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Let $A$ be a (non-commutative) $k$-algebra, where $k$ is an algebraically closed, characteristic zero field. Let $M$ be a finite-dimensional simple $A$-module. If $A/\operatorname{ann}(M)$ is commutative, then $M$ is $1$-dimensional. Is there a characterization (or at the very least, a name) for algebras in which this holds for all finite-dimensional simple modules? A simple example would be the quantum plane $\mathcal{O}_q(k^2)$ at $q \in k^\times$ a non-root of unity.

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Sorry, why do you need the hypothesis that $A$ is countably generated here? –  Qiaochu Yuan Aug 4 '12 at 15:57
    
Actually, that may not be necessary for the statement, but it does guarantee that $\ann(M)$ is a nonzero prime ideal. –  J. Gaddis Aug 4 '12 at 16:05
    
$M$ should be $1$-dimensional whenever $A/\text{ann}(M)$ is commutative. This follows from the fact that commuting operators have simultaneous eigenvectors. I don't see what $\text{ann}(M)$ being a nonzero prime ideal has to do with the property of $M$ being $1$-dimensional. –  Qiaochu Yuan Aug 4 '12 at 16:09
    
You're right. In the past I've proved this property for certain algebras by showing that $\ann(M)$ is a nonzero prime and then studying the prime spectrum of that ring. What I'm looking for is a more general approach to this problem. –  J. Gaddis Aug 4 '12 at 16:25

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