Primary ideals and localization of prime ideals

[Math Amateur]

I am reading Dummit and Foote Section 15.4 Localization.

Exercise 11 on page 727 reads as follows:

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Let \(\displaystyle R_P \) be the localization of R at the prime P. Prove that if Q is a P-primary idea of R then \(\displaystyle Q = ^c(^e Q) \) with respect to the extension and contraction of Q to \(\displaystyle R_P \).

Show the same result holds if Q is P'-primary for some P' contained in P.

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This exercise obviously uses concepts from Proposition 38, D&F Section 15.4 (see attached) and uses concepts from Section 15.2 - particularly those of primary ideal and P-primary ideal (see attachment).

I am somewhat intimidated by this exercise and have not made any real progress ... I would appreciate it if someone could give me a significant start on the problem ...

My basic understanding of the elements involved in the exercise follows.Since \(\displaystyle R_P \) be the localization of R at the prime P we have, in the notation of D&F Proposition 38, that P is a prime ideal of R, D = R - P and we have a mapping \(\displaystyle \pi : \ R \to R_P = D^{-1}R \) where \(\displaystyle \pi (r) = r/1 \).

The mapping \(\displaystyle \pi : \ R \to R_P = D^{-1}R \) constitutes the localization.
[? is this correct or is the localization actually the ring \(\displaystyle R_P = D^{-1}R \) ?]

Q is a P-primary ideal which implies that P is a prime ideal such that P = rad Q ( or \(\displaystyle \sqrt Q \) ) (see attachment page 682)

A primary ideal is defined as follows: (see attachment page 681)

Definition. A proper ideal Q in the commutative ring R is called primary if whenever \(\displaystyle ab \in Q \) and \(\displaystyle a \notin Q \) then \(\displaystyle b^n \in Q \) for some positive integer n. Equivalently if \(\displaystyle ab \in Q\) and \(\displaystyle a \notin Q \) then \(\displaystyle b \in \) rad Q.

Further to the above: rad \(\displaystyle Q = \{ a \in R \ | \ a^k \in Q \) for some \(\displaystyle k \ge 1 \} \).

Now the extension of Q to \(\displaystyle R_P = D^{-1}R \) is \(\displaystyle ^eQ = \pi (Q) R_P = \pi (Q)D{-1}R \)

and the contraction of this extension is \(\displaystyle ^c(^eQ) = \pi^{-1}(\pi(Q)R_P \).

I have also uploaded a sketch of my view of the structure of the elements of the exercise ... BUT ...

... as mentioned above, I have made no significant progress on the problem and would appreciate help in making a significant start ... ...

Peter

 

[jvicens]

,

Thank you for your question. I can understand why this exercise may seem intimidating at first, but with a little bit of guidance, I believe you can make progress on it.

First, let's clarify the notation used in the exercise. In the context of localization, the notation ^c(^eQ) refers to the contraction of the extension of Q to R_P. This means that we are taking the ideal ^eQ, which is an ideal in the localized ring R_P, and mapping it back to an ideal in the original ring R using the inverse map \pi^{-1}. In other words, ^c(^eQ) = \pi^{-1}(\pi(Q)R_P).

Now, let's take a closer look at the definition of a P-primary ideal. As you mentioned, a P-primary ideal is an ideal Q such that P = rad Q, or equivalently, if ab \in Q and a \notin Q, then b \in rad Q. We can rewrite this as follows: if ab \in Q and a \notin Q, then b^k \in Q for some positive integer k. This is because b \in rad Q means that b^k \in Q for some positive integer k.

Now, let's consider the extension and contraction of Q to R_P. We have ^eQ = \pi(Q)R_P, which means that any element in ^eQ can be written as a fraction of the form q/d, where q \in Q and d \in D. Similarly, ^c(^eQ) = \pi^{-1}(\pi(Q)R_P) means that any element in ^c(^eQ) can be written as a fraction of the form \pi^{-1}(q/d), where q \in Q and d \in D. But since \pi^{-1} is the inverse of \pi, we have \pi^{-1}(q/d) = q, so any element in ^c(^eQ) is in fact an element of Q.

Now, we need to show that every element in Q can be written as an element of ^c(^eQ). Let q \in Q, then q/d \in ^eQ for any d \in D. But since q \in Q, we have q/d \in ^c(^eQ) by definition. Therefore, every element in Q can be written as an element of ^c