Results 1 to 3 of 3

Math Help - automorphism problem

  1. #1
    Newbie
    Joined
    Jun 2008
    Posts
    1

    automorphism problem

    I need help with the following proof:



    Thanks
    Last edited by helpcules; June 24th 2008 at 11:19 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by helpcules View Post
    I need help with the following proof:

    show that the only ring automorphism of the real numbers is the identity mapping

    Thanks
    suppose f \in \text{Aut}(\mathbb{R}). obviously: f(0)=0, \ f(1)=1, \ f(-1)=-1. thus f(n)=n, \ \forall n \in \mathbb{Z}.

    hence for all m \in \mathbb{N}, \ n \in \mathbb{Z}: \ mf(\frac{n}{m})=f(n)=n. thus: f(\frac{n}{m})=\frac{n}{m}, i.e. f(r)=r,\ \forall r \in \mathbb{Q}.

    next we show that f is (strictly) increasing: let x, y be two real numbers and x<y. then for

    some z \neq 0 : y-x=z^2 thus: f(y)-f(x)=f(z^2)=(f(z))^2 > 0. so: f(x) < f(y), i.e. f is

    (strictly) increasing.

    now let x \in \mathbb{R}, and n \in \mathbb{N}. choose r \in \mathbb{Q} such that \frac{-1}{n} < x - r < \frac{1}{n}. call this (1). since f is

    increasing, by (1): \ \frac{-1}{n}=f(\frac{-1}{n}) < f(x-r)<f(\frac{1}{n})=\frac{1}{n}. but f(x-r)=f(x)-r. thus:

    \frac{-1}{n} < f(x) - r < \frac{1}{n}. call this (2). now (1) and (2) give us: |f(x)-x| < \frac{2}{n}. since n is arbitrary,

    we must have f(x)=x. \ \ \ \square
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Here is another way. If we can show f is continous then let x be any real number and x_n be a sequence of rationals converging to x. Then by continuity x_n\to x \implies \lim f(x_n) = f(x) \implies x = f(x). Let us show that f is continous WLOG at x_0>0. Let \epsilon > 0 and choose \delta = \epsilon. Then if 0<x<\delta we can find a rational r so that 0 < x < r < \delta and then |f(x) - f(0)| = f(x)< f(r) = r < \epsilon by using the strict property as NonCommAlg shown.

    A more interesting question is to find \text{Aut}(\mathbb{C}).
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. automorphism
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: May 18th 2009, 09:51 PM
  2. Automorphism
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: January 21st 2009, 12:48 PM
  3. AutoMorphism
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: July 31st 2008, 06:28 AM
  4. Automorphism
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: March 19th 2008, 08:03 AM
  5. automorphism
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: March 1st 2008, 08:11 PM

Search Tags


/mathhelpforum @mathhelpforum