Show that this submodule is not completely reducible

Question

Let $M = \begin{bmatrix} \mathbb{C} \\ \mathbb{C} \end{bmatrix}$.

Show that $M$ is not completely reducible as a left module over $R = \begin{bmatrix} \mathbb{C} & 0 \\ \mathbb{C} & \mathbb{C} \end{bmatrix}$.

I'm not sure how to tackle this? I can spot that $N = \begin{bmatrix} 0 \\ \mathbb{C} \end{bmatrix}$ is a submodule of $M$, but how do I show that it has no complement in $M$?

I thought about checking if $\begin{bmatrix} \mathbb{C} \\ 0 \end{bmatrix}$ is not a submodule of $M$, but this isn't sufficient as there are plenty of other submodules.

Answer 1

The subspaces of $M$ are of the form$$\left\{\begin{bmatrix}az\\bz\end{bmatrix}\,\middle|\,z\in\mathbb{C}\right\},$$for some complex numbers $a$ and $b$. It is easy to see that, unless $a=0$, none of them is a submodule.

Answer 2

Let $K$ be a nonzero submodule of $M$ and suppose $$ x=\begin{bmatrix} p \\ q \end{bmatrix} \in K $$ with $x\ne0$. Then $$ \begin{bmatrix} 0 & 0 \\ \beta & \gamma \end{bmatrix} \begin{bmatrix} p \\ q \end{bmatrix}= \begin{bmatrix} 0 \\ \beta p+\gamma q \end{bmatrix} $$ and we can obviously choose $\beta$ and $\gamma$ such that $\beta p+\gamma q$ is every complex number we fix. This shows $N\subseteq K$.