Field Automorphisms of the Real Numbers

Here, we try to solve the following problem (and some related ones):

Problem. Prove that if $f\mathbb{:R \rightarrow R}$ such that $f(x + y) = f(x) + f(y)$ and $f(xy) = f(x)f(y)$ , then either $f(x) = 0$ or $f(x) = x$ .

Solution. We assume there exists $u$ such that $f(u) \neq 0$ . Then we have $f(q) = q$ for $q\mathbb{\in Q}$ since $f$ satisfies Cauchy’s functional equation. (That is an elementary proof, done for example in Art of Problem Solving’s algebra class.)

The rest is suggested by abstract algebra – Is an automorphism of the field of real numbers the identity map? – Mathematics Stack Exchange, where I now do the steps.

Well, first, if $f(x) = 0$ but $x \neq 0$ , then $f(x)f\left( \frac{1}{x} \right) = f(1) = 1 = 0$ , contradiction.

Then, if $x > 0$ , then $\sqrt{x}\mathbb{\in R}$ and $f\left( \sqrt{x} \right)f\left( \sqrt{x} \right) = f(x) = \left( f\left( \sqrt{x} \right) \right)^{2}$ , and if this is zero then $x = 0$ , contradiction, so $f(x) > 0$ .

Then, if $x > y$ , $x = z + y,z > 0$ , $f(x) = f(z + y) = f(z) + f(y) > f(y)$ . So $f$ is increasing.

Then, we show that $\lim_{x \rightarrow 0}{f(x)} = 0$ . For any $\epsilon > 0$ , we may select rational $q > 0$ with $q < \epsilon$ ; let $\delta = q$ , then

(The absolute value comes from $f( - x) = - f(x)$ .)

Then, we show that $f$ is continuous. Indeed, we have

which is exactly what we showed above.

Then, it follows quickly from Cauchy’s functional equation for continuous functions (use the fact that irrationals are arbitrarily approximated by rationals.)

Corollary. The only field automorphism of $\mathbb{R}$ is the identity, where an automorphism must have $f(1) = 1$ .

Derivation of Cauchy’s Functional Equation Solution for Continuous Functions. We have $f(q) = cq$ clearly for $q\mathbb{\in Q}$ . Assume $f$ is continuous. Let’s derive $f(x) = cx$ .

Let $a$ be irrational. We have

$\displaystyle 0 < |x - a| < \delta \rightarrow \left| f(x) - f(a) \right| < \epsilon.$

We can choose $x$ rational. (This requires arbitrary-precision approximation of irrationals by rationals.) Then

$\displaystyle 0 < |x - a| < \delta \rightarrow \left| x - f(a) \right| < \epsilon.$

Then

$\displaystyle \left| a - f(a) \right| < \delta + \epsilon.$

But we can choose $\delta,\epsilon$ arbitrarily small, so $a = f(a)$ , as desired.

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Question. Let $G$ be a subfield of $F$ . If the only automorphism of $F$ is the identity, can we show that the only automorphism of $G$ is the identity?

Solution. Assume we had two automorphisms $a_{1},a_{2}$ of $G$ . We can get contradiction if we can enlarge an automorphism of $G$ into one of $F$ .

Well, $F$ is a vector space over $G$ . By linear algebra theorems, this vector space must have a basis, say $B = \{ e_{1},\ldots\}$ . Given an automorphism $g$ of $G$ , we simply map $c_{1}e_{1} + \ldots \rightarrow g\left( c_{1} \right)e_{1} + \ldots$ . Since the basis elements are independent, the automorphisms generated in this way are different for different choices of $g$ . (In fact, we can say this for arbitrary homomorphisms into any field as well.)

First written February 9, 2021.

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