Mixing
irreducible polynomial has all its roots in $F_s$ and every element in $F_s$ satisfies such a polynomial
$begingroup$
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
edited Jan 3 at 20:06
Bernard
119k740113
asked Jan 3 at 19:35
RukiaKuchikiRukiaKuchiki
337211
$endgroup$
add a comment |
$begingroup$
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
edited Jan 3 at 20:06
Bernard
119k740113
asked Jan 3 at 19:35
RukiaKuchikiRukiaKuchiki
337211
$endgroup$
add a comment |
$begingroup$
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
edited Jan 3 at 20:06
Bernard
119k740113
asked Jan 3 at 19:35
RukiaKuchikiRukiaKuchiki
337211
$endgroup$
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
edited Jan 3 at 20:06
Bernard
119k740113
asked Jan 3 at 19:35
RukiaKuchikiRukiaKuchiki
337211
edited Jan 3 at 20:06
Bernard
119k740113
asked Jan 3 at 19:35
RukiaKuchikiRukiaKuchiki
337211
edited Jan 3 at 20:06
Bernard
119k740113
edited Jan 3 at 20:06
Bernard
119k740113
edited Jan 3 at 20:06
Bernard
119k740113
119k740113
asked Jan 3 at 19:35
RukiaKuchikiRukiaKuchiki
337211
asked Jan 3 at 19:35
RukiaKuchikiRukiaKuchiki
337211
asked Jan 3 at 19:35
RukiaKuchikiRukiaKuchiki
337211
337211
add a comment |
add a comment |
1 Answer
1
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$begingroup$
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 $mathbb{F}_{q^s}$.
answered Jan 3 at 20:24
MikeMike
3,477411
$endgroup$
1
$begingroup$
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 at 13:31
add a comment |
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1 Answer
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1 Answer
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$begingroup$
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 $mathbb{F}_{q^s}$.
answered Jan 3 at 20:24
MikeMike
3,477411
$endgroup$
1
$begingroup$
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 at 13:31
add a comment |
$begingroup$
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 $mathbb{F}_{q^s}$.
answered Jan 3 at 20:24
MikeMike
3,477411
$endgroup$
1
$begingroup$
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 at 13:31
add a comment |
$begingroup$
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 $mathbb{F}_{q^s}$.
answered Jan 3 at 20:24
MikeMike
3,477411
$endgroup$
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 $mathbb{F}_{q^s}$.
answered Jan 3 at 20:24
MikeMike
3,477411
edited Jan 3 at 20:59
answered Jan 3 at 20:24
MikeMike
3,477411
answered Jan 3 at 20:24
MikeMike
3,477411
answered Jan 3 at 20:24
MikeMike
3,477411
3,477411
1
$begingroup$
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 at 13:31
add a comment |
1
$begingroup$
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 at 13:31
1
1
$begingroup$
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 at 13:31
$begingroup$
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 at 13:31
add a comment |
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