I want to know how hany monic irreducible polynomials of degree $3$ there are in a field $\mathbb{F}_q$.

The whole number of monic polynomials of degree three is $q^3$. Now I want to find out how many reducible polynomials there are among them. We can split any reducible $f$ in one and only one of the following ways:

$$ f \ = \ (X-\alpha)(X^2+bX+c) \qquad \text{or} \qquad f \ = \ (X-\alpha)(X-\beta)(X-\gamma) $$

  1. In the first case, we take the number of monic irreducible polynomials of degree $2$ and multiply them by $q$, which gives us $\frac12(q^3-q^2)$.
  2. To count the polynomials of the other kind, I thought it would be helpful to split up this set as $$ \{(X-\alpha)^3 \ : \ \alpha \in \mathbb{F}_q \} \quad \sqcup \quad \{ (X-\alpha)^2(X-\beta) \ : \ \alpha \neq \beta \ \} \quad \sqcup \quad \{ (X-\alpha)(X-\beta)(X- \gamma) \ : \ \alpha, \beta, \gamma \text{ all differ }\} $$ This would give us the following number of reducible polynomials of this shape: $$ q + \frac12q(q-1)+\frac16q(q-1)(q-2) $$

Would you think that I will get the right answer if I'd calculate the value below? $$ q^3 - \frac12(q^3-q^2) - \left( q + \frac12q(q-1)+\frac16q(q-1)(q-2) \right) $$

  • $\begingroup$ Have you tried your formula out on some small cases where you can do the count manually? $\endgroup$ – coffeemath Aug 25 '14 at 9:58
  • $\begingroup$ You seem to be counting monic polynomials. Easy fix. $\endgroup$ – André Nicolas Aug 25 '14 at 10:04
  • $\begingroup$ @André Nicolas, yes, that's right. I have added the word "monic" a few times to what I wrote. I hope that that makes things more insightful. $\endgroup$ – Koenraad van Duin Aug 25 '14 at 10:12
  • $\begingroup$ The general idea will work. There are some problems of detail. Check the number of monic irreducibles of degree $2$. For product of $3$ distinct, need to divide by $6$ not $3$. $\endgroup$ – André Nicolas Aug 25 '14 at 10:46
  • $\begingroup$ As @AndréNicolas points out, I think the number of monic irreducibles of degree 2 is still wrong, but after fixing that you'll have the right answer. A simpler route is to use the fact that every monic irreducible of degree 3 is the minimal polynomial of exactly three elements of $\mathbb{F}_{q^3}$ that generate $\mathbb{F}_{q^3}$ as an extension of $\mathbb{F}_q$. $\endgroup$ – Jeremy Rickard Aug 25 '14 at 13:58

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