The lattice of annihilator ideals of a ring The question is about an exercise from the book "Lattice-ordered rings and modules" from Stuart A. Steingberg. This is the exercise 7 from chapter 1, section 2.
Let $R$ a ring with no nonzero nilpotent ideals (equivalently, no nonzero nilpotent one-sided ideals). For each ideal $A$ of $R$, let $A'=\{x\in R:xA=0\}=\{x\in R:Ax=0\}$ be its annihilator, and let $\mathrm{Ann}(R)$ be the set of annihilator ideals of $R$. Show that $(\mathrm{Ann}(R),\subseteq,')$ is a complete Boolean algebra.
I do not understand why, in a ring without nonzero nilpotent ideals, every nilpotent one-sided ideal is zero, and the left and right annihilators of an ideal are the same.
 A: (1) If $R$ has a nonzero left ideal that is nilpotent, then it has one satisfying $L^2=0$.
(2)  If $L$ is a nonzero left ideal of $R$ satisfying $L^2=0$, then $N=\sum_{r\in R} Lr$ is nonzero a $2$-sided ideal that satisfies $N^2=0$. (This just uses $LrLs\subseteq LLs\subseteq 0s=0$.)
(3) If $A$ is an ideal, then $A'=\{x\in A\;|\;xA=0\}$ is also an ideal, and for any $x\in A'$ the set $Ax$ is a left ideal that squares to zero. If $R$ has no nonzero nilpotent ideals, then by (2) we have $Ax=0$. That is, $xA=0$ implies $Ax=0$.
A: A somewhat different description:

why, in a ring without nonzero nilpotent ideals, every nilpotent one-sided ideal is zero,

You can show that a ring has no nonzero nilpotent ideals iff $aRa=\{0\}$ implies $a=0$.
Suppose that $a$ belongs to a left ideal that squares to zero:  then $aRa\subseteq RaRa = \{0\}$, which implies $a=0$.
Now note that these two conditions are the same: a) $R$ has no nonzero nilpotent left ideals; b) $R$ has no nonzero left ideals squaring to zero.
The proof is the same as if you were proving "$R$ has no nonzero nilpotent elements iff it has no elements squaring to zero."

the left and right annihilators of an ideal are the same.

Just notice that $I\ell(I)$ is an ideal that squares to zero, so it is necessarily zero.  By the foregoing, $I\ell(I)=\{0\}$ so that $\ell(I)\subseteq r(I)$. By a symmetric argument, the other containment is proven and we have equality.
