How do I prove that $x^6+x^5+x^4+x^3+x^2+x+1$ is irreducible over $\mathbb{Z}_3[x]$?

Question

I am not really sure that the polynomial $x^6+x^5+x^4+x^3+x^2+x+1$ is really irreducible over $\mathbb{Z}_3[x]$, but all my attempts to factor it failed so far. I know I can't use Eisenstein for this so what are my options?

score 5 · Accepted Answer · answered Oct 11 '18 at 17:35

This is the seventh cyclotomic polynomial, so its roots are the primitive seventh roots of unity. An $n$-th root of unity generates an extension of degree $k$ in $\Bbb F_p$ where $k$ is the least number with $p^k\equiv 1\pmod n$.

So to show that this polynomial is irreducible you need to prove the least positive solution of $3^k\equiv1\pmod 7$ is $k=6$.

score 2 · Answer 2 · answered Oct 11 '18 at 18:38

This polynomial is $$p(x)=\frac{x^7-1}{x-1}$$

In particular $p(x)$ divides $x^7-1$. Suppose it is reducible. Then it has an irreducible factor of degree at most $3$, say $f(x)$.

You can easily see that $p(x)$ has no roots in $\Bbb Z_3$, hence $f(x)$ has degree $2$ or $3$.

Now, there is a result on irreducibles on $\Bbb Z_p[x]$:

All irreducible polynomials of degree at most $d$ divide $$x^{p^d}-x$$

In particular $f(x)$ divides $$\frac{x^{27}-x}{x}=x^{26}-1$$

Hence, $f(x)$ divides $\gcd(x^{26}-1, x^7-1)$. However, $$\gcd(x^{26}-1, x^7-1) = x^{\gcd(26,7)}-1=x-1$$ hence $f(x)$ cannot exist. We have to conclude that $p(x)$ is irreducible.

How do I prove that $x^6+x^5+x^4+x^3+x^2+x+1$ is irreducible over $\mathbb{Z}_3[x]$?

2 Answers2