# Thread: Degree of irreducible polynomial dividing composite of two polynomials.

1. ## Degree of irreducible polynomial dividing composite of two polynomials.

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.

2. ## Re: Degree of irreducible polynomial dividing composite of two polynomials.

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

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

3. ## Re: Degree of irreducible polynomial dividing composite of two polynomials.

Makes perfect sense, thanks NonCommAlg.