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

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 $$\mathbb F_2$$:

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

• Commented Dec 4, 2018 at 1:25
• Your factorization over $\Bbb{F}_2$ is incomplete. The octic is a product of two quartics. See here. Commented Dec 4, 2018 at 4:25

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 $$f_1(X),f_2(X),\ldots,f_k(X)\in\mathbb{F}_q[X]$$ 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 $$\mathbb{F}_4\setminus\mathbb{F}_2$$.
Therefore, you have to find $$6$$ polynomials of the form $$X^2+aX+b\,,$$ with $$a,b\in\mathbb{F}_4=\{0,1,t,t+1\}$$, 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 $$c\in\mathbb{F}_4$$, and $$X^2+aX+b=(X+c)^2$$. Therefore, $$a\neq 0$$.
If $$a=1$$, then $$b\in\{t,t+1\}$$, and both choices work. If $$a=t$$, then $$b\in\{1,t\}$$ and both choices work. If $$a=t+1$$, then $$b\in\{1,t+1\}$$ 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+1\big)\,\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+1\big)\,.\end{align}
• @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. Commented Dec 4, 2018 at 2:43