Define the square matrix $A$ over $\Bbb C[x]$ by $$ A:= \begin{pmatrix} x & -2x \\ -x^{2} & x^{3}-x \end{pmatrix}. $$
Prove that the submodule $AV=\{ Av \mid v \in V \}$ of the $\Bbb C[x]$-module $V= \left\{ \begin{pmatrix} f_{1} \\ f_{2} \end{pmatrix} \ \middle|\ f_{1},f_{2} \in \Bbb C[x] \right\}$ is a free module over $\Bbb C[x]$ and compute a basis for it.
I have worked on it as follows: $$A\begin{pmatrix} 1 \\ 0 \end{pmatrix}=\begin{pmatrix} x \\ -x^2 \end{pmatrix}=:v_1, \quad A\begin{pmatrix} 0 \\ 1 \end{pmatrix}=\begin{pmatrix} -2x \\ x^{3}-x \end{pmatrix}:=v_2, \quad A\begin{pmatrix} f_1 \\ f_2 \end{pmatrix}=f_1v_1+f_2v_2,$$ thus $v_1,v_2 \in AV$ and $v_1,v_2$ span $AV$. Therefore, $AV$ is a free module over $\Bbb C[x]$ once we prove $v_1,v_2$ are linearly independent over $\Bbb C[x]$. Now assume $f_1v_1+f_2v_2=0$ for some $f_1,f_2 \in\Bbb C[x]$. This gives a system of equations: $x(f_1-2f_2)=0$ and $-x^2f_1+(x^3-x)f_2=0$ for every $x$. Solving this system gives: $f_1=f_2=0$.
Are we done? I am not sure whether I understand the action of $\Bbb C[x]$ on $V$ correctly.
Any help would be much appreciated.
Let $R$ be a commutative ring and $A\in M_n(R)$. Consider the submodule $M$ of $R^n$ given by $\{Av:v\in R^n\}$. This is nothing but the submodule of $R^n$ generated by the columns of $A$.
Obviously, if the columns of $A$ are linearly independent over $R$ then $M$ is free.
Moreover, whenever $R$ is a PID (as it is $\mathbb C[x]$) we have that $M$ is free, and one can find a basis by using the Smith Normal Form of $A$.