TITLE: $I\cap J = P$ prime ideal, then $P=I$ or $P=J$
QUESTION [8 upvotes]: Question:
Prove that if $I,J$ are ideals and $I\cap J=P$ is a prime ideal, then either $P=I$ or $P=J$.
My proof:
Suppose $P\ne I$. Then $I\cap J=P\subsetneq I$ and $\exists i\in I\setminus P$. Now choose any $j\in J$. Then $ij\in I$ and $ij\in J$ as they are both ideals. So $ij\in I\cap J=P$. Since $P$ is prime, either $i\in P$ or $j\in P$. But we chose $i\notin P$, hence we must have $j\in P$, i.e. $J\subset P$. As $P\subset J$ (by definition $I\cap J=P$) we conclude $P=J$.
Our lecturer only provided a proof by contradiction (assume that $P\ne I$ and $P\ne J$ ...), but I was interested in a direct proof. Is it correct?

REPLY [2 votes]: I realize you are using the ordinary commutative defintion of prime ideals, but it might interest you to know that there's a noncommutative version of prime ideals (it reduces to the commutative definition if the ring is commutative.) It's as simple as you could imagine: $P$ is a prime ideal if for any two ideals $I,J$ of $R$, $IJ\subseteq P$ implies either $I$ or $J$ is a subset of $P$.
So, there is a very direct proof of your problem for general rings: Since $IJ\subseteq I\cap J$, either $I\subseteq I\cap J$ or $J\subseteq I\cap J$. Thus $I\subseteq J\cap I\subseteq I$ or $J\subseteq I\cap J \subseteq J$, showing that one of them must be equal to $I\cap J$.
I guess I would assign you this bonus work: show that the noncommutative definition of prime reduces to the commutative definition in commutative rings.