Dimension of a module (as vector space) I am reading the proof of the following  (part of a) lemma:

Let $A$ be a K-algebra. If $A$ is finite, then any nonzero module contains a simple submodule.

The proof of this is where I am stuck:

Proof:  Any submodule of minimal (finite) dimension as a K-vector space will be simple.

I have not yet familiarized myself with the use of modules that much. Could anyone explain to me 


*

*what is the dimension of a module as a K-vector space

*why is it so obviously simple?

 A: If $R$ is a ring then $R$ by definition acts on any $R$-module. If $S\subseteq R$ is a subring, then $S$ also acts on these $R$-modules. In this way, any $R$-module is also an $S$-module, as it has an action of $S$ on it.
If $A$ is a $K$-algebra, where $K$ is a field, then $K$ acts on any $A$-module $M$. In particular this means that $M$ is a vector space over $K$. Since it's a vector space over $K$, we can speak of its dimension.
Pick any $m\in M$ nonzero and consider the $A$-submodule $Am$. Since $Am$ is a quotient module of $A$ (there is an onto map $A\to Am$ given by $a\mapsto am$), it must have dimension less than or equal to that of $A$ as a vector space over $K$. In particular, it must be finite-dimensional if $A$ is. Therefore, if $A$ as a vector space over $K$ is finite-dimensional, then any $A$-module has an $A$-submodule that is also a finite-dimensional vector space over $K$.
Thus, it makes sense to speak of an $A$-submodule of $M$ of minimal nonzero dimension over $K$; one must exist. If the submodule $N$ is such, then can $N$ contain a proper nonzero $A$-submodule?
