Editing Galois theory (section)

==Inverse Galois problem==
{{main|Inverse Galois problem}}

The ''inverse Galois problem'' is to find a field extension with a given Galois group.

As long as one does not also specify the [[ground field]], the problem is not very difficult, and all finite groups do occur as Galois groups.
For showing this, one may proceed as follows. Choose a field {{math|''K''}} and a finite group {{math|''G''}}.  [[Cayley's theorem]] says that {{math|''G''}} is (up to isomorphism) a subgroup of the [[symmetric group]] {{math|''S''}} on the elements of {{math|''G''}}. Choose indeterminates {{math|{''x''<sub>''α''</sub><nowiki>}</nowiki>}}, one for each element {{math|''α''}} of {{math|''G''}}, and adjoin them to {{math|''K''}} to get the field {{math|''F'' {{=}} ''K''({''x''<sub>''α''</sub>})}}. Contained within {{math|''F''}} is the field {{math|''L''}} of symmetric [[rational function]]s in the {{math|{''x''<sub>''α''</sub><nowiki>}</nowiki>}}.  The Galois group of {{math|''F''/''L''}} is {{math|''S''}}, by a basic result of Emil Artin.  {{math|''G''}} acts on {{math|''F''}} by restriction of action of {{math|''S''}}. If the [[fixed field]] of this action is {{math|''M''}}, then, by the [[fundamental theorem of Galois theory]], the Galois group of {{math|''F''/''M''}} is {{math|''G''}}.

On the other hand, it is an open problem whether every finite group is the Galois group of a field extension of the field {{math|'''Q'''}} of the rational numbers. [[Igor Shafarevich]] proved that every solvable finite group is the Galois group of some extension of {{math|'''Q'''}}. Various people have solved the inverse Galois problem for selected non-Abelian [[simple group]]s. Existence of solutions has been shown for all but possibly one ([[Mathieu group]] {{math|''M''<sub>23</sub>}}) of the 26 sporadic simple groups. There is even a polynomial with integral coefficients whose Galois group is the [[Monster group]].