An ideal $I$ in a ring $R$ is said to satisfy property $A$ if the following holds: For $a,b\in\Bbb{R}$, if $a\cdot b\in I$, then either $a\in I$ or $b\in I$. Let $I_1, I_2$ be two distinct ideal in a ring $R$ such that both satisfy property $A$ and none is contained in the other. Prove that $I_1\cap I_2$ doesn't satisfy property $A$.
As much as I have understood I need to show, that there exists $a\cdot b\in I_1\cap I_2$, such that $a\in I_1$ and $b\in I_2$, then neither $a$ nor $b$ will belong to their intersection. But why should that be true. Please help.
You already assume that none of $I_1$ and $I_2$ contains the other. Thus you can take $a\in I_1\setminus I_2$ and $b\in I_2\setminus I_1$.