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- Jun 22, 2012

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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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... ...

(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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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

Proposition 19 Part (5) reads as follows:

----------------------------------------------------------------------------------------------------------------------------

**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

------------------------------------------------------------------------------------------------------------------------------

The proof of (5) above reads as follows:

-------------------------------------------------------------------------------------------------------------------------------

*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

--------------------------------------------------------------------------------------------------------------------------------

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

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