# Primary Ideals, prime ideals and maximal ideals.

• MHB
• Math Amateur
In summary, to prove Proposition 19 part 4, we first pass to the quotient ring $R/Q$ and use part 2 to show that every zero-divisor in this ring is nilpotent. This leads us to the situation where $Q$ is the zero ideal and $M = \text{rad}(Q)$ is the unique maximal ideal of $R/Q$. From here, we use the fact that maximal ideals are contained in every prime ideal to show that $M$ is the unique prime ideal of $R/Q$. Combining this with the fact that every maximal ideal is a prime ideal, we conclude that $M$ is the unique maximal ideal of $R/Q$. Finally, we show that any zero-div
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)

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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

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The proof of (4) above proceeds as follows:

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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)

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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.]

I think there is some unfortunate symbol-overloading, here. Let's look at (4):

Suppose $Q$ is an ideal whose radical is maximal. Consider the quotient ring $R/Q$. In this new ring, $Q$ is the "zero", so the nilpotent elements of $R/Q$ are the elements of $\text{rad}(Q) = \text{rad}(0)$.

Since this ideal ($M$, the nilpotent elements of $R/Q$) is a nilradical, by an earlier result (Proposition 12) it is a maximal ideal.

Now since (also by Proposition 12) this ideal is contained in EVERY prime ideal, it must be the UNIQUE prime ideal of $R/Q$. To see this, suppose we had 2 prime ideals $I \neq J$ with:

$M \subseteq I, M \subseteq J$.

Since we have some $y \in J - I$ (or vice-versa, in which case switch the letters), it follows from the maximality of $M$ that $J = R/Q$, contradicting the fact that $J$ is a prime ideal (prime ideals cannot be the entire ring).

Since maximal ideals are necessarily prime, this means that $M$ is the unique prime ideal of $R/Q$, and thus the unique maximal ideal of $R/Q$ (make sure you follow this!).

Ok, now let $d$ be any zero-divisor in $R/Q$, and consider the principal ideal generated by $d$. Since zero-divisors ARE NOT UNITS, this must be a PROPER ideal of $R/Q$. As such, it is contained in some MAXIMAL ideal (which might be, perhaps, itself). But...we just have the ONE maximal ideal, $M$. This shows that any zero-divisor must lie within $M$, that is: is nilpotent in $R/Q$. Since we now have satisfied the conditions of part (2), we see that $Q$ is primary.

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

Primary ideals are ideals in a ring that contain no zero divisors and are closed under multiplication. Prime ideals are special cases of primary ideals that are also closed under addition. Maximal ideals are the largest possible ideals in a ring, meaning that they cannot be properly contained in any other ideal.

## 2. What is the significance of primary, prime, and maximal ideals in algebraic geometry?

Primary, prime, and maximal ideals play a crucial role in algebraic geometry as they allow for the study of the geometric properties of algebraic varieties through the use of ideals. Primary ideals correspond to irreducible components of an algebraic variety, prime ideals correspond to subvarieties, and maximal ideals correspond to points on the variety.

## 3. Can a primary ideal be a prime ideal?

Yes, a primary ideal can be a prime ideal. In fact, every prime ideal is a primary ideal, but the converse is not always true. This means that not all primary ideals are prime ideals.

## 4. How do primary, prime, and maximal ideals relate to the concept of a radical ideal?

A radical ideal is an ideal where the nth power of any element in the ideal is also in the ideal. Primary ideals are always radical ideals, and prime ideals are radical ideals if and only if they are primary. Maximal ideals are always radical ideals, but the converse is not always true.

## 5. What applications do primary, prime, and maximal ideals have in other branches of mathematics?

Primary, prime, and maximal ideals have applications in a variety of branches of mathematics, including commutative algebra, algebraic geometry, number theory, and algebraic topology. They are also used in the construction of quotient rings and in the proof of the fundamental theorem of algebra.

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