$R/AB$ reduced implies $A\cap B= AB$ I would like to show that if $A$ and $B$ are ideals in some ring $R$ and if $R/AB$ has no nilpotent elements, then $A\cap B=AB$. 
Now,  $x\in AB\implies x=ab\implies x\in A$ and $x\in B$ as $A$ and $B$ are ideals. I am having trouble with the other direction. 
Suppose $x\in A\cap B$ and $x\notin AB$. I am guessing that $x^k+AB=AB$ for some $k\in \mathbb{Z}_{>0}$, but I am having trouble showing it.
Any help would be great. 

If $x \in A \cap B$, then $x^2 = xx \in AB$, so $(x+AB)^2 = AB \implies x \in AB$ by assumption.
