Question

Prove that p(x) is irreducible in F(x) if and only if for every $g\left(x\right)\in F\left(x\right)$, either p(x)|g(x) or p(x) is relatively prime to g(x).

Short answer

It is proved that p(x) is irreducible in F(x).

Step-by-step solution

Determine p(x) and g(x) are relatively prime

Consider p(x) is irreducible.

For any g(x), lets $g\left(x\right)=\prod _{i-1}^m{q}_i{\left(x\right)}^{\alpha }$ is the factorization of irreducible. Then, p(x)|g(x ) if and only if p(x)|q_i(x) for $0\le i\le m$.

If p(x)|g(x) then, p(x)=q_i(x) for$0\le i\le m$ and thus p(x) and g(x) are relatively prime.

Determine p(x) is irreducible in F(x) 

Now, assume that for all g(x)$\in$F(x) either$p\left(x\right)\left|g\right(x)$ or p(x) is relatively prime to g(x).

Let’s p(x)=$\prod _{i-1}^m{r}_i{\left(x\right)}^{\alpha }$ is the factorization into irreducible of p(x). Then, p(x) is not relatively prime with any $r_i\left(x\right)$thus, p(x)|r_i(x).

As$r_i\left(x\right)|p\left(x\right)$ it proved that p(x) is associate with all the$r_i\left(x\right)$ then, this is only possible when there is a unique irreducibler(x) in the factorization.