# Primary Ideals, prime ideals and maximal ideals - D&F Section 15.2

• Math Amateur
In summary, the proof of Proposition 19 Part (5) on page 682 shows that if M is a maximal ideal and Q is an ideal with M^n \subseteq Q \subseteq M for some n \ge 1 , then Q is a primary ideal with rad Q = M. To prove this, the definition of a maximal ideal is used to show that rad M cannot be equal to R, resulting in a contradiction. Thus, rad M = M and Q is a primary ideal with rad Q = M.
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I am studying Dummit and Foote Section 15.2. I am trying to understand the proof of Proposition 19 Part (5) on page 682 (see attachment)

Proposition 19 Part (5) reads as follows:
----------------------------------------------------------------------------------------------------------------------------

Proposition 19.

... ...

(5) Suppose M is a maximal ideal and Q is an ideal with $M^n \subseteq Q \subseteq M$ for some $n \ge 1$.

Then Q is a primary idea, with rad Q = M

----------------------------------------------------------------------------------------------The proof of (5) above reads as follows:----------------------------------------------------------------------------------------------

Proof.

Suppose $M^n \subseteq Q \subseteq M$ for some $n \ge 1$ where M is a maximal idea.

Then $Q \subseteq M$ so $rad \ Q \subseteq rad \ M = M$.

... ... etc

----------------------------------------------------------------------------------------------

My problem is as follows:

Why can we be sure that rad M = M?

I know that M is maximal and so no ideal in R can contain M. We also know that $M \subseteq rad \ M$

Thus either rad M = M (the conclusion D&F use) or rad M = R?

How do we know that $rad \ M \ne R$?

Would appreciate some help.

Peter

#### Attachments

• Page 682 of Dummit and Foote.pdf
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Try to use contradiction. Assume that ##\textrm{rad}(M) = R##. Then ##1\in \text{rm}(M)##. Now use the definition of the radical.

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Thanks R136a1!

But just thinking over this ...

Is the following thinking along the right track ...?

$rad \ M = \{a \in R \ | \ a^k \in M$ for some $k \ge 1$

So then $1 \in rad \ M \Longrightarrow 1^k \in M$ for some $k \ge 1$

$\Longrightarrow \ 1 \in M$

$\Longrightarrow \ M = R$

Thus would mean $M^n = R$ so $Q = R$ also ...

This result is not a contradiction but it leads to the collapse of the conditions of the Proposition to triviality ...

Can you confirm my reasoning ... or indeed point out errors/inadequacies in my thinking

Peter

Just thinking further ... maybe in my reasoning in the last post I have indeed achieved a contradiction since my reasoning (if correct!) establishes that M = R .,,, where of course M is a maximal ideal by assumption ... but by D&F's definition of a maximal ideal, this is not possible ... so contradiction!

Can someone confirm that this is correct?

Note: Definition of maximal ideal, Dummit and Foote, page 253:

"An ideal M in an arbitrary ring R is called a maximal ideal if $M \ne R$ and the only ideals containing M are M and R."

Yes, it is a contradiction because ##M=R## is not possible by definition.

## 1. What is the difference between a primary ideal and a prime ideal?

A primary ideal is a proper ideal in a commutative ring that is not equal to the whole ring, and its elements have the property that if a product of two elements is in the ideal, then one of the elements must be in the ideal. A prime ideal, on the other hand, is a proper ideal in a commutative ring that is not equal to the whole ring, and its elements have the property that if a product of two elements is in the ideal, then at least one of the elements is in the ideal. In other words, a primary ideal is a special case of a prime ideal, where the property is stricter.

## 2. How are primary ideals and prime ideals related to each other?

As mentioned before, a primary ideal is a special case of a prime ideal. This means that every primary ideal is also a prime ideal, but not every prime ideal is a primary ideal. In other words, primary ideals are a more specific type of prime ideals.

## 3. What is the significance of primary ideals in commutative rings?

Primary ideals are important in commutative rings because they help to classify the factors of ideals. In particular, they help to identify the prime factors of an ideal. This is useful in many areas of mathematics, including algebraic geometry and number theory.

## 4. How do maximal ideals differ from primary and prime ideals?

A maximal ideal is a proper ideal in a commutative ring that is not contained in any other proper ideal. This means that it is the largest possible ideal in the ring. Unlike primary and prime ideals, which have specific properties, maximal ideals are defined by their position in the hierarchy of ideals in a commutative ring.

## 5. Can a primary ideal or prime ideal also be a maximal ideal?

Yes, a primary ideal or prime ideal can also be a maximal ideal. In fact, every maximal ideal is also a prime ideal, and some are also primary ideals. However, not every primary or prime ideal is a maximal ideal. It is possible for a commutative ring to have multiple maximal ideals, but only one maximal ideal can be the whole ring.

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