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

#### Peter

##### Well-known member
MHB Site Helper
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 [TEX] M^n \subseteq Q \subseteq M [/TEX] for some [TEX] n \ge 1 [/TEX].

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 [TEX] M^n \subseteq Q \subseteq M [/TEX] for some [TEX] n \ge 1 [/TEX] where M is a maximal idea.

Then [TEX] Q \subseteq M [/TEX] so [TEX] rad \ Q \subseteq rad \ M = M [/TEX].

... ... 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 [TEX] M \subseteq rad \ M [/TEX]

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

How do we know that [TEX] rad \ M \ne R [/TEX]?

Would appreciate some help.

Peter

Last edited:

#### Siron

##### Active member
Like you said, we have $$M \subseteq \mbox{Rad} \ M$$ which implies that $$M = \mbox{Rad}\ M$$ as $$M$$ is a maximal ideal. Why is $$\mbox{Rad} \ M \neq R$$? Suppose that $$\mbox{Rad} \ M = R$$ then $$M=R$$ but that's impossible by definition of a maximal ideal.

#### Peter

##### Well-known member
MHB Site Helper
Like you said, we have $$M \subseteq \mbox{Rad} \ M$$ which implies that $$M = \mbox{Rad}\ M$$ as $$M$$ is a maximal ideal. Why is $$\mbox{Rad} \ M \neq R$$? Suppose that $$\mbox{Rad} \ M = R$$ then $$M=R$$ but that's impossible by definition of a maximal ideal.

Thanks for the helpful post, Siron

Peter