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











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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 $mathbb F_2$:



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






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    For reference.
    – Git Gud
    Dec 4 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 at 4:25















up vote
2
down vote

favorite












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 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 at 4:25













up vote
2
down vote

favorite









up vote
2
down vote

favorite











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













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edited Dec 4 at 1:26









Git Gud

28.6k104899




28.6k104899










asked Dec 4 at 1:16









Devilo

11718




11718








  • 1




    For reference.
    – Git Gud
    Dec 4 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 at 4:25














  • 1




    For reference.
    – Git Gud
    Dec 4 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 at 4:25








1




1




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




For reference.
– Git Gud
Dec 4 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 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 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 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 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 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 at 2:43















up vote
3
down vote













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 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 at 2:43













up vote
3
down vote










up vote
3
down vote









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 at 1:58

























answered Dec 4 at 1:37









Batominovski

33.2k23292




33.2k23292












  • How do you know in advance that they should factorise into quadratics?
    – Devilo
    Dec 4 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 at 2:43


















  • How do you know in advance that they should factorise into quadratics?
    – Devilo
    Dec 4 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 at 2:43
















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




How do you know in advance that they should factorise into quadratics?
– Devilo
Dec 4 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 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 at 2:43


















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Cabo Verde

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  Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º)
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Gyllenstierna

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