# irreducible polynomial has all its roots in $F_s$ and every element in $F_s$ satisfies such a polynomial

I'm currently trying to prove a certain statement( I will not post the whole statement because there is only one part, which is not clear to me).

Let us say that we have a finite field $$F$$ with $$q$$ elements. I was able to prove that $$x^{q^s} - x$$ is the product of all monic irreducible polynomials of degree dividing $$s$$. but now I have to follow that every such irreducible polynomial has all its roots in $$F_s$$ (finite field with $$q^s$$ elements) and conversely that every element in $$F_s$$ satisfies such a polynomial. But I don't see why. Is there an easy way to prove both directions?

For each $$y \in \mathbb{F}_{q^s}$$, note that $$y^{q^s} = y$$ [make sure you see why]. There are $$q^s$$ such $$y$$ (which is the degree of the polynomial $$x^{q^s}-x$$), so
1. those are precisely the $$q^s$$ roots of $$x^{q^s}-x$$.
2. let us write $$x^{q^s}-x= \prod_i p_i(x)$$ where $$p_i(x)$$ are irreducible in $$\mathbb{F}_q[x]$$. Suppose there exists a $$y \in \mathbb{F}_{q^s}$$ such that $$p_i(y) \not = 0$$ for all such $$i$$. Then this would imply that $$\prod_i p_i(y) = y^{q^s}-y \not = 0$$, which contradicts what we observed already above: $$y^{q^s} = y$$ for each $$y \in \mathbb{F}_{q^s}$$.
So 2 gives at least one of what you are trying to show, every $$y \in \mathbb{F}_{q^s}$$ satisfies $$p(y)=0$$ for some monic irreducible polynomial $$p(x)$$ dividing $$x^{q^s}-x$$; eqquivalently [as you had shown already] $$p \in \mathbb{F}_q[x]$$ irreducible and deg$$(p)|s$$.
On the other hand: You had already shown that every monic irreducible polynomial $$p \in \mathbb{F}_q[x]$$ satisfying deg$$(p) | s$$, divides the polynomial $$x^{q^s}-x$$. As $$x^{p^s}-x$$ factors completely in $$\mathbb{F}_{q^s}$$ (i.e., 1. above) it follows that $$p$$ must also factor completely in $$\mathbb{F}_{q^s}$$.
• Thank you for your answer, Mike. :) I proved now that for each $y \in F_{q^s}$ it holds that $y^{q^s} = y$. Now I will deal with the rest of your answer. – RukiaKuchiki Jan 5 '19 at 13:31