Any Automorphism of the Field of Real Numbers Must be the Identity Map

Field theory problems and solution in abstract algebra

Problem 507

Prove that any field automorphism of the field of real numbers $\R$ must be the identity automorphism.

 
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Proof.

We prove the problem by proving the following sequence of claims.

Let $\phi:\R \to \R$ be an automorphism of the field of real numbers $\R$.

  1. Claim 1. For any positive real number $x$, we have $\phi(x)>0$.
  2. Claim 2. For any $x, y\in \R$ such that $x>y$, we have $\phi(x) > \phi(y)$.
  3. Claim 3. The automorphism $\phi$ is the identity on positive integers.
  4. Claim 4. The automorphism $\phi$ is the identity on rational numbers.
  5. Claim 5. The automorphism $\phi$ is the identity on real numbers.

Let us now start proving the claims.
Let $\phi:\R \to \R$ be an automorphism of the field of real numbers $\R$.

Claim 1. For any positive real number $x$, we have $\phi(x)>0$.

Since $x$ is a positive real number, we have $\sqrt{x}\in \R$ and
\[\phi(x)=\phi\left(\sqrt{x}^2\right)=\phi(\sqrt{x})^2 \geq 0.\]

Note that since $\phi(0)=0$ and $\phi$ is bijective, $\phi(x)\neq 0$ for any $x\neq 0$.
Thus, it follows that $\phi(x) > 0$ for each positive real number $x$.
Claim 1 is proved.

Claim 2. For any $x, y\in \R$ such that $x>y$, we have $\phi(x) > \phi(y)$.

Since $x > y$, we have $x-y > 0$ and it follows from Claim 1 that
\[0<\phi(x-y)=\phi(x)-\phi(y).\] Hence, $\phi(x)> \phi(y)$.

Claim 3. The automorphism $\phi$ is the identity on positive integers.

Let $n$ be a positive integer. Then we have
\begin{align*}
\phi(n)=\phi(\underbrace{1+1+\cdots+1}_{\text{$n$ times}})=\underbrace{\phi(1)+\phi(1)+\cdots+\phi(1)}_{\text{$n$ times}}=n
\end{align*}
since $\phi(1)=1$.

Claim 4. The automorphism $\phi$ is the identity on rational numbers.

Any rational number $q$ can be written as $q=\pm m/n$, where $m, n$ are positive integers.
Then we have
\begin{align*}
\phi(q)=\phi\left(\, \pm \frac{m}{n} \,\right)=\pm \frac{\phi(m)}{\phi(n)}=\pm \frac{m}{n}=q,
\end{align*}
where the third equality follows from Claim 3.

Claim 5. The automorphism $\phi$ is the identity on real numbers.

In this claim, we finish the proof of the problem.

Let $x$ be any real number.
Seeking a contradiction, assume that $\phi(x)\neq x$.

There are two cases to consider:
\[x < \phi(x) \text{ or } x > \phi(x).\]

First, suppose that $x < \phi(x)$. Then there exists a rational number $q$ such that \[x< q < \phi(x).\] Then we have \begin{align*} \phi(x) &< \phi(q) && \text{by Claim 2 since $x < q$}\\ &=q && \text{by Claim 4 since $q$ is rational}\\ &<\phi(x) && \text{by the choice of $q$}, \end{align*} and this is a contradiction. Next, consider the case when $x > \phi(x)$.
There exists a rational number $q$ such that
\[\phi(x) < q < x.\] Then by the same argument as above, we have \[\phi(x) < q =\phi(q) < \phi(x),\] which is a contradiction. Thus, in either case we reached a contradiction, and hence we must have $\phi(x)=x$ for all real numbers $x$. This proves that the automorphism $\phi: \R \to \R$ is the identity map.


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