Let $R$ be a ring and $I\neq R$ an ideal. Suppose $I$ is not prime. Prove there exist ideals $J,K$ such that $I\subsetneq J$, $I\subsetneq K$ and $JK\subseteq I$.
I saw this question where the OP mentions the result I'm trying to prove but I couldn't find any actual proof on this site or any books online.
I have proven a similar result, which goes as follows:
Let $J,K$ be ideals and let $I$ be a prime ideal such that $JK\subseteq I$. Then, $J\subseteq I$ or $K\subseteq I$.
Proof:
Suppose $J\subsetneq I$. There exists $y\in J, y\notin I$. For every $x\in K$, $xy\in JK\subseteq I$. Since $I$ is prime, either $x\in I$ or $y\in I$. Hence, $x\in I$ and we have $K\subseteq I$. $\square$
It somehow feels like one is a contraposition of the other (idk if I'm using the correct term as english is not my main language), but I can't exactly see how to make the jump.
Is there a way to prove the first statement directly without needing to lean on the second one or use a contraposition argument?
Thank you
Let $x$ and $y$ be such that $xy \in I$ with neither $x$ or $y$ in I. Since I is not prime then such an $x$ and $y$ must exist. Let $J = (I, x)$ and $K = (I, y)$. Then $JK \subset I$.