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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.

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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. $n\mid [F(\alpha):F]=\deg p(x)$.

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