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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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    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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This follows easily from a result known as "Zariski's Lemma": if $k$ is a field, k' an extension which is a finitely generated $k$-algebra, then k' is a finite extension of $k$. So for any maximal ideal $I \subset k[x_1, \dots, x_n]$, the ring $k[x_1, \dots, x_n]/I$ is a finite field extension of $k$. For a short proof of this due to McCare, see section 7.5 of http://people.fas.harvard.edu/~amathew/CRing.pdf (in particular the exercise at the end for a really quick one, not using the machinery of the Noether normalization theorem).

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    @Heidi: Dear Heidi, I'm not completely sure about what you mean by your notation, but you are correct that $R/I$ is a finitely generated as an algebra over $\mathbb{Q}$ (as a homomorphic image of the polynomial ring $R$). Furthermore, *when $I$ is maximal,* it is in addition a field; Zariski's lemma then implies that it is algebraic, so the set $x, x^2, \dots$ becomes linearly dependent modulo $I$.2011-01-30