Why isn't $\{0\}$ being prime ideal is not maximal in $\mathbb{Z}$?

Question

I am confused with the following arguments :-

$\mathbb{Z}$ is a Euclidean Domain with the evaluation map $\phi(r)=|r|$ and so it is a PID.

The ideal $\{0\}$ is a prime ideal in $\mathbb{Z}$ since $ab=0$ implies either $a=0$ or $b=0$

I know the theorem that in a PID an ideal is maximal iff it is prime.

So this should give $\{0\}$ ideal as maximal! , which is obviously false since $\{0\}\subset p\mathbb{Z}\subset \mathbb{Z}$, where $p$ is prime.

This may be a naive question but where am I wrong. Please help.

Answer 1

I know the theorem that in a PID an ideal is maximal iff it is prime.

Then you know a misquoted version of that theorem. The real theorem says “In a PID a nonzero ideal is prime iff maximal.”