Proof.
Without loss of generality, it will be assumed that
$L$ is either
$\mathbb{Q}$ or
$\mathbb{Z}/c\mathbb{Z}$ .
Since $\varphi$ is a field homomorphism, $\varphi(0)=0$ , $\varphi(1)=1$ , and, for every $x \in F$ , $\varphi(-x)=-\varphi(x)$ .
Let $n \in \mathbb{Z}$ and $c$ be the characteristic of $F$ . Then
| $\varphi(n)$ |
$\equiv \varphi(\operatorname{sign}(n)|n|) \operatorname{mod} c$ , where $\operatorname{sign}$ denotes the signum function |
| |
$\displaystyle \equiv \operatorname{sign}(n)\varphi(|n|) \operatorname{mod} c$ |
| |
$\displaystyle \equiv \operatorname{sign}(n)\varphi\left(\sum_{j=1}^{|n|} 1\right) \operatorname{mod} c$ |
| |
$\displaystyle \equiv \operatorname{sign}(n)\sum_{j=1}^{|n|} \varphi(1) \operatorname{mod} c$ |
| |
$\displaystyle \equiv \operatorname{sign}(n)\sum_{j=1}^{|n|} 1 \operatorname{mod} c$ |
| |
$\equiv \operatorname{sign}(n)|n| \operatorname{mod} c$ |
| |
$\equiv n \operatorname{mod} c$ . |
This completes the proof in the case that $c$ is prime.
Now consider $c=0$ . Let $x \in \mathbb{Q}$ . Then there exist $a,b \in \mathbb{Z}$ with $b>0$ such that $\displaystyle x=\frac{a}{b}$ . Thus, $\displaystyle b\varphi(x)=\sum_{j=1}^b\varphi\left(\frac{a}{b}\right)=\varphi\left(\sum_{j=1}^b \frac{a}{b}\right)=\varphi(a)=a$ . Therefore, $\displaystyle \varphi(x)=\frac{a}{b}=x$ . Hence, $\varphi$ fixes $\mathbb{Q}$ . 
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