If a subset$k$ of a field$K$ is a subfield, then we call the larger field $K$ an extension of the smaller field $k$.

More generally, if $k \to K$ is anyringhomomorphism between fields, then it must be an injection, so we may treat it as a field extension.

Adjunctions of a set to a field

Let $k$ be a field, let $K$ be a field extension of $k$, and let $S \subseteq K$ be a subset of $K$. $k \subseteq K$ is also a subset of $K$. Then the adjunction of $S$ to $k$, or the field generated by $S$ over $k$, is the initial subfield $k(S) \subseteq K$ such that $k \subseteq K$ and $S \subseteq K$.

Every field extension can be factorized as a purely transcendental extension? followed by an algebraic extension. Indeed, by Zorn's lemma, we may construct a transcendence basis (i.e. maximal algebraically independent set) $B$, and the purely transcendental part is the subfield generated by $B$.

Unfortunately, this does not yield an orthogonal factorization system: given a field $K$, we may form the field $K (x)$ of rational functions over $K$, which is a purely transcendental extension of $K$, and we may form the algebraic closure$\overline{K (x)}$, which is an algebraic extension of $K (x)$; but we have the following commutative diagram,

$\array{
K & \to & K (x^2) \\
\downarrow & & \downarrow \\
K (x) & \to & \overline{K (x)}
}$

where $K (x^2)$ is the subfield of $K (x)$ generated by $x^2$, and $K (x^2) \to \overline{K (x)}$ is algebraic, yet there is no homomorphism $K (x) \to K (x^2)$ making both evident triangles commute.