prime ideal as the intersection of ideals

I am studying for an upcoming exam, and I came across the following statement that I am struggling to prove:

If $P$ is a prime ideal, then $P$ cannot be the intersection of two ideals that properly contain $P$.

So far, I have: If $P$ is properly contained in ideals $I$ and $J$, then we know that $P \subset I\cap J$. So I want to show that $I\cap J$ is not in $P$. Also, $P$ is prime $\implies$ if $ab\in P$ then at least $a \in P$ or $b \in P$. So, I have been trying to show that there is an element $x \in I \cap J$ such that $x$ is not in $P$, but I keep getting stuck at this step. Any help would be greatly appreciated!

If $P = I \cap J$, then $IJ \subseteq I \cap J \subseteq P$, hence $I \subseteq P$ or $J \subseteq P$.

• +1 for a solution that works even if the ring isn't commutative! Above and beyond what the OP probably aimed for, but as economical as possible. – rschwieb Mar 7 '14 at 12:38

Hint $\$ Reducing mod $P$ we may assume the ring $R$ is a domain, and we need to show that $I,J\ne 0\,\Rightarrow\,I\cap J\ne 0.$ But $I\cap J \supseteq IJ \ne 0,\,$ since $R$ is a domain.

Given that $P\subset I$ and $P\subset J$ so $P\subset I\cap J$

For $P\neq I\cap J$ we need to see that $I\cap J\not\subset P$

$P\subsetneq I$ so i have $a\in I$ and $a\notin P$

$P\subsetneq J$ so i have $b\in J$ and $b\notin P$

$a\in I$ and $b\in R$ so we would say $ab\in ???$

$b\in J$ and $a\in R$ so we would say $ab\in ???$

But then ????