Let $A$ be the $\mathbb{C}$-subalgebra of $M_4(\mathbb C)$ consisting of matrices of the form \begin{pmatrix} * & * & * & *\\ * & * & * &*\\ 0 & 0 & *&0 \\ 0 & 0&* &* \end{pmatrix} Then how would I find the Jacobson radical $J(A)$ of $A$?
I don't really understand finding the Jacobson radical of $\mathbb{C}$-subalgebras.
Any help on this would be greatly appreciated, thank you!
Let $R=M=M_2(\Bbb C)$ and $S=L_2(\Bbb C)$ be the ring of $2\times2$ lower triangular matrices over $\Bbb C$. Call your ring $T$.
Then $M$ is an $(R,S)$ bimodule in the obvious way, where $R$ acts on the left and $S$ acts on the right by matrix multiplication.
We can reexpress your ring as the ring $\begin{bmatrix}R&M\\0&S\end{bmatrix}$. Let's start looking for nilpotent ideals.
Work to show that $I=\begin{bmatrix}0&M\\0&0\end{bmatrix}$ is a nilpotent ideal.
Note that $T/I\cong R\oplus S$. If $I$ were the Jacobson radical, this should have radical zero. But it doesn't! There are still more nonzero nilpotent ideals left.
Show that $J=\{\begin{bmatrix}0&0\\x&0\end{bmatrix}\mid x\in \Bbb C\}$ is a nilpotent ideal of $S$, and that $I'=\begin{bmatrix}0&M\\0&J\end{bmatrix}$ is a nilpotent ideal of $T$.
Now $T/I'\cong R\oplus (S/J)\cong R\oplus\Bbb C\oplus\Bbb C$, and this is clearly semisimple (since $R$ and $\Bbb C$ are).
Conclude that $\mathrm{rad}(T)=I'$.
This is the thought process I followed to deduce the solution.
Now in general, you won't be able to get the answer solely through nilpotent ideals, as I did here. This was only possible because the ring is Artinian, and so its radical had to be nilpotent. You can try to show that in general, $\mathrm{rad}(\begin{bmatrix}R&M\\0&S\end{bmatrix})=\begin{bmatrix}\mathrm{rad}(R)&M\\0&\mathrm{rad}(S)\end{bmatrix}$ if you encounter a harder problem of the same type.