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I want to compute the Jacobson radical, the right socle, and the left socle of the ring of $3\times 3$ upper triangular matrices with entries in $\mathbb Z_4$ and such that the entries on the main diagonal are equal.

I know that the Jacobson radical of the ring of $3\times 3$ upper triangular matrices is comprised of those with zero main diagonal, and that, when the ring $R$ is Artinian, the right (left) socle of $R$ is the left (right) annihilator of the Jacobson radical of $R$. Thanks for any suggestion and/or help!

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Well, you have that $\begin{bmatrix}0&b&c\\ 0& 0& d\\ 0&0&0\end{bmatrix}$ is a nilpotent ideal, and that $\begin{bmatrix}2&0&0\\ 0& 2& 0\\ 0&0&2\end{bmatrix}$ is a central nilpotent, so the Jacobson radical must contain at least $\begin{bmatrix}2a&b&c\\ 0& 2a& d\\ 0&0&2a\end{bmatrix}$ for any choice of $a,b,c\in \mathbb Z_4$.

The quotient by this ideal is exactly $\mathbb Z_2$, so apparently we have already found the Jacobson radical.

I trust you can compute the annihilators after knowing this? Both socles have just four elements apiece.

I know that the Jacobson radical of the ring of $3\times 3$ upper triangular matrices is comprised of those with zero main diagonal.

That is not really correct. If you have $R$ and take the quotient in $T_n(R)$ by the strictly upper diagonal matrices, you get $R^n$, which is not Jacobson semisimple unless $R$ already was. So the radical is strictly bigger than the strictly upper triangular matrices, and contains some elements on the main diagonal (namely the elements in $J(R)$.

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  • $\begingroup$ I think the two socles are exactly the Jacobson radical of the ring. Am I right? $\endgroup$ – karparvar May 6 '17 at 3:11
  • $\begingroup$ @karparvar No, that would imply that $J^2=\{0\}$, but that does not hold... $\endgroup$ – rschwieb May 6 '17 at 13:20
  • $\begingroup$ @karparvar The way I did my computation was to consider what it meant to annihilate matrices like $\begin{bmatrix}0&1&0\\ 0&0&0\\ 0&0&0\end{bmatrix}$, $\begin{bmatrix}0&0&1\\ 0&0&0\\ 0&0&0\end{bmatrix}$ and $\begin{bmatrix}0&0&0\\ 0&0&1\\ 0&0&0\end{bmatrix}$ and $\begin{bmatrix}2&0&0\\ 0&2&0\\ 0&0&2\end{bmatrix}$, since these generate the radical. $\endgroup$ – rschwieb May 6 '17 at 19:43

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