# automorphism problem

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• Jun 24th 2008, 03:58 PM
helpcules
automorphism problem
I need help with the following proof:

http://i27.tinypic.com/dr9hzb.jpg

Thanks
• Jun 24th 2008, 05:45 PM
NonCommAlg
Quote:

Originally Posted by helpcules
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 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) 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 $f(x)=x. \ \ \ \square$
• Jun 24th 2008, 07:38 PM
ThePerfectHacker
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 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 $\text{Aut}(\mathbb{C})$.