proving that a ring is not semisimple.
A question asks me explicitly that the ring of matrices
$M_{a,b}=\begin{pmatrix} a & b \\ 0 & a \end{pmatrix}$ is not semisimple by showing that the submodule $M_{0, b}$ doesn't have a complement.
I proceed as follows.
Assume $M_{a,b}= M_{0, b} \oplus P$ where $P$ is an $M_{a,b}$-submodule. Choose any element from $M_{a,b}$, say $$\begin{pmatrix} a & b \\ 0 & a \end{pmatrix}=\begin{pmatrix} 0 & c \\ 0 & 0 \end{pmatrix}$$ where $\begin{pmatrix} 0 & b \\ 0 & 0 \end{pmatrix}\in M_{0,b}$. Then $$\begin{pmatrix} a & b-c\\ 0 & a \end{pmatrix} \in 0$$ and since $P$ is an ideal implies
$$\begin{pmatrix} -1 & 1\\0 & -1 \end{pmatrix} \begin{pmatrix} a & b-c \\ 0 & a \end{pmatrix} =\begin{pmatrix} -a & b-c+a\\0 & -a \end{pmatrix}\in P$$
Thus $$\begin{pmatrix} 0 & 2b-2c+a\\0 & 0 \end{pmatrix}=\begin{pmatrix} a & b-c \\ 0 & a \end{pmatrix}+\begin{pmatrix} -a & b-c+a \\ 0 & -a \end{pmatrix} \in P$$
Thus $M_{0,b} \cap P \neq 0$ which is a contradiction. Does this look correct?
Of course, as a vector space, $M_{0,b}$ does have a complement! Namely, $M_{a,0}$. However, $M_{a,0}$ is not an ideal. Why not? Because $M_{a,0}$ contains the identity matrix, but $M_{a,0}$ is not equal to the entire ring $M_{a,b}$.
$M_{a,0}$ is not the only complement of $M_{0,b}$ as a vector subspace of $M_{a,b}$, but we can try to extend this reasoning to see that $M_{0,b}$ has no complementary ideal whatsoever.
As you suggested, we should start by supposing for contradiction that $M_{a,b} = M_{0,b} \oplus P$ for some ideal $P \trianglelefteq M_{a,b}$. Now the identity matrix $1 \in M_{a,b}$ can be written as $$1 = X + Y$$ for some $X \in M_{0,b}$ and $Y \in P$. Write $X = \begin{pmatrix} 0 & x \\ 0 & 0 \end{pmatrix}$. Then $Y = \begin{pmatrix} 1 & -x \\ 0 & 1 \end{pmatrix}$. Now since $Y \in P$ and $P$ is an ideal, we have $$1 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & x \\ 0 & 1 \end{pmatrix}Y \in P.$$ Then, since $P$ is an ideal, $P = M_{a,b}$. Now $$M_{0,b} = M_{0,b} \cap M_{a,b} = M_{0,b} \cap P = 0,$$ contradiction!