Question Number 213208 by issac last updated on 01/Nov/24
![Let f(x)∈Q[x] irreducible of degree n and K it′s Splitting Field over Q Prove that if Gal(K\Q) is Abeilan then ∣Gal(K\Q)∣=n How can i prove this???](Q213208.png)
$$\mathrm{Let}\:{f}\left({x}\right)\in\mathbb{Q}\left[{x}\right]\:\mathrm{irreducible}\:\mathrm{of}\:\mathrm{degree}\:{n} \\ $$$$\mathrm{and}\:{K}\:\mathrm{it}'\mathrm{s}\:\mathrm{Splitting}\:\mathrm{Field}\:\mathrm{over}\:\mathbb{Q} \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{if}\:\mathrm{Gal}\left({K}\backslash\mathbb{Q}\right)\:\mathrm{is}\:\mathrm{Abeilan} \\ $$$$\mathrm{then}\:\mid\mathrm{Gal}\left({K}\backslash\mathbb{Q}\right)\mid={n} \\ $$$$\mathrm{How}\:\mathrm{can}\:\mathrm{i}\:\mathrm{prove}\:\mathrm{this}??? \\ $$
Answered by MrGaster last updated on 01/Nov/24
![f(x)∈ Q[x],deg(f)=n K/Q is the splitting field of f(x)=G=Gal(K/Q) G is abelian α_1 ,α_2 ,…,α_n are roots of f(x)in K K=Q(α_1 ,α_2 ,…,α_n ) ∣G∣=[K:Q] σ ∈ G,σ(α_i )=α_j ,σ(α_j )=α_i ,σ(α_κ )=α_κ ,k≠i,j στ=τσ,∀σ,τ ∈ G σ(α_i )α_j ⇒σ(α_j )=α_i or α_j σ(α_i )=α_j ,τ(α_i )=α_κ ,σ(α_κ )=α_κ στ(α_i )=σ(α_κ )=α_κ τσ(α_i )=τ(α_j )=α_κ σ(α_j )=α_j for all σ ∈ G σ(α_i )=α_j ⇒α_i and α_j are conjugate over Q α_1 ,α_2 ,…,α_n are distinct and conjugate o ∣G∣=n](Q213218.png)
$${f}\left({x}\right)\in\:\mathbb{Q}\left[{x}\right],\mathrm{deg}\left({f}\right)={n} \\ $$$${K}/\mathbb{Q}\:\mathrm{is}\:\mathrm{the}\:\mathrm{splitting}\:\mathrm{field}\:\mathrm{of}\:{f}\left({x}\right)={G}={G}\mathrm{al}\left({K}/\mathbb{Q}\right) \\ $$$${G}\:\mathrm{is}\:\mathrm{abelian} \\ $$$$\alpha_{\mathrm{1}} ,\alpha_{\mathrm{2}} ,\ldots,\alpha_{{n}} \mathrm{are}\:\mathrm{roots}\:\mathrm{of}\:{f}\left({x}\right)\mathrm{in}\:{K} \\ $$$${K}=\mathbb{Q}\left(\alpha_{\mathrm{1}} ,\alpha_{\mathrm{2}} ,\ldots,\alpha_{{n}} \right) \\ $$$$\mid{G}\mid=\left[{K}:\mathbb{Q}\right] \\ $$$$\sigma\:\in\:{G},\sigma\left(\alpha_{{i}} \right)=\alpha_{{j}} ,\sigma\left(\alpha_{{j}} \right)=\alpha_{{i}} ,\sigma\left(\alpha_{\kappa} \right)=\alpha_{\kappa} ,{k}\neq{i},{j} \\ $$$$\sigma\tau=\tau\sigma,\forall\sigma,\tau\:\:\in\:{G} \\ $$$$\sigma\left(\alpha_{{i}} \right)\alpha_{{j}} \Rightarrow\sigma\left(\alpha_{{j}} \right)=\alpha_{{i}} \mathrm{or}\:\alpha_{{j}} \\ $$$$\sigma\left(\alpha_{{i}} \right)=\alpha_{{j}} ,\tau\left(\alpha_{{i}} \right)=\alpha_{\kappa} ,\sigma\left(\alpha_{\kappa} \right)=\alpha_{\kappa} \\ $$$$\sigma\tau\left(\alpha_{{i}} \right)=\sigma\left(\alpha_{\kappa} \right)=\alpha_{\kappa} \\ $$$$\tau\sigma\left(\alpha_{{i}} \right)=\tau\left(\alpha_{{j}} \right)=\alpha_{\kappa} \\ $$$$\sigma\left(\alpha_{{j}} \right)=\alpha_{{j}} \mathrm{for}\:\mathrm{all}\:\sigma\:\in\:{G} \\ $$$$\sigma\left(\alpha_{{i}} \right)=\alpha_{{j}} \Rightarrow\alpha_{{i}} \mathrm{and}\:\alpha_{{j}} \mathrm{are}\:\mathrm{conjugate}\:\mathrm{over}\:\mathbb{Q}\:\alpha_{\mathrm{1}} ,\alpha_{\mathrm{2}} ,\ldots,\alpha_{{n}} \:\mathrm{are}\:\mathrm{distinct}\:\mathrm{and}\:\mathrm{conjugate}\:\mathrm{o} \\ $$$$\mid{G}\mid={n} \\ $$