1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member


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

    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
    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 =)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook