Noetherian ring and prime ideal contained in an invertible maximal ideal. 
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: 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.  
A: We have $P=(PM^{-1})M$. We also have $PM^{-1}\subseteq MM^{-1}=R$, so $PM^{-1}$ is an ideal of $R$ and obviously $PM^{-1}\ne R$. Moreover, since $1\in M^{-1}$ we have $P\subseteq PM^{-1}$. Now use that $P$ is prime (different from $M$) and get (from $P=(PM^{-1})M$) that $P\supseteq PM^{-1}$, so $P=PM^{-1}$ or equivalently $P=PM$. 
Remark. As the OP noticed, now it's easy to prove that $P$ must be in fact $0$. (Actually this part is the subject of another topic.)
