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?
 A: 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 


*

*those are precisely the $q^s$ roots of $x^{q^s}-x$.

*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  $\mathbb{F}_{q^s}$.
