User:Ragnar Schroder/Sandbox: Difference between revisions

Latest revision as of 03:34, 22 November 2023

The account of this former contributor was not re-activated after the server upgrade of March 2022.

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

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 Testing my sandbox:
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-===The Galois group of a polynomial - a basic example===
-As an example,  let us look at the second-degree polynomial <math>x^2-5</math>, 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. <math>x^2-5 = (x-r_0)(x-r_1), r_0, r_1 \in  C</math>.  From direct inspection of the polynomial we also realize that <math>r_0 = -r_1</math>.
-We now look for the smallest subfield of C that contains Q and both <math>r_0</math> and <math>-r_0</math>, which is L = { <math>   a+b r_0, a,b \in Q  </math>  }.  Since <math>r_0^2 = 5 \in Q</math>,  all products and sums are well defined.  This field is then the smallest extension of Q by the roots of <math>\alpha</math>.
-Now, in order to find the Galois group,  we need to look at all possible automorphisms of L that leave every element of Q alone.
-The only such automorphisms are the null automorphism and the map <math>a+b r_0   \rightarrow a - b r_0</math>.
-Under composition of automorphisms,  these two automorphisms together are isomorphic to the group <math>S_2</math>,  the group of permutations of two objects.
-The sought for Galois group is therefore <math>S_2</math>.

User:Ragnar Schroder/Sandbox: Difference between revisions

Latest revision as of 03:34, 22 November 2023

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