Let $k$ be a field and $R=k[x]$, the polynomial ring. An $R$-module $M$ is equivalent to a pair $(V,A)$, where $V$ is a $k$-vector space, $A$ is a linear map $A\colon V\to V$.
(i) For $k[x]$ modules $M$, $N$ with corresponding pairs $(V,A)$, $(W,B)$, prove that a homomorphism $f\colon M \to N$ corresponds to a linear map $F\colon V \to W$ such that $B \circ F=F \circ A$.
(ii) Generalise the above to the case $R=k[t_1,\dots,t_n]$.
I am a little stuck getting started. I think I want to use the property that $f(rm)=rf(m)$ for all $r\in R$, $m\in M$ from the definition of module homomorphisms, but am unsure what to do with it. Any help would be appreciated!
The $R$-module structure on $V$ is defined by $$ (a_0+a_1x+\dots+a_mx^m)v=a_0v+a_1A(v)+\dots+a_mA^m(v). $$ It should be obvious that a homomorphism $f\colon M\to N$ defines a $k$-linear map $F\colon V\to W$, by simply doing $F(v)=f(v)$.
Now we must have $f(xv)=xf(v)$, which means $$ F(A(v))=B(F(v)), $$ for all $v\in V$. This is exactly $F\circ A=B\circ F$. So the condition is necessary. It is also sufficient, because by induction we can prove $$ F\circ A^m=B^m\circ F $$ and checking $$ F(a_0+a_1x+\dots+a_mx^m)v)=(a_0+a_1x+\dots+a_mx^m)F(v) $$ is easy.
What about $R=k[t_1,\dots,t_n]$? An $R$-module is a $k$-vector space where we need to define an action of $t_i$, which, as before, means to give a linear map $A_i$. However, $t_it_j=t_jt_i$, so we also need $A_i\circ A_j=A_j\circ A_i$.
The compatibility condition for homomorphisms is like before: $F\circ A_i=B_i\circ F$.