Is an automorphism of the field of real numbers the identity map? Is an automorphism of the field of real numbers $\mathbb{R}$ the identity map?
If yes, how can we prove it?
Remark An automorphism of $\mathbb{R}$ may not be continuous.
 A: Here's a detailed proof based on the hint given by lhf.
Let $\phi$ be an automorphism of the field of real numbers.
Let $x \gt 0$ be a positive real number.
Then there exists $y$ such that $x = y^2$.
Hence $\phi(x) = \phi(y)^2 \gt 0$.
If $a \lt b$, then $b - a \gt 0$.
Hence $\phi(b) - \phi(a) = \phi(b - a) \gt 0$ by the above.
Hence $\phi(a) \lt \phi(b)$.
This means that $\phi$ is strictly increasing.
If $n$ is a natural number, it can be written in the form $1 + \ldots + 1$, so $\phi(n) = n$. Now, any rational number is of the form $r = (a - b)c^{-1}$, for $a, b, c$ natural numbers, so it follows that $\phi(r) = r$ for any rational number.
Let $x$ be a real number.
Let $r, s$ be rational numbers such that $r \lt x \lt s$.
Then $r \lt \phi(x) \lt s$.
Since $s - r$ can be arbitrarily small, $\phi(x) = x$.
This completes the proof.
A: Hint: Let $\phi$ be a field automorphism of $\mathbb R$. Then prove: 


*

*$\phi$ sends positive numbers to positive numbers

*$\phi$ is increasing

*$\phi$ is continuous

*$\phi$ is the identity on $\mathbb Q$

*$\phi$ is the identity on $\mathbb R$.
A: I have never liked the proofs of this that use analysis more than necessary.  Once you get that $\phi$ is order-preserving and the identity map on the rationals, take an arbitrary real number $a$.  If $\phi(a) \neq a$, then there is a rational $q \in \mathbb{Q}$ between $a$ and $\phi(a)$.  If $a \leq q \leq \phi(a)$, then $\phi(a) \leq \phi(q) = q$, so $\phi(a) \leq q$ and $q \leq \phi(a)$, so $\phi(a) = q = \phi(q)$, so $a = q$, contradicting the fact that $\phi(a) \neq a$.  Similarly if $\phi(a) \leq q \leq a$.    
A: I think the correct way to use density of rationals in this problem is this: 
Let $a$ be a real number and suppose $\phi(a) \neq a$. Then, there exists a rational number $q$ such that $a<q<\phi(a)$. Hence, $q < \phi(a)$ and $q>\phi(a)$. This is a contradiction since $\phi$ is a bijection.
A: Since $f$ preserves addition, $f(0+0)=f(0)+f(0)$, and so $f(0)=0$. Since $f$ preserves mutliplication, for all $x\in\mathbb R$ we have $f(x)=f(1\cdot x)=f(1)\cdot f(x)$. Therefore, either $f(1)=1$ or $f$ is identically $0$; the second possibility is ruled out by the fact that $f$ maps $\mathbb R$ onto $\mathbb R$. Let $n\in\mathbb N$. It can be easily proven by induction that if $a_1,\dots,a_n$ is any list of numbers, then
$$
f\left(\sum_{i=1}^{n}a_i\right)=\sum_{i=1}^{n}f(a_i) \, ,
$$
and so in particular
$$
f(n)=f\left(\sum_{i=1}^{n}1\right)=\sum_{i=1}^{n}f(1)=\sum_{i=1}^{n}1=n \, .
$$
Since $f(n)+f(-n)=f(0)=0$, we have $f(-n)=-f(n)=-n$. This shows that $f$ is the identity map on $\mathbb Z$. Let $a,b\in\mathbb Z$, with $b\neq0$. We have $f(1/b)\cdot f(b)=f(1)$, so $f(1/b)=f(1)/f(b)=1/b$. Hence $f(a/b)=f(a)\cdot f(1/b)=a/b$. This shows that $f$ equals the identity on $\mathbb Q$.
For the final stage of the proof, we utilise three facts that rest upon $\mathbb R$ being a complete ordered field:

*

*Every positive real has one positive square root.

*The square of a nonzero real number is positive.

*$\mathbb Q$ is dense in $\mathbb R$: between any two distinct real numbers there is a rational number in between them.

Let $x>0$. Then $f(x)=f(\sqrt x)f(\sqrt x)>0$. This means that if $x>y$, then as $x-y>0$,
$$
f(x-y)=f(x)-f(y)>0\implies f(x)>f(y) \, .
$$
Therefore, $f$ preserves order. Now, let $z$ be irrational. If it were the case that $f(z)<z$, then there would an $r\in\mathbb Q$ such that $f(z)<r<z$. But as $f(r)=r$, this contradicts the fact that $f$ is order-preserving. The case $f(z)>z$ similarly leads to a contradiction. Hence $f(z)=z$ and so $f$ is the identity on $\mathbb R\setminus\mathbb Q$, completing the proof.
A: For related but slightly stronger results, see $\S$ 16.7 of these field theory notes.
Highlights:
(i) Every Archimedean ordered field $K$ admits a unique homomorphism of ordered fields
$K \hookrightarrow \mathbb{R}$.
(ii) Let $(F,<)$ be an ordered field in which every positive element is a square (e.g. any real-closed field, e.g. $\mathbb{R}$).  Then the ordering $<$ is unique, so every homomorphism of fields between two such fields is necessarily a homomorphism of ordered fields.  Thus:
The identity map on $\mathbb{R}$ is the unique field homomorphism from $\mathbb{R}$ to $\mathbb{R}$: "$\mathbb{R}$ is strongly rigid".
(In the Lemma that occurs just before the "Main Theorem on Archimedean Ordered Fields" -- currently numbered Lemma 192 and on p. 106, but both of these are subject to change -- where it says "topological rings", I think it should say "Hausdorff topological rings".)
A: Here is a quick formulation of the proof which bypasses explicitly showing the ordering is preserved.  Let $f:\mathbb{R}\to\mathbb{R}$ be an automorphism.  We know $f$ fixes all rational numbers.  Suppose $f$ is not the identity; say $f(x)\neq x$ for some $x$.  We may assume $f(x)>x$ (if $f(x)<x$, then $f(-x)>-x$, so we can replace $x$ by $-x$).  Now let $q$ be a rational number such that $x<q<f(x)$ and let $y=\sqrt{q-x}$.  Note that $$f(q)=f(x+y^2)=f(x)+f(y)^2\geq f(x).$$ But $f(q)=q$ since $q$ is rational, so this contradicts the fact that $q<f(x)$.  Thus $f$ must be the identity.
