Automorphisms of the Rational Field and the Real Field

Proposition 1. The only automorphism of the rational field $\mathbb{Q}$ is the identity automorphism.

Proof: Let $f$ be an automorphism of the field $\mathbb{Q}$. Then

$$
f(1)=f(1\cdot 1)=f(1)^2.
$$

Hence

$$
f(1)=0 \quad \text{or} \quad 1.
$$

But since $f$ is injective, we have

$$
1\notin \ker f,
$$

so

$$
f(1)\neq 0.
$$

Therefore,

$$
f(1)=1.
$$

Now for any integer $n\in\mathbb{Z}$, we have

$$
f(n)=f(1+1+\cdots+1)=nf(1)=n.
$$

For any $x\in\mathbb{Q}$, there exist $m,n\in\mathbb{Z}$ with $n\neq 0$ such that

$$
x=\frac{m}{n}.
$$

Then

$$
m=f(m)=f\left(n\cdot \frac{m}{n}\right)=f(n),f\left(\frac{m}{n}\right)=n,f\left(\frac{m}{n}\right).
$$

Hence

$$
f\left(\frac{m}{n}\right)=\frac{m}{n}.
$$

So for every $x\in\mathbb{Q}$, we have

$$
f(x)=x.
$$

Therefore $f$ is the identity automorphism.

Proposition 2. The only automorphism of the real field $\mathbb{R}$ is the identity automorphism.

Proof: Let $f$ be an automorphism of the field $\mathbb{R}$.

By Proposition 1, the restriction of $f$ to the rational field $\mathbb{Q}$ is the identity automorphism.

Now let $x,y\in\mathbb{R}$ with $x\ge y$. Then

$$
f(x)-f(y)=f(x-y)=f\bigl(\sqrt{x-y}\bigr)^2\ge 0.
$$

Thus $f$ is monotone increasing.

Therefore, for any $r\in\mathbb{R}$, there exist two sequences of rational numbers $\{a_n\}$ and $\{b_n\}$ such that

$$
a_n\le r\le b_n
$$

for every $n\in\mathbb{N}^*$, and

$$
\lim_{n\to\infty} a_n=r=\lim_{n\to\infty} b_n.
$$

Since $f$ is increasing and fixes every rational number, we have

$$
a_n=f(a_n)\le f(r)\le f(b_n)=b_n.
$$

Letting $n\to\infty$, we obtain

$$
f(r)=r.
$$

Hence $f$ is the identity automorphism.

The cover image of this article was taken at Lake Baikal in Russia.