$E/F$ is a and a separable extension.

normal extension

$E$ is a of a separable polynomial with coefficients in $F.$

splitting field

$|\!\operatorname {Aut} (E/F)|=[E:F],$ that is, the number of automorphisms equals the of the extension.

degree

An important theorem of Emil Artin states that for a finite extension $E/F,$ each of the following statements is equivalent to the statement that $E/F$ is Galois:

Other equivalent statements are:

An infinite field extension $E/F$ is Galois if and only if $E$ is the union of finite Galois subextensions $E_{i}/F$ indexed by an (infinite) index set $I$ , i.e. $E=\bigcup _{i\in I}E_{i}$ and the Galois group is an inverse limit $\operatorname {Aut} (E/F)=\varprojlim _{i\in I}{\operatorname {Aut} (E_{i}/F)}$ where the inverse system is ordered by field inclusion $E_{i}\subset E_{j}$ .^[4]

Take any field $E$ , any finite subgroup of $\operatorname {Aut} (E)$ , and let $F$ be the fixed field.

Take any field $F$ , any separable polynomial in $F[x]$ , and let $E$ be its .

splitting field

There are two basic ways to construct examples of Galois extensions.

Adjoining to the rational number field the square root of 2 gives a Galois extension, while adjoining the cubic root of 2 gives a non-Galois extension. Both these extensions are separable, because they have characteristic zero. The first of them is the splitting field of $x^{2}-2$ ; the second has normal closure that includes the complex cubic roots of unity, and so is not a splitting field. In fact, it has no automorphism other than the identity, because it is contained in the real numbers and $x^{3}-2$ has just one real root. For more detailed examples, see the page on the fundamental theorem of Galois theory.

An algebraic closure ${\bar {K}}$ of an arbitrary field $K$ is Galois over $K$ if and only if $K$ is a perfect field.