Galois described field extension and intermediate field.
F=coefficient, K=field extension, M=intermediate field. G=Galois group.
Therefore
F⊂M⊂K
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G(K/M)
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G(K/F)
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∴
G(K/M)⊂G(K/F)
There is cubic equation.
x^3+ax+b=(x-α)(x-β)(x-γ)=0
α+β+γ=0, αβ+βγ+αγ=a, αβγ=-b
G | e | f1 | f2 | g1 | g2 | g3 |
α | α | β | γ | β | α | γ |
β | β | γ | α | α | γ | β |
γ | γ | α | β | γ | β | α |
e⊂H⊂G
H1=(e,g1)
H2=(e,g2)
H3=(e,g3)
H=(e,f1,f2)
H | e | f1 | f2 |
α | α | β | γ |
β | β | γ | α |
γ | γ | α | β |
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I define
F⊂M⊂K
andG(K/M)⊂G(K/F)
Therefore, G=F and H=M.
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K⊂M⊂F
This is upside-down.
∴
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