Exercise A0.V.1.08

Proposition

Let $R$ be a Noetherian ring and $I$ be an ideal of $R.$ Then $I$ contains a finite product of prime ideals.

Proof

Let $\mathscr{F}$ be the family of ideals that do not contain finite products of prime ideals. For contradiction we assume that $\mathscr{F}$ is non-empty, and thus has a maximal element $M$ (by Noetherianity of $R$). Note that $M$ is not prime, so there exist $a, b \not\in M$ such that $ab \in M.$

Suppose $aR + M = R.$ Then $bR = baR + bM \subseteq M$ and this contradicts $b \not\in M,$ so $aR + M \neq R.$ By the same argument, $bR + M \neq R.$

Since $M$ is maximal in $\mathscr{F}$, and since $aR + M$ and $bR + M$ both strictly contain $M,$ they most both contain a finite product of prime ideals. But all elements of $(aR + M)(bR + M)$ are contained in $M,$ making it also contain a finite product of prime ideals. This is a contradiction.

$\blacksquare$