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Primary Ideals, prime ideals and maximal ideals.

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
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.]
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
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.