Originally Posted by

**AshleyLin** I've recently been looking at the relationships between divisibility of polynomials and their degrees. This following idea has eluded me for a while.

Suppose $\displaystyle f$ and $\displaystyle g$ are polynomials in $\displaystyle F[X]$ for $\displaystyle F $ a field, with $\displaystyle f $ irreducible. If $\displaystyle p $ is another irreducible polynomial that divides $\displaystyle f(g(X))$, then how can you conclude that the degree of $\displaystyle f $ divides the degree of $\displaystyle p$?

I know if $\displaystyle f $ has degree $\displaystyle n$ and $\displaystyle g$ has degree $\displaystyle m$ then $\displaystyle f(g(X))$ has degree $\displaystyle nm$, I'm not sure if that immediately relates to the degree of $\displaystyle p$. Thanks.

let $\displaystyle p(x)$ be an irreducible factor of $\displaystyle f(g(x)).$ let $\displaystyle \mathfrak{m}$ and $\displaystyle \mathfrak{n}$ be the ideals of $\displaystyle F[x]$ generated by $\displaystyle f(x)$ and $\displaystyle p(x),$ respectively. let $\displaystyle E=F[x]/\mathfrak{m}$ and $\displaystyle L = F[x]/\mathfrak{n}.$ since both $\displaystyle f(x)$ and $\displaystyle p(x)$ are irreducible, $\displaystyle E$ and $\displaystyle L$ are field extensions of $\displaystyle F.$ now, define the map $\displaystyle \varphi : E \longrightarrow L$ by

$\displaystyle \varphi(h(x) + \mathfrak{m})=h(g(x)) + \mathfrak{n},$

for all $\displaystyle h(x) \in F[x].$ the claim is that $\displaystyle \varphi$ is well-defined. to see this, suppose that $\displaystyle h(x) \in \mathfrak{m}.$ then $\displaystyle h(x)=f(x)u(x)$ for some $\displaystyle u(x) \in F[x]$ and so $\displaystyle h(g(x)) = f(g(x))u(g(x)) \in \mathfrak{n}$ because $\displaystyle p(x) \mid f(g(x)).$ clearly $\displaystyle \varphi$ is a ring homomorphism and, since $\displaystyle E$ is a field, $\displaystyle \varphi$ is one-to-one. so we may assume that $\displaystyle F \subseteq E \subseteq L$ and thus

$\displaystyle \deg p(x) = [L:F]=[L:E][E:F]=(\deg f(x))[L:E]. \ \Box$