Results 1 to 2 of 2

Math Help - Algebra, Problems For Fun (26)

  1. #1
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7

    Algebra, Problems For Fun (26)

    Let F be a field and \sigma \in \text{Aut}(F) with o(\sigma)=2. Prove that A=\{a^{-1}\sigma(a): \ 0 \neq a \in F \} is the solution set of the equation x\sigma(x)=1.


    Remark: In general, for any integer n \geq 1 and any \sigma \in \text{Aut}(F) of order n, the solution set of the equation x\sigma(x)\sigma^2(x) \cdots \sigma^{n-1}(x)=1 is still A. (David Hilbert)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Bruno J.'s Avatar
    Joined
    Jun 2009
    From
    Canada
    Posts
    1,266
    Thanks
    1
    Awards
    1
    Finally I got it!

    Let S be the solution set. Clearly A \subset S; we want to show A \supset S.

    Let x \in S; suppose x \neq -1.
    Clearly then y=(1+x)^{-1}\sigma(1+x) \in A; but we have

    y(1+x)=\sigma(1+x) = 1+\sigma(x)
    y(1+x)x=x+x\sigma(x) = x+1

    hence yx=1; i.e. y=x^{-1} \in A, and clearly A is closed under taking inverses so x \in A.

    To show -1 \in A (clearly -1 \in S), take any y \in F such that z=y-\sigma(y) \neq 0; then

    \sigma(z) = \sigma(y)-y = -z

    hence z^{-1}\sigma(z) = -1 \in A.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Algebra 2 problems.
    Posted in the Algebra Forum
    Replies: 3
    Last Post: July 29th 2009, 11:50 AM
  2. Algebra, Problems For Fun (21)
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: June 22nd 2009, 07:57 AM
  3. Algebra, Problems For Fun (22)
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: June 21st 2009, 12:59 AM
  4. Algebra, Problems For Fun (13)
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: June 5th 2009, 03:53 PM
  5. Need Help With All these Algebra Problems.
    Posted in the Algebra Forum
    Replies: 10
    Last Post: November 26th 2006, 06:42 PM

Search Tags


/mathhelpforum @mathhelpforum