Editing Inverse Galois problem (section)

==Symmetric and alternating groups==
[[David Hilbert|Hilbert]] showed that all symmetric and [[alternating group|alternating]] groups are represented as Galois groups of polynomials with rational [[coefficient]]s.

The polynomial {{math|''x<sup>n</sup>'' + ''ax'' + ''b''}} has discriminant

:<math>(-1)^{\frac{n(n-1)}{2}} \!\left( n^n b^{n-1} + (-1)^{1-n} (n-1)^{n-1} a^n \right)\!.</math>

We take the special case

:{{math|1=''f''(''x'', ''s'') = ''x<sup>n</sup>'' − ''sx'' − ''s''}}.

Substituting a prime integer for {{mvar|s}} in {{math|''f''(''x'', ''s'')}} gives a polynomial (called a '''specialization''' of {{math|''f''(''x'', ''s'')}}) that by [[Eisenstein's criterion]] is [[irreducible polynomial|irreducible]]. Then {{math|''f''(''x'', ''s'')}} must be irreducible over <math>\mathbb{Q}(s)</math>. Furthermore, {{math|''f''(''x'', ''s'')}} can be written

:<math>x^n - \tfrac{x}{2} - \tfrac{1}{2} - \left( s - \tfrac{1}{2} \right)\!(x+1)</math>

and {{math|''f''(''x'', 1/2)}} can be factored to:

:<math>\tfrac{1}{2} (x-1)\!\left( 1+ 2x + 2x^2 + \cdots + 2 x^{n-1} \right)</math>

whose second factor is irreducible (but not by Eisenstein's criterion). Only the reciprocal polynomial is irreducible by Eisenstein's criterion. We have now shown that the group {{math|Gal(''f''(''x'', ''s'')/'''Q'''(''s''))}} is [[doubly transitive]].

We can then find that this Galois group has a transposition. Use the scaling {{math|1=(1 − ''n'')''x'' = ''ny''}} to get

:<math> y^n - \left \{ s \left ( \frac{1-n}{n} \right )^{n-1} \right \} y - \left \{ s \left ( \frac{1-n}{n} \right )^n \right \}</math>

and with

:<math> t = \frac{s (1-n)^{n-1}}{n^n},</math>

we arrive at:

:{{math|1=''g''(''y'', ''t'') = ''y<sup>n</sup>'' − ''nty'' + (''n'' − 1)''t''}}

which can be arranged to

:{{math|''y<sup>n</sup>'' − ''y'' − (''n'' − 1)(''y'' − 1) + (''t'' − 1)(−''ny'' + ''n'' − 1)}}.

Then {{math|''g''(''y'', 1)}} has {{math|1}} as a [[double root|double zero]] and its other {{math|''n'' − 2}} zeros are [[simple zero|simple]], and a transposition in {{math|Gal(''f''(''x'', ''s'')/'''Q'''(''s''))}} is implied. Any finite [[doubly transitive permutation group]] containing a transposition is a full symmetric group.

[[Hilbert's irreducibility theorem]] then implies that an infinite set of rational numbers give specializations of {{math|''f''(''x'', ''t'')}} whose Galois groups are {{math|''S<sub>n</sub>''}} over the rational field {{nowrap|<math>\mathbb{Q}</math>.}} In fact this set of rational numbers is dense in {{nowrap|<math>\mathbb{Q}</math>.}}

The discriminant of {{math|''g''(''y'', ''t'')}} equals

:<math> (-1)^{\frac{n(n-1)}{2}} n^n (n-1)^{n-1} t^{n-1} (1-t),</math>

and this is not in general a perfect square.

===Alternating groups===
Solutions for alternating groups must be handled differently for [[parity (mathematics)|odd]] and [[parity (mathematics)|even]] degrees.

====Odd degree====
Let

:<math>t = 1 - (-1)^{\tfrac{n(n-1)}{2}} n u^2</math>

Under this substitution the discriminant of {{math|''g''(''y'', ''t'')}} equals

:<math>\begin{align}
(-1)^{\frac{n(n-1)}{2}} n^n (n-1)^{n-1} t^{n-1} (1-t)
&= (-1)^{\frac{n(n-1)}{2}} n^n (n-1)^{n-1} t^{n-1} \left (1 - \left (1 - (-1)^{\tfrac{n(n-1)}{2}} n u^2 \right ) \right) \\
&= (-1)^{\frac{n(n-1)}{2}} n^n (n-1)^{n-1} t^{n-1} \left ((-1)^{\tfrac{n(n-1)}{2}} n u^2 \right ) \\
&= n^{n+1} (n-1)^{n-1} t^{n-1} u^2 
\end{align}</math>

which is a perfect square when {{mvar|n}} is odd.

====Even degree====
Let:

:<math>t = \frac{1}{1 + (-1)^{\tfrac{n(n-1)}{2}} (n-1) u^2}</math>

Under this substitution the discriminant of {{math|''g''(''y'', ''t'')}} equals:

:<math>\begin{align}
(-1)^{\frac{n(n-1)}{2}} n^n (n-1)^{n-1} t^{n-1} (1-t)
&= (-1)^{\frac{n(n-1)}{2}} n^n (n-1)^{n-1} t^{n-1} \left (1 - \frac{1}{1 + (-1)^{\tfrac{n(n-1)}{2}} (n-1) u^2} \right ) \\
&= (-1)^{\frac{n(n-1)}{2}} n^n (n-1)^{n-1} t^{n-1} \left ( \frac{\left ( 1 + (-1)^{\tfrac{n(n-1)}{2}} (n-1) u^2 \right ) - 1}{1 + (-1)^{\tfrac{n(n-1)}{2}} (n-1) u^2} \right ) \\
&= (-1)^{\frac{n(n-1)}{2}} n^n (n-1)^{n-1} t^{n-1} \left ( \frac{(-1)^{\tfrac{n(n-1)}{2}} (n-1) u^2}{1 + (-1)^{\tfrac{n(n-1)}{2}} (n-1) u^2} \right ) \\
&= (-1)^{\frac{n(n-1)}{2}} n^n (n-1)^{n-1} t^{n-1} \left (t (-1)^{\tfrac{n(n-1)}{2}} (n-1) u^2 \right ) \\
&= n^n (n-1)^n t^n u^2
\end{align}</math>

which is a perfect square when {{mvar|n}} is even.

Again, Hilbert's irreducibility theorem implies the existence of infinitely many specializations whose Galois groups are alternating groups.