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Determine the number of irreducible monic polynomials of degree 3 in $\mathbb F_p[x]$ where $p$ is a prime.

I'd like to start off by acknowledging that I know there are many posts relating to similar problems both on here and mathoverflow, however none address my problem.

In class, my professor went over the monic polynomials of degree 2 by reasoning that there are $p^2$ distinct polynomials, the reducible ones look like $(x - a)^2$ and $(x - a)(x - b)$ for the first one there are $p$ of them and for the second one there are $p(p-1)/2$ of them, subtracting the reducible ones from the total ones gives $$p^2 - p - \frac{p(p - 2)}{2} = \frac{2p^2 - 2p - p(p-1)}{2} = \frac{2p^2 - 2p - p^2 + p}{2} = \frac{p^2 - p}{2}.$$

I'm trying to use similar reasoning for polynomials of degree three using the fact that there are $p$ irreducible monic linear polynomials and $p(p-1)/2$ irreducible monic quadratic polynomials. But I can never get the result shown here and here.

My reasoning is as follows: Observe that reducible polynomials would look like $(x - a)(x - b)(x - c)$, $(x - a)^2(x - b)$, $(x - c)^3$ and $(x^2 + ax + b)(x - c)$ formed from irreducible linear and quadratic monic polynomials.

  1. For the first type where all three are distinct we can surmise that there are $p$ choices for the first, $p-1$ choices for the second and $p - 2$ choices for the third. So we have $p(p-1)(p-2)$. (is it $p$ choose $3$? Do I have to account for order not mattering?)
  2. For the second type there are $p$ choices for the square, and $p-1$ choices for the linear polynomial, so we have $p(p-1)/2$.
  3. For the third type, we have shown that there are $p(p-1)/2$ choices for the quadratic polynomials and there are $p$ choices for the linear polynomial. So we have $p^2(p-1)/2$.

Subtracting these values from $p^3$ should give me what I'm looking for, which is $p(p^2 - 1)/3$. But I'm not getting that.

Can someone tell me where my reasoning is wrong? Or tell me how to go about counting the number of irreducibles? (I know there are other methods of getting the answer, even more general ones, but I'm looking for a simple counting method that can help me better understand where my reasoning is wrong and help me correct it.)

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For step 1, you do need $\binom{p}{3}$. For step 2, you've incorrectly divided by $2$: here the order does matter, since $(x-a)^2(x-b) \ne (x-b)^2(x-a)$. Finally, you forgot the perfect cubes, of which there are $p$. This should give you the right answer. –  Paul VanKoughnett Feb 13 '13 at 4:42
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1 Answer 1

up vote 4 down vote accepted

There are, as you pointed out, four types.

For the three distinct roots, it is $\dfrac{p(p-1)(p-2)}{3!}$. Whatever order you pick the three distinct roots, you get the same cubic.

For the two identical, and one different, it is $p(p-1)$. (There should not have been division by $2$: a double root of $a$ and a single of $b$ is different from single $a$, double $b$.)

For all identical, it is of course $p$.

And you are right about the quadratic times linear, it is $\dfrac{p^2(p-1)}{2}$.

Calculate. We get the right answer.

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