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simple question. I am looking at a presentation of the ring $k[x,yx^{-1}]$ in the following form $k[a,b]/I$ where $I$ is an ideal.

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Note that $x$ and $y/x$ are algebrically independent since the field of fraction of $k[x.y/x]$ is $k(x.y)$. If there is map from $k[a,b]$ to $k[x,y/x]$ sending $a$ to $x$ and $b$ to $y/x$ then the kernel $I$ is definitely a prime ideal as $k[x.y/x]$ is a domain. Now equating transcendence degree of $k[a.b]/I$ and $k[x.y/x]$ over $k$ we can conclude that the kernel is zero.

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    @Mariano Suarez Alvarez don't the alg. independence of $x$ and $\frac{y}{x}$ is an immediate consequence of the independence of $x,y$???2011-03-22