Primary Ideals, prime ideals and maximal ideals.

• Math Amateur
This is a contradiction, so there cannot be other maximal ideals than ##M##. Therefore, ##M## is the unique maximal ideal.
Math Amateur
Gold Member
MHB
I have a problem in understanding the proof to Dummit and Foote Section 15.2, Proposition 19 regarding primary ideals. I hope someone can help.

My problem is with Proposition 19 part 4 - but note that part 4 relies on part 2 - see attachment.

The relevant sections of Proposition 19 read as follows: (see attachment)

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

Proposition 19.

Let R be a commutative ring with 1

... ...

(2) The ideal Q is primary if and only if every zero divisor in R/Q is nilpotent.

... ...

(4) If Q is an ideal whose radical is a maximal ideal. then Q is a primary ideal

... ... etc

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

The proof of (4) above proceeds as follows:

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

Proof. (see attachment)

... ...

To prove (4) we pass to the quotient ring R/Q: by (2) it suffices to show that every zero divisor in this quotient ring is nilpotent.

We are reduced to the situation where Q = (0) and M = rad Q = rad (0), which is the nilradical, is a maximal ideal.

Since the nilradical is contained in every prime ideal (Proposition 12), it follows that M is the unique prime ideal, so also the unique maximal ideal.

... ... etc (see attachment)

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

I have two problems with the proof above.

(1) I do not completely follow the statement:

"We are reduced to the situation where Q = (0) and M = rad Q = rad (0), which is the nilradical, is a maximal ideal."

I know this is something to do with zero divisors in R/Q, but why/how do we end up discussing Q = (0) exactly? Can someone please clarify? ( I suspect it is simple but I cannot see the connection :-( )

(2) I am puzzled by the statement "it follows that M is the unique prime ideal, so also the unique maximal ideal"

This statement appears to indicate that in R we have that prime ideals are maximal ideals.

I thought that we could only assume that maximal ideals were prime ideals (D&F Corollary 14 page 256) but not the converse? Can someone please clarify.

Would appreciate some help.

Peter

[Note: D&F Corollary 14, page 256 reads as follows:

Corollary 14. Assume R is commutative. Every maximal ideal is a prime ideal.]

Attachments

• Dummit and Foote - Ch 15 - page 682.pdf
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Math Amateur said:
I have a problem in understanding the proof to Dummit and Foote Section 15.2, Proposition 19 regarding primary ideals. I hope someone can help.

My problem is with Proposition 19 part 4 - but note that part 4 relies on part 2 - see attachment.

The relevant sections of Proposition 19 read as follows: (see attachment)

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

Proposition 19.

Let R be a commutative ring with 1

... ...

(2) The ideal Q is primary if and only if every zero divisor in R/Q is nilpotent.

... ...

(4) If Q is an ideal whose radical is a maximal ideal. then Q is a primary ideal

... ... etc

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

The proof of (4) above proceeds as follows:

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

Proof. (see attachment)

... ...

To prove (4) we pass to the quotient ring R/Q: by (2) it suffices to show that every zero divisor in this quotient ring is nilpotent.

We are reduced to the situation where Q = (0) and M = rad Q = rad (0), which is the nilradical, is a maximal ideal.

Since the nilradical is contained in every prime ideal (Proposition 12), it follows that M is the unique prime ideal, so also the unique maximal ideal.

... ... etc (see attachment)

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

I have two problems with the proof above.

(1) I do not completely follow the statement:

"We are reduced to the situation where Q = (0) and M = rad Q = rad (0), which is the nilradical, is a maximal ideal."

I know this is something to do with zero divisors in R/Q, but why/how do we end up discussing Q = (0) exactly? Can someone please clarify? ( I suspect it is simple but I cannot see the connection :-( )

You need to prove that every zero divisor is nilpotent in ##R/Q##. This is equivalent to saying that ##\{0\}## is primary, since ##(R/Q)/\{0\} \cong R/Q##.
Now, you know that the radical of ##Q## is maximal in ##R##. You can say that this implies that the radical of ##\{0\}## is maximal in ##R/Q##. So you know the radical of ##\{0\}## is maximal, and you need to prove ##\{0\}## is primary. This is exactly the theorem with ##Q=\{0\}##.

(2) I am puzzled by the statement "it follows that M is the unique prime ideal, so also the unique maximal ideal"

This statement appears to indicate that in R we have that prime ideals are maximal ideals.

I thought that we could only assume that maximal ideals were prime ideals (D&F Corollary 14 page 256) but not the converse? Can someone please clarify.

Assume that there are other maximal ideals than ##M##. Then these other maximal ideals are also prime ideals. Thus there are other prime ideals than ##M##. But ##M## was the unique prime ideal.

1 person

1. What is the difference between primary ideals, prime ideals, and maximal ideals?

Primary ideals, prime ideals, and maximal ideals are all types of ideals in abstract algebra. A primary ideal is an ideal in a commutative ring where if a product of two elements is in the ideal, then one of the elements is also in the ideal. A prime ideal is an ideal where if a product of two elements is in the ideal, then one of the elements is also in the ideal. A maximal ideal is an ideal that is not contained in any other proper ideal. In other words, a maximal ideal is as "big" as an ideal can be without being the whole ring.

2. How do primary ideals, prime ideals, and maximal ideals relate to one another?

Primary ideals are a special case of prime ideals. All primary ideals are prime ideals, but not all prime ideals are primary ideals. Similarly, all maximal ideals are prime ideals, but not all prime ideals are maximal ideals. In general, primary ideals are "smaller" than prime ideals, and prime ideals are "smaller" than maximal ideals.

3. What is the significance of primary ideals, prime ideals, and maximal ideals in abstract algebra?

Ideals are important in abstract algebra because they allow us to study the structure of a ring by looking at its "substructures." Primary ideals, prime ideals, and maximal ideals are important because they give us information about the properties of the elements in a ring and the relationships between them. For example, if a ring has a unique maximal ideal, it is called a local ring and has special properties.

4. Can a ring have multiple maximal ideals?

Yes, a ring can have multiple maximal ideals. For example, the ring of integers has infinitely many maximal ideals, each corresponding to a different prime number. However, a field, which is a special type of ring, can only have one maximal ideal (the zero ideal).

5. How are primary ideals, prime ideals, and maximal ideals used in applications?

Primary ideals, prime ideals, and maximal ideals are used in many applications, particularly in algebraic geometry and number theory. In algebraic geometry, these types of ideals help us understand the structure and properties of algebraic varieties. In number theory, they are used to study number rings and their properties. They also have applications in coding theory and cryptography.

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