In most sources, the field of definition of the modular curve $X(n)_\mathbb{C}$ (quotient of the upper half plane by the subgroup $\Gamma(n)$ of $SL_2(\mathbb{Z})$ congruent to $I\mod n$) is $\mathbb{Q}(\mu_n)$, where $\mu_n$ is the group of $n$th roots of unity.

However, in Deligne Rapoport V.4.4 (http://publications.ias.edu/sites/default/files/Number22.pdf), they define $\mathcal{G}[1/n]$ to be the stack over $\mathbb{Z}[1/n]$ classifying generalized elliptic curves $E/S$ whose singular geometric fibers are $n$-gons, equipped with an isomorphism of group schemes $$E[n]\cong\mu_n\times\mathbb{Z}/n$$ "of determinant 1'', by which they mean that the isomorphism is compatible with the weil pairing on $E[n]$ and the alternating pairing $\Lambda^2(\mu_n\times\mathbb{Z}/n)\rightarrow\mu_n$ defined by $\alpha\wedge k\mapsto \alpha^k$ if $\alpha\in\mu_n$ and $k\in\mathbb{Z}/n$.

It seems like by requiring that the level structures have ``determinant 1'', we've singled out a component of the usual moduli space for $\Gamma(n)$-structures, and our component based changed to $\mathbb{C}$ should be isomorphic to $X(n) := \Gamma(n)\backslash\overline{\mathcal{H}}$. However, this would seem to imply that $X(n)$ is defined over $\mathbb{Q}$ (or even over $\mathbb{Z}[1/n]$), which seems to contradict the usual wisdom that $X(n)$ is only defined over $\mathbb{Q}(\mu_n)$.

In fact, on the second page of Ogg's paper "Rational Points on Certain Elliptic Modular Curves", he mentions a ``canonical model of Shimura, as communicated to me by Casselman", which is a $\mathbb{Q}$-rational model of $X(n)$, such that the cusps are defined over $\mathbb{Q}(\mu_n)$.

Am I reading Deligne-Rapoport correctly? Is $X(n)$ actually definable over $\mathbb{Q}$? (I wonder if Ogg's 'canonical model of shimura' somehow coincides with DR's point of view)

Edit: Also does anyone know what it means for a morphism of stacks to be "localement représentable" (eg, right after definition IV.3.2)? I know the translation is locally representable, but that sounds like relatively representable (fiber products with a scheme is also a scheme).

  • 3
    $\begingroup$ It's a question of whether you are working with a geometrically connected curve or not. $\endgroup$
    – anon
    Jan 5, 2015 at 4:47
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    $\begingroup$ For $n>2$, $G=\mu_n\times(\mathbf{Z}/n\mathbf{Z})$ and $H=(\mathbf{Z}/n\mathbf{Z})^2$ over $\mathbf{Q}$, $G$ has a $\mu_n$-valued symplectic form and the $\mathbf{Z}/(n)$-valued symplectic form on $H$ is isomorphic to the one on $G$ only over $\mathbf{Q}(\zeta_n)$ (uses a choice of $\zeta_n$). Moduli of $E\rightarrow S$ equipped with a symplecto-morphism $E[n]\simeq G_S$ is represented by a geometrically connected curve over $\mathbf{Q}$. The $H$-analogue is represented by a geometrically connected curve over $\mathbf{Q}(\zeta_n)$, where moduli defines $\zeta_n$ via pairing of $(1,0),(0,1)$! $\endgroup$
    – user74230
    Jan 5, 2015 at 6:53
  • $\begingroup$ @user74230 Can you explain your last sentence? What do you mean by "moduli defines $\zeta_n$ via pairing of (1,0),(0,1)", and why does that imply that the $H$-analogue is represented by a geometrically connected curve over $\mathbb{Q}(\zeta_n)$? $\endgroup$
    – Will Chen
    Jan 5, 2015 at 18:26
  • $\begingroup$ @oxeimon: are there local experts in algebraic geometry where you are? If so, please discuss it with them. $\endgroup$
    – user74230
    Jan 6, 2015 at 6:45
  • $\begingroup$ @user74230 I don't think there are any that are experts on Deligne-Rapoport's paper... $\endgroup$
    – Will Chen
    Jan 10, 2015 at 7:06

1 Answer 1


Deligne-Rapoport proved that $\mathcal{G}[1/N]$ is an algebraic stack proper and smooth over $\mathbb{Z}[1/N]$. If $N>2$, for any scheme $S$ over $\mathbb{Z}[1/N]$, the objects of $\mathcal{G}[1/N](S)$ do not have a non trivial automorphism, thus $\mathcal{G}[1/N]$ is an algebraic space over $\mathbb{Z}[1/N]$ and since it is regular, it is necessarily a scheme.

Remark: The finite flat group scheme $\mu_N$ is defined over $\mathbb{Z}$ and it is étale over $\mathbb{Z}[1/N]$.


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