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Conjecture: the primes for which the polynomial $x^3 - x^2 - 2 x + 1$ splits mod $p$ are the primes $\equiv 1$ or $6$ mod $7$ (OEIS sequence A045472). Is this correct?

Robert Israel
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  • You may need to include $p=7$ because $x^3 - x^2 - 2 x + 1 = (x+2)^3 \bmod 7$. – lhf Jun 14 '17 at 18:26
  • see final example in https://www.impan.pl/en/publishing-house/journals-and-series/acta-arithmetica/all/28/3/75096/representability-by-certain-norm-forms-over-algebraic-number-fields – Will Jagy Jun 14 '17 at 18:41
  • also see page 6 in Reuschle, https://books.google.com/books?id=wt7lgfeYqMQC&pg=PR1&lpg=PR1&dq=reuschle++tafeln+complexer+primzahlen&source=bl&ots=VGZFPrfUBn&sig=MlQ667PqXaQ9rAvLWkG3_F1rwsk&hl=en&sa=X&ved=0ahUKEwiIwtSvm9TQAhUJ-2MKHXJIA_kQ6AEIODAE#v=onepage&q=reuschle%20%20tafeln%20complexer%20primzahlen&f=false – Will Jagy Jun 14 '17 at 18:50
  • Related. – Jyrki Lahtonen Jun 27 '20 at 03:56

2 Answers2

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If $\alpha$ is a root of $x^3-x^2-2x+1$ (in any field) then the latter splits as $$(x-\alpha)(x-\alpha^2+\alpha+1)(x+\alpha^2-2).$$ Also if $\alpha \neq -2$ and $\beta$ is a root of $$x^2+\alpha x + 1$$ then $\beta$ is a primitive seventh root of unity. Combining all this for the field $\mathbb{F}_p$ where $p\neq 7$: If $x^3-x^2-2x+1$ has a root in $\mathbb{F}_p$ then $\mathbb{F}_{p^2}$ contains a primitive seventh root of unity and so $p^2 \equiv 1 \pmod 7$. The other way around: If $p^2\equiv 1 \pmod 7$ then $\mathbb{F}_{p^2}$ contains a primitive seventh root of unity $\beta$ and $\alpha = -\beta-\beta^{-1}$ is a root of $x^3-x^2-2x+1$. Moreover $\beta^p = \beta$ (if $p\equiv 1 \bmod 7$) or $\beta^p = \beta^{-1}$ (if $p \equiv -1 \bmod 7)$. Either way $\alpha^p = \alpha$ so $\alpha \in \mathbb{F}_p$. For $p=7$ $$x^3-x^2-2x+1\equiv (x+2)^3 \pmod 7$$ as pointed out by @lhf in a comment.

WimC
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  • Do you mind explaining why if $x^3 - x^2 - 2x + 1$ has a root in $\Bbb{F}p$, then $\Bbb{F}{p^2}$ contains a primitive seventh root of unity? – Clement Yung Jun 26 '20 at 16:20
  • @ClementYung If $\alpha_1$, $\alpha_2$, $\alpha_3$ are the roots in $\mathbb{F}p$ (as provided in the answer) then $$x^6+x^5+x^4+x^3+x^2+x+1 = \prod{k=1}^3(x^2 + \alpha_k x + 1)$$ – WimC Jun 26 '20 at 16:51
  • Sorry, but I still don't understand. The polynomial you gave contains all the primitive roots. What's next? And where does $\Bbb{F}_{p^2}$ comes in? – Clement Yung Jun 26 '20 at 17:26
  • @ClementYung The right hand side is a product of quadratics over $\mathbb{F}_p$ so any root has degree $\leq 2$ over $\mathbb{F}_p$. – WimC Jun 26 '20 at 17:31
  • Why does the degree of the root matter here? Apologies but I'm afraid I'm completely lost. Do you mind explaining in more detail? I'm still relatively new to number theory. – Clement Yung Jun 26 '20 at 18:35
  • In particular, I'm not sure how we can assure they the polynomial you mentioned has a root. – Clement Yung Jun 26 '20 at 18:55
  • I have made a question on my queries here. If you are available, feel free to take a look. – Clement Yung Jun 27 '20 at 01:30
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Yes, this is because the zeros of the polynomial is $-(\zeta^j+\zeta^{-j})$ for $\zeta=\exp(2\pi i/7)$ and $j\in\{1,2,3\}$. So it splits iff $\zeta+\zeta^{-1}\in\Bbb F_p$ for $\zeta$ now a primitive seventh root of unity in an extension of $\Bbb F_p$. This is the case iff $(\zeta+\zeta^{-1})^p =\zeta+\zeta^{-1}$ in characteristic $p$, and that is the case iff $p\equiv\pm1\pmod7$.

Angina Seng
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