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)} 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


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Question Number 214098 by issac last updated on 28/Nov/24
$Let F be Field of characteristic 0 L_i (i=1,2) be two algebraic extension of F , and L_1 L_2 be a field in F^� (where F^� is the algebraic closure of F) defined by {l_1 l_2 ∣l_i ∈L_i (i=1,2)} 1. show that if L_1 and L_2 are galois over F then L_1 L_2 is also Galois over F 2. show that if G(L_1 /F^ ) and G(L_2 /F^ ) are Solvable , then Gal(L_1 L_2 /F^ ) is also Solvable$
$Let F be Field of characteristic 0$ $L_{i} (i = 1, 2) be two algebraic extension$ $of F, and L_{1} L_{2} be a field in \bar{F}$ $(where \bar{F} is the algebraic closure of F)$ $defined by {l_{1} l_{2} ∣ l_{i} \in L_{i} (i = 1, 2)}$ $1 . show that if L_{1} and L_{2} are galois over F$ $then L_{1} L_{2} is also Galois over F$ $2 . show that if G (L_{1} / F^{}) and G (L_{2} / F^{})$ $are Solvable, then Gal (L_{1} L_{2} / F^{}) is also$ $Solvable$
	Answered by MrGaster last updated on 24/Dec/24
	$∀σ ∈ Gal(F^_ /F) σ(L_1 )=L_1 and σ(L_2 )=L_2 _(Since L_1 ,L_2 are Galois over F) ⇒σ(L_1 L_2 )=σ(L_1 )σ(L_2 ).>=L_1 L_2 ∴L_1 L_2 is Galois over F Let G_i =Gal(L_i /F),(i=1,2) Consider ϕ:Gal(L_1 L_2 /F)→G_1 ×G_2 σ (σ∣L_1 ,σ∣L_2 ) ker(ϕ)={σ ∈ Gal(L_1 L_2 /F):σ∣L_1 =id_L_1 and σ∣_L_2 =id_L_2 } ⇒ker(ϕ)={id_(L_1 L_2 ) }⇒ϕ is injective ∀(τ_1 ,τ_2 )∈ G_1 ×G_2 ∃!σ ∈ Gal(L_1 L_2 /F) such that σ∣_L_1 =τ_1 and σ∣_L_2 =τ_2 _(By the universal property of compositum) ⇒ϕ is surjectice ∴ϕ is an isomorphism G_1 ×G_2 is solvable if G_1 and G_2 are solvable ⇒Gal(L_1 L_2 /F)≅G_1 ×G_2 is solvable$
	$\forall σ \in G al (\overline{F} / F)$ $\underset{Since L_{1}, L_{2} are G alois over F}{\underset{⏟}{σ (L_{1}) = L_{1} and σ (L_{2}) = L_{2}}}$ $\Rightarrow σ (L_{1} L_{2}) = σ (L_{1}) σ (L_{2}) . >= L_{1} L_{2}$ $∴ L_{1} L_{2} is G alois over F$ $L et G_{i} = G al (L_{i} / F), (i = 1, 2)$ $C onsider φ : G al (L_{1} L_{2} / F) \to G_{1} \times G_{2}$ $σ (σ ∣ L_{1}, σ ∣ L_{2})$ $\ker (φ) = {σ \in G al (L_{1} L_{2} / F) : σ ∣ L_{1} = {i d}_{L_{1}} and σ ∣_{L_{2}} = {i d}_{L_{2}}}$ $\Rightarrow \ker (φ) = {{i d}_{L_{1} L_{2}}} \Rightarrow φ is injective$ $\forall (τ_{1}, τ_{2}) \in G_{1} \times G_{2} \exists! σ \in Gal (L_{1} L_{2} / F)$ $\underset{By the universal property of compositum}{\underset{⏟}{such that σ ∣_{L_{1}} = τ_{1} and σ ∣_{L_{2}} = τ_{2}}}$ $\Rightarrow φ is surjectice$ $∴ φ is an isomorphism$ $G_{1} \times G_{2} is solvable if G_{1} and G_{2} are solvable$ $\Rightarrow Gal (L_{1} L_{2} / F) ≅ G_{1} \times G_{2} is solvable$
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