User:Ragnar Schroder/Sandbox

Testing my sandbox:

1995: 0,28 grader 1997: 0,36 grader 1998: 0,52 grader

2001: 0,40 grader 2002: 0,46 grader 2003: 0,46 grader 2004: 0,43 grader 2005: 0,48 grader 2006: 0,42 grader 2007: 0,41 grader

ÆØÅ æøå

The Galois group of a polynomial - a basic example

As an example, let us look at the second-degree polynomial $x^{2}-5$, with the coefficients {-5,0,1} viewed as elements of Q.

This polynomial has no roots in Q. However, from the fundamental theorem of algebra we know that it has exacly two roots in C, and can be written as the product of two first-degree polynomials there - i.e. $x^{2}-5=(x-r_{0})(x-r_{1}),r_{0},r_{1}\in C$. From direct inspection of the polynomial we also realize that $r_{0}=-r_{1}$.

L = $\lbrace a+br_{0},a,b\in Q\rbrace$ is the smallest subfield of C that contains Q and both $r_{0}$ and $-r_{0}$.

The are exactly 2 automorphisms of L that leave every element of Q alone: the do-nothing automorphism $\phi _{0}:a+br_{0}\rightarrow a+br_{0}$ and the map $\phi _{1}:a+br_{0}\rightarrow a-br_{0}$.

Under composition of automorphisms, these two automorphisms together form a group isomorphic to $S_{2}$, the group of permutations of two objects.

The sought for Galois group is therefore $S_{2}$, which has no nontrivial subgroups.

The 3 requirements (explained below) for the Galois correspondence to exist happen to be fullfilled, so we we conclude from the subgroup structure of $S_{2}$ that there is no intermediate field extension containing K and also roots of the polynomial.