Question

(b) Let F be a field and $\mathit{p}\left(x\right)\mathbf{\in }\mathit{F}\left[x\right]$. Prove that$\mathit{p}\left(x\right)$ is irreducible if and only if the idealrole="math" localid="1653368960356" $\left(p\left(x\right)\right)$ is maximal in$\mathit{F}\left[x\right]$ .

Short answer

It is proved that$p\left(\mathrm{x}\right)$ is maximal if and only if$p\left(\mathrm{x}\right)$ is irreducible.

Step-by-step solution

Determine p(x) is maximal

Consider that, $p\left(\mathrm{x}\right)$is irreducible and $\left(\mathrm{p}\left(\mathrm{x}\right)\right)$is not maximal.

But $q\left(\mathrm{x}\right)\in F\left[\mathrm{x}\right]$so that $\left(\mathrm{p}\left(\mathrm{x}\right)\right)⊑\left(\mathrm{q}\left(\mathrm{x}\right)\right)$and here $\left(\mathrm{q}\left(\mathrm{x}\right)\right)\ne F\left[\mathrm{x}\right]$, that means

deg$\left(\mathrm{q}\left(\mathrm{x}\right)\right)>0$and $q\left(\mathrm{x}\right)|p\left(\mathrm{x}\right)$, this contradicts the fact that $p\left(\mathrm{x}\right)$is irreducible. This implies that $p\left(\mathrm{x}\right)$is maximal.

Hence it is proved that$p\left(\mathrm{x}\right)$ is maximal when$p\left(\mathrm{x}\right)$ is irreducible.

Determine p(x) is irreducible

As$p\left(\mathrm{x}\right)$ is maximal, then by corollary, that is, In a commutative ring R with identity, every maximal ideal is prime, since$p\left(\mathrm{x}\right)$ is a prime ideal, this implies that$p\left(\mathrm{x}\right)$ is irreducible.

Hence it is proved that$p\left(\mathrm{x}\right)$is maximal if and only if$p\left(\mathrm{x}\right)$ is irreducible.