field homomorphism FieldHomomorphism 2003-08-29 16:47:32 2007-01-04 06:47:23 Definition ring homomorphism field homomorphism field isomorphism field map % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsthm} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here \newtheorem{thm}{Theorem} \newtheorem*{defn}{Definition} \newtheorem{prop}{Proposition} \newtheorem*{lemma}{Lemma} \newtheorem{cor}{Corollary} % Some sets \newcommand{\Nats}{\mathbb{N}} \newcommand{\Ints}{\mathbb{Z}} \newcommand{\Reals}{\mathbb{R}} \newcommand{\Complex}{\mathbb{C}} \newcommand{\Rats}{\mathbb{Q}} Let $F$ and $K$ be fields. \begin{defn} A {\em field homomorphism} is a function $\psi\colon F \to K$ such that: \begin{enumerate} \item $\psi(a+b) = \psi(a)+\psi(b)$ for all $a,b \in F$ \item $\psi(a\cdot b) = \psi(a) \cdot \psi(b)$ for all $a,b \in F$ \item $\psi(1)=1,\quad \psi(0)=0$ \end{enumerate} If $\psi$ is injective and surjective, then we say that $\psi$ is a \emph{field isomorphism}. \end{defn} \begin{lemma} Let $\psi\colon F\to K$ be a field homomorphism. Then $\psi$ is injective. \end{lemma} \begin{proof} Indeed, if $\psi$ is a field homomorphism, in particular it is a ring homomorphism. Note that the kernel of a ring homomorphism is an ideal and a field $F$ only has two ideals, namely $\{0\}, F$. Moreover, by the definition of field homomorphism, $\psi(1)=1$, hence $1$ is not in the kernel of the map, so the kernel must be equal to $\{0\}$. \end{proof} {\bf Remark}: For this reason the terms ``field homomorphism'' and ``field monomorphism'' are synonymous. Also note that if $\psi$ is a field monomorphism, then $$\psi(F)\cong F, \quad \psi(F)\subseteq K$$ so there is a ``copy'' of $F$ in $K$. In other words, if $$\psi\colon F\to K$$ is a field homomorphism then there exist a subfield $H$ of $K$ such that $H\cong F$. Conversely, suppose there exists $H\subset K$ with $H$ isomorphic to $F$. Then there is an isomorphism $$\chi \colon F \to H$$ and we also have the inclusion homomorphism $$\iota\colon H \hookrightarrow K$$ Thus the composition $$\iota \circ \chi\colon F \to K$$ is a field homomorphism. {\bf Remark}: Let $\psi : F \to K$ be a field homomorphism. We claim that the characteristic of $F$ and $K$ must be the same. Indeed, since $\psi(1_F)=1_K$ and $\psi(0_F)=0_K$ then $\psi(n\cdot 1_F)=n\cdot 1_K$ for all natural numbers $n$. If the characteristic of $F$ is $p>0$ then $0=\psi(p\cdot 1)=p\cdot 1$ in $K$, and so the characteristic of $K$ is also $p$. If the characteristic of $F$ is $0$, then the characteristic of $K$ must be $0$ as well. For if $p\cdot 1=0$ in $K$ then $\psi(p\cdot 1)=0$, and since $\psi$ is injective by the lemma, we would have $p\cdot 1=0$ in $F$ as well.