In a commutative ring with 1, every proper ideal is contained in a maximal ideal.

and we prove it using Zorn's lemma, that is, $I$ is an ideal, $P=\{I\subset A\mid A\text{is an ideal}\}$, then by set inclusion, every totally ordered subset has a bound, then $P$ has a maximal element $M$.

My question is why $M$ must contain $I$?

and we prove it using Zorn's lemma, that is, $I$ is an ideal, $P=\{I\subset A\mid A\text{is an ideal}\}$, then by set inclusion, every totally ordered subset has a bound, then $P$ has a maximal element $M$.

My question is why $M$ must contain $I$?