Let $R=\begin{pmatrix} \mathbb{C} & \mathbb{C} & \mathbb{C}\\ 0 & \mathbb{C} & \mathbb{C}\\ 0 & \mathbb{C} & \mathbb{C} \end{pmatrix}$. I want to find the nilradical: $$N(R)=\sum\{I: I \text{ is a nilpotent ideal of } R\}.$$
The observation is that $N:=\begin{pmatrix} 0 & \mathbb{C} & \mathbb{C}\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{pmatrix}$ is a nilpotent set. Then define map $$\phi: R \rightarrow \begin{pmatrix} \mathbb{C} & 0 & 0\\ 0 & \mathbb{C} & \mathbb{C}\\ 0 & \mathbb{C} & \mathbb{C} \end{pmatrix} \cong\mathbb{C} \oplus M_2(\mathbb{C}), \begin{pmatrix} a & b & c \\ 0 & d & e\\ 0 & f & g\end{pmatrix} \mapsto \left(a, \begin{pmatrix} d & e\\ f & g\end{pmatrix} \right).$$ Indeed, $\phi$ is a surjective ring homomorphism and clearly ker$(\phi)=N$ and so by the First Isomorphism Theorem for Rings, $R/N\cong\text{Im}(\phi) =\mathbb{C} \oplus M_2(\mathbb{C}).$
I do not understand the next steps in the argument which are that this must be semisimple and so $N(R) \subseteq N$. Furthermore, $N$ is nilpotent so $N \subseteq N(R)$, hence $N=N(R)$.
I would appreciate any hints here. I am not sure if this follows from the Artin-Wedderburn Theorem or otherwise.