17. Let f(x)∈F[x] be irreducible of degree n, and let g(x)∈F[x]. Prove that every irreducible factor of f(g(x)) has degree divisible by n.
Proof: (16) It suffices to show that every element r∈R has a multiplicative inverse. Since r∈K is algebraic over F, we observe r−1∈F(r)⊆R.
(17) Let h(x)∣f(g(x)) be irreducible of degree k, and let α be a solution to h(x), so that h(α)=0 and thus f(g(α))=0. Since f(x) is irreducible, we must have g(α) is of degree n. We thus havedeg h(x)=k=[F(α) : F]=[F(α) : F(g(α))]⋅[F(g(α)) : F]=[F(α) : F(g(α))]⋅nso that n∣k. ◻
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