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Could someone explain the proof of the following statement to me:

Let $A$ be a $K$-algebra. $A$ is finite $\implies$ a simple module over $A$ has finite dimension over $K$

Proof: If $M$ is simple, then $mA\subset M$ is a sub-module, hence we see our statment holds.

($A$ is finite means $\dim_KA<\infty$)

The proof is a bit to direct for me. I have a gut feeling that $mA\subset M$ ${\small (\text{implying } mA=M\text{ due to simpleness})}$, however I can't seem to make it solid. The same thing goes for the part on the dimension.

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gebruiker
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1 Answers1

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$mA$ is a submodule: $ma_1+ma_2=m(a_1+a_2), (ma_1)a_2=m(a_1a_2), 0\in mA$. So the point is that for $M$ simple $mA=M$ unless $m=0$.

Kevin Carlson
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  • But why is it contained in $M$? I can see this for $m\in M$, but my source never mentions that it is... – gebruiker Jun 05 '14 at 20:12
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    Where else could $m$ be from? Regardless, that's certainly what's intended. If $m$ were from, I don't know, $A$, then clearly $ma$ wouldn't be in $M$. Besides, if $m$ were from $A$ it'd be named $a$. – Kevin Carlson Jun 05 '14 at 20:40