This question refers to WimC's answer to this question. Consider the cubic congruence problem: $$ f(x) := x^3 - x^2 - 2x + 1 \equiv 0 \pmod{p} $$ We want to know for which $p$ does $f(x)$ splits. The answer to this is $p \equiv 0,1,6 \pmod{7}$, and when proving this WimC made the following claim:
If $f(x)$ has a root in $\Bbb{F}_p$, then $\Bbb{F}_{p^2}$ contains a primitive seventh root of unity.
I fail to see why this is the case. As discussed in his answer/comments, if $\alpha$ is a solution then we can split $f(x) \equiv (x - \alpha_1)(x - \alpha_2)(x - \alpha_3)$ in any field, where $\alpha_1 = \alpha$, $\alpha_2 = \alpha^2 - \alpha - 1$, $\alpha_3 = -\alpha^2 + 2$. Furthermore, if $\beta$ is a root to $x^2 + \alpha x + 1$, then $\beta$ is a primitive seventh root of unity. I can understand these parts.
He then further pointed out that: $$ \Phi_7(x) = x^6 + x^5 + x^4 + x^3 + x^2 + x + 1 = \prod_{i=1}^3 (x^2 + \alpha_ix + 1) $$ and any root of $\Phi_7(x)$ must have degree $\leq 2$. I'm lost at this part.
- Is he making the claim that $\Phi_7(x)$ has a root in $\Bbb{F}_{p^2}$? If so, why is this true?
- Why does the degree of roots matter here?
I'm relatively new to number theory with minimal exposure to Galois theory, so any beginner-friendly explanation would be appreciated.
Any root of $\Phi_7(x)$ must be a root of one of the quadratics $x^2 + \alpha_i x + 1$. If the quadratic is reducible, then clearly the root lies in $\mathbb{F}_p$; otherwise, the roots of an irreducible quadratic must lie in $\mathbb{F}_{p^2}$. This is a special case of the general fact that the field $\mathbb{F}_{p^n}$ contains roots for all irreducible polynomials of degree $n$ over $\mathbb{F}_p$, which can be proven from the uniqueness of the field of order $p^n$.