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Math Help - Prove x^4 - x^2 + 1 is reducible over F_p for all p

  1. #1
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    Prove x^4 - x^2 + 1 is reducible over F_p for all p

    Prove that x^4 - x^2 + 1 is reducible over \mathbb{F}_p for every prime p.

    I know x^4 - x^2 + 1 is the 12th cyclotomic polynomial and know a few properties of such polynomials, but I'm not sure if that really helps here. I can also show that p^2 -1 is divisible by 12 for p>3 but can't seem to make use of that (although have been given a hint that this would be useful).
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  2. #2
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    Quote Originally Posted by Boysilver View Post
    Prove that x^4 - x^2 + 1 is reducible over \mathbb{F}_p for every prime p.

    I know x^4 - x^2 + 1 is the 12th cyclotomic polynomial and know a few properties of such polynomials, but I'm not sure if that really helps here. I can also show that p^2 -1 is divisible by 12 for p>3 but can't seem to make use of that (although have been given a hint that this would be useful).

    == In \mathbb{F}_2\,,\,\,x^4-x^2+1=(x^2-x+1)^2 , and in \mathbb{F}_3\,,\,\,x^4-x^2+1=\left(x^2+1\right)^2 . All the following is based on Gauss' Quadratic Reciprocity Law:

    == Sometimes it helps to factor out these things over the reals: in \mathbb{R}\,,\,\,x^4-x^2+1=\left(x^2+\sqrt{3}\,x+1\right)\left(x^2-\sqrt{3}\,x+1\right) , so in every field in which 3 is a quadratic

    residue we get the above factorization , and this happens in \mathbb{F}_p\,,\,\,with\,\,p=1\!\!\!\pmod 3\,,\,\,or\,\,p=2\!\!\!\pmod 3\,\,and\,\,also\,\,p=3\!\!\!\pmod 4. This last kind of primes

    may be expressed more simply by p=3+4k\,,\,\,k=2\!\!\!\mod 3. Examples: p=11\,,\,23\,,\,47 , etc.


    So the primes that haven't been considered are those that are  1\!\!\!\pmod 4 but not  1\!\!\!\pmod 3 , like  17\,,\,29\,,\,41 , etc., and also primes =3\!\!\!\pmod 4 which

    are =1\!\!\!\pmod 3 , like 19\,,\,43 , etc., but...but then nothing: unless I made some mistake somewhere, the pol. is irreducible over \mathbb{Z}_{17} since (1) 3 is not a quadratic

    residue modulo 17 (and thus the pol. doesn't have a actorization in two quadratic pol's. as above), and (2) it's easy to check (this pol. is an even function...) that

    no element in \mathbb{Z}_{17} is a root of the pol.

    Check this.

    Tonio
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  3. #3
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    I didn't really follow the argument completely, but in any case the polynomial does factor over \mathbb{F}_{17} as (x^2+4x-1)(x^2-4x+1)...
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    Quote Originally Posted by Boysilver View Post
    I didn't really follow the argument completely, but in any case the polynomial does factor over \mathbb{F}_{17} as (x^2+4x-1)(x^2-4x+1)...

    No, it doesn't: (x^2+4x-1)(x^2-4x+1)=x^4-16x^2-1\neq x^4-x^2+1\!\!\!\!\pmod{17}

    Tonio
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  5. #5
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    Sorry, you're right. I meant (x^2+4x-1)(x^2-4x-1).
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  6. #6
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    Quote Originally Posted by Boysilver View Post
    Sorry, you're right. I meant (x^2+4x-1)(x^2-4x-1).

    Now this is true, and it's exactly the case I missed to check! In the decomposition in two quadratics  x^4-x^2+1=\left(x^2+\sqrt{3}\,x+1\right)\left(x^2-\sqrt{3}\,x+1\right) I missed the case when the two free coefficients are -1, so it can also be in general

    x^4-x^2+1=(x^2+ax-1)(x^2-ax-1) , where a^2=-1, and thus we already covered all the primes =1\!\!\!\!\pmod 4 (among them, of course, 17...!)

    Resuming, we have that our pol. is reducible \mod p if:

    0) p=2\,,\,3 ;

    1) p=1\!\!\!\!\pmod 4 ;

    2) p=1\!\!\!\!\pmod 3 ;

    3) p=2\!\!\!\!\pmod 3\mbox{ and also }=3\!\!\!\!\pmod 4 .

    And that covers all the possible cases for a prime, because if p=3\!\!\!\!\pmod 4\mbox{ but }p\neq 2\!\!\!\!\pmod 3\mbox{ then }p=1\!\!\!\!\pmod 3 , which is already covered in case (2), so Q.E.D.!

    Great eye-opener example with p=17, boysilver: thanx.

    Tonio
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