Must $\operatorname{Soc}(R)^2 =\operatorname{Soc}(R)^3$ (for a ring $R$)? I am struggling with the following problem. Show that the right socle of a ring $E := \operatorname{Soc}(R)$ has $E^2 = E^3$. I know that $E$ is a two sided ideal and so $E^3 \subset E^2$.
I am also asked for an example where $E \neq E^2$, and not sure where to start with this.
Thanks. :)
 A: I think this has everything to do with this lemma, which says minimal right ideals square to zero or else are idempotent.
If $A$ and $B$ are minimal right ideals and $AB=A$, then for some $a\in A$, $aB=A$, so that $a$ induces an isomorphism between $B$ and $A$.  We can conclude that $AB=\{0\}$ when $A$ and $B$ are nonisomorphic.
If $A^2=\{0\}$ and $B$ is isomorphic to $A$, they have the same annihilator, which contains both $A$ and $B$, so they annihilate each other also.
So when you square the socle, the only things that contribute are products of pairwise isomorphic minimal ideals, all generated by idempotents.
This suggests the following proposition:

$Soc(R_R)^2=$  the sum of all minimal right ideals generated by idempotents.

By the discussion above, the only products in $Soc(R_R)^2$ that contribute anything nonzero happen between pairs of minimal right ideals generated by idempotents. If $A$ and $B$ are any two minimal right ideals generated by idempotents, then $AB=A$ or $AB=\{0\}$, so the left hand side is contained in the right hand side. In the other direction, if $A$ is generated by an idempotent, then $A^2=A$, so the right hand side is contained in the left hand side.
Now you can see how an argument will go, which you can fill in, that $Soc(R_R)^3=Soc(R_R)^2$. Basically squaring the socle eliminated the "nilpotent part", and all that's left is the idempotent part.

Now it is also clear how one would produce an example with $Soc(R_R)\neq Soc(R_R)^2$: you just have to produce something with a nilpotent socle.
An easy example would be $\mathbb Z/4\mathbb Z$, which has socle $2\mathbb Z/4\mathbb Z$, which squares to $4\mathbb Z/4\mathbb Z$.
A: To show $E^2 \subset E^3$, it is enough to show that the product of two minimal right ideals is in $E^3$.
Given minimal ideals $A,B$, then if $AB \neq 0$, $AB = A$.
If $B^2 = 0$, then $A = 0$, which can't happen.
So $B^2 = B$ and $AB = ABB \subset E^3$.
Thus $E^2 = E^3$.
