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Let $R$ be a commutative ring with 1, and $P$ be a prime ideal in $R$. Let $D = R$ \ $P$. Show that $R_P := D^{-1}R$ has only one maximal ideal.

Problem 2b in this link http://math.arizona.edu/~cais/371Page/homework/371s3.pdf is very similar to my problem, but I don't understand the explanations. I tried to derive my solution. Typically, I want to show that under the canonical map $\varphi: R \rightarrow R_P$, $\quad$ $\varphi (P)$ is the maximal ideal in $R_P$, and it is unique.

I already knew that any ideal in $D^{-1}R$ is of the form $D^{-1}I$, where $I$ is an ideal of $R$. I assume this is true for prime and maximal ideals. Am I correct?

I tried to prove $R_P/\varphi(P)$ is a field, but I'm stuck. I also tried to assume that $D^{-1}M$ is a maximal ideal and show that any other ideals $K$ must be contained in $D^{-1}M$, but it didn't work. Could anyone please give me some hints?

To me, showing a prime ideal seems to be easier than showing a maximal ideal. I'm really weak in abstract algebra. This is my first algebra course, and I'm surprised to see how others use "facts" that are absolutely not obvious to me and which means I have to prove. Any suggestions?

Thank you very much.

dh16
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  • I think the answer is clearly visible. The primes of $R_{\mathfrak{p}}$ are in a bijective correspondence with those primes of $R$ which $\mathfrak{q}\cap (R\setminus \mathfrak{p})=\emptyset$ this means that the $\mathfrak{q}\subseteq \mathfrak{p}$ and $\mathfrak{p}$ would be put above all other primes of $R_{\mathfrak{p}}$ in the chain and hence is the only maximal ideal. – Arvin Rasoulzadeh Apr 08 '19 at 09:38

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