FieldHomomorphismsFixPrimeSubfields 2006-10-15 01:09:40 2006-10-18 12:16:14 Theorem field homomorphism \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} \usepackage{psfrag} \usepackage{graphicx} \usepackage{amsthm} \usepackage{xypic} \newtheorem*{thm*}{Theorem} \begin{thm*} 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 $L$. \end{thm*} \begin{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 \begin{center} \begin{tabular}{rl} $\varphi(n)$ & $\equiv \varphi(\operatorname{sign}(n)|n|) \operatorname{mod} c$, where $\operatorname{sign}$ denotes the signum function \\ & $\displaystyle \equiv \operatorname{sign}(n)\varphi(|n|) \operatorname{mod} c$ \\ & $\displaystyle \equiv \operatorname{sign}(n)\varphi\left(\sum_{j=1}^{|n|} 1\right) \operatorname{mod} c$ \\ & $\displaystyle \equiv \operatorname{sign}(n)\sum_{j=1}^{|n|} \varphi(1) \operatorname{mod} c$ \\ & $\displaystyle \equiv \operatorname{sign}(n)\sum_{j=1}^{|n|} 1 \operatorname{mod} c$ \\ & $\equiv \operatorname{sign}(n)|n| \operatorname{mod} c$ \\ & $\equiv n \operatorname{mod} c$. \end{tabular} \end{center} This \PMlinkescapetext{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 $\mathbb{Q}$. \end{proof}