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For some reason I am just stuck on this question, although my intuition insists it's easy - I'd appreciate anyone telling me any obvious fact I'm overlooking here.

Suppose we have an infinite-dimensional algebra $A$ over some field, and we have some finite-dimensional $A$-module $M$ - dimension always meaning as a vector space. How does the codimension of $\text{Ann}(M)$ - that is, $\dim (A/\text{Ann}(M))$ - compare to $\dim (M)$? They're both finite, but are they always the same?

My guess is that they're not, and in particular that $\dim (M) \geq \dim (A/\text{Ann}(M))$, but I can't seem to argue this without getting tangled up. If it is true, I'll expand my question - in what case does equality hold? Is it when the module is irreducible? Something else?

We can restate the question via quotienting out by the annihilator, in case this is easier: if we have a finite-dimensional faithful module over an algebra, what can its dimension be? What conditions force its dimension to be equal to that of the algebra?

In case it matters, my algebras are Noetherian, noncommutative, and Hopf.

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Is $A/Ann(M)$ always finite? take $A=k[x]$ and $M=(x)$ then $Ann(M)=0$ so $A/Ann(M)$ is isomorphic to $A$ which is infinite dimensional. –  user10 Apr 11 '12 at 8:26
I think it is when $M$ is finite-dimensional, which is the case I'm considering (and $(x)$ wouldn't be). After all, you can map $A$ into $\text{End}_k(M)$, which is finite-dimensional, and the kernel of that will be $\text{Ann}(M)$. –  Astrid Apr 11 '12 at 8:58
oh sorry, misread. –  user10 Apr 11 '12 at 9:03

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