field adjunctionFieldAdjunction2004-05-30 17:58:182008-01-11 16:46:08Definitionring adjunctionadjunction% this is the default PlanetMath preamble. as your knowledge
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\newtheorem*{thmplain}{Theorem}Let $K$ be a field and $E$ its extension field.\, If $\alpha \in E$, then the smallest subfield of $E$, that contains $K$ and $\alpha$, is denoted by $K(\alpha)$.\, We say that $K(\alpha)$ is obtained from the field $K$ by {\em adjoining} the element $\alpha$ to $K$ via {\em field adjunction}.
\begin{thmplain}
\, $K(\alpha)$ is identical with the quotient field $Q$ of $K[\alpha]$.
\end{thmplain}
{\em Proof.} (1) Because $K[\alpha]$ is an integral domain (as a subring of the field $E$), all possible quotients of the elements of $K[\alpha]$ belong to $E$. So we have
$$K\cup\{\alpha\} \subseteq K[\alpha] \subseteq Q \subseteq E,$$
and because $K(\alpha)$ was the smallest, then \,$K(\alpha) \subseteq Q.$
(2) $K(\alpha)$ is a subring of $E$ containing $K$ and $\alpha$, therefore also the whole ring $K[\alpha]$, that is, \,$K[\alpha] \subseteq K(\alpha)$. \,And because $K(\alpha)$ is a field, it must contain all possible quotients of the elements of $K[\alpha]$, i.e., \,$Q \subseteq K(\alpha)$.
In \PMlinkescapetext{addition} to the adjunction of one single element, we can adjoin to $K$ an arbitrary subset $S$ of $E$:\, the resulting field $K(S)$ is the smallest of such subfields of $E$, i.e. the intersection of such subfields of $E$, that contain both $K$ and $S$.\, We say that $K(S)$ is obtained from $K$ by adjoining the set $S$ to it.\, Naturally,
$$K \subseteq K(S) \subseteq E.$$
The field $K(S)$ contains all elements of $K$ and $S$, and being a field, also all such elements that can be formed via addition, subtraction, multiplication and division from the elements of $K$ and $S$.\, But such elements constitute a field, which therefore must be equal with $K(S)$.\, Accordingly, we have the
\begin{thmplain}
\, $K(S)$ constitutes of all rational expressions formed of the elements of the field $K$ with the elements of the set $S$.
\end{thmplain}
\textbf{Notes.}\\
1. $K(S)$ is the union of all fields $K(T)$ where $T$ is a finite subset of $S$.\\
2. $K(S_1 \cup S_2) = K(S_1)(S_2)$.\\
3. If, especially, $S$ also is a subfield of $E$, then one may denote\, $K(S) = KS$.
\begin{thebibliography}{9}
\bibitem{vdW}{\sc B. L. van der Waerden:} {\em Algebra. Erster Teil}.\, Siebte Auflage der {\em Modernen Algebra}. Springer-Verlag; Berlin, Heidelberg, New York (1966).
\end{thebibliography}