Let-F-be-Field-of-characteristic-0-L-i-i-1-2-be-two-algebraic-extension-of-F-and-L-1-L-2-be-a-field-in-F-where-F-is-the-algebraic-closure-of-F-defined-by-l-1-l-2-l-i-L-i-i-1-2-

Question Number 214098 by issac last updated on 28/Nov/24

Let F be Field of characteristic 0

L_{i} (i = 1, 2) be two algebraic extension

of F, and L_{1} L_{2} be a field in \bar{F}

(where \bar{F} is the algebraic closure of F)

defined by {l_{1} l_{2} ∣ l_{i} \in L_{i} (i = 1, 2)}

1 . show that if L_{1} and L_{2} are galois over F

then L_{1} L_{2} is also Galois over F

2 . show that if G (L_{1} / F^{}) and G (L_{2} / F^{})

are Solvable, then Gal (L_{1} L_{2} / F^{}) is also

Solvable

Answered by MrGaster last updated on 24/Dec/24

\forall σ \in G al (\overline{F} / F)

\underset{Since L_{1}, L_{2} are G alois over F}{\underset{⏟}{σ (L_{1}) = L_{1} and σ (L_{2}) = L_{2}}}

\Rightarrow σ (L_{1} L_{2}) = σ (L_{1}) σ (L_{2}) . >= L_{1} L_{2}

∴ L_{1} L_{2} is G alois over F

L et G_{i} = G al (L_{i} / F), (i = 1, 2)

C onsider φ : G al (L_{1} L_{2} / F) \to G_{1} \times G_{2}

σ (σ ∣ L_{1}, σ ∣ L_{2})

\ker (φ) = {σ \in G al (L_{1} L_{2} / F) : σ ∣ L_{1} = {i d}_{L_{1}} and σ ∣_{L_{2}} = {i d}_{L_{2}}}

\Rightarrow \ker (φ) = {{i d}_{L_{1} L_{2}}} \Rightarrow φ is injective

\forall (τ_{1}, τ_{2}) \in G_{1} \times G_{2} \exists! σ \in Gal (L_{1} L_{2} / F)

\underset{By the universal property of compositum}{\underset{⏟}{such that σ ∣_{L_{1}} = τ_{1} and σ ∣_{L_{2}} = τ_{2}}}

\Rightarrow φ is surjectice

∴ φ is an isomorphism

G_{1} \times G_{2} is solvable if G_{1} and G_{2} are solvable

\Rightarrow Gal (L_{1} L_{2} / F) ≅ G_{1} \times G_{2} is solvable

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