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>.
-L = <math>\lbrace a+b r_0, a,b \in Q  \rbrace </math> is the smallest subfield of C that contains Q and both <math>r_0</math> and <math>-r_0</math>.
-Now, in order to find the Galois group,  we need to look at all possible .
-The are exactly 2 automorphisms of L that leave every element of Q alone: 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 form a group  isomorphic to <math>S_2</math>,  the group of permutations of two objects.
-The sought for Galois group is therefore <math>S_2</math>, which has no nontrivial subgroups.
-This group has no nontrivial subgroups.
-Two requirements (known as "seperability" and "normality") need to be satisfied in order for the Galois correspondence to work.  Both happen to be fulfilled in this case,  so we may invoke the Galois correspondence,  and conclude that no intermediate field extension exists.

User:Ragnar Schroder/Sandbox: Difference between revisions

Latest revision as of 03:34, 22 November 2023

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