Question

Question: If p(x) and q(x) are non associate irreducible in F(x) , prove that arep(x) and q(x) relatively prime.

Short answer

Answer:

Thus, the given statement is proved

Step-by-step solution

To prove:

The objective is to prove that p(x) and q(x) are relatively prime if p(x) and q(x) are non associate irreducible.

Obtaining p(x) and q(x)  are relatively prime

Let’s consider that d(x)= p(x) q(x)

As per definition, d(x)/p(x) and d(x)is monic thus, either d(x) =1 or d(x) is an associate of p(x). Also,d(x)/p(x) and d(x) is monic thus, either d(x) =1 or d(x) is an associate of q(x).

Thus, if d(x) $\ne$1 then, d(x) is an associate of both p(x) and q(x) which would imply p(x) and q(x) are associate, which is a contradiction.

Therefore, (p(x) ,q(x)) thus p(x) and q(x) arerelatively prime.