Let $\tau$ be a CM point of discriminant $D$. Assume that $D$ is not divisible by $3$. Then $j(\tau)$ is an algebraic integer of degree equal to the class number $h(D)$. Let $ \gamma_2(\tau)=j(\tau)^{1/3}$, the cube root being chosen in such a way that $\gamma_2(\tau)$ is positive on the imaginary axis.

Weber had shown that the degree of $ \gamma_2(\tau) $ is $h(D)$ instead of the expected $3h(D)$. In other words, $\mathbb Q(\gamma_2(\tau))$=$\mathbb Q(j(\tau))$.

The modular function $\gamma_2$ has level $3$, and the exact subgroup under which it is invariant is $$\Gamma(\gamma_2)=\bigg \lbrace \begin{pmatrix}a&b\\c&d\end{pmatrix}\in SL_2(\mathbb Z) :\begin{pmatrix}a&b\\c&d\end{pmatrix}\equiv\begin{pmatrix}0&*\\*&0\end{pmatrix}\text{or}\begin{pmatrix}*&b\\b&*\end{pmatrix}\text{mod } 3\bigg \rbrace.$$


  1. Is there an intuitive reason why $\gamma_2(\tau)$ has degree $h(D)$?
  2. What is the connection between Cartan subgroups of $ GL_2(\mathbb Z/N\mathbb Z)$ and this problem?
  3. Can someone please explain to me what a non-split Cartan subgroup is?


Jeremy Booher, "Modular curves and the class number one problem", Theorem $36$ and Definition $45$.

Serre: Lectures on the Mordell-Weil Theorem page 196.

Serre, "Proprietes galoisiennes des points d'ordre fini des courbes elliptiques", Section 5.3.b

I would be especially interested in some comments on the third referenced article (in connection to the problem at hand). Serre writes that this is an elementary proof that $j$ is a cube.

  • $\begingroup$ The 3rd one p.47 mentions that $\Delta^{1/3}$ appears in the coefficients of the equation for $x_P, P \in E[3]$, it could help to write how this happens exactly (when $E: y^2= \prod_{j=1}^3(x-e_j)$) in which case it suffices that those $x_P \in \mathbb{Q}(\tau,j)$ ie. that $\mathbb{Q}(\tau,x_P)/\mathbb{Q}(\tau)$ is unramified (from the CM you know its Artin map) $\endgroup$
    – reuns
    Dec 6, 2018 at 14:48
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    $\begingroup$ In the particular case of $GL_2(\mathbb{F}_p)$ and characteristic not $2$, a split Cartan is a subgroup conjugate to the diagonal matrices, and a nonsplit Cartan is a subgroup conjugate to $\mathbb{F}_{p^2}^{\times}$, embedded by identifying $\mathbb{F}_{p^2}$ with $\mathbb{F}_p^{2}$ as $\mathbb{F}_p$ vector spaces. $\endgroup$ Dec 13, 2018 at 16:04
  • $\begingroup$ @DavidESpeyer, I think your description also works perfectly well in characteristic 2. (Note also that we don't have to specify a conjugate of $\mathbb F_{p^2}^\times$; you had to choose a basis of $\mathbb F_{p^2}$ as an $\mathbb F_p$-vector space, and varying the choice of basis covers all the conjugates anyway.) $\endgroup$
    – LSpice
    May 29, 2019 at 1:39
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    $\begingroup$ @LSpice I think you are right. I no longer remember why I thought I had to be careful in characteristic $2$. $\endgroup$ May 29, 2019 at 1:40

1 Answer 1


There is a map from the $\mathbb P^1$ with coordinate $\gamma_2$ to the $\mathbb P^1$ with coordinate $j$ given by $j= \gamma_2^3$. We want to check that the fiber over $j$ has a $\mathbb Q(j)$ rational point. Because this is cubic covering, it can't gain a rational point over a quadratic extension if it didn't have one already, so for simplicity we can check that the fiber has a $\mathbb Q( j, \tau , \sqrt{-3})$-rational point.

Over $\mathbb Q(\mu_3)$, the $3$-torsion points define an $SL_2(\mathbb Z/3)$-covering of the modular curve, and this map is simply the map associated by the Galois correspondence to the subgroup $\Gamma(\gamma_2)$ of $SL_2(\mathbb Z/3)$. So by the Galois correspondence, the fiber has a rational point if and only if the action of the absolute Galois group of $\mathbb Q(j, \tau, \sqrt{-3})$ on the fiber factors through $\Gamma(\gamma_2)$.

To check this, we split into two cases depending on whether $3$ is split or inert in $\tau$. In either case, there is a natural action of the ring of integers of $\mathbb Q(\tau)$ on the $3$-torsion points which the Galois group commutes with. The action necessarily factors through the ring of integers mod $3$. If $3$ is split, the ring of integers mod $3$ is $\mathbb F_3 \times \mathbb F_3$, acting diagonalizably, and so the Galois group consists of elements that commute with it, which must lie in $\mathbb F_3^\times \times \mathbb F_3^\times$ (or the determinant $1$ elements of it). If $3$ is inert, the ring is $\mathbb F_9$, and so the Galois group must lie in $\mathbb F_9^\times$ (or the determinant $1$ elements of it). No matter how one of these two subgroups is conjugated, they must lie in the subgroup you have written down (for instance because it contains all the elements of order a power of $2$).

The group $\mathbb F_3^\times \times \mathbb F_3^\times$ here is a split Cartan, as are all its conjugates, and $\mathbb F_9^\times$ and all its conjugates are non-split Cartans.

  • $\begingroup$ Split or non-split Cartan where? (You mention $\mathrm{SL}_2(\mathbb F_3)$, but neither $\mathbb F_3^\times \times \mathbb F_3^\times$ nor $\mathbb F_9^\times$ embeds in it.) $\endgroup$
    – LSpice
    Dec 11, 2018 at 20:55
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    $\begingroup$ @LSpice Those are the split and non-split Cartans in $GL_2$. For the split and non-split Cartan's in $SL_2$, you take the subgroup of norm $1$ elements from each. $\endgroup$
    – Will Sawin
    Dec 11, 2018 at 22:38
  • $\begingroup$ You write $\mathbb{Q}(j, \tau, \sqrt{-3})$ a few times. To check my understanding, this is a typo for $\mathbb{Q}(j(\tau), \sqrt{-3})$, right? Nice answer! $\endgroup$ Dec 13, 2018 at 16:02
  • $\begingroup$ @DavidESpeyer Thanks! I just wanted to also adjoin $\tau$ (which is a quadratic irrational) so the endomorphisms will also be defined over the base field. Otherwise we would have to work with the normalizers of the endomorphism rings, which would be slightly more complicated. $\endgroup$
    – Will Sawin
    Dec 13, 2018 at 16:17
  • $\begingroup$ Re mathoverflow.net/questions/316958/… : Oh, I see; I missed the "or the determinant 1 elements of it" addenda, and was wondering how we'd got to $\mathrm{GL}_2$ from the poster's question which seemed to be about $\mathrm{SL}_2$. $\endgroup$
    – LSpice
    Dec 13, 2018 at 17:40

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