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












2














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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  • 1




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










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
















2














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




















  • 1




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










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














2












2








2







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















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






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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









Devilo

11718




11718








  • 1




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










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














  • 1




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










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








1




1




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




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












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




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










1 Answer
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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























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






  • 1




    @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.
    – Batominovski
    Dec 4 '18 at 2:43











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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























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






  • 1




    @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.
    – Batominovski
    Dec 4 '18 at 2:43
















3














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























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






  • 1




    @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.
    – Batominovski
    Dec 4 '18 at 2:43














3












3








3






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














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









Batominovski

33.9k33292




33.9k33292












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






  • 1




    @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.
    – Batominovski
    Dec 4 '18 at 2:43


















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






  • 1




    @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.
    – Batominovski
    Dec 4 '18 at 2:43
















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




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




1




1




@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.
– Batominovski
Dec 4 '18 at 2:43




@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.
– Batominovski
Dec 4 '18 at 2:43


















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Sphinx de Gizeh

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Pour les articles homonymes, voir Sphinx. Sphinx de Gizeh.mw-parser-output .entete.egypte-antique{background-image:url("//upload.wikimedia.org/wikipedia/commons/c/cc/Picto_infobox_ancient_Egypt.png")} Site d'Égypte antique Le visage du sphinx de Gizeh. Noms en arabe أبو الهول , Abou al-Hôl Localisation Région Basse-Égypte Coordonnées 29° 58′ 31″ nord, 31° 08′ 16″ est Géolocalisation sur la carte : Égypte Sphinx de Gizeh modifier   Le sphinx de Gizeh est la statue thérianthrope qui se dresse devant les grandes pyramides du plateau de Gizeh, en Basse-Égypte. Sculpture monumentale monolithique la plus grande du monde [ 1 ] avec 73,5 mètres de longueur, 14 mètres de largeur et 20,22 mètres de hauteur, elle représente un sphinx couchant. Réalisée vers -2500, elle serait attribuée à Djédefrê, l'un des pharaons de la IV e  dynastie, représentant son père le pharaon Khéops. Sommaire 1 Étymologie 2 Desc...
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Different font size/position of beamer's navigation symbols template's content depending on regular/plain...

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3 Let's suppose I'd like to divert the navigation symbols template (e.g. in order replace them by the frame numbers, see this bug report ;) I would be facing troubles regarding the font size/position, depending on regular/plain frames, as shown by the following MCE. How could I get rid of these troubles? documentclass{beamer} usetheme{PaloAlto} setbeamertemplate{navigation symbols}{insertframenumber/inserttotalframenumber} begin{document} begin{frame} frametitle{A regular frame} end{frame} begin{frame}[plain] frametitle{A plain frame} end{frame} end{document} beamer header-footer share | improve this question asked Dec 10 '18 a...
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