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:

Proposition. 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

$$\{x_1,\mathrm{\dots },x_n\}\subset M.$$

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

$$\{x_1,\mathrm{\dots },x_n\}\subset M$$

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

$$\{y_1,\mathrm{\dots },y_t\}\subset A$$

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

$$m=\sum _{i=1}^n{a}_i\cdot x_i=\sum _{i=1}^n\left(\sum _{j=1}^t{\lambda }_{ij}\cdot y_j\right)\cdot x_i=$$

$$=\sum _{i=1}^n\sum _{j=1}^t{\lambda }_{ij}\cdot (y_j\cdot x_i)$$

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. $\mathrm{\square }$