Editing Inverse Galois problem (section)

==A simple example: cyclic groups==
It is possible, using classical results, to  construct explicitly a [[polynomial]] whose Galois group over <math>\mathbb{Q}</math> is the [[cyclic group]] {{math|'''Z'''/''n'''''Z'''}} for any positive [[integer]] {{mvar|n}}. To do this, choose a [[prime number|prime]] {{mvar|p}} such that {{math|''p'' ≡ 1 (mod ''n'')}}; this is possible by [[Dirichlet's theorem on arithmetic progressions|Dirichlet's theorem]]. Let {{math|'''Q'''(''μ'')}} be the [[Cyclotomic field#Cyclotomic fields|cyclotomic extension]] of <math>\mathbb{Q}</math> generated by {{mvar|μ}}, where {{mvar|μ}} is a primitive {{math|''p''}}-th [[root of unity]]; the Galois group of {{math|'''Q'''(''μ'')/'''Q'''}} is cyclic of order {{math|''p'' − 1}}.

Since {{mvar|n}} [[divisor|divides]] {{math|''p'' − 1}}, the Galois group has a cyclic [[subgroup]] {{mvar|H}} of order {{math|(''p'' − 1)/''n''}}. The [[fundamental theorem of Galois theory]] implies that the corresponding fixed field, {{math|1=''F'' = '''Q'''(''μ'')<sup>''H''</sup>}}, has Galois group {{math|'''Z'''/''n'''''Z'''}} over <math>\mathbb{Q}</math>. By taking appropriate sums of conjugates of {{mvar|μ}}, following the construction of [[Gaussian period]]s, one can find an element {{mvar|α}} of {{mvar|F}} that generates {{mvar|F}} over {{nowrap|<math>\mathbb{Q}</math>,}} and compute its [[Minimal polynomial (field theory)|minimal polynomial]].

This method can be extended to cover all finite [[abelian group]]s, since every such group appears in fact as a quotient of the Galois group of some cyclotomic extension of <math>\mathbb{Q}</math>. (This statement should not though be confused with the [[Kronecker–Weber theorem]], which lies significantly deeper.)

===Worked example: the cyclic group of order three===
For {{math|1=''n'' = 3}}, we may take {{math|1=''p'' = 7}}. Then {{math|Gal('''Q'''(''μ'')/'''Q''')}} is cyclic of order six. Let us take the generator {{mvar|η}} of this group which sends {{mvar|μ}} to {{math|''μ''<sup>3</sup>}}. We are interested in the subgroup {{math|1=''H'' = {1, ''η''<sup>3</sup>}}} of order two. Consider the element {{math|1=''α'' = ''μ'' + ''η''<sup>3</sup>(''μ'')}}. By construction, {{mvar|α}} is fixed by {{mvar|H}}, and only has three conjugates over <math>\mathbb{Q}</math>:

: {{math|1=''α'' = ''η''<sup>0</sup>(''α'') = ''μ'' + ''μ''<sup>6</sup>}}, 
: {{math|1=''β'' = ''η''<sup>1</sup>(''α'') = ''μ''<sup>3</sup> + ''μ''<sup>4</sup>}}, 
: {{math|1=''γ'' = ''η''<sup>2</sup>(''α'') = ''μ''<sup>2</sup> + ''μ''<sup>5</sup>}}.

Using the identity:

:{{math|1=1 + ''μ'' + ''μ''<sup>2</sup> + ⋯ + ''μ''<sup>6</sup> = 0}},

one finds that

: {{math|1=''α'' + ''β'' + ''γ'' = −1}},
: {{math|1=''αβ'' + ''βγ'' + ''γα'' = −2}},
: {{math|1=''αβγ'' = 1}}.

Therefore {{mvar|α}} is a [[root of a polynomial|root]] of the polynomial

:{{math|1=(''x'' − ''α'')(''x'' − ''β'')(''x'' − ''γ'') = ''x''<sup>3</sup> + ''x''<sup>2</sup> − 2''x'' − 1}},

which consequently has Galois group {{math|'''Z'''/3'''Z'''}} over <math>\mathbb{Q}</math>.