I have to show that if $M$ is an $R$-module where $R$ is a field then: $M$ is artinian if and only if $M$ is finite dimensional.
$\Rightarrow:$ Assume for sake of contradiction that $M$ is not finite dimensional. Then let $B=\{x_i\}_I$ be a basis for $M$ where $I$ must be an infinite index set. Let $\{y_n:n\in \mathbb{N}\}$ be a subset of the basis.
Then I consider $M_0=M$ and $M_1=\langle B\backslash y_1\rangle$ and in general $M_n=\langle B\backslash\{y_1,\ldots,y_n\}\rangle$. I just furnished an infinite descending chaing of submodules contradiction the assumption that $M$ was artinian.
$\Leftarrow:$ It is clear since if $M_1$ properly contains $M_2$ then the dimension of $M_1$ must be greater than that of $M_2$, and the result follows since we cannot have an infinite descending chain of submodules since the dimension we started with was finite.
My question goes: I am assuming that $M$ has a basis (i.e., $M$ is free), is it true in general that if we are over a field our modules must be free? or does the way the question was phrased implies that we are assuming that we have a basis (when the say finite dimensional)?
Thanks.