Factorising $X^{16}- X$ over $mathbb F_4$.
$begingroup$
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
edited Dec 4 '18 at 1:26
Git Gud
28.7k1050100
asked Dec 4 '18 at 1:16
DeviloDevilo
11718
$endgroup$
add a comment |
$begingroup$
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
edited Dec 4 '18 at 1:26
Git Gud
28.7k1050100
asked Dec 4 '18 at 1:16
DeviloDevilo
11718
$endgroup$
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
add a comment |
$begingroup$
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
edited Dec 4 '18 at 1:26
Git Gud
28.7k1050100
asked Dec 4 '18 at 1:16
DeviloDevilo
11718
$endgroup$
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
edited Dec 4 '18 at 1:26
Git Gud
28.7k1050100
asked Dec 4 '18 at 1:16
DeviloDevilo
11718
edited Dec 4 '18 at 1:26
Git Gud
28.7k1050100
asked Dec 4 '18 at 1:16
DeviloDevilo
11718
edited Dec 4 '18 at 1:26
Git Gud
28.7k1050100
edited Dec 4 '18 at 1:26
Git Gud
28.7k1050100
edited Dec 4 '18 at 1:26
Git Gud
28.7k1050100
28.7k1050100
asked Dec 4 '18 at 1:16
DeviloDevilo
11718
asked Dec 4 '18 at 1:16
DeviloDevilo
11718
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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,
$$begin{align}X^4+X^3+X^2+X+1&=X^2,(Y^2+Y+1)=X^2(Y+t)(Y+t+1)\&=(X^2+tX+1),big(X^2+(t+1)X+1big),.end{align}$$
answered Dec 4 '18 at 1:37
BatominovskiBatominovski
1
$endgroup$
$begingroup$
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
add a comment |
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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,
$$begin{align}X^4+X^3+X^2+X+1&=X^2,(Y^2+Y+1)=X^2(Y+t)(Y+t+1)\&=(X^2+tX+1),big(X^2+(t+1)X+1big),.end{align}$$
answered Dec 4 '18 at 1:37
BatominovskiBatominovski
1
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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,
$$begin{align}X^4+X^3+X^2+X+1&=X^2,(Y^2+Y+1)=X^2(Y+t)(Y+t+1)\&=(X^2+tX+1),big(X^2+(t+1)X+1big),.end{align}$$
answered Dec 4 '18 at 1:37
BatominovskiBatominovski
1
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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,
$$begin{align}X^4+X^3+X^2+X+1&=X^2,(Y^2+Y+1)=X^2(Y+t)(Y+t+1)\&=(X^2+tX+1),big(X^2+(t+1)X+1big),.end{align}$$
answered Dec 4 '18 at 1:37
BatominovskiBatominovski
1
$endgroup$
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,
$$begin{align}X^4+X^3+X^2+X+1&=X^2,(Y^2+Y+1)=X^2(Y+t)(Y+t+1)\&=(X^2+tX+1),big(X^2+(t+1)X+1big),.end{align}$$
answered Dec 4 '18 at 1:37
BatominovskiBatominovski
1
edited Dec 4 '18 at 1:58
answered Dec 4 '18 at 1:37
BatominovskiBatominovski
1
answered Dec 4 '18 at 1:37
BatominovskiBatominovski
1
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
add a comment |
$begingroup$
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
add a comment |
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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