Is an automorphism of the field of real numbers the identity map?

Question

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.

Answer 1

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.

Answer 2

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 $\mathbb R$.

Answer 3

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 $\phi(a) \leq q \leq a$.

Answer 4

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".)

Answer 5

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.

Answer 6

Let $f:\mathbb R\to\mathbb R$ be a field automorphism. Since $f$ preserves addition, $f(0+0)=f(0)+f(0)$, and so $f(0)=0$. Similarly, since $f$ preserves multiplication, $f(1\cdot 1)=f(1)\cdot f(1)$, and so either $f(1)=1$ or $f(1)=0$; the second possibility is ruled out by the fact that $f$ is injective, and $f(0)=0$.

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 $x-y>0$, so $f(x)-f(y)=f(x-y)>0$, so $f(x)>f(y)$. Hence, $f$ preserves the usual ordering of the reals.

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.

Answer 7

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.