# Primary ideals and localization of prime ideals

• MHB
• Math Amateur
In summary: Q), which means that Q \subseteq ^c(^eQ).On the other hand, let x \in ^c(^eQ). Then x = \pi^{-1}(q/d) for some q \in Q and d \in D. But this means that \pi(x) = q/d, which means that x \in ^eQ. Therefore, every element in ^c(^eQ) is also an element of ^eQ, which means that ^c(^eQ) \subseteq ^eQ.In summary, we have shown that Q \subseteq ^c(^eQ) and ^c(^eQ) \subseteq ^eQ, which means that Q = ^c(^eQ
Math Amateur
Gold Member
MHB
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

Last edited:
,

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

## 1. What are primary ideals?

Primary ideals are a type of ideal in commutative algebra that have a special relationship with prime ideals. They are defined as ideals in a ring whose radical (set of all elements that raise to a power in the ideal) is a prime ideal. In other words, primary ideals are those whose elements have no nontrivial divisors that are zero divisors.

## 2. How are primary ideals different from prime ideals?

While primary ideals have a special relationship with prime ideals, they are not the same. Prime ideals are those that are maximal with respect to not containing zero divisors, while primary ideals are maximal with respect to not containing nontrivial zero divisors. Additionally, primary ideals can be decomposed into a product of prime ideals, but the converse is not always true.

## 3. What is the localization of a prime ideal?

The localization of a prime ideal is the process of extending a ring by formally adding inverses for a set of elements. In other words, it is the process of "localizing" the ring at a particular prime ideal, making it possible to invert elements that were previously not units. The resulting ring is called the localized ring.

## 4. Why is the localization of prime ideals important?

The localization of prime ideals is important because it allows us to study the properties of a ring at a specific prime ideal. This can be useful in understanding the behavior of the ring at a local level, as well as in applications such as algebraic geometry and number theory.

## 5. How are primary ideals and localization of prime ideals related?

Primary ideals and localization of prime ideals are closely related because localization can be used to define primary ideals. In fact, primary ideals are the prime ideals of the localized ring. Additionally, localization can help in studying the primary decomposition of an ideal, which is the process of breaking it down into a product of primary ideals.

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