It is known (by Gauss) that for a prime $p \equiv 1 \pmod 3$ there is a "unique" writing of $4p=A^2+27B^2$ where $A=1+p-M_p$ and $M_p$ is the number of solutions of $X^3+Y^3+Z^3=0$ in the projective plane $\mathbb{P^2(\mathbb F_p)}$.

It's also known that the points on the cubic $y^3=x^3+1$ are in 1-1 correspondence with the points on the elliptic curve $y^2=x^3-432$.

I noted that $A$ is the same (up to its sign) of the Frobenius trace in the Weil conjectures for elliptic curves; and the number $A^2-4p$ is actually the discriminant of the characteristic polynomial of the map the Frobenius automorphism induces on the Tate module.

So, I've the following question to you:

Is there a proof of $p \equiv 1 \pmod 3 \Rightarrow 4p=A^2+27B^2, \ A=1+p-M_p$ using Weil conjectures?

  • $\begingroup$ What does "using Weil conjectures" mean? That statement follows from identifying Frobenius in the ring of endomorphisms of the elliptic curve, which is an order in the field of cube roots of one. $\endgroup$ Sep 23, 2011 at 11:49
  • $\begingroup$ The problem is that I do not know how to show that Frobenius is in $\mathbb Z[\zeta_3]$. $\endgroup$
    – user18019
    Sep 23, 2011 at 12:00
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    $\begingroup$ The elliptic curve has an automorphism of order 3, $x \mapsto \zeta_3 x$, so the ring of endomorphisms contains $\zeta_3$. The condition $p \equiv 1 \mod 3$ implies that the curve is ordinary so the endomorphism ring is an order in a quadratic field. So the quadratic fiels has to be generated by $\zeta_3$ and you are done. But the "Weil conjectures" are not used anywhere. $\endgroup$ Sep 23, 2011 at 12:32
  • $\begingroup$ I've read of this. I found this proof in many books, but the point it that I would like to know if or not a prove by WCs exists. Thank you anyway!! $\endgroup$
    – user18019
    Sep 23, 2011 at 13:03
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    $\begingroup$ Just becouse I would like -if possible of course!- to put in my thesis this fact as an application of the Weil conjectures. $\endgroup$
    – user18019
    Sep 23, 2011 at 14:00

1 Answer 1


I am going to interpret "the Weil conjectures" as a synecdoche for "thinking about counting points in terms of eigenvalues of Frobenius operators." In this case, the answer is a resounding yes! However, the specific facts about eigenvalues of Frobenius which which are being used here are easier than the ones discussed in the Weil conjectures. Silverman's The Arithmetic of Elliptic Curves might be a good reference for you.

Let $E$ be the genus one curve $X^3+Y^3+Z^3$ over $\mathbb{F}_p$. This curve has an automorphism $\Phi: (x:y:z) \mapsto (z^p:y^p:z^p)$. The map $\Phi$ sends every point with values in $\mathbb{F}_p$ to itself. However, points with values in $\mathbb{F}_{p^n}$ for higher values of $n$ are permuted. Our goal is to find the number of fixed points of $\Phi$.

The Weil conjecture way to do this is to associate to $E$ certain vector spaces $H^0(E)$, $H^1(E)$ and $H^2(E)$. The map $\Phi$ acts functorially on cohomology, and we have $$\#(\mbox{Fixed points of }\Phi) = Tr(\Phi^*:H^0 \to H^0) - Tr(\Phi^*:H^1 \to H^1) + Tr(\Phi^*:H^2 \to H^2).$$ Now, in this case, you can really get quite explicit about what these vector spaces are. $H^0$ and $H^2$ are both $1$-dimensional, and $\Phi$ acts by $1$ and $p$ respectively. $H^1$ the Tate module of $E$ (Silverman III.7); a two dimensional space.

So the number of fixed points is $p+1-a$ where $a$ is the trace of $\Phi^*$ acting on the Tate module.

One can show that the endomorphism ring of $E$ is $\mathbb{Z}[\omega]$ where $\omega$ is a primitive cube root of unity (see Silverman V.3). So we must have $\Phi = c+d \omega$ for some integers $c$ and $d$. The fact that $\Phi$ has degree $p$ (see Silverman II.2) means that $N(c \omega +d) = c^2 - cd + d^2 =p$. And one computes that $a = Tr(\Phi) = 2c - d$.

So, in short, the number of points on $E$ is $p+1-a$, where $a=2c-d$ with $(c,d)$ such that $N(c+d \omega)=p$. Converting that into your statement, and getting the signs right, is left to you.

Focusing on the more specific question of using the Weil conjectures. The fact that $N(c+d \omega)$ is $p$ comes from the Riemann hypothesis part of the Weil conjectures. But you also need to know that $\mathbb{Z}[\omega]$ is the full ring of automorphisms, which is not particularly related to the Weil conjectures. And you need to figure out which of the $6$ elements of $\mathbb{Z}[\omega]$ which have norm $p$ is the correct one, which is also not a Weil computation. So I would say that this is not specifically a Weil conjecture result, although definitely can be proven using the same sort of ideas as in the Weil conjectures.

  • $\begingroup$ I found it! One can write the number of solution of $y^2=x^3-432$ in term of a Jacobi sum $S \in \mathbb Z[\zeta_3]$, namenly $\# E(\mathbb F_p)=1+p+S+\overline S$. It has absolute value $p$. Furthermore, from what I know from Weil Conjectures, $S+\overline S$ is equal to minus the norm of Frobenius. Then Frobenius is $-S$ and hence it is in the form $c+d\zeta_3$. The rest comes from simple aritmetics calculations and the fact that $3$ (and $9$ too) divides $\# E(\mathbb F_p)$. Thak you very much David, you've been very usefull to me! $\endgroup$
    – user18019
    Sep 25, 2011 at 7:25
  • $\begingroup$ Isn't $H^1$ dual (rather than equal) to the tate module? $\endgroup$ Oct 3, 2011 at 21:54

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