14.4: Algebraic lemma

The following lemma was used in the proof of Proposition 14.3.1.

Lemma $\PageIndex{1}$

Assume $f:\mathbb{R}\to\mathbb{R}$ is a function such that for any $x,y\in\mathbb{R}$ we have

1. $f(1)=1$,
2. $f(x+y)=f(x)+f(y)$,
3. $f(x\cdot y)=f(x)\cdot f(y)$.

Then $f$ is the identity function; that is, $f(x)=x$ for any $x\in \mathbb{R}$.

Note that we do not assume that $f$ is continuous.

A function $f$ satisfying these three conditions is called a field automorphism. Therefore, the lemma states that the identity function is the only automorphism of the field of real numbers. For the field of complex numbers, the conjugation $z\mapsto \bar{z}$ (see Section 14.3) gives an example of a nontrivial automorphism.

Proof

By (b) we have

By (a),

$f(0)+1=1;$

whence

$$f(0)=0.$$

Applying (b) again, we get that

$0=f(0)=f(x)+f(-x).$

Therefore,

$$f(-x)=-f(x) \ \ \ \ \text{for any} \ \ \ \ x\in \mathbb{R}.$$

Applying (b) recurrently, we get that

$\begin{array} {l} {f(2) = f(1) + f(1) = 1 + 1 = 2;} \\ {f(3) = f(2) + f(1) = 2 + 1 = 3;} \\ {\ \ \ \ ...} \end{array}$

Together with 14.4.2, the latter implies that

$f(n)=n \ \ \ \text{for any integer}\ \ \ n.$

By (c)

Therefore

$f(\dfrac{m}{n})=\dfrac{m}{n}$

for any rational number $\dfrac{m}{n}$.

Assume $a\ge 0$. Then the equation $x\cdot x=a$ has a real solution $x = \sqrt{a}$. Therefore, $[f(\sqrt{a})]^2=f(\sqrt{a})\cdot f(\sqrt{a})=f(a)$. Hence $f(a)\ge 0$. That is,

$$a\ge 0 \ \ \ \Longrightarrow\ \ \ f(a)\ge 0.$$

Applying 14.3.2, we also get

$$a\le 0 \ \ \ \Longrightarrow \ \ \ f(a) \le 0.$$

Now assume $f(a)\ne a$ for some $a\in\mathbb{R}$. Then there is a rational number $\dfrac{m}{n}$ that lies between $a$ and $f(a)$; that is, the numbers

have opposite signs.

By 14.4.3

$\begin{array} {rcl} {y + \dfrac{m}{n}} & = & {f(a) =} \\ {} & = & {f(x + \dfrac{m}{n}) =} \\ {} & = & {f(x) + f(\dfrac{m}{n}) =} \\ {} & = & {f(x) + \dfrac{m}{n};} \end{array}$

that is, $f(x)=y$. By 14.4.4 and 14.4.5 the values $x$ and $y$ cannot have opposite signs — a contradiction.