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I was interested in counting (and more generally having somehow an interesting expression) the numbers of solution of cubic equations modulo a prime $p$.

So here are my thoughts. Let take a cubic polynomial $P$, $L$ its decomposition field over $\mathbf{Q}$ and $G$ its Galois group. We denote by $N_p(P)$ the number of solutions of $P(x) = 0$ modulo $p$.

$N_p(P)$ is the number of $z\in Z$ fixed by $\sigma_p$, the frobenius substitution corresponding to $p$.

To pursue the study of the action of $\sigma_p$, two cases occurs: either $G$ is isomorphic to $\mathfrak{S}_3$ either to $\mathfrak{A}_3$.

Let focus on $G = \mathfrak{S}_3$. Then the field $L$ is a cubic extension of the quadratic field $K = \mathbf{Q}(\sqrt{\Delta})$. ($\Delta$ is the discriminant of $P$).

The cases where $N_p(P)$ is 1 or 2 are not difficult and depends on the value of the Legendre of $\genfrac(){}{0}{p}{-\Delta}$. Problems arises for determining if $N_p(P) = 0$ or $3$, both cases falling under the case where $\genfrac(){}{0}{p}{-\Delta} = 1$, the prime $(p)$ is therefore splited in $K$: we decompose it in $\mathfrak{p}\bar{\mathfrak{p}}$.

Let now suppose that $L$ is a Hilbert class field of $K$.
When $\mathfrak{p}$ is a principal ideal, $N_p(P) = 3$, whereas, when it is not principal, $N_p(P) = 0$. But the principality of this ideal is directly related to the representability of $p$ by quadratic forms: (p) splits completely if it can be represented by the form $x^2+\Delta y^2$.

So one can discuss the number of solutions by looking at the representation of $p$.

But what happens when $L$ is not a Hilbert class field of $K$?

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  • $\begingroup$ In this case L is an ordinary class field and there are a few ramified primes. $\endgroup$ Jan 9, 2016 at 13:02
  • $\begingroup$ I agree, but how can you found out generically which one are ramified? $\endgroup$
    – user70925
    Jan 15, 2016 at 12:23
  • $\begingroup$ Write disc $L = \Delta \cdot f^2$; the primes dividing $f$ ramify. $\endgroup$ Feb 7, 2016 at 17:50

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