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This question originates from Pinter's Abstract Algebra, Chapter 25 Exercise C4.

How many irreducible cubics are there over a field of $n$ elements?

  1. Total number of monic cubics $x^3 + ax + bx + c$: $n^3$.

  2. For monic reducible cubics in the form of $(x+a)(x+b)(x+c)$, either all, only two, or none of $a, b, c$ are the same. Therefore the total number of such cases is $\displaystyle {n\choose 1} + 2{n\choose 2} + {n\choose 3}$.

  3. Per this discussion, the number of monic irreducible quadratics is $n(n-1)/2$. This suggests for monic reducible cubics in the form of $(x^2+ax+b)(x+c)$ where $x^2+ax+b$ is irreducible, there are $n^2(n-1)/2$ cases.

Hence, the total number of

Monic reducible cubics:

$\displaystyle\quad R = {n\choose 1} + 2{n\choose 2} + {n\choose 3} + \frac{n^2(n-1)}{2}$

Monic irreducible cubics:

$\displaystyle\quad n^3 - R$

Irreducible cubics:

$\displaystyle\quad(n-1)\left(n^3 - R\right)$.

Correct?

user26857
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hchar
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  • I think everything you wrote is correct. I can't find any mistakes. – user729424 Feb 09 '20 at 21:01
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    Do you know that with $a(d)$ the number of monic irreducibles $\in \Bbb{F}q[x]$ of degree $d$, by unique factorization $$\prod{d\ge 1}\frac1{(1-T^d)^{a(d)}}=\prod_{f \in F_q[x] \ monic \ irreducible}(1+\sum_{k\ge 1} T^{\deg(f^k)})=\sum_{g\in F_q[x]\ monic} T^{\deg(g)}= \sum_{d\ge 0} q^d T^d$$ From which $a(d)= \frac1d \sum_{l | d} \mu(l) q^{d/l}$ – reuns Feb 09 '20 at 21:18
  • This and related questions have been asked and answered on our site many times. The two that I selected contain a slight generalization (one of the answers has the most general formula also) as well as an approach like yours for the particular case of cubics. If those leave anything unclear, please comment, and we can reconsider. Do also check other threads linked to the chosen ones (see the right margins for a list). Plenty of information there. – Jyrki Lahtonen Feb 10 '20 at 05:37
  • And +1 for showing your work, making it easier to select appropriate duplicate targets. – Jyrki Lahtonen Feb 10 '20 at 05:38

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