field adjunction

Let $K$ be a field and $E$ an extension field of $K$.  If $\alpha \in E$, then the smallest subfield of $E$, that contains $K$ and$\alpha$, is denoted by $K(\alpha )$.  We say that $K(\alpha )$ isobtained from the field $K$ by adjoining the element $\alpha$to $K$ via field adjunction.

Theorem.  $K(\alpha )$ is identical with the quotient field $Q$ of $K[\alpha ]$.

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 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

Theorem.  $K(S)$ constitutes of all rational expressions formed of the elements of the field $K$ with the elements of the set $S$.

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  $K(S)=KS$.