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[Abstract Algebra] GCD and Relatively Prime Proof

  1. Feb 20, 2016 #1

    RJLiberator

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    Gold Member

    1. The problem statement, all variables and given/known data
    If [itex]gcd(f(x),g(x)) = 1[/itex] and m,n ∈ ℕ, show that [itex]gcd(f(x)^m, g(x)^n) = 1[/itex].

    2. Relevant equations


    3. The attempt at a solution

    So I had previously proved this for non-polynomials:

    gcd(a,b)=1
    then gcd(a^n,b^n)=1

    Proof: a = p1*p2*...*pn
    b = p1*p2*...*pm

    then
    a^n = p1^n*p2^n*...*pn^n
    b^n = p1^n*p2^n*...*pm^n

    Since a and a^n have the same prime factors and b and b^n have the same prime factors and a and b are relatively prime then a^n and b^n are relatively prime.



    I guess, I am looking at this question from the same point of view.

    Is there a difference here with polynomials? Why would there be?
     
  2. jcsd
  3. Feb 20, 2016 #2

    fresh_42

    Staff: Mentor

    As long as the underlying ring or field is of characteristic zero there shouldn't be a difference. (Not quite sure about other characteristics.)
    However, your proof with the numbers uses the fact that you have a unique integer factorization into primes. You need to assure this for your polynomial ring if you want to adapt the same proof. As far as I can remember this is true for an integral domain. Another proof could be with contradiction: assume there is an irreducible ##p=p(x)## with ##p|f^n## and ##p|g^m##.
     
  4. Feb 20, 2016 #3

    RJLiberator

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    How is this proof:

    Since f and g are relatively prime, by Bezout's Identity there exists polynomials p and q such that fp+gq = 1.
    Take the (2n-1)th power of both sides.
    Any term of the binomial expansion of (fp+gq)^(2n-1) is divisible by one of the f^n or g^n.
    Therefore:
    1 = (fp+gq)^(2n-1) = f^nP+g^nQ
    for some polynomials P and Q. It follows that f^n and g^n are relatively prime.

    But does this hold for f^n and g^m ? (different degrees) hm...

    I don't think so, right? Since we use (2n-1) as the exponential.
     
  5. Feb 20, 2016 #4

    RJLiberator

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    Perhaps this is cleaner / better:

    Suppose p is a prime that divides gcd(f^n,g^m)
    then p divides f^n and hence divides f. Similarly for g^m. Since gcd(f,g)=1 no such p can exist.

    This is proven by:
    If p divides f^n, than p divides f. for n>1
    Proof using induction: true for n = 1
    Assume it is true for n.
    Then for n+1 p divides f^(n+1) = f*f^n implies that p divides f or it divides f^n. By hypothesis, p divides f.

    However, not sure how this incorporates polynomials or if i even need to (the question doesn't necessarily make any statement about polynomials just is the question stated)
     
  6. Feb 20, 2016 #5

    fresh_42

    Staff: Mentor

    Your first proof is wrong because not every term of the binomial expansion is divided by ##f^n##.
    Let's say ##p## is irreducible instead.
    ##n \geq 1##
    What is the underlying ring or field for the polynomials? ##ℤ##?
    This here is the crucial point: ##p|f \cdot f^n ⇒ p|f ∨ p|f^n##?
    Here I think you need the polynomial ring be an integral domain.
    E.g. if ##p(x) = x+1## and ##f(x)=2x## both in ##ℤ_4[x]## then ##p \nmid f## but ##p|f^2=4x^2=0##.
    However, I suppose ##ℤ[x]## or a polynomial ring over field of characteristic zero ##(ℚ[x],ℝ[x],ℂ[x])## is meant.
    Then everything looks ok.
     
  7. Feb 21, 2016 #6

    RJLiberator

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    Indeed. Thank you for analysis here. (as always)
    fresh_42, I bow to you!
     
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