Maximal ideal in $\mathbb Z[\sqrt 2]$

Question

$\mathbb Z[\sqrt 2]/(\sqrt 2)$ is isomorphic to set of all integers, which is not a field. But $(\sqrt 2)$ is maximal ideal in $\mathbb Z[\sqrt 2]$ (since in a P.I.D every prime ideal is maximal). So the quotient has to be a field. But why does it contradict the fact that R/I is a field if and only if I is a maximal ideal in ring R.

Answer 1

$\Bbb Z[\sqrt2]/(\sqrt2)\cong \Bbb Z/2\Bbb Z$ is a field (note that $2\in (\sqrt2)$, which makes this case different from $\Bbb Z[x]/(x)\cong \Bbb Z$ since $(x)$ doesn't contain any non-zero integers).