# Primary Ideals, prime ideals and maximal ideals.

1. Nov 8, 2013

### Math Amateur

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.]

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2. Nov 9, 2013

### R136a1

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 $Q=\{0\}$.

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.