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Math Help - Polynomial Rings Over Fields - Dummit and Foote

  1. #1
    Super Member Bernhard's Avatar
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    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:

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

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

     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  f_i(x) are distinct.

    Then we have the following isomorphism of rings:

     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}

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

    The proof reads as follows:

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

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

    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  {f_i{(x)}}^{n_i} and  {f_j{(x)}}^{n_j} are comaximal.

    That is, why are  {f_i{(x)}}^{n_i} and {f_j{(x)}}^{n_j} comaximal - how exactly does it follow from  f_i(x) and f_j(x) being relatively prime in the Euclidean Domain F[x]?

    Any help would be very much appreciated.

    Peter
    Last edited by Bernhard; June 1st 2013 at 03:28 AM.
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    Re: Polynomial Rings Over Fields - Dummit and Foote

    Because all the divisors of f_i(x)^{n_i} are powers of f_i(x) and all the divisors of f_j(x)^{n_j} are powers of f_j(x), their gcd is 1, so the ideals are comaximal.
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    Super Member Bernhard's Avatar
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    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  {f_i{(x)}}^{n_i} and  {f_j{(x)}}^{n_j} having a gcd of 1 that the to ideals are comaximal?

    Peter
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    Re: Polynomial Rings Over Fields - Dummit and Foote

    Quote Originally Posted by Bernhard View Post
    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  {f_i{(x)}}^{n_i} and  {f_j{(x)}}^{n_j} having a gcd of 1 that the to ideals are comaximal?

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

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

    Conversely, if gcd(a,b)=1, then ra+sb=1 for some r,s\in R. Therefore 1 \in (a)+(b), so (a)+(b)=R
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    Super Member Bernhard's Avatar
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    Re: Polynomial Rings Over Fields - Dummit and Foote

    Quote Originally Posted by Gusbob View Post
    I'll speak in general terms. Let R be a principal ideal domain (it is certainly the case for you).

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

    Conversely, if gcd(a,b)=1, then ra+sb=1 for some r,s\in R. Therefore 1 \in (a)+(b), so (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
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    Re: Polynomial Rings Over Fields - Dummit and Foote

    I had a feeling we went through this before

    Comaximal Ideas in a Principal Ideal Domain

    Actually I think that answers both your questions.
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    Super Member Bernhard's Avatar
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    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
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    Super Member Bernhard's Avatar
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    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)?

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

    By definition (Dummit and Foote page 251)  (a, b) = \{r_1a + r_2b  \ | \ r_1, r_2 \in R \}

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

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

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

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

    so we are missing the 'addition' terms  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

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

    Working, then, with this definition we have

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

     (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  \{a_1, a_2, ... ... a_n \}

    do there always (necessarily?) exist ideals  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
    Last edited by Bernhard; June 1st 2013 at 05:37 PM.
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    Re: Polynomial Rings Over Fields - Dummit and Foote

    Quote Originally Posted by Bernhard View Post

    If the above is correct then seemingly for an ideal generated by the set  A = \{ a_1, a_2, ... ... a_n \} we have that

     (a_1, a_2, ... ... a_n) = (a_1) + (a_2) + ... ... (a_n)

    Is this correct?
    In the setting of PID, I believe so.

    Just another vaguely connected question.

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

    do there always (necessarily?) exist ideals  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.
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