Factorising $X^{16}- X$ over $mathbb F_4$.












2












$begingroup$


I need to factorise $X^{16}- X$ over $mathbb F_4$. How might I go about this? I have factorised over $mathbb F_2$ and I know the quadratic must split but I'm not sure about the quartic and octic. Is there any theory that can help me here or is bruteforce the only way?



Here is the factorisation over $mathbb F_2$:



$$X(X+1)(X^2+X+1)(X^4+X^3+X^2+X+1)(X^8+X^7+X^5+X^4+X^3+X+1)$$






polynomials finite-fields factoring







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$endgroup$








  • 1




    $begingroup$
    For reference.
    $endgroup$
    – Git Gud
    Dec 4 '18 at 1:25










  • $begingroup$
    Your factorization over $Bbb{F}_2$ is incomplete. The octic is a product of two quartics. See here.
    $endgroup$
    – Jyrki Lahtonen
    Dec 4 '18 at 4:25
















2












$begingroup$


I need to factorise $X^{16}- X$ over $mathbb F_4$. How might I go about this? I have factorised over $mathbb F_2$ and I know the quadratic must split but I'm not sure about the quartic and octic. Is there any theory that can help me here or is bruteforce the only way?



Here is the factorisation over $mathbb F_2$:



$$X(X+1)(X^2+X+1)(X^4+X^3+X^2+X+1)(X^8+X^7+X^5+X^4+X^3+X+1)$$






polynomials finite-fields factoring







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    For reference.
    $endgroup$
    – Git Gud
    Dec 4 '18 at 1:25










  • $begingroup$
    Your factorization over $Bbb{F}_2$ is incomplete. The octic is a product of two quartics. See here.
    $endgroup$
    – Jyrki Lahtonen
    Dec 4 '18 at 4:25














2












2








2





$begingroup$


I need to factorise $X^{16}- X$ over $mathbb F_4$. How might I go about this? I have factorised over $mathbb F_2$ and I know the quadratic must split but I'm not sure about the quartic and octic. Is there any theory that can help me here or is bruteforce the only way?



Here is the factorisation over $mathbb F_2$:



$$X(X+1)(X^2+X+1)(X^4+X^3+X^2+X+1)(X^8+X^7+X^5+X^4+X^3+X+1)$$






polynomials finite-fields factoring







share|cite|improve this question











$endgroup$




I need to factorise $X^{16}- X$ over $mathbb F_4$. How might I go about this? I have factorised over $mathbb F_2$ and I know the quadratic must split but I'm not sure about the quartic and octic. Is there any theory that can help me here or is bruteforce the only way?



Here is the factorisation over $mathbb F_2$:



$$X(X+1)(X^2+X+1)(X^4+X^3+X^2+X+1)(X^8+X^7+X^5+X^4+X^3+X+1)$$






polynomials finite-fields factoring




polynomials finite-fields factoring






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 4 '18 at 1:26









Git Gud

28.7k1050100




28.7k1050100










asked Dec 4 '18 at 1:16









DeviloDevilo

11718




11718








  • 1




    $begingroup$
    For reference.
    $endgroup$
    – Git Gud
    Dec 4 '18 at 1:25










  • $begingroup$
    Your factorization over $Bbb{F}_2$ is incomplete. The octic is a product of two quartics. See here.
    $endgroup$
    – Jyrki Lahtonen
    Dec 4 '18 at 4:25














  • 1




    $begingroup$
    For reference.
    $endgroup$
    – Git Gud
    Dec 4 '18 at 1:25










  • $begingroup$
    Your factorization over $Bbb{F}_2$ is incomplete. The octic is a product of two quartics. See here.
    $endgroup$
    – Jyrki Lahtonen
    Dec 4 '18 at 4:25








1




1




$begingroup$
For reference.
$endgroup$
– Git Gud
Dec 4 '18 at 1:25




$begingroup$
For reference.
$endgroup$
– Git Gud
Dec 4 '18 at 1:25












$begingroup$
Your factorization over $Bbb{F}_2$ is incomplete. The octic is a product of two quartics. See here.
$endgroup$
– Jyrki Lahtonen
Dec 4 '18 at 4:25




$begingroup$
Your factorization over $Bbb{F}_2$ is incomplete. The octic is a product of two quartics. See here.
$endgroup$
– Jyrki Lahtonen
Dec 4 '18 at 4:25










1 Answer
1






active

oldest

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3












$begingroup$

Let $q:=4$. You are factorizing $X^{q^2}-X$ over $mathbb{F}_q$. Thus, you should expect that $$X^{q^2}-X=(X^q-X),f_1(X),f_2(X),cdots,f_k(X)$$
where $f_1(X),f_2(X),ldots,f_k(X)inmathbb{F}_q[X]$ are all the irreducible monic quadratic polynomials over $mathbb{F}_q$ (whence $k=dfrac{q^2-q}{2}$, which equals $6$ when $q=4$). You are right that the quadratic polynomial $X^2+X+1$ factors into linear terms over $mathbb{F}_q=mathbb{F}_4$, namely,
$$X^q-X=X^4-X=X(X+1)(X^2+X+1)=X(X+1)(X+t)(X+t+1),,$$
where $t$ is an element of $mathbb{F}_4setminusmathbb{F}_2$.



Therefore, you have to find $6$ polynomials of the form
$$X^2+aX+b,,$$
with $a,binmathbb{F}_4={0,1,t,t+1}$, that are not divisible by $X$, $X+1$, $X+t$, or $X+t+1$. If $a=0$, then note that every element of $mathbb{F}_4$ is a square ($0=0^2$, $1=1^2$, $t=(t+1)^2$, and $t+1=t^2$), so $b=c^2$ for some $cinmathbb{F}_4$, and $X^2+aX+b=(X+c)^2$. Therefore, $aneq 0$.



If $a=1$, then $bin{t,t+1}$, and both choices work. If $a=t$, then $bin{1,t}$ and both choices work. If $a=t+1$, then $bin{1,t+1}$ and both choices work. This shows that
$$begin{align}X^{16}-X&=X(X+1)(X+t)big(X+(t+1)big),(X^2+X+t),big(X^2+X+(t+1)big)\&phantom{abcf}(X^2+tX+1),(X^2+tX+t),big(X^2+(t+1),X+1big),big(X^2+(t+1),X+(t+1)big),.end{align}$$



I cannot offer a good way to directly deal with the octic $X^8+X^7+X^5+X^4+X^3+X+1$, except from the work above. However, there is an easy way to factor the quartic polynomial $X^4+X^3+X^2+X+1$. Let $Y:=X+dfrac{1}{X}$. Then,
$$begin{align}X^4+X^3+X^2+X+1&=X^2,(Y^2+Y+1)=X^2(Y+t)(Y+t+1)\&=(X^2+tX+1),big(X^2+(t+1)X+1big),.end{align}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    How do you know in advance that they should factorise into quadratics?
    $endgroup$
    – Devilo
    Dec 4 '18 at 2:36






  • 1




    $begingroup$
    @Devilo It follows from a theorem which states that $mathbb{F}_{q}$ is the splitting field of $X^{q}-X$, where $q$ is a prime power. Therefore, $mathbb{F}_{16}$ is the splitting field of $X^{16}-X$. Since $[mathbb{F}_{16}:mathbb{F}_4]=2$, it follows that the irreducible factors of $X^{16}-X$ over $mathbb{F}_4$ are either linear or quadratic.
    $endgroup$
    – Batominovski
    Dec 4 '18 at 2:43











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$begingroup$

Let $q:=4$. You are factorizing $X^{q^2}-X$ over $mathbb{F}_q$. Thus, you should expect that $$X^{q^2}-X=(X^q-X),f_1(X),f_2(X),cdots,f_k(X)$$
where $f_1(X),f_2(X),ldots,f_k(X)inmathbb{F}_q[X]$ are all the irreducible monic quadratic polynomials over $mathbb{F}_q$ (whence $k=dfrac{q^2-q}{2}$, which equals $6$ when $q=4$). You are right that the quadratic polynomial $X^2+X+1$ factors into linear terms over $mathbb{F}_q=mathbb{F}_4$, namely,
$$X^q-X=X^4-X=X(X+1)(X^2+X+1)=X(X+1)(X+t)(X+t+1),,$$
where $t$ is an element of $mathbb{F}_4setminusmathbb{F}_2$.



Therefore, you have to find $6$ polynomials of the form
$$X^2+aX+b,,$$
with $a,binmathbb{F}_4={0,1,t,t+1}$, that are not divisible by $X$, $X+1$, $X+t$, or $X+t+1$. If $a=0$, then note that every element of $mathbb{F}_4$ is a square ($0=0^2$, $1=1^2$, $t=(t+1)^2$, and $t+1=t^2$), so $b=c^2$ for some $cinmathbb{F}_4$, and $X^2+aX+b=(X+c)^2$. Therefore, $aneq 0$.



If $a=1$, then $bin{t,t+1}$, and both choices work. If $a=t$, then $bin{1,t}$ and both choices work. If $a=t+1$, then $bin{1,t+1}$ and both choices work. This shows that
$$begin{align}X^{16}-X&=X(X+1)(X+t)big(X+(t+1)big),(X^2+X+t),big(X^2+X+(t+1)big)\&phantom{abcf}(X^2+tX+1),(X^2+tX+t),big(X^2+(t+1),X+1big),big(X^2+(t+1),X+(t+1)big),.end{align}$$



I cannot offer a good way to directly deal with the octic $X^8+X^7+X^5+X^4+X^3+X+1$, except from the work above. However, there is an easy way to factor the quartic polynomial $X^4+X^3+X^2+X+1$. Let $Y:=X+dfrac{1}{X}$. Then,
$$begin{align}X^4+X^3+X^2+X+1&=X^2,(Y^2+Y+1)=X^2(Y+t)(Y+t+1)\&=(X^2+tX+1),big(X^2+(t+1)X+1big),.end{align}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    How do you know in advance that they should factorise into quadratics?
    $endgroup$
    – Devilo
    Dec 4 '18 at 2:36






  • 1




    $begingroup$
    @Devilo It follows from a theorem which states that $mathbb{F}_{q}$ is the splitting field of $X^{q}-X$, where $q$ is a prime power. Therefore, $mathbb{F}_{16}$ is the splitting field of $X^{16}-X$. Since $[mathbb{F}_{16}:mathbb{F}_4]=2$, it follows that the irreducible factors of $X^{16}-X$ over $mathbb{F}_4$ are either linear or quadratic.
    $endgroup$
    – Batominovski
    Dec 4 '18 at 2:43
















3












$begingroup$

Let $q:=4$. You are factorizing $X^{q^2}-X$ over $mathbb{F}_q$. Thus, you should expect that $$X^{q^2}-X=(X^q-X),f_1(X),f_2(X),cdots,f_k(X)$$
where $f_1(X),f_2(X),ldots,f_k(X)inmathbb{F}_q[X]$ are all the irreducible monic quadratic polynomials over $mathbb{F}_q$ (whence $k=dfrac{q^2-q}{2}$, which equals $6$ when $q=4$). You are right that the quadratic polynomial $X^2+X+1$ factors into linear terms over $mathbb{F}_q=mathbb{F}_4$, namely,
$$X^q-X=X^4-X=X(X+1)(X^2+X+1)=X(X+1)(X+t)(X+t+1),,$$
where $t$ is an element of $mathbb{F}_4setminusmathbb{F}_2$.



Therefore, you have to find $6$ polynomials of the form
$$X^2+aX+b,,$$
with $a,binmathbb{F}_4={0,1,t,t+1}$, that are not divisible by $X$, $X+1$, $X+t$, or $X+t+1$. If $a=0$, then note that every element of $mathbb{F}_4$ is a square ($0=0^2$, $1=1^2$, $t=(t+1)^2$, and $t+1=t^2$), so $b=c^2$ for some $cinmathbb{F}_4$, and $X^2+aX+b=(X+c)^2$. Therefore, $aneq 0$.



If $a=1$, then $bin{t,t+1}$, and both choices work. If $a=t$, then $bin{1,t}$ and both choices work. If $a=t+1$, then $bin{1,t+1}$ and both choices work. This shows that
$$begin{align}X^{16}-X&=X(X+1)(X+t)big(X+(t+1)big),(X^2+X+t),big(X^2+X+(t+1)big)\&phantom{abcf}(X^2+tX+1),(X^2+tX+t),big(X^2+(t+1),X+1big),big(X^2+(t+1),X+(t+1)big),.end{align}$$



I cannot offer a good way to directly deal with the octic $X^8+X^7+X^5+X^4+X^3+X+1$, except from the work above. However, there is an easy way to factor the quartic polynomial $X^4+X^3+X^2+X+1$. Let $Y:=X+dfrac{1}{X}$. Then,
$$begin{align}X^4+X^3+X^2+X+1&=X^2,(Y^2+Y+1)=X^2(Y+t)(Y+t+1)\&=(X^2+tX+1),big(X^2+(t+1)X+1big),.end{align}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    How do you know in advance that they should factorise into quadratics?
    $endgroup$
    – Devilo
    Dec 4 '18 at 2:36






  • 1




    $begingroup$
    @Devilo It follows from a theorem which states that $mathbb{F}_{q}$ is the splitting field of $X^{q}-X$, where $q$ is a prime power. Therefore, $mathbb{F}_{16}$ is the splitting field of $X^{16}-X$. Since $[mathbb{F}_{16}:mathbb{F}_4]=2$, it follows that the irreducible factors of $X^{16}-X$ over $mathbb{F}_4$ are either linear or quadratic.
    $endgroup$
    – Batominovski
    Dec 4 '18 at 2:43














3












3








3





$begingroup$

Let $q:=4$. You are factorizing $X^{q^2}-X$ over $mathbb{F}_q$. Thus, you should expect that $$X^{q^2}-X=(X^q-X),f_1(X),f_2(X),cdots,f_k(X)$$
where $f_1(X),f_2(X),ldots,f_k(X)inmathbb{F}_q[X]$ are all the irreducible monic quadratic polynomials over $mathbb{F}_q$ (whence $k=dfrac{q^2-q}{2}$, which equals $6$ when $q=4$). You are right that the quadratic polynomial $X^2+X+1$ factors into linear terms over $mathbb{F}_q=mathbb{F}_4$, namely,
$$X^q-X=X^4-X=X(X+1)(X^2+X+1)=X(X+1)(X+t)(X+t+1),,$$
where $t$ is an element of $mathbb{F}_4setminusmathbb{F}_2$.



Therefore, you have to find $6$ polynomials of the form
$$X^2+aX+b,,$$
with $a,binmathbb{F}_4={0,1,t,t+1}$, that are not divisible by $X$, $X+1$, $X+t$, or $X+t+1$. If $a=0$, then note that every element of $mathbb{F}_4$ is a square ($0=0^2$, $1=1^2$, $t=(t+1)^2$, and $t+1=t^2$), so $b=c^2$ for some $cinmathbb{F}_4$, and $X^2+aX+b=(X+c)^2$. Therefore, $aneq 0$.



If $a=1$, then $bin{t,t+1}$, and both choices work. If $a=t$, then $bin{1,t}$ and both choices work. If $a=t+1$, then $bin{1,t+1}$ and both choices work. This shows that
$$begin{align}X^{16}-X&=X(X+1)(X+t)big(X+(t+1)big),(X^2+X+t),big(X^2+X+(t+1)big)\&phantom{abcf}(X^2+tX+1),(X^2+tX+t),big(X^2+(t+1),X+1big),big(X^2+(t+1),X+(t+1)big),.end{align}$$



I cannot offer a good way to directly deal with the octic $X^8+X^7+X^5+X^4+X^3+X+1$, except from the work above. However, there is an easy way to factor the quartic polynomial $X^4+X^3+X^2+X+1$. Let $Y:=X+dfrac{1}{X}$. Then,
$$begin{align}X^4+X^3+X^2+X+1&=X^2,(Y^2+Y+1)=X^2(Y+t)(Y+t+1)\&=(X^2+tX+1),big(X^2+(t+1)X+1big),.end{align}$$






share|cite|improve this answer











$endgroup$



Let $q:=4$. You are factorizing $X^{q^2}-X$ over $mathbb{F}_q$. Thus, you should expect that $$X^{q^2}-X=(X^q-X),f_1(X),f_2(X),cdots,f_k(X)$$
where $f_1(X),f_2(X),ldots,f_k(X)inmathbb{F}_q[X]$ are all the irreducible monic quadratic polynomials over $mathbb{F}_q$ (whence $k=dfrac{q^2-q}{2}$, which equals $6$ when $q=4$). You are right that the quadratic polynomial $X^2+X+1$ factors into linear terms over $mathbb{F}_q=mathbb{F}_4$, namely,
$$X^q-X=X^4-X=X(X+1)(X^2+X+1)=X(X+1)(X+t)(X+t+1),,$$
where $t$ is an element of $mathbb{F}_4setminusmathbb{F}_2$.



Therefore, you have to find $6$ polynomials of the form
$$X^2+aX+b,,$$
with $a,binmathbb{F}_4={0,1,t,t+1}$, that are not divisible by $X$, $X+1$, $X+t$, or $X+t+1$. If $a=0$, then note that every element of $mathbb{F}_4$ is a square ($0=0^2$, $1=1^2$, $t=(t+1)^2$, and $t+1=t^2$), so $b=c^2$ for some $cinmathbb{F}_4$, and $X^2+aX+b=(X+c)^2$. Therefore, $aneq 0$.



If $a=1$, then $bin{t,t+1}$, and both choices work. If $a=t$, then $bin{1,t}$ and both choices work. If $a=t+1$, then $bin{1,t+1}$ and both choices work. This shows that
$$begin{align}X^{16}-X&=X(X+1)(X+t)big(X+(t+1)big),(X^2+X+t),big(X^2+X+(t+1)big)\&phantom{abcf}(X^2+tX+1),(X^2+tX+t),big(X^2+(t+1),X+1big),big(X^2+(t+1),X+(t+1)big),.end{align}$$



I cannot offer a good way to directly deal with the octic $X^8+X^7+X^5+X^4+X^3+X+1$, except from the work above. However, there is an easy way to factor the quartic polynomial $X^4+X^3+X^2+X+1$. Let $Y:=X+dfrac{1}{X}$. Then,
$$begin{align}X^4+X^3+X^2+X+1&=X^2,(Y^2+Y+1)=X^2(Y+t)(Y+t+1)\&=(X^2+tX+1),big(X^2+(t+1)X+1big),.end{align}$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 4 '18 at 1:58

























answered Dec 4 '18 at 1:37









BatominovskiBatominovski

1




1












  • $begingroup$
    How do you know in advance that they should factorise into quadratics?
    $endgroup$
    – Devilo
    Dec 4 '18 at 2:36






  • 1




    $begingroup$
    @Devilo It follows from a theorem which states that $mathbb{F}_{q}$ is the splitting field of $X^{q}-X$, where $q$ is a prime power. Therefore, $mathbb{F}_{16}$ is the splitting field of $X^{16}-X$. Since $[mathbb{F}_{16}:mathbb{F}_4]=2$, it follows that the irreducible factors of $X^{16}-X$ over $mathbb{F}_4$ are either linear or quadratic.
    $endgroup$
    – Batominovski
    Dec 4 '18 at 2:43


















  • $begingroup$
    How do you know in advance that they should factorise into quadratics?
    $endgroup$
    – Devilo
    Dec 4 '18 at 2:36






  • 1




    $begingroup$
    @Devilo It follows from a theorem which states that $mathbb{F}_{q}$ is the splitting field of $X^{q}-X$, where $q$ is a prime power. Therefore, $mathbb{F}_{16}$ is the splitting field of $X^{16}-X$. Since $[mathbb{F}_{16}:mathbb{F}_4]=2$, it follows that the irreducible factors of $X^{16}-X$ over $mathbb{F}_4$ are either linear or quadratic.
    $endgroup$
    – Batominovski
    Dec 4 '18 at 2:43
















$begingroup$
How do you know in advance that they should factorise into quadratics?
$endgroup$
– Devilo
Dec 4 '18 at 2:36




$begingroup$
How do you know in advance that they should factorise into quadratics?
$endgroup$
– Devilo
Dec 4 '18 at 2:36




1




1




$begingroup$
@Devilo It follows from a theorem which states that $mathbb{F}_{q}$ is the splitting field of $X^{q}-X$, where $q$ is a prime power. Therefore, $mathbb{F}_{16}$ is the splitting field of $X^{16}-X$. Since $[mathbb{F}_{16}:mathbb{F}_4]=2$, it follows that the irreducible factors of $X^{16}-X$ over $mathbb{F}_4$ are either linear or quadratic.
$endgroup$
– Batominovski
Dec 4 '18 at 2:43




$begingroup$
@Devilo It follows from a theorem which states that $mathbb{F}_{q}$ is the splitting field of $X^{q}-X$, where $q$ is a prime power. Therefore, $mathbb{F}_{16}$ is the splitting field of $X^{16}-X$. Since $[mathbb{F}_{16}:mathbb{F}_4]=2$, it follows that the irreducible factors of $X^{16}-X$ over $mathbb{F}_4$ are either linear or quadratic.
$endgroup$
– Batominovski
Dec 4 '18 at 2:43


















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Ellipse (mathématiques)

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Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
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Quarter-circle Tiles

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up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
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Mont Emei

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Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
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