Show that $\text {Hom}_{\mathbb R[x]}(M,N)=\{0\}$ where $M,N=\mathbb R^2$ are $\mathbb R[x]$-modules where $X$ acts as $A$ in $M$ and $B$ in $N$ Let $A=\begin{pmatrix}2&0\\0&3\end{pmatrix}$ and $B=\begin{pmatrix}1&0\\0&0\end{pmatrix}$
Show that $\text {Hom}_{\mathbb R[x]}(M,N)=\{0\}$ where $M,N=\mathbb R^2$ are $\mathbb R[X]$-modules and $X$ acts as $A$ in $M$ and $B$ in $N$.
I've completed the first part of this question which asked to calculate $(1-3X+X^2)\cdot\begin{pmatrix}1\\2\end{pmatrix}$ in both $M$ and $N$ for which I got $\begin{pmatrix}-1\\2\end{pmatrix}$ in both $M$ and $N$ (which is probably important in showing the part above that I can't do.)
However I'm struggling to use the fact that both answers were the same (if that's even relevant) to arrive at $\text {Hom}_{\mathbb R[x]}(M,N)=\{0\}$. I've tried fiddling around with the properties of homomorphisms of modules like $\varphi(rm)=r\varphi(m)$ for $r\in\mathbb R[X]$ and $m\in M$. I also don't really understand how we can define homomorphisms like these when $M$ and $N$ are essentially modules over different rings $\mathbb R[A]$ and $\mathbb R[B]$. I know the ring they're over is $\mathbb R[X]$ but I hope you get what I mean.
I couldn't find much about modules over polynomial rings involving a matrix online so I'm hoping someone here can help me. Thanks
EDIT: Corrected wrong entry in $B$
 A: Use the fact that $\varphi$ is a homomorphism and your first part to deduce
$$ \varphi((1,2))-2\varphi(e_1) = \varphi((-1,2)) = \varphi((1-3A+A^2) (1,2)) = (1-3B+B^2) \varphi((1,2)). $$
This implies
$$ 2\varphi(e_1) = (3B-B^2) \varphi((1,2)). $$
We write now $\varphi((1,2)) = (a,b)$ and obtain
$$ 2\varphi(e_1) = (2a,0). $$
Hence,
$$ \varphi(e_1) = (a,0). $$
Then we obtain
$$ \varphi((0,2)) = \varphi((1,2)-e_1) = (a,b) - (a,0) = (0,b). $$
Then we get
$$ (0,0) = B(0,b) = B \varphi(e_2) = \varphi(Ae_2) = \varphi(3e_2), $$
which implies $b=0$. On the other hand we have
$$ (a,0) = B(a,0) = B \varphi(e_1)= \varphi(Ae_1) = \varphi(2e_1) = (2a, 0). $$
Thus, $a=0$ and so we get $\varphi(e_1) = (0,0) = \varphi(e_2)$. Hence, we have $\varphi=0$.
A: One way to do this is as follows. First, the one dimensional modules for your ring are given by $\mathbb R_\lambda$ where $p(X)$ acts by $p(\lambda)$. These are pairwise non-isomorphic, and this is immediate. Because they are one dimensional, in fact this means there are no morphisms between them at all.
Your first module is $\mathbb R_2\oplus \mathbb R_3$ and the second one is $\mathbb R_1\oplus \mathbb R_0$. Moreover, $\hom(\mathbb R_2\oplus \mathbb R_3,\mathbb R_1\oplus \mathbb R_0)$ is just the sum of the four possible $\hom(\mathbb R_a,\mathbb R_b)$, which are all zero.
