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Primary Ideals, prime ideals and maximal ideals.

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.

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(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.]

Re: Primary Ideals, prime ideals and maximal ideals.

Hello once again :)

In the first statement, you have probably seen statements like:

$\displaystyle P$ is a prime ideal of $\displaystyle R$ if and only if $\displaystyle R/P$ is an integral domain.

This is because the condition [TEX]ab\in P[TEX] implies $\displaystyle a\in P$ or $\displaystyle b\in P$

becomes $\displaystyle ab=0$ implies $\displaystyle a=0$ or $\displaystyle b=0$ in $\displaystyle R/P$, as we are modding out by elements of P.

Similarly the primary condition becomes the condition that all zero divisors are nilpotent and it is a similar proof to the statement above. By passing to the quotient ring R/P, and using the ideal correspondence theorem, we see P corresponds to the zero ideal in R/P. So it suffices to prove the statement for the zero ideal in R/P.

Now the radical of the zero ideal, actually consists of all nilpotent elements, by definition.

For the second point, we appeal to the statement that requires Zorn's Lemma: Any proper ideal is contained in some maximal ideal. So P must be contained in some maximal ideal Q. All maximal ideals are prime, so Q is prime and Q contains P. As P is the only prime it follows that P=Q and so P is maximal.