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Homework Help: Relatively Prime Polynomials in Extension Fields

  1. Jul 22, 2008 #1
    The problem statement, all variables and given/known data
    Recall that two polynomials f(x) and g(x) from F[x] are said to be relatively prime if there is no polynomial of positive degree in F[x] that divides both f(x) and g(x). Show that if f(x) and g(x) are relatively prime in F[x], they are relatively prime in K[x], where K is an extension of F.

    The attempt at a solution
    I'm guessing this will be a proof by contradiction where the contradiction will be that f(x) and g(x) are not relatively prime in F[x]:

    Let p(x) be an irreducible factor of the divisor of f(x) and g(x) in K[x]. How can I show that p(x) divides f(x) and g(x) in F[x]?
     
  2. jcsd
  3. Jul 23, 2008 #2
    You can't do what you ask for in your last sentence: p(x) isn't even necessarily in F[x]. But the question doesn't ask you to show that.
     
  4. Jul 23, 2008 #3
    Is there another way of getting a contradiction? I can't think of anything else.
     
  5. Jul 23, 2008 #4
    Why have you stopped what you're doing? The only thing wrong was your assumption that any factor p(x) in K[x] must be in F[x].

    Consider an example if it helps you. E.g. x^2+1 and x^4-1 over Q[x]. These are not relatively prime, but you don't prove that by considering *one* irreducible linear factor in C[x], since x-i and x+i are not in Q[x].
     
  6. Jul 23, 2008 #5
    I got the impression that my whole approach was wrong.

    So are you saying I have to consider all the irreducible factors in K[x]? In what way? Perhaps the product of some of the irreducible linear factors of the divisor is a member of F[x], e.g. (x - i)(x + i) = x2 + 1 in your example?
     
  7. Jul 23, 2008 #6
    Let's suppose that I have a polynomial f(x) with rational coefficients. And suppose that, passing to the complex numbers, I know x-i is a root of f. What other linear factor must be a root of f(x)?

    What do you know about things like Aut(K:F)? The field automorphisms of K that fix F. Such as complex conjugation in Aut(C:Q).
     
  8. Jul 23, 2008 #7
    I'm guessing x + i. I don't really know. To find another linear factor I would try to find a zero of f(x)/(x - i).

    I've never seen that notation before.

    My problem is from a chapter on extension and splitting fields.
     
  9. Jul 23, 2008 #8
    Yes, that makes sense - you can assume that K is a splitting field for f(x).

    You do know the first part: complex roots of a real polynomial occur in complex conjugate pairs.
     
  10. Jul 23, 2008 #9
    That's an interesting fact. So in the case of Q and C, the contradiction works. But in general? Does p(x) have a conjugate?
     
  11. Jul 23, 2008 #10
    I was told that since f(x) and g(x) are relatively prime in F[x], then there are s(x) and t(x) in F[x] with f(x)s(x) + g(x)t(x) = 1. This equation is also valid in K[x] and this supposedly implies that f(x) and g(x) are relatively prime in K[x]. I don't understand this implication. Why is this?
     
  12. Jul 23, 2008 #11
    It's just euclid's algorithm. Just think of Z rather than a polynomial algebra.

    You can do it with a contradiction for any extension. But you don't need to.
     
  13. Jul 23, 2008 #12
    Does f(x)s(x) + g(x)t(x) = 1 imply that f(x) and g(x) are relatively prime? Isn't it the other way around?
     
  14. Jul 24, 2008 #13
    It is exactly the same as it is for the integers. You should be able to answer your own question.
     
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