k[x,y,z]/(p)$ is isomorphic to a polynomial ring, again with certain elements inverted. This will show that the quotient field of $\displaystyle T$ is isomorphic to the field of rational functions in some (should be 2) variables (which is the ACTUAL goal).

    My first thought was to consider the ring $\displaystyle T[\frac{1}{y}]$, where the relationship $\displaystyle z^3=y(y^2-x^2)(x-1)$ can be rewritten as $\displaystyle (\frac{z}{y})^3=(1-(\frac{x}{y})^2)(x-1)$. Then do a change of variables, say $\displaystyle w=z/x,v=y/x$, and then solve for $\displaystyle x$. That relation should then let me define a map to some localization of $\displaystyle k[v,w]$.

    I could be doing it right, but the equations seem to get way too complicated to check whether or not I have an isomorphism. Anyone have an idea?
    Last edited by topspin1617; Oct 22nd 2011 at 07:25 PM.
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    Re: Quotient of polynomial ring is a localization of a polynomial ring

    Quote Originally Posted by topspin1617 View Post
    Let $\displaystyle p(x,y,z)=z^3-y(y^2-x^2)(x-1)\in k[x,y,z]$, where $\displaystyle k$ is an algebraically closed field. The goal is to show that some ring obtained by inverting certain elements of $\displaystyle T=k[x,y,z]/(p)$ is isomorphic to a polynomial ring, again with certain elements inverted. This will show that the quotient field of $\displaystyle T$ is isomorphic to the field of rational functions in some (should be 2) variables (which is the ACTUAL goal).

    My first thought was to consider the ring $\displaystyle T[\frac{1}{y}]$, where the relationship $\displaystyle z^3=y(y^2-x^2)(x-1)$ can be rewritten as $\displaystyle (\frac{z}{y})^3=(1-(\frac{x}{y})^2)(x-1)$. Then do a change of variables, say $\displaystyle w=z/x,v=y/x$, and then solve for $\displaystyle x$. That relation should then let me define a map to some localization of $\displaystyle k[v,w]$.

    I could be doing it right, but the equations seem to get way too complicated to check whether or not I have an isomorphism. Anyone have an idea?
    let $\displaystyle S = k[x,y,z],$ where $\displaystyle x,y$ are algebraically independent and $\displaystyle z^3=y(y^2-x^2)(x-1)$. clearly $\displaystyle T \cong S.$ let $\displaystyle u = \frac{y}{x}$ and $\displaystyle v = \frac{z}{x}$. then $\displaystyle x = 1 + \frac{v^3}{u(u^2-1)}$ and thus the quotient field of $\displaystyle S$ is $\displaystyle Q(S) = k(u,v).$
    so the only thing you need to prove is that $\displaystyle u$ and $\displaystyle v$ are algebraically independent.
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