Exercise A0.V.1.09

Proposition

Let $R$ be a Noetherian ring and $I$ be an ideal of $R.$ Then the set of minimal primes of $I$ is finite.

Proof

Let $\mathscr{F}$ be the family of ideals that have infinitely many minimal primes. For contradiction we assume that $\mathscr{F}$ is non-empty, and thus has a maximal element $M$ (by Noetherianity of $R$).

If $M$ were prime, then it would be it’s own minimal prime, since any other ideal containing $M$ must, well, contain $M.$ In that case $M$ is the only minimal prime, and hence cannot be in $\mathscr{F}.$

So $M$ is not prime. Thus, 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 have a finite set of minimal primes.

Any prime ideal $p$ containing $M$ must contain $ab.$ Hence, it must contain $a$ or $b.$ Hence it must contain either $aR + M$ or $bR + M.$ This makes it a minimal prime of one of these larger ideals. Since both have finitely many minimal primes, so does $M.$ This is a contradiction.

$\blacksquare$