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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

$$\mathrm{Let}\:{F}\:\mathrm{be}\:\:\mathrm{Field}\:\mathrm{of}\:\mathrm{characteristic}\:\mathrm{0} \\ $$$${L}_{{i}} \:\left({i}=\mathrm{1},\mathrm{2}\right)\:\mathrm{be}\:\mathrm{two}\:\mathrm{algebraic}\:\mathrm{extension} \\ $$$$\mathrm{of}\:{F}\:,\:\mathrm{and}\:{L}_{\mathrm{1}} {L}_{\mathrm{2}} \:\mathrm{be}\:\mathrm{a}\:\mathrm{field}\:\mathrm{in}\:\bar {{F}}\: \\ $$$$\left(\mathrm{where}\:\bar {{F}}\:\:\mathrm{is}\:\mathrm{the}\:\mathrm{algebraic}\:\mathrm{closure}\:\:\mathrm{of}\:{F}\right) \\ $$$$\mathrm{defined}\:\mathrm{by}\:\left\{{l}_{\mathrm{1}} {l}_{\mathrm{2}} \mid{l}_{{i}} \in{L}_{{i}} \:\left({i}=\mathrm{1},\mathrm{2}\right)\right\} \\ $$$$\mathrm{1}.\:\mathrm{show}\:\mathrm{that}\:\mathrm{if}\:{L}_{\mathrm{1}} \:\mathrm{and}\:{L}_{\mathrm{2}} \:\mathrm{are}\:\mathrm{galois}\:\mathrm{over}\:{F} \\ $$$$\mathrm{then}\:{L}_{\mathrm{1}} {L}_{\mathrm{2}} \:\mathrm{is}\:\mathrm{also}\:\mathrm{Galois}\:\mathrm{over}\:{F} \\ $$$$\mathrm{2}.\:\mathrm{show}\:\mathrm{that}\:\mathrm{if}\:{G}\left({L}_{\mathrm{1}} /{F}^{\:} \right)\:\mathrm{and}\:{G}\left({L}_{\mathrm{2}} /{F}^{\:} \right) \\ $$$$\mathrm{are}\:\mathrm{Solvable}\:,\:\mathrm{then}\:\mathrm{Gal}\left({L}_{\mathrm{1}} {L}_{\mathrm{2}} /{F}^{\:} \right)\:\mathrm{is}\:\mathrm{also} \\ $$$$\mathrm{Solvable} \\ $$

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

$$\forall\sigma\:\in\:{G}\mathrm{al}\left(\overset{\_} {{F}}/{F}\right) \\ $$$$\underset{\mathrm{Since}\:{L}_{\mathrm{1}} ,{L}_{\mathrm{2}} \:\mathrm{are}\:{G}\mathrm{alois}\:\mathrm{over}\:{F}} {\underbrace{\sigma\left({L}_{\mathrm{1}} \right)={L}_{\mathrm{1}} \mathrm{and}\:\sigma\left({L}_{\mathrm{2}} \right)={L}_{\mathrm{2}} }} \\ $$$$\Rightarrow\sigma\left({L}_{\mathrm{1}} {L}_{\mathrm{2}} \right)=\sigma\left({L}_{\mathrm{1}} \right)\sigma\left({L}_{\mathrm{2}} \right).>={L}_{\mathrm{1}} {L}_{\mathrm{2}} \\ $$$$\therefore{L}_{\mathrm{1}} {L}_{\mathrm{2}} \:\mathrm{is}\:{G}\mathrm{alois}\:\mathrm{over}\:{F} \\ $$$${L}\mathrm{et}\:{G}_{{i}} ={G}\mathrm{al}\left({L}_{{i}} /{F}\right),\left({i}=\mathrm{1},\mathrm{2}\right) \\ $$$${C}\mathrm{onsider}\:\varphi:{G}\mathrm{al}\left({L}_{\mathrm{1}} {L}_{\mathrm{2}} /{F}\right)\rightarrow{G}_{\mathrm{1}} ×{G}_{\mathrm{2}} \\ $$$$\sigma \left(\sigma\mid{L}_{\mathrm{1}} ,\sigma\mid{L}_{\mathrm{2}} \right) \\ $$$$\mathrm{ker}\left(\varphi\right)=\left\{\sigma\:\in\:{G}\mathrm{al}\left({L}_{\mathrm{1}} {L}_{\mathrm{2}} /{F}\right):\sigma\mid{L}_{\mathrm{1}} ={id}_{{L}_{\mathrm{1}} } \mathrm{and}\:\sigma\mid_{{L}_{\mathrm{2}} } ={id}_{{L}_{\mathrm{2}} } \right\} \\ $$$$\Rightarrow\mathrm{ker}\left(\varphi\right)=\left\{{id}_{{L}_{\mathrm{1}} {L}_{\mathrm{2}} } \right\}\Rightarrow\varphi\:\mathrm{is}\:\mathrm{injective} \\ $$$$\forall\left(\tau_{\mathrm{1}} ,\tau_{\mathrm{2}} \right)\in\:{G}_{\mathrm{1}} ×{G}_{\mathrm{2}} \exists!\sigma\:\in\:\mathrm{Gal}\left({L}_{\mathrm{1}} {L}_{\mathrm{2}} /{F}\right) \\ $$$$\underset{\mathrm{By}\:\mathrm{the}\:\mathrm{universal}\:\mathrm{property}\:\mathrm{of}\:\mathrm{compositum}} {\underbrace{\mathrm{such}\:\mathrm{that}\:\sigma\mid_{{L}_{\mathrm{1}} } =\tau_{\mathrm{1}} \mathrm{and}\:\sigma\mid_{{L}_{\mathrm{2}} } =\tau_{\mathrm{2}} }} \\ $$$$\Rightarrow\varphi\:\mathrm{is}\:\mathrm{surjectice} \\ $$$$\therefore\varphi\:\mathrm{is}\:\mathrm{an}\:\mathrm{isomorphism} \\ $$$${G}_{\mathrm{1}} ×{G}_{\mathrm{2}} \mathrm{is}\:\mathrm{solvable}\:\mathrm{if}\:{G}_{\mathrm{1}} \mathrm{and}\:{G}_{\mathrm{2}} \mathrm{are}\:\mathrm{solvable} \\ $$$$\Rightarrow\mathrm{Gal}\left({L}_{\mathrm{1}} {L}_{\mathrm{2}} /{F}\right)\cong{G}_{\mathrm{1}} ×{G}_{\mathrm{2}} \mathrm{is}\:\mathrm{solvable} \\ $$

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