Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm just now beginning to learn about descending the curves $X(N)$ to $\mathbb{Q}$, and I have a few questions:

  1. Does $X(N)$ have a $\mathbb{Q}$-point for every $N$?

  2. What is $\operatorname{Aut}(X(N))$? I know that $\operatorname{PSL}_2(\mathbb{F}_p)\leq \operatorname{Aut}(X(p))$ for $p$ prime, but can we say more? (for example, is this an equality for $p$ prime; what can we say if $p$ isn't prime?)

  3. Does the map $X(N)\rightarrow \mathbb{P}^1_{\mathbb{C}}$ together with the Galois action descend to $\mathbb{Q}$? How about without it?

  4. Do the curves $X_0(N)$ (and curves given by congruence groups in general) also descend to $\mathbb{Q}$? If so, I ask the above questions for those as well.

share|cite|improve this question

For $X(N)$, one can give a $\mathbb{Q}$-model, but there's a sense in which doing so is "cheating".

The moduli interpretation of $X(N)$ only makes sense over $\mathbb{Q}(\zeta_N)$; if $R$ is an algebra over this field, then $X(N)$ classifies triples $(E, P_1, P_2)$ where $E$ is an elliptic curve over $R$ and $P_1, P_2$ are $R$-points of $E$ order $N$ which pair to $\zeta_N$ under the Weil pairing.

You can choose a model for $X(N)$ over $\mathbb{Q}$ in various ways, but you lose the moduli interpretation above. The usual way of doing this is such that the rational functions defined over $\mathbb{Q}$ are those whose $q$-expansions at the cusp $\infty$ have coefficients in $\mathbb{Q}$. This is done in, for instance, Stevens' book "Arithmetic on modular curves". With this choice, it is essentially tautological that the cusp $\infty$ is a $\mathbb{Q}$-point, and that the $j$-invariant map $X(N) \to X(1) \cong \mathbf{P}^1$ is defined over $\mathbb{Q}$. But this $\mathbb{Q}$-structure is rather artificial, and doesn't interact well with the moduli interpretation; e.g. sometimes one encounters modular curves $X$ which are quotients of $X(N)$, and which have a natural $\mathbb{Q}$-structure which is compatible with the moduli space interpretation of $X$, but where the map $X(N) \to X$ isn't defined over $\mathbb{Q}$ when we give $X(N)$ the $\mathbb{Q}$-structure described above. (See e.g. Elkies' paper on mod 3 and mod 9 representations of elliptic curves.)

I hope this at least partially answers your questions (1) and (3). I don't know the answer to (2); but for $N < 6$ we have $X(N) \cong \mathbf{P}^1$, so the automorphism group of $X(N)$ is infinite in these cases (in particular it is much larger than $PSL_2(\mathbf{Z} / N)$).

For $X_0(N)$ and $X_1(N)$ life is much easier than $X(N)$: the moduli space interpretations of these curves make sense over $\mathbb{Q}$, and they have rational models compatible with this moduli space structure, and the cusp $\infty$ is always a $\mathbb{Q}$-point.

(EDIT: Actually the last sentence is wrong, sorry! This works for $X_0(N)$ but not for $X_1(N)$. The canonical $\mathbb{Q}$-structure on $X_1(N)$ is the one that makes $\mathbb{Q}$-expansions at the cusp 0, not the cusp $\infty$, rational.)

share|cite|improve this answer
Are you saying $X(N)\rightarrow \mathbb{P}^1$ is always defined together with the group action over $\mathbb{Q}$? This surprises me. I thought $PSL_2(\mathbb{F}_p)$ is not known (for a general prime $p$) to be realized as a Galois group over $\mathbb{Q}$. I think Shih proved it was just for some of the primes.... Are you sure you didn't mean defined without the group action over $\mathbb{Q}$? That would make more sense. (P.S. Thanks a lot for the answer) – Nicole Sep 4 '11 at 20:55
I'm not sure I quite understand your question -- what do you mean by "defined without the group action"? – David Loeffler Sep 6 '11 at 11:40
The cover is defined, but it is not nec. Galois. – Nicole Sep 13 '11 at 0:11

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.