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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!

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    Since $2\mathbb{F}_4=0$, your statement for the radical is technically correct: it is $J(R)=\begin{pmatrix} 0 & \mathbb{F}_4 \\ 0 & 0 \end{pmatrix}$. The one for $Soc(R_R)$ seems wrong, however: note that the right $R$-module $\begin{pmatrix} 0 & 0 \\ 0 & \mathbb{F}_4 \end{pmatrix}$ does not have non-trivial $R$-submodules. The same holds for $\begin{pmatrix} 0 & \mathbb{F}_4 \\ 0 & 0 \end{pmatrix}$. This is why $Soc(R_R) = \begin{pmatrix} 0 & \mathbb{F}_4 \\ 0 & \mathbb{F}_4 \end{pmatrix}$. – 2017-01-22

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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.

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    Nice easy example! – 2017-01-17
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    @karparvar Here, $\mathbb{F}_4$ is the field with four elements. It is a field of characteristic $2$, and as such, $2\mathbb{F}_4=0$. It is not to be confused with the cyclic group $\mathbb{Z}/4\mathbb{Z}$. For this reason, I still think my answer is correct. – 2017-01-22
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    The matrices $\begin{bmatrix}0&\mathbb F_4\\0&0\end{bmatrix}$ and $\begin{bmatrix}0&0\\0&\mathbb F_4\end{bmatrix}$ are two minimal right ideals of this ring, and every minimal right ideal must be contained in $\begin{bmatrix}0&\mathbb F_4\\0&\mathbb F_4\end{bmatrix}$: this can all be learned in [this post](http://math.stackexchange.com/a/137526/29335). I do not see any problems with $Soc(R_R)$ here. – 2017-01-22