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Difficult polynomial questions

  1. Dec 8, 2007 #1
    I'm trying to prepare for finals, and these questions have me completely stumped.

    1. The problem statement, all variables and given/known data
    1) For what primes p is x^2 + 1 a factor of x^3 + x^2 + 22x + 15 in F_p[x]? (F_p = finite field with p elements)

    2) F a field. Let x^m - 1 have m distinct roots in F, suppose k divides m. Show x^k - 1 has k distinct roots in K.

    2. The attempt at a solution

    1) Obviously if x^2 + 1 is a factor the other factor must be linear of the form ax + b with coefficients a and b in F_p, a not zero. The only thing I could think of doing is setting up some nasty congruence relations on a and b but they got me nowhere.

    2) I don't even know where to start. I don't really see how divisibility plays a role.
     
  2. jcsd
  3. Dec 8, 2007 #2
    I'm not sure, but...

    Okay, so for the first one, I'd think you'd just apply the division algorithm for polynomials and then choose p so that your remainder is 0. If this way is right, I think there's only one answer.

    For the second one, I'm not sure. Maybe write m as nk for some n and see what happens?
     
  4. Dec 8, 2007 #3
    I'm not sure but I think for #2 you could write m=ka and then x^ka - 1= 0 has ka distinct roots and then write (x^k)^a - 1 = 0 has ka distinct roots. then we have (x^k)^a=1 and then taking to power a we get x^k=1 and then x^k -1 =0 for k distinct roots. I'm not sure I think it might be the right direction.
     
  5. Dec 8, 2007 #4
    First off it should be k roots in F, not K.

    Mystic I don't see how I find p that way, after all I have no way of figuring out the coefficients of the remainder.

    Buzz thanks for your help but that doesn't work.
     
  6. Dec 8, 2007 #5
    Sure you do. It's just normal polynomial long division. I just didn't want to outright say it.

    You know, this seems to be a pattern in my answers lately.
     
  7. Dec 9, 2007 #6
    Um how am I supposed to do long division if p is arbitrary???
     
  8. Dec 9, 2007 #7

    Office_Shredder

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    Can't you just do it like a polynomial over the integers and then at the end see which p gives you a zero remainder?
     
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