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

[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

------------------------------------------------------------------------------------------------------------------------------The proof of (5) above reads as follows:-------------------------------------------------------------------------------------------------------------------------------

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

 

[Siron]

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.

 

[Math Amateur]

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