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This is part of a larger question where I'm supposed to determine if $x^6+x^3+1$ is irreducible over $\mathbb{F}_2$, $\mathbb{F}_3$, $\mathbb{F}_{19}$ and $\mathbb{Q}$.

For $\mathbb{F}_3$ and $\mathbb{F}_{19}$ it's easy enough to just find roots and factorise, while over $\mathbb{Q}$ I can substitute and then apply Eisenstein.

I'm struggling to think of how to show it's irreducible over $\mathbb{F}_2$(I'm pretty sure it is). Clearly it doesn't have any roots in $\mathbb{F}_2$ , and so all that's left to show is that no polynomials of degree 2 or 3 divide it, but I have no idea how to go about doing that.

Any help you could offer would be really appreciated.

med
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    See https://math.stackexchange.com/questions/32197/find-all-irreducible-monic-polynomials-in-mathbbz-2x-with-degree-equal for a list of possible factors. There are only a few to test. – lhf Nov 01 '19 at 16:14

3 Answers3

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Let's look for factors of degree $3$. Suppose (over $\mathbb F_2$) $$x^6 + x^3 + 1 = (x^3 + a_2 x^2 + a_1 x + a_0)(x^3 + b_2 x^2 + b_1 x + b_0)$$ Since $a_0 b_0 = 1$, we must have $a_0 = b_0 = 1$. By looking at the $x^5$ and $x^1$ terms, you must have $a_2=b_2$ and $a_1=b_1$. But then the $x^3$ term on the right is $0$.

The case of a factor of degree $2$ is similar.

Robert Israel
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$f(x)=x^6+x^3+1=(x^9-1)/(x^3-1)$ is the ninth cyclotomic polynomial. Its zeros are the primitive ninth roots of unity.

As long as the characteristic of $F$ is not three, $f$ has six distinct zeros in an extension field of $f$. If $\alpha$ is one then $\alpha^9=1$ and $\alpha,\alpha^2,\ldots,\alpha^9$ are all distinct. By Lagrange's theorem, if $\alpha$ lies in a finite field, of order $q$ then $9\mid(q-1)$.

Let $K=\Bbb F_2(\alpha)$ have order $q$. Then $q$ is a power of $2$, and indeed $q=2^k$ where $k\le 6$ (as the degree of $f$ is $6$). But $2^k-1$ must be a multiple of $9$, and the smallest $k$ with $9\mid(2^k-1)$ is $k=6$. Therefore $K$ is a degree $6$ extension of $\Bbb F_2$ so that $f$ must be irreducible over $\Bbb F_2$.

By similar considerations one can find the degree of $\Bbb F_p(\alpha)$ for any prime $p$, and so determine the degrees of the irreducible factors of $f(x)$ over $\Bbb F_p$.

Angina Seng
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This answer will assume that you know some field theory.

Consider the irreducible polynomial

$g(X) = X^2+X+1 \in \mathbb{F}_2[X]$

and the polynomial

$f(X) = X^6+X^3+1 \in \mathbb{F}_2[X]$

Let $\alpha_1,\alpha_2$ be the roots of $g$ in a fixed algebraic closure $\overline{\mathbb{F}_2}$ of $\mathbb{F}_2$. Let $\beta_{11},\beta_{12},\beta_{13}$ be the cube roots of $\alpha_1$ and let $\beta_{21},\beta_{22},\beta_{23}$ be the cube roots of $\alpha_2$. Since $g(X^3) = f(X)$ we have that the $\beta_{ij}$'s are all the roots of $f$ in $\overline{\mathbb{F}_2}$.

Can you argue why adjoining any one of these roots to $\mathbb{F}_2$ necessarily gives a degree $6$ extension of $\mathbb{F}_2$?

Gunnar Sveinsson
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