1

I am completely in the dark about how to approach a problem like this. I know that over the splitting field, the polynomial will have 24 roots, and it will look like $$ (x - \gamma_1)(x - \gamma_2) \dots (x - \gamma_{24}) $$

maritsm
  • 607
  • 2
    Hmm, $11$ is not a root. $P(x)=x^{24}+x+1$ or $P(x)=x^{23}+x+1$? – Martín Vacas Vignolo Oct 31 '21 at 08:32
  • 2
    Using the Frobenius automorphism $F(x)=x^{23}$ you can show that if $\alpha$ is a zero of this polynomial, then $F(F(F(\alpha)))=\alpha$. It follows that $\alpha$ is an element of $\Bbb{F}_{23^3}$. Along the way you will find that a smaller field will not work. – Jyrki Lahtonen Oct 31 '21 at 08:43
  • 1
    @JyrkiLahtonen Yes, I recognized $F$ looking like an element of the anharmonic group, but I got sidetracked by the $11$ thing. :-) – anon Oct 31 '21 at 08:49
  • 1
    $P(x) = x^{24} + x + 1$. I erroneously thought that 11 was a root because the calculator I used interpreted the input "11^24 mod 23" as "11^(24 mod 23)".... I'm sorry for the confusion – maritsm Oct 31 '21 at 08:50
  • 4
    Now I know why this felt familiar. I have given a bounty to an older question generalizing this. – Jyrki Lahtonen Oct 31 '21 at 08:53
  • 1
    Also sprach Mathematica: Factor[x^24+x+1,Modulus->23] outputs $$\left(x^3+2 x^2+22 x+22\right) \left(x^3+3 x^2+22\right) \left(x^3+4 x^2+x+22\right) \left(x^3+5 x^2+2 x+22\right) \left(x^3+8 x^2+5 x+22\right) \left(x^3+11 x^2+8 x+22\right) \left(x^3+16 x^2+13 x+22\right) \left(x^3+20 x^2+17 x+22\right)$$ as a totally different verification for the fact that all the roots reside in the cubic extension. – Jyrki Lahtonen Oct 31 '21 at 10:03

1 Answers1

1

Recall the Frobenius endomorphism defined by $x \mapsto x^{p}$ where $p$ is the characteristic of the field. We need to determine what field the roots are in. We know that in a field with $p^n$ elements, applying the Frobenius endomorphism $n$ times will yield the identity function. (This is because all elements of the field satisfy $x^{p^n} - x = 0$)

So we take $F(x) = x^{23}$. From the polynomial, we see that is $\gamma$ is a root of the polynomial, then $\gamma^{23} = \frac{-\gamma - 1}{\gamma}$.

Using this and some elementary algebraic manipulations, you can see that $$ F^2(\gamma) = - \frac{1}{\gamma + 1}$$ and $$ F^3(\gamma) = \gamma$$ So in general, applying Frobenius three times gives back the root itself, so the root must be in $\mathbb F_{23^3}$.

maritsm
  • 607
  • Upvoted your answer as a nod of appreciation. But I think I will use my dupehammer and close this as a duplicate. My understanding is that this won't give you any trouble. If I'm wrong, please @-ping me, and explain the situation (or bring it up in meta so that more people can contribute to the discussion). – Jyrki Lahtonen Oct 31 '21 at 09:59
  • Many users here are sticklers for the guidelines of what makes a good question. The statement of your question is a bit short on context. Familiarizing yourself with those expectations will improve your future stay on the site, so I recommend that you take a look if you haven't already. – Jyrki Lahtonen Oct 31 '21 at 10:02