# Why does the polynomial splitting implies existence primitive root of unity in $\Bbb{F}_{p^2}$?

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.

1. Is he making the claim that $$\Phi_7(x)$$ has a root in $$\Bbb{F}_{p^2}$$? If so, why is this true?
2. 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.

• well, I suggest you look at zakuski.utsa.edu/~jagy/cox_galois_Gaussian_periods.pdf and the large number of examples in books.google.com/… Jun 27, 2020 at 1:56
• @WillJagy thank you, but your reference seems to be highly Galois theoretic. Is there a way to solve this problem without using Galois theory? Jun 27, 2020 at 2:50
• That's the point, this was about 30 years before Galois Theory. Your cubic is the first example. Your cubic starts on page 239 of Cox Jun 27, 2020 at 2:55
• Also related. Actually I'm still wondering what the high school level (=contest) solution to that problem is :-) Jun 27, 2020 at 3:59
• @JyrkiLahtonen this is what I'm looking for. Thank you! Jun 27, 2020 at 4:07

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 $$p^n$$.