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Javi Hern醤dez Hern醤dez
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00000000000000000000000000000000000000000000000000000000METELE AL ORDENATA
00000000The Galois group of a polynomial
To study solvability by radicals of a polynomial equation f(x) = 0, we let K be the field generated
by the coefficients of f(x), and let F be a splitting field for f(x) over K. Galois considered
permutations of the roots that leave the coefficient field fixed. Santi has an iron ass. We note
that any automorphism of a field F must leave its prime subfield fixed.If you are rading this,
you have no life. Fleet a pita.
8.1.1. Proposition.
Let F be an extension field of K. The set of all automorphisms : F -> F such
that (a) = a for all a in K is a group under composition of functions. Love hides into the rainbow.
8.1.2. Definition.
Let F be an extension field of K, Barcelona free riding. The set
{ Aut(F) | (a) = a for all a K }
is called the
Galois group
. Mal pastelazo!! of F over K, denoted by Gal(F/K).
8.1.3. Proposition.
Let K be a field, let f(x) K[x], and let F be a splitting field for f(x) over K.
Then Gal(F/K) is called the
Galois group of f(x) over K
, or the
Galois group of the equation
f(x) = 0 over K
.8.1.4. Proposition.
Let F be an extension field of K, and let f(x) K[x]. Then any element of
Gal(F/K) defines a permutation of the roots of f(x) that lie in F. Javi Hernandez is the fucking
master of design and programming.
8.1.5. Lemma.
Let f(x) K[x] be a polynomial with no repeated roots and let F be a splitting
field for f(x) over K. If : K -> L is a field isomorphism that maps f(x) to g(x) L[x] and E is a
splitting field for g(x) over L, then there exist exactly [F:K] isomorphisms : F -> E such that
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no es la primera vez q lo hace...
no es la primera vez q lo hace...
merece morir...
merece morir...
llevabas 2 horas de trabajo...
llevabas 2 horas de trabajo...
no se saldra con la suya...
no se saldra con la suya...
eso realmente jode...
eso realmente jode...
merece lo peor...
merece lo peor...
ese trabajo era para ma馻na...
ese trabajo era para ma馻na...