Quotient of polynomial ring is a localization of a polynomial ring

**topspin1617** · October 22nd 2011, 06:33 PM

Let $p(x,y,z)=z^3-y(y^2-x^2)(x-1)\in k[x,y,z]$ , where $k$ is an algebraically closed field. The goal is to show that some ring obtained by inverting certain elements of $T=k[x,y,z]/(p)$ is isomorphic to a polynomial ring, again with certain elements inverted. This will show that the quotient field of $T$ 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 $T[\frac{1}{y}]$ , where the relationship $z^3=y(y^2-x^2)(x-1)$ can be rewritten as $(\frac{z}{y})^3=(1-(\frac{x}{y})^2)(x-1)$ . Then do a change of variables, say $w=z/x,v=y/x$ , and then solve for $x$ . That relation should then let me define a map to some localization of $k[v,w]$ .

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?

**NonCommAlg** · October 23rd 2011, 06:36 AM

Originally Posted by topspin1617

Let $p(x,y,z)=z^3-y(y^2-x^2)(x-1)\in k[x,y,z]$ , where $k$ is an algebraically closed field. The goal is to show that some ring obtained by inverting certain elements of $T=k[x,y,z]/(p)$ is isomorphic to a polynomial ring, again with certain elements inverted. This will show that the quotient field of $T$ 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 $T[\frac{1}{y}]$ , where the relationship $z^3=y(y^2-x^2)(x-1)$ can be rewritten as $(\frac{z}{y})^3=(1-(\frac{x}{y})^2)(x-1)$ . Then do a change of variables, say $w=z/x,v=y/x$ , and then solve for $x$ . That relation should then let me define a map to some localization of $k[v,w]$ .

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 $S = k[x,y,z],$ where $x,y$ are algebraically independent and $z^3=y(y^2-x^2)(x-1)$ . clearly $T \cong S.$ let $u = \frac{y}{x}$ and $v = \frac{z}{x}$ . then $x = 1 + \frac{v^3}{u(u^2-1)}$ and thus the quotient field of $S$ is $Q(S) = k(u,v).$
so the only thing you need to prove is that $u$ and $v$ are algebraically independent.

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