Degree of an irreducible factor of composition of two polynomials

Posted: November 17, 2011 in Elementary Algebra; Problems & Solutions, Groups and Fields
Tags: ,

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

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