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

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
• Why do we know that $[F(g(\alpha)):F] =n$, particularly why do we know it is an equality? Shouldn't it be an inequality? Namely $[F(g(\alpha)):F] \leq n$ – user110320 May 12 '18 at 19:58
• @user110320: We know $[F(g(\alpha)):F] = n$ because $g(\alpha)$ is a root of $f(X)$, and all fields obtained by adjoining a root of $f(X)$ to $F$ are isomorphic to one another by sending a root $\alpha'$ in $F(\alpha')$, say, to $\alpha''$ in $F(\alpha'')$. In particular, $F(\alpha')$ and $F(\alpha'')$ have the same dimension as $F$-vector spaces. – Alex Ortiz Mar 17 at 21:56