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Localization - Bijections between prime ideals of R and D^-1R

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
I am reading Dummit and Foote, Section 15.4: Localization and am currently working on Proposition 38, part 3 (contraction bijection) - see attachments.

I am hoping that someone can demonstrate a proof of the following propostion (without - as D&F do - referring to or relying on translating the result of Exercise 13, Section 7.4)

c maps prime ideals of \(\displaystyle D^{-1}R \) to prime ideals P of R where \(\displaystyle P \cap D = \emptyset \)

Note: c is a contraction of ideals Q of \(\displaystyle D^{-1}R \) to R defined as folows:

\(\displaystyle c: \ D^{-1}R \to R \)

where

\(\displaystyle c(Q) = \Pi^{-1}(Q) \) where Q is an ideal of \(\displaystyle D^{-1}R \)

Hoping someone can help!

Peter
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
By an earlier exercise, if $I \cap D \neq \emptyset$, we have ${}^eI = D^{-1}R$, which is not a prime ideal of $D^{-1}R$.

Since $J = {}^e({}^cJ)$ we see that if $J$ is a prime ideal of $D^{-1}R$, it must be the case that ${}^cJ$ does not intersect $D$ (if $D$ is an ideal, this is often expressed as: " ${}^cJ$ does not meet $D$").

D&F do not give a very good characterization of which elements of $R$ are actually in ${}^cJ$.

I claim that ${}^cJ = S = \{a \in R: a/1 \in J\}$.

Suppose $a \in S$. Then $a/1 \in J$

Hence $a \in \pi^{-1}(J) = {}^cJ$.

On the other hand suppose $a \in {}^cJ$. Then $a \in \pi^{-1}(r/d)$ for some $r/d \in J$. From:

$\pi(a)= a/1 = \pi(\pi^{-1}(r/d)) = r/d$, we see that $a/1 \in J$.

It remains to be seen that if $J$ is prime, ${}^cJ$ is prime.

So suppose we have $ab \in {}^cJ$, with $a \not\in {}^cJ$.

This means that $(ab)/1 = (a/1)(b/1) \in J$. Since $a \not\in {}^cJ$,

$a/1 \not\in J$, and since $J$ is prime, $b/1 \in J$, so $b \in {}^cJ$.

I leave it to you to show that $c$ is a bijection of the two sets.

(Hint: show any prime ideal of $D^{-1}R$ is an extension of a prime ideal $I$ in $R$ with $I \cap D = \emptyset$).