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'field homomorphism'
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| Title of object: |
field homomorphism |
| Canonical Name: |
FieldHomomorphism |
| Type: |
Definition |
| Created on: |
2003-08-29 16:47:32 |
| Modified on: |
2006-11-04 17:05:59 |
| Classification: |
msc:12E99 |
| Keywords: |
field, map |
| Defines: |
field homomorphism, field isomorphism |
| Synonyms: |
field homomorphism=field monomorphism |
Revision comment (for changes between this and next version):
| Changes for correction #10832 ('remark clarification'). |
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Content:
Let $F$ and $K$ be fields.
\begin{defn}
A {\em field homomorphism} is a function $\psi\colon F \longrightarrow K$ 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 $\psi(1)=1$ and $\psi(0)=0$ then $\psi(n)=n$ 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. |
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