Projective, free and semisimple submodules of $R =\left(\begin{smallmatrix} K & K \\ 0 & K \end{smallmatrix}\right)$

Question

We consider $K$ a field and the ring $R = \begin{pmatrix} K & K \\ 0 & K \end{pmatrix}$. Recall that $e_{11},e_{22}$ are non-central idempotents. Prove that:

1. $e_{ii}R$ are projective and not free right $R$-modules.
2. $e_{22}R$ is simple but $e_{11}R$ is not semi-simple.

Could you give some hints to solve it?

My approach

1. $e_{11}R = \begin{pmatrix} K & K \\ 0 & 0 \end{pmatrix} \;$, $\; e_{22}R = \begin{pmatrix} 0 & 0 \\ 0 & K \end{pmatrix} \;$ , I assume that product by $R$ is normal product of matrices. Then, the operation is stable and indeed this sets form $R$-modules. I try to prove that is a direct summand of a free right $R$-module. I take as $R$-module $R$ itself. And it seems it is free since it has basis $e_{11},e_{12},e_{22}$. Then $e_{11}R \oplus e_{22}R = R$ so they are projective.But they are not free since $\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}\begin{pmatrix} 0 & 0 \\ 0 & K \end{pmatrix} = 0$ and $\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} K & K \\ 0 & 0 \end{pmatrix} = 0$ so there is a linear combination equal to zero to non-zero coefficients for any element that we put in the basis.

2. To show $e_{22}R$ is simple, I assume that $\begin{pmatrix} 0 & 0 \\ 0 & a \end{pmatrix}$ is in a eventual non-trivial submodule. But then, since $K$ is a field, there exists $a^{-1}$ and multiplication by $\begin{pmatrix} 0 & 0 \\ 0 & a^{-1} \end{pmatrix}$ gives $\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$ and if the module contains this matrix then it would contain the whole of $e_{22}R$.

But how can I proof that $e_{11}R$ is not semisimple?

Other comments:

$J(M) = MJ(R)$ if $M$ is projective.

$J(R) = \{x \in R:Mx = 0 \text{ for all simple right modules } M\}$

https://mathstrek.blog/2015/01/09/radical-of-module/

Answer 1

(For reference, these are both clearly projective because the are direct summands of a free module. They aren't free because they have dimension between $0$ and $3$, so you can't make them isomorphic to a $3n$ dimensional free module $R^n$. The one is obviously simple because it is $1$ dimensional over $K$.)

Each semisimple module is annihilated by the Jacobson radical, and here you can compute that the Jacobson radical is $J=\begin{bmatrix}0&K\\0&0\end{bmatrix}$

But if you check, $e_{11}R\begin{bmatrix}0&1\\0&0\end{bmatrix}\neq \{0\}$.