A commutative ring $R$ is said absolutely flat (or alternatively von Neumann regular) if every $R$-module is flat. This property is equivalent to $r \in (r^2)$ for all elements $r\in R$.
The ring $R$ is said semi-artinian (also called Loewy ring) if any non-zero R-module admits a non-zero simple submodule. This is easily seen equivalent to the socle of every non-zero module being not trivial. We can provide another characterization by means of the Loewy series, defined recursively using socles. If $s(N)$ denotes the socle of the module $N$, then for an $R$-module $M$ we define $s_0M=0$, $s_1M=s(M)$ and for a general ordinal $\alpha$ we set $s_{\alpha+1}M$ to be the unique submodule of $M$ such that $s_{\alpha+1}M/s_{\alpha}M=s(M/s_{\alpha}M)$. If $\lambda$ is a limit ordinal, we set $s_{\lambda}M=\bigcup_{\alpha<\lambda}s_{\alpha}M$. We have that $R$ is semi-artinian if and only if for any module $M$ there exists an ordinal $\delta$ such that $s_{\delta}M=M$.
The example of absolutely flat ring usually presented is the infinite product of copies of a field, $\prod_{\mathbb{N}}k$. This is not semi-artinian: we have $s_1(R)=\sum_{\mathbb{N}}k$ but $s_2(R)$ is not defined, since $\prod_{\mathbb{N}}k/\sum_{\mathbb{N}}k$ does not admit simple submodules. See my previous question Quotient of infinite product of fields which is semi-artinian for details.
My question is: can you provide a non-trivial explicit example of absolutely flat ring which is semi-artinian?
Let $R$ be the subring of $T=\prod_{i=1}^\infty F_2$ which is generated by the socle of $T$ (left or right, it doesn't matter, they match because it is a reduced ring) and the identity of $T$.
$R$ is semiartinian and von Neumann regular (it is even boolean.)
This is a special case of a construction given in
on page 365.