0

The following question is from my Field Theory assignments.

Prove that $Aut_{\mathbb{Q} } { \mathbb{R}} $ is identity group.

I took $f:\mathbb{R} \to \mathbb{R}$ be an automorphism. Now which result I should use to prove that all the automorphism which are $\mathbb{Q}$ -module homomorphisms are identity.

Let $f'$ be such an automorphism and q belongs to $\mathbb{Q}$. So, $f'(q) = qf'(1) =q$.but how to prove that $f'(q) =1$ .

Kindly guide.

Paul Frost
  • 76,394
  • 12
  • 43
  • 125

3 Answers3

1

First of all, note that the identity is not the map that maps alls real numbers to $1$; that is not even a homomorphism. Instead, it is the map that maps every real number to itself.

Hint to prove this:

  • $f$ is increasing. Hint: consider $b > a$ and look at $f(\left(\sqrt{b-a}\right)^2$).
  • $f$ is continuous. Hint: use (only) that $f$ is an increasing bijection.
  • $f = \mathop{id}_{\mathbb R}$. Hint: $f|_{\mathbb Q} = \mathop{id}_{\mathbb Q}$ and ${\mathbb Q}$ is dense in ${\mathbb R}$.
Magdiragdag
  • 15,049
0

For every $x$ $f(x)=f(x\cdot1)=f(x)\cdot(1)$. So if $f$ is not constant $=0$, $f(1)=1$.

Thomas
  • 7,470
0

First off, note that it would be enough to show that any such automorphism must be continuous. (and then since any element of $\Bbb R$ is a limit of elements in $\Bbb Q$ you are done)

To do this, note that for $a\in\Bbb R$, we have $f(a)=f(\sqrt a ^2)=f(\sqrt a)^2\geq0$, so $f$ must be increasing, and its continuity follows because if $f(a^+)>f(a^-)$ then the elements in the open interval $(f(a^-),f(a^+))$ don't appear in the image of $f$ (since it is strictly increasing).

We thus get that $f(a^+)=f(a^-)$ for any $a\in\Bbb R$ and thus $f$ is continuous, proving the claim using $f(q_n)=q_n\to r$ for $q_n$ a sequence of rationals converging to $r$.

NL1992
  • 972