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 adjoining the element $\alpha$ to $K$ via field adjunction.

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 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

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 $K(S) = KS$ .

Bibliography

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B. L. VAN DER WAERDEN: Algebra. Erster Teil. Siebte Auflage der Modernen Algebra. Springer-Verlag; Berlin, Heidelberg, New York (1966).