I'm trying to understand/solve an exercise problem in "Algebras of linear transformations" by Farenick. The following is the problem:
[exercises 4.6-5] Let $\mathfrak{A}$ be a finite-dimensional algebra that is not nilpotent. If $e\in \mathfrak{A}$ is a nonzero idempotent, then set $$ \mathfrak{L}_e=\{x\in\mathfrak{A}:xe=0\},\quad \mathfrak{R}_e=\{y\in \mathfrak{A}:ey=0\}. $$ a. Prove that $$(e\mathfrak{L}_e+\mathfrak{R}_e e+(\mathfrak{L}_e\cap \mathfrak{R}_e))\subseteq \text{Rad}\, \mathfrak{A}.$$
Considering a simple example $\mathfrak{A}=M_2(\mathbb{C}),\; e=\pmatrix{1&0\\0&0}$, I see that $$ \mathfrak{L}_e=\left\{\pmatrix{0&b\\0&d}:b,d\in \mathbb{C}\right\},\mathfrak{R}_e=\left\{\pmatrix{0&0\\c&d}:c,d\in \mathbb{C}\right\},\\ e\mathfrak{L}_e=\left\{\pmatrix{0&b\\0&0}:b\in \mathbb{C}\right\},\mathfrak{R}_e e=\left\{\pmatrix{0&0\\c&0}:c\in \mathbb{C}\right\}, \mathfrak{L}_e\cap \mathfrak{R}_e=\left\{\pmatrix{0&0\\0&d}:d\in \mathbb{C}\right\}\\ \therefore (e\mathfrak{L}_e+\mathfrak{R}_e e+(\mathfrak{L}_e\cap \mathfrak{R}_e))=\left\{\pmatrix{0&b\\c&d}:b,c,d\in \mathbb{C}\right\}\nsubseteq \text{Rad}\, \mathfrak{A}=\left\{\pmatrix{0&0\\0&0}\right\}. $$ (In the last line, I used the fact that the radical of $\mathfrak{A}$ is trivial since $\mathfrak{A}=M_2(\mathbb{C)}$ is a (central) simple algebra.) So, what am I missing or am I misunderstanding some concepts? If the problem should be corrected, what might be the intention or take-home message of this problem?
It looks to me that there is a typo, it should say $\supseteq$ instead of $\subseteq$. In that case, all you need to show is that $$ V_1\cap\text{Rad}\,\mathfrak A=\{0\}, $$ in the language of the proof of Theorem 4.6 (page 123).