# field homomorphisms fix prime subfields

###### Theorem.

Let $F$ and $K$ be fields having the same prime subfield $L$ and $\varphi\colon F\to K$ be a field homomorphism. Then $\varphi$ fixes $L$.

###### 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 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 $\mathbb{Q}$. ∎

Title field homomorphisms fix prime subfields FieldHomomorphismsFixPrimeSubfields 2013-03-22 16:19:54 2013-03-22 16:19:54 Wkbj79 (1863) Wkbj79 (1863) 10 Wkbj79 (1863) Theorem msc 12E99