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.

  • $\begingroup$ 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/… $\endgroup$
    – Will Jagy
    Jun 27, 2020 at 1:56
  • $\begingroup$ @WillJagy thank you, but your reference seems to be highly Galois theoretic. Is there a way to solve this problem without using Galois theory? $\endgroup$ Jun 27, 2020 at 2:50
  • 2
    $\begingroup$ 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 $\endgroup$
    – Will Jagy
    Jun 27, 2020 at 2:55
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    $\begingroup$ Also related. Actually I'm still wondering what the high school level (=contest) solution to that problem is :-) $\endgroup$ Jun 27, 2020 at 3:59
  • $\begingroup$ @JyrkiLahtonen this is what I'm looking for. Thank you! $\endgroup$ Jun 27, 2020 at 4:07

1 Answer 1


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 $p^n$.

  • $\begingroup$ That clears things up! Thanks! $\endgroup$ Jun 27, 2020 at 4:07

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