Why this argument does not prove powers of prime are always primary?

Question

I know this lemma

Let $P, Q$ be ideals in a commutative ring $R$, and suppose:

• $P$ is maximal

• $\sqrt Q = P$

Then $Q$ is $P$-primary.

Then, as $\sqrt{P^n} = P$, every such power is primary. On the other hand, I know that if $P$ is prime, $P^n$ needs not be primary, since counterexamples are known.

What's going on then?

Answer 1

The trouble here is that $P$ is assumed to be maximal, not just prime. Hence the argument does not apply to arbitrary prime ideals since maximality is a strictly stronger property than primality.