An equivalent condition with $\{0\}$ being the only nilpotent ideal

Question

In a ring $R$ prove that $\{0\}$ is the only nilpotent ideal if and only if for every ideals $A$ and $B$ from $R$, $AB=\{0\}$ implies $A\cap B=\{0\} $.

Answer 1

Suppose $\langle 0\rangle$ is the only nilpotent ideal and that $AB=\langle 0\rangle$. Let $x \in A \cap B$. Then $x^2=0$ so $\langle x\rangle^2=\langle 0\rangle$. Thus $\langle x\rangle=\langle 0\rangle$.

For the reverse direction, suppose that $I$ is a nilpotent ideal of $R$. Then $I^n=\langle 0\rangle$ for some $n>0$. If $n=1$, then $I=\langle 0\rangle$. If $n>1$, then $II^{n-1}=\langle 0\rangle$. Thus $I^{n-1}=I \cap I^{n-1}=\langle 0\rangle$. Continuing like this we get $I^2=\langle 0\rangle$, so $I=I\cap I=\langle 0\rangle$.