# $P$ is prime if and only if for every pair of ideals $I,J$ containing $P$ properly, we have $IJ\nsubseteq P$

I'm trying to show that if $P$ is an ideal of a ring $A$; $A$ not necessarly conmutative, then $P$ is prime if and only if for every pair of ideals $I,J$ containing $P$ properly, we have $IJ\nsubseteq P$. Recall that an ideal $P$ in a not necessarly commutative ring $A$ is prime if for any ideals $I,J$ of $A$ such that $IJ\subseteq P$, we must have that $I\subseteq P$ or $J\subseteq P$.

The "only if" part is clear, I don't know how to begin with the other direction, I'd like to have a hint; not a complete answer please.

Thanks

• Try to form the quotient ring, and see what this condition means. Recall that the quotient ring is then an integral domain if and only if the ideal is prime! – awllower Feb 23 '13 at 4:51
• ok, I just got it thanks – Camilo Arosemena-Serrato Feb 23 '13 at 4:52
• Is there any reason you want to delete your question? Please note that this site helps everyone online - even though you've got the answer now, if someone had the same question as you, we'd want them to be able to see the answer here as well. You're welcome to write up your own answer now that you understand the problem; this is explicitly encouraged on stackexchange sites. – Zev Chonoles Feb 23 '13 at 4:54

Hint: Let $I$ and $J$ be principal ideals.