# Question about proof of Corollary 2.18 from Eisenbud

I am reading Eisenbud's Commutative Algebra. The following is the proof I am trying to understand.  My question is the second sentence in the proof.

I understand that a power of $P_P$ annihilates $M_P$. However, to conclude $M_P$ is of finite length using Corollary 2.17, I need to know that $P_P$ is maximal in $R_P$. I really have no idea why $P_P$ is maximal in $R_P$. All I know is that it is a minimal prime containing $I_P$.

Do I miss something really obvious? Thanks!

EDIT: I found the answer, so I will put it below.

• I guess it's trivial that $PR_P$ is the maximal ideal of $R_P$. Jun 11, 2015 at 18:21
• @user26857 Commutative algebra is new to me. So, I guess I missed a lot of standard results and tricks.
– YYF
Jun 11, 2015 at 18:26

The thing I missed is the correspondence between primes in $R$ and primes in $R_P$.
Let $\varphi\colon R\to R_P$ be the natural map sending $r$ to $r/1$. The correspondence $I\mapsto\varphi^{-1}(I)$ is a bijection between primes of $R_P$ and the primes of $R$ contained in $P$. Also, the correspondence preserves inclusions. This is exactly the Proposition 2.2 in Eisenbud.
Now, suppose $PR_P\subset Q^*\subsetneq R_P$, where $Q^*$ is prime. By the correspondence, we have another prime $Q$ in $R_P$ such that $Q\subset P\subset R$. If we can show that $PR_P$ corresponds to $P$, we have $P\subset Q$, so that $P=Q$, proving that $P_P$ is maximal.
Clearly, $P\subset\varphi^{-1}(PR_P)$. If $r\in\varphi^{-1}(PR_P)$, then $r/1=r'/s'$ for some $r'\in P$ and $s'\notin P$. Therefore, $xs'r=xr'\in P$ for some $x\notin P$. Since $x,s'\notin P$, $r\in P$. Therefore, $\varphi^{-1}(PR_P)=P$.
This correspondence also shows that $PR_P$ is the only maximal ideal in $R_P$, so that $R_P$ is a local ring.