Equivalent Conditions For a Prime Ideal in a Commutative Ring

Problem 174

Let $R$ be a commutative ring and let $P$ be an ideal of $R$. Prove that the following statements are equivalent:

(a) The ideal $P$ is a prime ideal.

(b) For any two ideals $I$ and $J$, if $IJ \subset P$ then we have either $I \subset P$ or $J \subset P$.

Proof.

(a) $\implies$ (b)

Suppose that $P$ is a prime ideal. Let $I$ and $J$ be ideals such that $IJ \subset P$. Assume that
$I \not \subset P \text{ and } J \not \subset P.$ Then there exist
$a \in I \setminus P \text{ and } b\in J \setminus P.$

Then the element $ab$ is in both $I$ and $J$ since $I, J$ are ideals. Then we have
$ab \in IJ \subset P$ and this implies either $a \in P$ or $b\in P$ since $P$ is a prime ideal.

However, this contradicts the choice of the elements $a, b$.
Therefore, we must have
$I \subset P \text{ or } J \subset P.$

(b) $\implies$ (a)

Now we assume statement (b) is true.
Suppose that $ab \in P$, where $a, b \in R$.
Let $I=(a)$, $J=(b)$ be ideals generated by $a$ and $b$, respectively.

Then we have
$IJ=(ab)\subset P$ since $ab \in P$, and statement (b) implies that we have either $(a)=I\subset P$ or $(b)=J \subset P$.

Hence we have either $a \in P$ or $b\in P$.
Thus $P$ is a prime ideal.

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