# The degree of an irreducible factor of a composite polynomial

Dummit and Foote Ch 13.2 Q17

Let $$f(x)$$ be an irreducible polynomial of degree $$n$$ over a field $$F$$. Let $$g(x)$$ be any polynomial in $$F[x]$$. Prove that every irreducible factor of the composite polynomial $$f(g(x))$$ has degree divisible by $$n$$.

Following theorem may be useful

Let $$p(x)\in F[x]$$ be an irreducible polynomial of degree $$n$$ over the field $$F$$ and let $$K$$ be the field $$K=\frac{F[x]}{(p(x))}$$. Then the degree of the extension is $$n$$.

Let $$h_1,..., h_m$$ be irreducible factors of $$f(g(x))$$. By Chinese remainder theorem $$\frac{F[x]}{(f(g(x)))}\cong \frac{F[x]}{(h_1(x))} \times...\times \frac{F[x]}{(h_m(x))}$$. So the RHS talks about the degree of the irreducible factors. The LHS should incorporate the given information, i.e. $$f$$ is irreducible polynomial of degree $$n$$. Please give a hint on how to incorporate this given information in LHS.

Edit: If some other method works, please give a hint for that method.

Thanks.

Let $$p(x)$$ be an irreducible factor of $$f(g(x))$$ and $$\alpha$$ a root of $$p$$ in some extension field of $$F$$. Then the degree of $$F(\alpha)/F$$ is $$\deg p(x)$$. Since $$p(\alpha)=0$$ and $$p(x)\mid f(g(x))$$ we have $$f(g(\alpha))=0$$, i.e. $$g(\alpha)$$ is a root of $$f$$, furthermore $$g(\alpha)\in F(\alpha)$$. Now consider the degrees in the field tower $$F(\alpha)/F(g(\alpha))/F$$ Since $$f$$ is irreducible we have as above $$[F(g(\alpha)):F]=n$$ and hence $$[F(\alpha):F]=[F(\alpha):F(g(\alpha))]\cdot n$$, i.e. $$n\mid [F(\alpha):F]=\deg p(x)$$.