Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This should be quite easy, but somehow I can't find the proof. Let $P\neq Q$ be two maximal ideals in the commutative ring $R$. Then $P_Q=R_Q$.

($P_Q$ is the localisation of the R-module $P$ at $Q$ and $R_Q$ is the localisation of R at Q)

share|improve this question
    
Are you sure this is the question? What if R=Z and P, Q are <2>, <3>? The localizations are not equal, and I don't think they're isomorphic. –  Gadi A May 22 '11 at 16:05
    
Are they really distinct? an element of $\mathbb{Z}_{(3)}$ is a fraction $\frac{x}{y}$ with y coprime to 3. This element is equal to $\frac{2x}{2y}$, which is an element of $(2)_{(3)}$. –  Michalis May 22 '11 at 16:40
2  
Michalis, your excellent response to Gadi's objection is easy to generalize to a proof of the general statement: just use that $P$ is not contained in $Q$. –  Georges Elencwajg May 22 '11 at 17:10
    
@elgeorges: you're right :D I was a bit confused when I posed the question. –  Michalis May 22 '11 at 17:15
add comment

1 Answer

up vote 5 down vote accepted

Since $P$ and $Q$ are distinct maximal ideals, $P$ is not contained in $Q$ and thus there exists $x \in P \cap (R \setminus Q)$. This element becomes a unit in the localization, so the localized ideal contains a unit and is thus the entire localized ring $R_Q$.

This is a special case of basic results on pushing forward and pulling back ideals under a localization map: see e.g. $\S 7.2$ of my commutative algebra notes for more details. (Or see any other commutative algebra text, of course.)

share|improve this answer
    
Oh I shouldve seen that myself! thanks, I will definitely have a look at your notes. –  Michalis May 22 '11 at 17:14
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.