Let u be a root of f(x) in some splitting field. Since f(x) is an irreducible polynomial of degree m over K, we have [K(u): K] = m.

Consider the extension F(u) over F. Since f(x) is also a polynomial in F[x], we have [F(u): F] < m with equality if and only if f(x) is irreducible as a polynomial over F.

Note that [F(u): F]n = [F(u): K(u)]m. Since m and n are relatively prime, m divides [F(u): F]. Hence, m < [F(u): F].

Thus, [F(u): F] = m and so f(x) is irreducible as a polynomial over F.