I have been asked to prove the following, but I believe the statement is not correct.
We have a field $K$, and a subring $R$ of the 2x2 matrices over $K$ ($R<M_2(K)$), whose elements are $r=\begin{bmatrix} a & b \\\ 0 & c \end{bmatrix}\in R$ , with $a,b,c\in K$. We know (it was another exercise) that given a homomorphism of K-vector spaces $T:Y \rightarrow X$, the ring $R$ defines on $X \oplus Y$ a module, with the operation given by $\begin{bmatrix} a & b \\\ 0 & c \end{bmatrix}\begin{bmatrix} x \\\ y \end{bmatrix}=\begin{bmatrix} ax + bT(y) \\\ cy \end{bmatrix} $.
Now, he asked me to prove that every $R$-module is isomorphic to one of those. So he wants me to prove that $_RM\equiv _R(X \oplus Y)$, for a general group $M$. I don’t even know how to start. The action of $R$ on $M$ can be so crazy that I have begun to wonder whether it is true. I’m lacking the way the ring $R$ is giving $M$ the module structure. Any hint, link, or comment will be really helpful.
The statement is correct indeed. Let me show a possible way for proving it:
I will denote by $1$ the identity element of $R$ (i.e. the identity matrix) and I will denote by "$*$" the action of $R$ on $M$. Since I guess you are working with unital modules, I will assume $M$ to be unital, that is, $1*m=m$ $\forall m\in M$.
Let $e= \big(\begin{smallmatrix} 1 & 0\\ 0 & 0 \end{smallmatrix}\big)$. Then $e$ is an idempotent (i.e. $e^2=e$). Similarly, $f=1-e=\big(\begin{smallmatrix} 0 & 0\\ 0 & 1 \end{smallmatrix}\big)$ is another idempotent. We also have $0=fe=ef$.
Let $m\in M$. Then we have $m=1*m=(e+f)*m=e*m+f*m$. Hence, it follows that $M=e*M + f*M$. Furthermore, if $e*n=f*n'$ then $e*n=e^2 *n=e*(e*n)=e*(f*n)=ef*n'=0*n'=0$, and so $e*M\cap f*M=0$.
Now, after identifying every $a\in K$ with the matrix $1 a=\big(\begin{smallmatrix} a & 0\\ 0 & a \end{smallmatrix}\big)$ we have $ea=ae$, and $fa=af$. Then $e*M$ is a $K$-vector space via the action $a\cdot (e*m)=e*(a*m)$. Similarly, $f*M$ is a $K$-vector space. So it follows that $M=e*M\oplus f*M$ as $K$-vector spaces.
Now, let $g=\big(\begin{smallmatrix} 0 & 1\\ 0 & 0 \end{smallmatrix}\big)$. We have $gf=g=eg$. Define now the $K$-linear map $\phi: f*M\rightarrow e*M$ by $\phi(f*m)=g*m$. It is well-defined since $g=eg$ (and so the image of $\phi$ is in $e*M$), and since $f*m=0\implies g*m=g*(f*m)=g*0=0$.
It is not hard now to see that $M$ is isomorphic to the $R$-module $X=(e*M\oplus f*M, \phi)$.
I hope this helps.