User:Ragnar Schroder/Sandbox: Difference between revisions
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===The Galois group of a polynomial - a basic example=== | ===The Galois group of a polynomial - a basic example=== | ||
As an example, let us look at the | 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>. | 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>. | ||
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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>. | 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>. | ||
The are exactly 2 automorphisms of L that leave every element of Q alone: the do-nothing automorphism <math>\phi_0: a+b r_0 \rightarrow a + b r_0 </math> and the map <math>\phi_1 : a+b r_0 \rightarrow a - b r_0</math>. | |||
The are exactly 2 automorphisms of L that leave every element of Q alone: the | |||
Under composition of automorphisms, these two automorphisms together form a group isomorphic to <math>S_2</math>, the group of permutations of two objects. | Under composition of automorphisms, these two automorphisms together form a group isomorphic to <math>S_2</math>, the group of permutations of two objects. | ||
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The sought for Galois group is therefore <math>S_2</math>, which has no nontrivial subgroups. | The sought for Galois group is therefore <math>S_2</math>, which has no nontrivial subgroups. | ||
The 3 requirements (explained below) for the Galois correspondence to exist are fullfilled, so we we conclude from the subgroup structure of <math>S_2</math> that there is no intermediate field extension containing K and also roots of the polynomial. | |||
Revision as of 01:08, 20 December 2007
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 , 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. . From direct inspection of the polynomial we also realize that .
L = is the smallest subfield of C that contains Q and both and .
The are exactly 2 automorphisms of L that leave every element of Q alone: the do-nothing automorphism and the map .
Under composition of automorphisms, these two automorphisms together form a group isomorphic to , the group of permutations of two objects.
The sought for Galois group is therefore , which has no nontrivial subgroups.
The 3 requirements (explained below) for the Galois correspondence to exist are fullfilled, so we we conclude from the subgroup structure of that there is no intermediate field extension containing K and also roots of the polynomial.