# Primary Ideals, prime ideals and maximal ideals - D&F Section 15.2

1. Nov 9, 2013

### Math Amateur

I am studying Dummit and Foote Section 15.2. I am trying to understand the proof of Proposition 19 Part (5) on page 682 (see attachment)

Proposition 19 Part (5) reads as follows:
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Proposition 19.

... ...

(5) Suppose M is a maximal ideal and Q is an ideal with $M^n \subseteq Q \subseteq M$ for some $n \ge 1$.

Then Q is a primary idea, with rad Q = M

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The proof of (5) above reads as follows:

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Proof.

Suppose $M^n \subseteq Q \subseteq M$ for some $n \ge 1$ where M is a maximal idea.

Then $Q \subseteq M$ so $rad \ Q \subseteq rad \ M = M$.

... ... etc

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My problem is as follows:

Why can we be sure that rad M = M?

I know that M is maximal and so no ideal in R can contain M. We also know that $M \subseteq rad \ M$

Thus either rad M = M (the conclusion D&F use) or rad M = R?

How do we know that $rad \ M \ne R$?

Would appreciate some help.

Peter

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2. Nov 9, 2013

### R136a1

Try to use contradiction. Assume that $\textrm{rad}(M) = R$. Then $1\in \text{rm}(M)$. Now use the definition of the radical.

3. Nov 10, 2013

### Math Amateur

Thanks R136a1!

But just thinking over this ...

Is the following thinking along the right track ...?

$rad \ M = \{a \in R \ | \ a^k \in M$ for some $k \ge 1$

So then $1 \in rad \ M \Longrightarrow 1^k \in M$ for some $k \ge 1$

$\Longrightarrow \ 1 \in M$

$\Longrightarrow \ M = R$

Thus would mean $M^n = R$ so $Q = R$ also ....

This result is not a contradiction but it leads to the collapse of the conditions of the Proposition to triviality ....

Can you confirm my reasoning ... or indeed point out errors/inadequacies in my thinking

Peter

4. Nov 10, 2013

### Math Amateur

Just thinking further ... maybe in my reasoning in the last post I have indeed achieved a contradiction since my reasoning (if correct!) establishes that M = R .,,, where of course M is a maximal ideal by assumption ... but by D&F's definition of a maximal ideal, this is not possible ... so contradiction!

Can someone confirm that this is correct?

Note: Definition of maximal ideal, Dummit and Foote, page 253:

"An ideal M in an arbitrary ring R is called a maximal ideal if $M \ne R$ and the only ideals containing M are M and R."

5. Nov 10, 2013

### R136a1

Yes, it is a contradiction because $M=R$ is not possible by definition.