I am looking at the last paragraph of this lemma. Basically given a finite group $G$ acting faithfully on the field $F$, it claimed that $\left[F:\operatorname{Fix}(G)\right] \geq |G|$, or using primitive element theorem, $$[\operatorname{Fix}(G)[\alpha]:\operatorname{Fix}(G)]\geq|G|.$$
Equivalently, the degree of the minimal polynomial $p_\alpha\in \operatorname{Fix}(G)[x]$ cannot be smaller than $|G|$. I understand $p_\alpha$ is a factor of the polynomial $$P=\prod_{\beta\in \alpha\cdot G}(x-\beta)\in\operatorname{Fix}(G)[x]$$ but how does the faithfulness imply $P$ is irreducible?