## Degree of an irreducible factor of composition of two polynomials

Posted: November 17, 2011 in Elementary Algebra; Problems & Solutions, Groups and Fields
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Problem. Let $F$ be a field and suppose that $f(x),g(x), p(x)$ are three polynomials in $F[x].$ Prove that if both $f(x)$ and $p(x)$ are irreducible and $p(x) \mid f(g(x)),$ then $\deg f(x) \mid \deg p(x).$

Solution. Let $\mathfrak{m}$ and $\mathfrak{n}$ be the ideals of $F[x]$ generated by $f(x)$ and $p(x),$ respectively. Let $E=F[x]/\mathfrak{m}$ and $L = F[x]/\mathfrak{n}.$ Since both $f(x)$ and $p(x)$ are irreducible, $E$ and $L$ are field extensions of $F.$ Now, define the map $\varphi : E \longrightarrow L$ by $\varphi(h(x) + \mathfrak{m})=h(g(x)) + \mathfrak{n},$ for all $h(x) \in F[x].$ We first show that $\varphi$ is well-defined. To see this, suppose that $h(x) \in \mathfrak{m}.$ Then $h(x)=f(x)u(x)$ for some $u(x) \in F[x]$ and hence $h(g(x)) = f(g(x))u(g(x)) \in \mathfrak{n},$ because $p(x) \mid f(g(x)).$ So $\varphi$ is well-defined. Now $\varphi$ is clearly a ring homomorphism and, since $E$ is a field and $\ker \varphi$ is an ideal of $E,$ we must have $\ker \varphi = \{0\}.$ Therefore we may assume that $F \subseteq E \subseteq L$ and hence $\deg p(x) = [L : F] = [L: E][E:F]=(\deg f(x))[L : E]. \ \Box$