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Polynomials and Ideals

  1. Feb 9, 2012 #1
    1. The problem statement, all variables and given/known data

    (see attachment)

    2. Relevant equations



    3. The attempt at a solution

    I have been attempting a proof by contradiction (for the last statement) for a while now, but I can't seem to reach a contradiction from these premises:

    1 ≤ deg(f) ≤ deg(g) (without loss)
    N ≠ F[x]
    both f and g are irreducible over F
    I also know that since F is a field, N is principal (by a lemma)
     

    Attached Files:

  2. jcsd
  3. Feb 9, 2012 #2

    SammyS

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    https://www.physicsforums.com/attachment.php?attachmentid=43713&d=1328829169

    Having the image appear directly in this thread, may help someone to answer this.
     
  4. Feb 9, 2012 #3

    micromass

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    If f(x) is irreducible, what can you say about (f(x)) in relation to N?? Are they equal?? (use that N is principal)
     
  5. Feb 9, 2012 #4
    So since f(x) is irreducible over F[x], <f(x)> is maximal. Is this what you're getting at micromass?
     
  6. Feb 9, 2012 #5

    micromass

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    Yes. So <f(x)>=N, right?? But if g(x) is irreducible, then also <g(x)>=N.

    So <f(x)>=<g(x)>!! Try to write out what that means.
     
  7. Feb 9, 2012 #6
    I may be blanking, but how can we be sure that <f(x)> = N?
     
  8. Feb 9, 2012 #7

    micromass

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    <f(x)> is maximal, N is an ideal that contains <f(x)>...
     
  9. Feb 9, 2012 #8
    Foolish of me. Thank you.

    So since <f(x)> = <g(x)>, {f(x)s(x) | s(x) in F[x]} = {g(x)r(x) | r(x) in F[x]}, which, for s(x), r(x) = 1, implies that f(x) = g(x), a contradiction since we assumed that f and g differ in degree.
     
  10. Feb 9, 2012 #9

    micromass

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    No, that is not correct. You can't conclude that f(x)=g(x) because it is simply not true.

    You now that f(x)=g(x)r(x) for some r. You also now that there is an s such that g(x)=f(x)s(x). Take the degrees of both equations.
     
  11. Feb 9, 2012 #10
    Let deg(r) = j and let deg(s) = k.

    deg(f) = m = deg(g) + deg(r) = n + j
    deg(g) = n = deg(f) + deg(s) = m + k

    So by substitution from above, n = (n + j) + k, which implies that j + k = 0. But the only way this is true is if j = k = 0, which is shows that m = n; a contradition.
     
  12. Feb 9, 2012 #11

    micromass

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    Good!!

    Another way to prove it is to manipulate the equations directly to prove that r(x)s(x)=1 and therefore r(x) and s(x) are invertible and thus be an element of F

    So now you know that if (f(x))=(g(x)) then there must be a c in F such that f(x)=cg(x)!!

    Well done!
     
  13. Feb 9, 2012 #12
    One question remains: How do we know for sure that <f(x)> is not equal to {0}?
     
  14. Feb 9, 2012 #13

    micromass

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    That would mean that f(x)=0. But f(x) is supposed to be irreducible, and 0 (by definition) isn't.
     
  15. Feb 9, 2012 #14
    *embarrassed. I don't know why i keep wasting your time tonight micromass... I think too much problem solving for one day is drying me out. Much thanks =)
     
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