$R/\mathrm{Soc}(R_R)$ is a Boolean ring, but $R/\mathrm{Soc}( _RR)$ is not > Can anybody give me an example of a ring $R$ such that $R/\mathrm{Soc}(R_R)$ is a Boolean ring, but $R/\mathrm{Soc}( _RR)$ is not? The ring $R = \begin{pmatrix} \mathbb{F}_2 & \mathbb{F}_4 \\\ 0 & \mathbb{F}_4 \end{pmatrix}$ suggested, first, by Pierre-Guy Plamondon has the Jacobson radical $J(R) = \begin{pmatrix} 0 & \mathbb{F}_4 \\\ 0 & 2\mathbb{F}_4 \end{pmatrix}$, and hence $Soc(R_R) = \begin{pmatrix} 0 & 2\mathbb{F}_4 \\\ 0 & 2\mathbb{F}_4 \end{pmatrix}$ (which is exactly the left annihilator of $J(R)$). So, $R/Soc(R_R)$ would not be Boolean, as for the (nilpotent) element $x=\begin{pmatrix} 0 & 1 \\\ 0 & 0 \end{pmatrix}$, we have $x-x^2=x\notin Soc(R_R)$. Thanks for any leading answer!

Take $R = \begin{pmatrix} \mathbb{F}_2 & \mathbb{F}_4 \\\ 0 & \mathbb{F}_4 \end{pmatrix}$.

Then $Soc(R_R) = \begin{pmatrix} 0 & \mathbb{F}_4 \\\ 0 & \mathbb{F}_4 \end{pmatrix}$, so $R/Soc(R_R) \cong \mathbb{F}_2$ is boolean, but $Soc(_RR) = \begin{pmatrix} \mathbb{F}_2 & \mathbb{F}_4 \\\ 0 & 0 \end{pmatrix}$, so $R/Soc(_RR) \cong \mathbb{F}_4$ is not.