Let $R$ be a ring. Prove that for any two ideals $I$ and $J$ of $R$, the following conditions are equivalent:
$(1)$ $IJ = (0) \implies I\cap J = (0) $;
$(2)$ $(0)$ is the only nilpotent ideal of $R$.
I am able to prove $1 \implies 2$.
Let us assume that there exist another ideal $K \neq (0) $ and $K$ is nil-potent i.e. $K^n = (0)$ for some n. Then $K^{n-1}K= (0)$ and using the statement $(1)$ we have $K^{n-1} \cap K =(0)$ and since $K$ being an ideal $K^{n-1} \subset K$. Thus $K^{n-1} = (0)$. In the same manner after going $n-1$ steps we can show that $K = (0)$. Thus $(0)$ is only the nil-potent ideal of $R$.
But I am not able to proof $2\implies 1 $. Please help!!
Thank You.
Suppose $IJ=(0)$ and $(0)$ is the only nilpotent ideal.
Clearly $I \cap J \subseteq I$ and $I \cap J \subseteq J$, so $$(I\cap J)^2 \subseteq IJ = (0) \Rightarrow I \cap J = (0) \Rightarrow I \cap J \ \text{is nilpotent} \Rightarrow I \cap J = (0)$$
This shows that $2 \Rightarrow 1$.