field homomorphisms fix prime subfields

Theorem.

Let $F$ and $K$ be fields having the same prime subfield $L$ and $\phi \mathrm{:}F\mathrm{\to }K$ be a field homomorphism. Then $\phi$ fixes $L$.

Proof.

Without loss of generality, it will be assumed that $L$ is either $\mathbb{Q}$ or $\mathbb{Z}/c\mathbb{Z}$.

Since $\phi$ is a field homomorphism, $\phi (0)=0$, $\phi (1)=1$, and, for every $x\in F$, $\phi (-x)=-\phi (x)$.

Let $n\in \mathbb{Z}$ and $c$ be the characteristic of $F$. Then

$\phi (n)$	$\equiv \phi (\mathrm{sign}(n)|n|)\mathrm{mod}c$, where $\mathrm{sign}$ denotes the signum function
	$\equiv \mathrm{sign}(n)\phi (|n|)\mathrm{mod}c$
	$\equiv \mathrm{sign}(n)\phi \left({\displaystyle \sum _{j=1}^{|n|}}1\right)\mathrm{mod}c$
	$\equiv \mathrm{sign}(n){\displaystyle \sum _{j=1}^{|n|}}\phi (1)\mathrm{mod}c$
	$\equiv \mathrm{sign}(n){\displaystyle \sum _{j=1}^{|n|}}1\mathrm{mod}c$
	$\equiv \mathrm{sign}(n)|n|\mathrm{mod}c$
	$\equiv n\mathrm{mod}c$.

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 $x={\displaystyle \frac{a}{b}}$. Thus, $b\phi (x)={\displaystyle \sum _{j=1}^b}\phi \left({\displaystyle \frac{a}{b}}\right)=\phi \left({\displaystyle \sum _{j=1}^b}{\displaystyle \frac{a}{b}}\right)=\phi (a)=a$. Therefore, $\phi (x)={\displaystyle \frac{a}{b}}=x$. Hence, $\phi$ fixes $\mathbb{Q}$. ∎

Title	field homomorphisms fix prime subfields
Canonical name	FieldHomomorphismsFixPrimeSubfields
Date of creation	2013-03-22 16:19:54
Last modified on	2013-03-22 16:19:54
Owner	Wkbj79 (1863)
Last modified by	Wkbj79 (1863)
Numerical id	10
Author	Wkbj79 (1863)
Entry type	Theorem
Classification	msc 12E99