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I am trying to solve the following question.

Let $A$ be the algebra over $\mathbb{C}$ consisting of matrices of the form

\begin{pmatrix} * & * & 0 & 0 \\ * & * & 0 & 0 \\ * & * & * & 0 \\ * & * & * & * \\ \end{pmatrix}

Find the Jacobson radical of $A$.

To be honest I don't really have any idea how to do this so any help would very much be appreciated!

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Basically the same as this other question, although not exactly the same ring. –  rschwieb Dec 3 at 18:34

1 Answer 1

up vote 1 down vote accepted

We can use that post again here!

The construction there uses an $R-S$ bimodule to create the ring: $T=\begin{pmatrix} R & M \\ 0 & S \\ \end{pmatrix}$ with formal matrix multiplication as the product. We can use the same thing here, except we can use an $S-R$ bimodule in the lower left corner: $T=\begin{pmatrix} R & 0 \\ M & S \\ \end{pmatrix}$. In your case, $R=M=M_2(\mathbb{C})$ and $S$ is the subring of lower triangular matrices of $R$.

Applying (analogous) info from "that post", we have $rad(T)=\begin{pmatrix} rad(R) & 0 \\ M & rad(S) \\ \end{pmatrix}$.

Can you take it from here?

Extra tools to keep in mind when doing things like this:

  • if $I$ is a nilpotent (one or two sided) ideal, then $I \subseteq rad(R)$.
  • if $R/I$ is semisimple, then $rad(R)\subseteq I$.

Check and see what you get when you compute $T/rad(T)$ for your example! :)

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Thanks again for another brilliant answer! I am pretty sure I have understood everything you have said in this post! –  Alex Kite May 24 '12 at 20:07
    
@AlexKite How about you write up your solution, and whatever else you found useful, here in another answer? –  rschwieb May 24 '12 at 20:21

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