TITLE: Modules over ring of polynomials in several indeterminates.
QUESTION [5 upvotes]: I read this proposition and its proof in the Dummit's book (It's not exactly like this, but this is the idea of the proposition).

Let $F$ be a field, $V$ a vector space over $F[x]$ and $T\in \mathcal{L}(V)$. Then $V$ can be seen as a $F[x]-$module defining 
  $$\left(\sum_{i=0}^na_ix^i\right)\cdot v=\sum_{i=0}^na_i T^i(v)$$
  (Let's call this extension for now $V(F[x],T)$)
  Furthermore, every module $M$ over $F[x]$ is a vector space over $F$, the transformation $T(v)=x\cdot v$ is a linear mapping of $M$ and, as a module, $M=M(F[x],T)$

This characterizes all the modules over $F[x]$ as vector spaces over $F$ with a linear mapping on them.
Is there any analogous result for modules over $F[x_1,\cdots,x_r]$?

REPLY [5 votes]: Defining a structure of module over $F[x_1,\dots,x_n]$ on $V$ (assuming the action of $F$ is preserved) is the same as specifying an $F$-algebra homomorphism
$$
\varphi\colon F[x_1,\dots,x_n]\to \operatorname{End}_F(V)
$$
(the codomain is the ring of $F$-endomorphisms of $V$). This is equivalent to specifying endomorphisms $T_1,\dots, T_n$ that are pairwise commuting, that is
$$
T_i\circ T_j=T_j\circ T_i,\qquad 1\le i\le n,\quad 1\le j\le n
$$
If $\varphi$ is given, define $T_i=\varphi(x_i)$; if the commuting endomorphisms are given, use the universal property of polynomial rings, together with the fact that the $F$-subalgebra of $\operatorname{End}_F(V)$ generated by $T_1,\dots,T_n$ is commutative.