I am reading Eisenbud's Commutative Algebra. The following is the proof I am trying to understand.

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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.

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

1 Answer 1


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.


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