Galois extension

In mathematics, a Galois extension is an algebraic field extension E/F that is normal and separable;^[1] or equivalently, E/F is algebraic, and the field fixed by the automorphism group Aut(E/F) is precisely the base field F. The significance of being a Galois extension is that the extension has a Galois group and obeys the fundamental theorem of Galois theory.^[a]

A result of Emil Artin allows one to construct Galois extensions as follows: If E is a given field, and G is a finite group of automorphisms of E with fixed field F, then E/F is a Galois extension.^[2]

The property of an extension being Galois behaves well with respect to field composition and intersection.^[3]

Characterization of Galois extensions

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

• ${\displaystyle E/F}$ is a normal extension and a separable extension.
• ${\displaystyle E}$ is a splitting field of a separable polynomial with coefficients in ${\displaystyle F.}$
• ${\displaystyle |\!\operatorname {Aut} (E/F)|=[E:F],}$ that is, the number of automorphisms equals the degree of the extension.

Other equivalent statements are:

• Every irreducible polynomial in ${\displaystyle F[x]}$ with at least one root in ${\displaystyle E}$ splits over ${\displaystyle E}$ and is separable.
• ${\displaystyle |\!\operatorname {Aut} (E/F)|\geq [E:F],}$ that is, the number of automorphisms is at least the degree of the extension.
• ${\displaystyle F}$ is the fixed field of a subgroup of ${\displaystyle \operatorname {Aut} (E).}$
• ${\displaystyle F}$ is the fixed field of ${\displaystyle \operatorname {Aut} (E/F).}$
• There is a one-to-one correspondence between subfields of ${\displaystyle E/F}$ and subgroups of ${\displaystyle \operatorname {Aut} (E/F).}$

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

Examples

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

• Take any field ${\displaystyle E}$, any finite subgroup of ${\displaystyle \operatorname {Aut} (E)}$, and let ${\displaystyle F}$ be the fixed field.
• Take any field ${\displaystyle F}$, any separable polynomial in ${\displaystyle F[x]}$, and let ${\displaystyle E}$ be its splitting field.

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 ${\displaystyle x^{2}-2}$; 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 ${\displaystyle x^{3}-2}$ has just one real root. For more detailed examples, see the page on the fundamental theorem of Galois theory.

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

Notes

Citations

References

Further reading