Suppose px is an element of Fx is irreducible of degree n Sh

Suppose p(x) is an element of F[x] is irreducible of degree n. Show that if a(x) an element of F[x]\\{0}, and n>deg(a(x)), then a(x) and p(x) are relatively prime.

Solution

Solution: (Irreducible polynomial over a field): Let F be a field and f(x) be a non-zero and non-unit polynomial in F[x] i.e., f(x) be a   polynomial of positive degree. Then f(x) is said to be irreduciable over F if it has no proper divisors in F[x], otherwise it is reducible.

Since p(x) is irreducible polynomial of F[x] of degree n, so it has no proper divisors in F[x] i.e., it can not be expressed as a product of two proper divisors of f(x). Also given that a(x) is non-zero polynomial of F[x] of degree less than n.

Case I. If a(x) is irreducible over F, then p(x) and a(x) are relatively prime, since these have no common divisors.

Case II. If a(x) is reducible, that is a(x) = g(x).h(x), for some polynomials g(x), h(x) of F[x] of degree less than n separtely. Also in this case a(x) and p(x) no common polynomial divisors except 1. Hence a(x) and p(x) are relatively prime.