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Let $R$ and $S$ be rings with unity and $_RM_S$ be a bimodule. Then $A=\begin{pmatrix} R & M \\ 0 & S \end{pmatrix}$ is a ring (with formal multiplication). I want to know what is $Soc(A_A)$, the socle of $A$ as a right $A$-module. I know, as stated in Lam's "A First Course in Noncommutative rings", that the right ideals of $A$ are of the form $J_1\oplus J_2$, where $J_1$ is a right ideal of $R$, and $J_2$ is a right $S$-submodule of $M\oplus S$ containing $J_1M$. To be aware of the right socle of $A$ one should know the minimal right ideals of $A$ to be summed up (from among them, $S$ is mentioned) .

Thanks for any cooperation!

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    leaving the whole reference could be useful for future readers (if anybody has it) – 2018-03-26

1 Answers 1

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This all flows rather straightforwardly from the characterization.

Suppose we have a minimal right ideal $J_1\oplus J_2$.

First suppose $J_1\neq \{0\}$. Then if $J_1M\neq \{0\}$, then $J_2\neq \{0\}$, and $\{0\}\oplus J_2$ is a nonzero right ideal strictly contained in our minimal right ideal, a contradiction. So if $J_1\neq\{0\}$, it follows $J_1M=\{0\}$. $J_2$ can then be replaced with $\{0\}$ to obtain the smallest ideal possible.

On the other hand if $J_1=\{0\}$ any $J_2$ will do.

So we can have as minimal right ideals:

  1. $\{0\}\oplus J_2$ for a simple right $S$ submodule $J_2\subseteq M\oplus S$
  2. $J_1\oplus \{0\}$ for some minimal right ideal $J_1\subseteq ann(_RM)$

This is apparently $soc(ann(_RM)_R)\oplus soc((M\oplus S)_S)$

Obviously if $_RM$ is faithful, you only have right ideals of the first type.