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

$\displaystyle g(x) = {f_1{(x)}}^{n_1} {f_2{(x)}}^{n_2}..... {f_k{(x)}}^{n_k} $

be its factorization into irreducibles, where all the $\displaystyle f_i(x) $ are distinct.

Then we have the following isomorphism of rings:

$\displaystyle F[x]/(g(x)) \cong F[x]/{f_1{(x)}}^{n_1} \ \times \ F[x]/{f_2{(x)}}^{n_2} \ \times \ ... ... \ \times \ F[x]/{f_k{(x)}}^{n_k} $

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The proof reads as follows:

Proof: This follows from the Chinese Remainder Theorem (Theorem 7.17), since the ideals $\displaystyle {f_i{(x)}}^{n_i} $ and $\displaystyle {f_j{(x)}}^{n_j} $ are comaximal if $\displaystyle f_i(x) $ and$\displaystyle f_j(x) $ 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 $\displaystyle {f_i{(x)}}^{n_i} $ and $\displaystyle {f_j{(x)}}^{n_j} $ are comaximal.

That is, why are$\displaystyle {f_i{(x)}}^{n_i} $ and $\displaystyle {f_j{(x)}}^{n_j}$ comaximal - how exactly does it follow from$\displaystyle f_i(x)$ and $\displaystyle f_j(x)$ 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 $\displaystyle f_i(x)^{n_i}$ are powers of $\displaystyle f_i(x)$ and all the divisors of $\displaystyle f_j(x)^{n_j}$ are powers of $\displaystyle f_j(x)$, 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 $\displaystyle {f_i{(x)}}^{n_i} $ and $\displaystyle {f_j{(x)}}^{n_j} $ 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 $\displaystyle {f_i{(x)}}^{n_i} $ and $\displaystyle {f_j{(x)}}^{n_j} $ having a gcd of 1 that the to ideals are comaximal?

Peter

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

Suppose $\displaystyle (a)+(b)=R=(1)$. In Section 8.2, proposition 6 of D&F, this means $\displaystyle gcd(a,b)=1$.

Conversely, if $\displaystyle gcd(a,b)=1$, then $\displaystyle ra+sb=1$ for some $\displaystyle r,s\in R$. Therefore $\displaystyle 1 \in (a)+(b)$, so $\displaystyle (a)+(b)=R$

Re: Polynomial Rings Over Fields - Dummit and Foote

Quote:

Originally Posted by

**Gusbob** I'll speak in general terms. Let $\displaystyle R$ be a principal ideal domain (it is certainly the case for you).

Suppose $\displaystyle (a)+(b)=R=(1)$. In Section 8.2, proposition 6 of D&F, this means $\displaystyle gcd(a,b)=1$.

Conversely, if $\displaystyle gcd(a,b)=1$, then $\displaystyle ra+sb=1$ for some $\displaystyle r,s\in R$. Therefore $\displaystyle 1 \in (a)+(b)$, so $\displaystyle (a)+(b)=R$

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) $\displaystyle (a, b) = \{r_1a + r_2b \ | \ r_1, r_2 \in R \} $

[Note (a, B) includes the terms $\displaystyle r_1a $ and $\displaystyle r_2b $ since $\displaystyle r_1 $ or $\displaystyle r_2 $ can equal 0.]

Also by definition we have $\displaystyle (a) = \{r_1a \ | \ r_1 \in R \} $ and $\displaystyle (b) = \{r_2b \ | \ r_2 \in R \} $

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

$\displaystyle (a) + (b) = \{r_1a, r_2b \ | \ r_1, r_2 \in R \} $

so we are missing the 'addition' terms $\displaystyle r_1a + r_2b $ 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

$\displaystyle X + Y = \{x + y \ | \ x \in X, y \in Y \} $

Working, then, with this definition we have

$\displaystyle (a) + (b) = \{ r_1a + r_2b \ | \ r_1, r_2 \in R \} $ 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 $\displaystyle A = \{ a_1, a_2, ... ... a_n \} $ we have that

$\displaystyle (a_1, a_2, ... ... a_n) = (a_1) + (a_2) + ... ... (a_n) $

Is this correct?

Just another vaguely connected question.

Given a ring R consisting of the elements $\displaystyle \{a_1, a_2, ... ... a_n \} $

do there always (necessarily?) exist ideals $\displaystyle A_1 = (a_1) , A_2 = (a_2), ... ... A_n = (a_n) $

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 $\displaystyle A = \{ a_1, a_2, ... ... a_n \} $ we have that

$\displaystyle (a_1, a_2, ... ... a_n) = (a_1) + (a_2) + ... ... (a_n) $

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 $\displaystyle \{a_1, a_2, ... ... a_n \} $

do there always (necessarily?) exist ideals $\displaystyle A_1 = (a_1) , A_2 = (a_2), ... ... A_n = (a_n) $

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