# f(g(x)) has a degree divisible by n

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

I don't know even how to begin. I really need help. Thanks

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Suppose $h$ is an irreducible factor of $f \circ g$, and $\alpha$ is a root of $h$ (in some extension field). Then $g(\alpha)$ is a root of $f$, and so, since $f$ is irreducible of degree $n$, $[F[g(\alpha)]:F]=n$. Thus $\deg(h)=[F[\alpha]:F]=[F[\alpha]:F[g(\alpha)]]\cdot[F[g(\alpha)]:F]$ is divisible by $n$.
thank you very much for the help. Just one question, why $f(g(\alpha))$ is well-defined? –  user42912 Nov 10 '12 at 0:52
@user42912: I'm having trouble seeing the problem. $\alpha$ is an element of an extension field of $F$. $f\circ g$ is a polynomial over $F$, hence defines a function on every extension field. I'm just applying that function to that element. –  Chris Eagle Nov 10 '12 at 9:56