Polynomial Rings Over Fields - Dummit and Foote

I am reading Dummit and Foote Section 9.5 Polynomial Rings Over Fields II.

I am trying to understand the proof of Proposition 16 which reads as follows:

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Proposition 16. Let F be a field. Let g(x) be a nonconstant monic polynomial of F[x] and let

be its factorization into irreducibles, where all the are distinct.

Then we have the following isomorphism of rings:

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The proof reads as follows:

Proof: This follows from the Chinese Remainder Theorem (Theorem 7.17), since the ideals and are comaximal if and are distinct (they are relatively prime in the Euclidean Domain F[x], hence the ideal generated by them is F[x]).

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My question: I can follow the reference to the Chinese Remainder Theorem but I cannot follow the argument that establishes the necessary condition that the and are comaximal.

That is, why are and comaximal - how exactly does it follow from and being relatively prime in the Euclidean Domain F[x]?

Any help would be very much appreciated.

Peter

Re: Polynomial Rings Over Fields - Dummit and Foote

Because all the divisors of are powers of and all the divisors of are powers of , their gcd is 1, so the ideals are comaximal.

Re: Polynomial Rings Over Fields - Dummit and Foote

Thanks Gusbob, but I still have a problem

In Dummit and Foote Section 7.6 The Chinese Remainder Theorem (page 265) we find the following definition of comaximal ideals

Definition. The ideals A and B of the ring R are said to be comaximal if A + B = R

How does it follow that and having a gcd of 1 that the to ideals are comaximal?

Peter

Re: Polynomial Rings Over Fields - Dummit and Foote

Quote:

Originally Posted by

**Bernhard** Thanks Gusbob, but I still have a problem

In Dummit and Foote Section 7.6 The Chinese Remainder Theorem (page 265) we find the following definition of comaximal ideals

Definition. The ideals A and B of the ring R are said to be comaximal if A + B = R

How does it follow that

and

having a gcd of 1 that the to ideals are comaximal?

Peter

I'll speak in general terms. Let be a principal ideal domain (it is certainly the case for you).

Suppose . In Section 8.2, proposition 6 of D&F, this means .

Conversely, if , then for some . Therefore , so

Re: Polynomial Rings Over Fields - Dummit and Foote

Quote:

Originally Posted by

**Gusbob** I'll speak in general terms. Let

be a principal ideal domain (it is certainly the case for you).

Suppose

. In Section 8.2, proposition 6 of D&F, this means

.

Conversely, if

, then

for some

. Therefore

, so

Gusbob,

You mention Proposition 6 in Section 8.2, but in this proposition we have that d is the generator for the principal ideal that is generated by a and b

That is (d) = (a,b) [Is this correct?)

Now you seem to be using (d) = (a) + (b) ... ... indeed the proof needs this ...

Can you show how (d) = (a, b) implies (d) = (a) + (b)

Peter

PS Note that Proposition 2 in Section 8.1 gives a similar finding for a Commutative ring R

Re: Polynomial Rings Over Fields - Dummit and Foote

I had a feeling we went through this before :)

http://mathhelpforum.com/advanced-al...al-domain.html

Actually I think that answers both your questions.

Re: Polynomial Rings Over Fields - Dummit and Foote

Thanks for the help Gusbob ... At first glance looks like you are correct ... I will definitely work through it agin

apologies ... My day job often distracts me from my mathematical adventures ... :-)

Peter

Re: Polynomial Rings Over Fields - Dummit and Foote

Just a further (very basic!) question:

Is the following argument - working from definitions - correct

Does (a) + (b) = (a,b)?

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By definition (Dummit and Foote page 251)

[Note (a, B) includes the terms and since or can equal 0.]

Also by definition we have and

Now if by '+' we mean the "addition" (union or putting together) of sets then we have

so we are missing the 'addition' terms of (a, B).

But if we take (as we probably should) the '+' to mean the sum of ideals - then the definition is (Dummit and Foote page 247) for ideals X and Y in R

Working, then, with this definition we have

and this is the same as the definition of (a, b) so (a) + (b) = (a, b) from the definitions.

**Is this correct?**

If the above is correct then seemingly for an ideal generated by the set we have that

Is this correct?

Just another vaguely connected question.

Given a ring R consisting of the elements

do there always (necessarily?) exist ideals

Help with confirming (or otherwise) my reasoning & clarifying the above issues would be much appreciated.

Peter

Re: Polynomial Rings Over Fields - Dummit and Foote

Quote:

Originally Posted by

**Bernhard**
If the above is correct then seemingly for an ideal generated by the set

we have that

Is this correct?

In the setting of PID, I believe so.

Quote:

Just another vaguely connected question.

Given a ring R consisting of the elements

do there always (necessarily?) exist ideals

I'm not sure what you mean here. You can always have an ideal generated by a single element.