Is $K:=\begin{pmatrix} 0 & \Bbb C \\ 0 & \Bbb C\end{pmatrix} \leq M_2(\Bbb C)$ simple in $M_2(\Bbb C)$?

Question

We take the ring $R:=M_2(\Bbb C)$ and we regard it as a $R$-module over itself.

If we take its left $R$-module $K:=\begin{pmatrix} 0 & \Bbb C \\ 0 & \Bbb C\end{pmatrix} \leq M_2(\Bbb C)$, is it simple?

I tried it by definition, but I failed. Any help please?

Answer 1

If $a$ is a nonzero element of $K$, can you show that $Ra = K$?

Then if $M \subseteq K$ is a nonzero submodule, it contains some nonzero $a$, and $K = Ra \subseteq M$.

Answer 2

$\newcommand{gae}[1]{\newcommand{#1}{\operatorname{#1}}}\gae{End}\gae{rk}$ Call $e_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and identify $R:=M_2(\Bbb C)=\End_{\Bbb C}(\Bbb C^2)$. Then $K=\{f\in R\,:\,e_1\in\ker f\}$ and the question is:

Given $f\in K\setminus\{0\}$ and another map $g\in K$, does there exist a linear map $h\in R$ such that $h\circ f=g$?

And this is clearly the case. In fact, the relevant linear algebra result is the following:

Let $T:V\to W$ and $S:V\to Z$ be linear maps of $k$-vector spaces. There is a linear map $H:W\to Z$ such that $HT=S$ if and only if $\ker S\supseteq \ker T$ and $\rk T\ge\rk S$.