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.

Solution 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.

Solution 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 $\mathbb R$.

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