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












1














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?






number-theory finite-fields







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    1














    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?






    number-theory finite-fields







    share|cite|improve this question



























      1












      1








      1







      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?






      number-theory finite-fields







      share|cite|improve this question















      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?






      number-theory finite-fields




      number-theory finite-fields






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 2 days ago









      Bernard

      118k639112




      118k639112










      asked 2 days ago









      RukiaKuchiki

      322211




      322211






















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






          share|cite|improve this answer



















          • 1




            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
            23 hours ago













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






          share|cite|improve this answer



















          • 1




            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
            23 hours ago


















          1














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






          share|cite|improve this answer



















          • 1




            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
            23 hours ago
















          1












          1








          1






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






          share|cite|improve this answer














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







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 2 days ago

























          answered 2 days ago









          Mike

          3,099211




          3,099211








          • 1




            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
            23 hours ago
















          • 1




            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
            23 hours ago










          1




          1




          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
          23 hours ago






          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
          23 hours ago




















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