Exercise 5.8.1 (Discriminants of cubic equations on $\mathbb{Z}/p\mathbb{Z}$)

Proof. Consider the affine curve (with cusp) on $\mathbb{Z}∕\mathit{p\mathbb{Z}}$,

(We write $0,1,2,\dots$ for the classes $\bar{0},\bar{1},\bar{2},\dots$ in $\mathbb{Z}∕\mathit{p\mathbb{Z}}$.)

We show that $𝒞$ has $p$ elements.

• If $p=2$, then

  $$𝒞=\{(a,b)\in {(\mathbb{Z}∕2\mathbb{Z})}^2\mid 0=b^2\}=\{(0,0),(1,0)\},$$

  thus $|𝒞|=2=p.$

• If $p=3$, then

  $$𝒞=\{(a,b)\in {(\mathbb{Z}∕3\mathbb{Z})}^2\mid a^3=0\}=\{(0,0),(0,1),(0,2)\},$$

  thus $|𝒞|=3=p.$

• Suppose now that $p\ne 2,p\ne 3$.

  Since $(0,0)$ is a point of multiplicity $2$, we search for every $\lambda \in {(\mathbb{Z}∕\mathit{p\mathbb{Z}})}^{\text{∗}}$ the point of intersection $M\ne (0,0)$ of the line $b=\mathit{\lambda a}$ with the curve $𝒞$. If $a\ne 0$,

  $$\begin{array}{llll}\hfill \{\begin{array}{cc}\hfill 4a^3& =27b^2\hfill \\ \hfill b& =\mathit{\lambda a}\hfill \end{array}& ⟺\{\begin{array}{cc}\hfill 0& =a^2(4a-27{\lambda }^2)\hfill \\ \hfill b& =\mathit{\lambda a}\hfill \end{array}& \hfill & \\ \hfill & ⟺\{\begin{array}{cc}\hfill a& =3^3\cdot 2^{-2}\cdot {\lambda }^2\hfill \\ \hfill b& =3^3\cdot 2^{-2}\cdot {\lambda }^3.\hfill \end{array}& \hfill & \\ \hfill & ⟺\{\begin{array}{cc}\hfill a& =k{\lambda }^2\hfill \\ \hfill b& =k{\lambda }^3.\hfill \end{array}& \hfill & \\ \hfill & & \hfill \end{array}$$

  where $k=3^3\cdot 2^{-2}\ne 0$.

  Consider the map

  $$\phi \{\begin{array}{ccc}\hfill \mathbb{Z}∕\mathit{p\mathbb{Z}}\hfill & \hfill \to \hfill & 𝒞\hfill \\ \hfill \lambda \hfill & \hfill \mapsto \hfill & (k{\lambda }^2,k{\lambda }^3).\hfill \end{array}$$

  Then

  • $\phi$ is injective (one-to-one): for all $\lambda ,\mu$ in $\mathbb{Z}∕\mathit{p\mathbb{Z}}$, since $k\ne 0$,

    $$\begin{array}{llll}\hfill \phi (\lambda )=\phi (\mu )& \Rightarrow (k{\lambda }^2,k{\lambda }^3)=(k{\mu }^2,k{\mu }^3)& \hfill & \\ \hfill & \Rightarrow {\lambda }^2={\mu }^2\text{ and }{\lambda }^3={\mu }^3& \hfill & \\ \hfill & \Rightarrow \lambda =\mu =0\text{ or }{\lambda }^3∕{\lambda }^2={\mu }^3∕\mu & \hfill & \\ \hfill & \Rightarrow \lambda =\mu .& \hfill & \end{array}$$
  • $\phi$ is surjective (onto). Let $(a,b)\in 𝒞$. If $a=0$, then $b=0$, so $(a,b)=\phi (0)$. If $a\ne 0$, put $\lambda =b{a}^{-1}$. Then $4a^3=27b^2$ and $b=\mathit{\lambda a}$. By the preceding equivalences, $(a,b)=(k{\lambda }^2,k{\lambda }^3)=f(\lambda )$.

  Therefore $f$ is bijective, so

  $$|𝒞|=|\mathbb{Z}∕\mathit{p\mathbb{Z}}|=p.$$

In any case, $|𝒞|=p.$

Therefore the complementary set in ${(\mathbb{Z}∕\mathit{p\mathbb{Z}})}^2$ has $p^2-p$ elements. This is equivalent to the waited result: the number of pairs $(A,B)$ of integers, $0\le A<p,0\le B<p$, for which $4A^3\not\equiv 27B^2(\mathit{mod}p)$ is exactly $p^2-p$. □