# finite dimensional modules over algebra

Assume that $k$ is a field, $A$ is a $k$-algebra and $M$ is a $A$-module over $k$. In particular $M$ is a $A$-module and a vector space  over $k$, thus we may speak about $M$ being finitely generated   as $A$-module and finite dimensional as a vector space. These two concepts are related as follows:

Assume that $A$ and $M$ are both unital and additionaly $A$ is finite dimensional. Then $M$ is finite dimensional vector space if and only if $M$ is finitely generated $A$-module.

Proof. ,,$\Rightarrow$” Of course if $M$ is finite dimensional, then there exists basis

 $\{x_{1},\ldots,x_{n}\}\subset M.$

Thus every element of $M$ can be (uniquely) expressed in the form

 $\sum_{i=1}^{n}\lambda_{i}\cdot x_{i}$

which is equal to

 $\sum_{i=1}^{n}(\lambda_{i}\cdot 1)\cdot x_{i}$

since $M$ and $A$ are unital. This completes   this implication  , because $\lambda_{i}\cdot 1\in A$ for all $i$.

,,$\Leftarrow$” Assume that $M$ is finitely generated $A$-module. In particular there is a subset

 $\{x_{1},\ldots,x_{n}\}\subset M$

such that every element of $M$ is of the form

 $\sum_{i=1}^{n}a_{i}\cdot x_{i}$

with all $a_{i}\in A$. Let $m\in M$ be with the decomposition as above. Now $A$ is finite dimensional, so there is a subset

 $\{y_{1},\ldots,y_{t}\}\subset A$

which is a $k$-basis of $A$. In particular for each $i$ we have

 $a_{i}=\sum_{j=1}^{t}\lambda_{ij}\cdot y_{j}$

with $\lambda_{ij}\in k$. Thus we obtain

 $m=\sum_{i=1}^{n}a_{i}\cdot x_{i}=\sum_{i=1}^{n}\big{(}\sum_{j=1}^{t}\lambda_{% ij}\cdot y_{j}\big{)}\cdot x_{i}=$
 $=\sum_{i=1}^{n}\sum_{j=1}^{t}\lambda_{ij}\cdot(y_{j}\cdot x_{i})$

which shows, that all $y_{j}\cdot x_{i}\in M$ together make a set of generators   of $M$ over $k$ (note that $y_{j}$ and $x_{i}$ are independent on $m$). Since it is finite, then $M$ is finite dimensional and the proof is complete. $\square$

Title finite dimensional modules over algebra FiniteDimensionalModulesOverAlgebra 2013-03-22 19:16:35 2013-03-22 19:16:35 joking (16130) joking (16130) 4 joking (16130) Definition msc 16S99 msc 20C99 msc 13B99