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Let $R=\mathbb{Q}[x,y,z]$, then every simple $R$-module $M$ is finite dimensional over $\mathbb{Q}$.

Had this been over $\mathbb{C}$ (complex field), it would have been rather easy. I have tried to use a theorem which says simple modules over $R$ is isomorphic to $R/I$, where I is a maximal regular ideal. But I don't understand regular ideals all that well. (For example, $Q[x,y]/(xy-1) \simeq Q(y)$, but $Q(y)$ is not finite dimensional over $Q$.)

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    Out of curiosity, how are you solving the problem with $\mathbb{Q}$ replaced by $\mathbb{C}$? Are you using Hilbert's Nullstellensatz?2011-01-30
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    if $R$ is a domain which is not a field, every nonzero ideal is regular (i.e., contains a non-zero divisor), so "maximal regular" = "maximal". The ideal $(xy-1)$ is not a maximal ideal of $\mathbb{Q}[x,y]$, and the quotient by that ideal is isomorphic neither to $\mathbb{Q}(t)$ nor $\mathbb{Q}[t]$ but to $\mathbb{Q}[t,t^{-1}]$. (You seem a bit confused...)2011-01-30
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    all maximal ideals of R= C[x,y,z] look like I=(x-a,y-b,z-c) by Nullstellensatz so then it looks like $\bar x$ is algebraic and etc. i was wrong about (xy-1) being maximal. once i made an homomorphism sending x to t and y to t^{-1} i assumed i had a field but that's wrong. thanks for what you have written.2011-01-30

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