Let $R$ be a Noetherian ring which is an integral domain, and $M$ be an invertible maximal ideal. Suppose that $P$ is a prime ideal and $P<M$. How to show then that $P=PM$?
I was trying to say something about $PM^{-1}$ ($M^{-1}$ is an inverse of $M$), but still stuck.
Actually, my goal is to show that $P=0$, but this is an easy corollary to Nakayama's Lemma if indeed $P=PM$.
A prime invertible ideal $P$ in a noetherian domain $R$ has height one: this is proved in our friend's Pete Clark's great commutative algebra notes as Proposition 19.22.
An immediate corollary is that a prime satisfying $P\subsetneq M$ must be zero and the step $P=PM$ is not needed.
If you know, and thus love!, algebraic geometry this is a consequence of the correspondence between line bundles on a scheme and divisors: the point $[\mathfrak m]$ corresponds to a divisor only if it has codimension one i.e. if $\mathfrak m$ has height one.