How can one interpret a module as a vector space? (in a specific circumstance) I'm reading a remark in Atiyah, Macdonald - Introduction to Commutative Algebra and I can't really see what is going on.
After giving the definition and a few examples of $R$-Modules (for a ring $R$), the following statement appears:
If $A = k[x]$, where $k$ is a field, then an $A$-module is a $k$-vector space with a linear transformation.
I'm not sure how to "precisify" this statement.  Certainly $A$ is a ring in this case so the concept of an $A$-module is well defined.
I'm certainly open to even a vague hint; I think I would benefit from working out the details and I need to rapidly get familiar these ideas.
 A: Let $M$ be a left $A$-module.  Then $M$ is also a module over the subring of constant polynomials, which is isomorphic to $k$.  This makes $M$ a vector space over $k$.  Now let $T:M\to M$ be defined by $T(m)=x\cdot m$.  Then $T$ is a linear transformation on $M$.  The action of $A$ on $M$ is given by $p(x)\cdot m =p(T)(m)$.
Conversely, if $V$ is a $k$-vector space and $T$ is a linear transformation on $V$, then defining $p(x)\cdot v=p(T)(v)$ makes $V$ an $A$-module.
A: First of all, if $S$ is a subring of $R$ and $M$ is an $R$-module, then $M$ is automatically an $S$-module. [To make this concrete just think of something like $\mathbb{R}^2$. It's an $\mathbb{R}$-module (a real vector space), but it's also a $\mathbb{Z}$-module (since you can scale by real numbers, you can also scale by integers).]
Thus any $k[x]$-module is a $k$-module (thus a $k$-vector space). Next, just consider module axioms. Can you show that the action of "$x$" is a linear transformation?
Conversely, if you take a vector space with linear transformation (call it "$T$"). Then let  $x$ act like $T$: $x \cdot x \cdot v =T(T(v))$ etc. Poof! You get a $k[x]$ action on your vector space.
