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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?

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

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    $\begingroup$ 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. $\endgroup$ – RukiaKuchiki Jan 5 '19 at 13:31

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