Field extension

A field extension $L/K$ is algebraic if every element of $L$ is an algebraic element over $K$. Every finite field extension is algebraic, but algebraic extensions can have infinite degree.

The degree $[L:K]$ of a field extension is the dimension of a vector space $L$ over $K$.

Algebraic independence means the absence of a nonzero polynomial relation over the base field. It extends the distinction between algebraic and transcendental elements to finite tuples.

A transcendence basis of a field extension $L/k$ is an algebraically independent set $T$ such that $L/k(T)$ is algebraic. Its cardinality is the transcendence degree. For finitely generated extensions it is finite, and the remaining algebraic extension is finite.

A transcendence basis $T$ for $L/k$ is separating if $L/k(T)$ is a separable field extension. In this case the $dt$, $t\in T$, form a basis of the Kähler differentials. In characteristic zero every transcendence basis is separating. In positive characteristic the condition on the chosen basis is essential.

For a finitely generated field extension $L/k$ in characteristic zero, a tuple $t_i$ is algebraically independent exactly when its differentials $d{t}_i$ are linearly independent in the Kähler differentials. A relation of minimum total degree gives a nontrivial differential relation; an independent tuple extends to a transcendence basis. In positive characteristic, derivatives of nonconstant relations can vanish identically.

The transcendence degree is the cardinality of a transcendence basis. It counts independent transcendental parameters. In a finitely generated separably generated field extension, it equals the dimension of the Kähler differentials as a vector space over the extension field.

Elements $x_1,\dots ,x_n$ of a field extension of $K$ are algebraically independent over $K$ when no nonzero polynomial in $K[X_1,\dots ,X_n]$ vanishes at $(x_1,\dots ,x_n)$.

The rational function field $K(x_1,\dots ,x_n)$ is the field of fractions of $K[x_1,\dots ,x_n]$ when the variables are algebraically independent over $K$.

If $f\in k[X]$ is irreducible, then $(f)$ is a maximal ideal andis a field extension of $k$. The class $\alpha =X+(f)$ is a root of $f$, so this quotient adjoins a root of $f$ to the base field.

For finite extensions $K\subseteq L\subseteq M$,

Inside a common overfield, the compositum $L_1{L}_2$ is the smallest field containing both $L_1$ and $L_2$.

Finite extensions $L_1/K$ and $L_2/K$ are linearly disjoint when products of basis elements form a basis of their compositum. In particular,Extensions of coprime finite degrees are linearly disjoint.

An algebraic closure of a field $K$ is an algebraic extension $\overline{K}/K$ for which $\overline{K}$ is algebraically closed.

For a field extension $L/K$, an intermediate field is a field $M$ satisfying $K\subseteq M\subseteq L$.

A quadratic extension is a field extension of degree two. It is normal because the other root of the minimal polynomial $T^2+bT+c$ of any non-base-field element is $-b-\alpha$ and lies in the same extension.

For a field of characteristic different from two, adjoining a square root of a nonsquare gives every quadratic extension. If two such extensions are isomorphic over $K$, write $\sqrt{a}=x+y\sqrt{b}$; the vanishing of its cross coefficient forces $x=0$, giving $a/b=y^2$. This explains why one subtracts the identity class when counting extensions from the square-class group.

For a p-adic field with odd residue characteristic, the square map is an isomorphism on $1+{\mathfrak{m}}_K$ by Hensel lemma. The parity of the valuation and the square class of the residue unit therefore determine the square-class group of a field. Its three nontrivial classes give the three quadratic extensions: one unramified extension and two totally ramified extensions.

If $L/K$ and $K/F$ are algebraic field extensions, then $L/F$ is algebraic. In particular, an element algebraic over an algebraic extension of $F$ is algebraic over $F$.