How to prove that $IJ\subseteq P\implies I\subseteq P \vee J \subseteq P$

Question

I'm having some trouble with the following exercise:

Let $R$ be a non trivial ring with unity, and let $P\neq 0$ be an ideal of $R$. Prove that if $R/P$ is a prime ring, then for all right Ideals $I,J$ in $R$, we have that $IJ\subseteq P\implies I\subseteq P \text{ or } J \subseteq P$.

Here I use the following definitions: An ideal $P$ of $R$ is prime, if for any ideals $I,J$, we have that $IJ\subseteq P\implies I\subseteq P \vee J \subseteq P$.

$R$ is a prime ring if $(0)$ is a prime ideal.

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This is what I've tried so far:

Let $I,J$ be right ideals of $R$ such that $IJ\subseteq P$. Then $(IJ)/P \subseteq 0 + P =(0)$. This implies that $(I/P)(J/P)\subseteq 0+P$. The problem now is that $I/P$ and $J/P$ are right Ideals of $R/P$, meaning that we cannot use the fact that $(0)_{R/P}$ is prime.

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how can I continue and conclude this proof?

Answer 1

Consider ideals $RI$ and $RJ$. Since $RIRJ=RIJ\subset RP\subset P$, $(RI/P)(RJ/P)=0$. It follows that $RI/P=0$ or $RJ/P=0$ and hence $I\subset RI\subset P$ or $J\subset RJ\subset P$.