Let \(f(x) \in F[x]\) be an irreducible degree-\(n\) polynomial, and \(g(x) \in F[x]\). Prove that every irreducible factor of the composite polynomial \(f(g(x))\) has degree divisible by \(n\).
If \(\alpha\) be some root of \(f(x)\), then the field extension \(F(\alpha)/F\) has degree \(n\). If \(\beta\) is a root of \(f(g(x))\), then \(\beta\) must also be a root of the polynomial \(g(x)-\alpha \in F(\alpha)[x]\). Thus the degree of the extension \(F(\alpha,\beta)/F\) is divisible by \(n\).
Now observe the following chain: \(F \subseteq_a F(\beta) \subseteq_b F(\beta)(\alpha)\). Note that \(\beta\) is a root of some irreducible factor \(p(x)\) of \(f(g(x))\), thus \(a = \text{deg}\ p\). Conversely for every irreducible factor \(p(x)\) we can find a root \(\beta\) such that \([F(\beta) : F] = \text{deg}\ p\). Suppose the degree of some irreducible factor \(p(x)\) is not divisible by \(n\). Then \(b\) divides \(n\), since \(ab\) divides \(n\) (as shown in the previous paragraph). Furthermore, \(b\) is the degree of the minimal polynomial of \(\alpha\) in \(F(\beta)[x]\).
The above statement implies that if we can exhibit a degree \(n\) polynomial in \(F(\beta)[x]\) with \(\alpha\) as a root, then it must be a unit-multiple of the minimal polynomial, meaning it is irreducible. But \(f(x)\) is precisely such a polynomial. However, since \(g(\beta) = \alpha\) is a root of \(f(x)\), \(f(x)\) has a linear factor \((x-g(\beta))\), contradicting its irreducibility (it is critical that we are working in \(F(\beta)[x]\) and not just \(F[x]\)). Thus \(b\) cannot divide \(n\) and the conclusion follows.
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