Prove that the identity is the only field automorphism of \(\mathbb R\).
Solution
Let \[ \sigma:\RR\to\RR \] be a field automorphism.
1. The automorphism \(\sigma\) fixes every rational number.
A field automorphism preserves \(0\) and \(1\). Hence, for every positive integer \(n\), \[ \sigma(n) = \sigma(\underbrace{1+\cdots+1}_{n\text{ terms}}) =n. \] It follows that \(\sigma(-n)=-n\). If \(m\in\ZZ\) and \(n\in\ZZ\) is nonzero, then \[ \sigma\left(\frac{m}{n}\right) = \frac{\sigma(m)}{\sigma(n)} = \frac{m}{n}. \] Thus \(\sigma(q)=q\) for every \(q\in\QQ\).
2. If \(x>0\), then\[ \sigma(x)>0. \]
Every positive real number is a nonzero square: \[ x=y^2 \] for some \(y\neq0\). Since \(\sigma\) is injective, \[ \sigma(y)\neq0. \] Therefore \[ \sigma(x) = \sigma(y)^2 >0. \]
3. The automorphism \(\sigma\) preserves the usual order on \(\RR\):\[ x<y \quad\Longrightarrow\quad \sigma(x)<\sigma(y). \]
If \(x<y\), then \(y-x>0\). By step 2, \[ \sigma(y)-\sigma(x) = \sigma(y-x) >0. \] Hence \(\sigma(x)<\sigma(y)\).
4. Every \(x\in\RR\) satisfies\[ \sigma(x)=x. \]
Suppose first that \[ x<\sigma(x). \] By density of \(\QQ\) in \(\RR\), choose \(q\in\QQ\) with \[ x<q<\sigma(x). \] Step 3 applied to \(x<q\), together with step 1, gives \[ \sigma(x)<\sigma(q)=q, \] contradicting \(q<\sigma(x)\).
If instead \[ \sigma(x)<x, \] choose \(q\in\QQ\) with \[ \sigma(x)<q<x. \] Step 3 applied to \(q<x\) gives \[ q=\sigma(q)<\sigma(x), \] contradicting \(\sigma(x)<q\).
Both strict inequalities are impossible, so \(\sigma(x)=x\).
5. Therefore\[ \boxed{\sigma=\operatorname{id}_{\RR}}. \]
Step 4 shows that \(\sigma\) fixes every real number.
6. Q.E.D.
Since \(\sigma\) was an arbitrary field automorphism, step 5 proves that the identity is the only one.