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. It can be easily proven by induction that for all $n\in\mathbb N$, if $a_1,\dots,a_n$ is a list of real numbers, then $$ f\left(\sum_{i=1}^{n}a_i\right)=\sum_{i=1}^{n}f(a_i) \, ; $$ 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 $f$ fixes all integers. 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 $f$ fixes all the rational numbers.

So far, we have not used any deep results about the real numbers, and indeed, with some very mild adjustments, the above argument shows that any homomorphism of fields of characteristic zero preserves $\mathbb Q$. For the final stage of the proof, however, we need to use the following facts (which rest upon the order-theoretic properties of $\mathbb R$, and in particular its Dedekind-completeness):

• Every positive real $x$ has a 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 we can find a rational number.

Let $x>0$ and let $t>0$ be such that $t^2=x$. Then $f(x)=f(t)f(t)>0$. 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.

Finally, let $z$ be a real number. 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, $f$ fixes all real numbers, 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.