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Ideals and GCDs in k[x] ... ... Cox et al ...

  1. Jul 2, 2016 #1
    1. The problem statement, all variables and given/known data

    I am reading the undergraduate introduction to algebraic geometry entitled "Ideals, Varieties and Algorithms: An introduction to Computational Algebraic Geometry and Commutative Algebra (Third Edition) by David Cox, John Little and Donal O'Shea ... ...

    I am currently focused on Chapter 1, Section 5: Polynomials of One Variable ... ... and need help with the proof of Proposition 8, part 3 ...

    Proposition 8 of Chapter 1 (including Definition 7 which is relevant) reads as follows:


    ?temp_hash=0292a2083050a3ef6d81c413a30bf337.png



    In the above text from Cox et al we read the following:


    " ... ... To prove part (iii), let ##h \ = \ GCD(f_2, \ ... \ ... \ , f_s)##. We leave it as an exercise to show that

    ##<f_1, h> \ = \ <f_1, \ ... \ ... \ , f_s>##

    ... ... "


    I need help to show that ##<f_1, h> \ = \ <f_1, \ ... \ ... \ , f_s>## ... ...


    2. Relevant equations

    These are all introduced in context, in 3 below ... ...



    3. The attempt at a solution


    Work so far ...

    We need to show that ##<f_1, h> \ \subset \ <f_1, \ ... \ ... \ , f_s>## ... and also that

    ##<f_1, \ ... \ ... \ , f_s> \ \subset \ <f_1, h>##


    So to show ##<f_1, h> \ \subset \ <f_1, \ ... \ ... \ , f_s>## we start with


    Let ##l \in <f_1, h>## ...

    Then, by definition of ##<f_1, h>##, we have that ##l = f_1 t_1 + h t_2 where t_1, t_2 \in k[x]## ...


    Now we have that ##h = GCD(f_2, \ ... \ ... \ , f_s)## ... BUT ... how do we use this in the proof?


    Note that we also have

    ##(1) \ h = GCD(f_2, \ ... \ ... \ , f_s) \ \Longrightarrow \ h \text{ divides } f_2, \ ... \ ... \ , f_s##

    ## \ \Longrightarrow \ h = f_2 u_2, h \ = \ f_3 u_3, \ ... \ ... \ , h \ = \ f_s u_s##

    for some ##u_2, \ ... \ ... \ , u_s \in k[x]## ...


    (2) ##k[x]## is a PID so that:

    ##<f_1, h > \ = \ <v>## for some polynomial ##v \in k[x]## ... ...


    ... but, how do we use (1) and (2) in the required proof ...



    Can someone please help me to complete the proof of ##<f_1, h> \ = \ <f_1, \ ... \ ... \ , f_s>## ...

    Help will be appreciated ...

    Peter
     

    Attached Files:

  2. jcsd
  3. Jul 2, 2016 #2

    fresh_42

    Staff: Mentor

    The easy part is ##I := <f_1, \dots , f_s> \;⊆ \;<h>\; ⊆\; <h,f_1>\;## because every ##f_i## can be written ##f_i = g_i \cdot h## by definition of ##h##. Thus any ##f = \sum a_i f_i ∈ I## gets ##f = \sum a_i g_i h ∈ \; <h>.##
    The other way, your first, is essentially Bézout's Lemma. I'm too lazy to type it from Wikipedia. Maybe you get along with what is written there:
    https://en.wikipedia.org/wiki/Bézout's_identity

    It might as well be the quoted proposition 6 or corollary 4. It means that you can write ##h_2 = p_1 f_1 + p_2 f_2## if ##h_2## is the greatest common divisor of ##f_1## and ##f_2## and then by induction ##h = p_1 f_1 + \dots + p_s f_s## which is what we need for ##h \in I.##

    It applies to PID, so you may substitute "integers" by "polynomials". There is a written proof and a link to the polynomial version (without proof as far as I could see): https://en.wikipedia.org/wiki/Polyn...9zout.27s_identity_and_extended_GCD_algorithm

    If you have further questions to it or about the Wiki entry, please let me know.
     
  4. Jul 3, 2016 #3

    Thanks for the help fresh_42 ...

    Still reflecting on your post ... but basically followed that ...

    Thanks again ... most helpful

    Peter
     
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