Results 1 to 2 of 2

Math Help - Splitting field of x^p-x-a, a neq b^p-b

  1. #1
    Junior Member
    Joined
    Aug 2012
    From
    Sweden
    Posts
    37
    Thanks
    1

    Splitting field of x^p-x-a, a neq b^p-b

    Hi again guys. I have the following assignment:

    Let k be a field of characteristic p and a in k an element which cannot be written b^p-b for any b in k. Compute the Galois group of the splitting field of f(x)=x^p-x-a.

    In a previous thread here you helped me to determine that the above polynomial either splits in k or is irreducible there. It is shown here Math Forum - Ask Dr. Math how to do. The ansatz is simply what you guys taught me before:

    Suppose [; \alpha ;] is a root, i.e. [; \alpha^p - \alpha - a = 0 ;]. Then so is [; \alpha+j ;], because [; (\alpha+j)^p-(\alpha+j)-a = \alpha^p - \alpha + j^p - j - a = \alpha^p - \alpha - a = 0 ;], hence yada yada.

    My problem is with two things:
    1) Alright, suppose [; \alpha ;] is a root... well, it's not. Wouldn't it follow from Fermat's little theorem that [; \alpha^p - \alpha - a = -a ;], which is not zero?

    2) Fine, let's accept that [; \alpha ;] is indeed a root, that my objection in (1) is wrong. Then why is j^p-j = 0? Is it because j belongs to k while [; \alpha ;] does not?

    Thanks in advance!


    (Bonus question: Determine all intermediate fields of [; Q(\sqrt{2+\sqrt{2}}) ;]. The intermediate fields correspond to the subgroups of the Galois group, in this case the cyclic group on four elements, so wouldn't there be only one intermediate field, corresponding to the (up to isomorphism) unique group of two elements, namely [; Q(\sqrt{2}) ;] ?)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jan 2008
    Posts
    588
    Thanks
    87

    Re: Splitting field of x^p-x-a, a neq b^p-b

    Quote Originally Posted by spudwish View Post
    My problem is with two things:
    1) Alright, suppose [; \alpha ;] is a root... well, it's not. Wouldn't it follow from Fermat's little theorem that [; \alpha^p - \alpha - a = -a ;], which is not zero?
    Fermat's little theorem only holds for integers. This root \alpha is something that lives in the splitting field of f(x), and is probably not an integer, or even an element of Z/pZ. However, since you know there exists a splitting field for every polynomial f(x), you can assume the existence of a root.


    2) Fine, let's accept that [; \alpha ;] is indeed a root, that my objection in (1) is wrong. Then why is j^p-j = 0? Is it because j belongs to k while [; \alpha ;] does not?
    In this instance j is necessary an integer (or more precisely, an element of Z/pZ). Fermat's little theorem holds in this case.

    (Bonus question: Determine all intermediate fields of [; Q(\sqrt{2+\sqrt{2}}) ;]. The intermediate fields correspond to the subgroups of the Galois group, in this case the cyclic group on four elements, so wouldn't there be only one intermediate field, corresponding to the (up to isomorphism) unique group of two elements, namely [; Q(\sqrt{2}) ;] ?)
    Yes.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Splitting Field
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: April 10th 2011, 11:57 AM
  2. Splitting Field of a Polynomial over a Finite Field
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: April 1st 2011, 03:45 PM
  3. Splitting field
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: September 2nd 2009, 11:33 PM
  4. Splitting Field
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: August 26th 2009, 09:14 AM
  5. Field of char p>0 & splitting field
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: April 22nd 2009, 12:20 AM

Search Tags


/mathhelpforum @mathhelpforum