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For a positive integer $N$, we define $$\Gamma(N)=\big \{ \begin{bmatrix} a & b \newline c & d\end{bmatrix}\in \operatorname{SL}_2(\mathbb{Z}): \begin{bmatrix} a & b \newline c & d\end{bmatrix}\equiv \begin{bmatrix} 1 & 0 \newline 0 & 1\end{bmatrix}(\operatorname{mod}N)\big\}$$

Then, [Diamond, Shurman] defines the set of equivalence classes $S(N)$ of enhanced elliptic curves associated to $\Gamma(N)$ to be the set of equivalence classes of all triples $(E,P,Q)$, where $E/\mathbb{C}$ is an elliptic curve, and $P,Q\in E(\mathbb{C})[N]$ such that the Weil pairing $e_N(P,Q)=e^{2\pi i/N}$ .

Defining a Weil pairing:

Let $P_1,P_2\in E(\mathbb{C})[N]$ (possibly equal). Firstly, fix $E(\mathbb{C})=\mathbb{C}/\Lambda$ for some lattice $\Lambda=\mathbb{Z}\omega_1+\mathbb{Z}\omega_2\subset \mathbb{C}$. Now, there is a matrix $\gamma\in M_{2\times 2}(\mathbb{Z}/N\mathbb{Z})$ such that $\begin{bmatrix} P_1 \\ P_2\end{bmatrix}=\gamma \begin{bmatrix} \omega_1/N+\Lambda \\ \omega_2/N+\Lambda\end{bmatrix}$. We define $e_N(P_1,P_2)=e^{2\pi i\operatorname{det}(\gamma)/N}$.

Now, I know that there exists a smooth affine curve $X_{\widetilde{\Gamma}(N)}/\operatorname{Spec}\mathbb{Z}[1/N]$ which represents the moduli problem (assume $N$ is large enough): $$X_{\widetilde{\Gamma}(N)}(S)=\{(E,\alpha):E/S \operatorname{\ is\ an\ elliptic\ curve\ }, \alpha:\underline{(\mathbb{Z}/N\mathbb{Z})^{\oplus 2}}_S\xrightarrow{\cong}E[N]\}.$$

At the face of it, it seems like $X_{\widetilde{\Gamma}(N)}(\mathbb{C})=S(N)$. But, the problem I am having is that an isomorphism $\alpha:(\mathbb{Z}/N\mathbb{Z})^{\oplus 2}\xrightarrow{\cong}E(\mathbb{C})[N]$ gives two points $P,Q\in E(\mathbb{C})[N]$ such that $e_N(P,Q)=e^{2\pi i x/N}$ for some $x\in (\mathbb{Z}/N\mathbb{Z})^\times$, which is not the same as a point in $S(N)$

Am I missing something? Are these two the same?

Reference:

Diamond, Fred; Shurman, Jerry, A first course in modular forms, Graduate Texts in Mathematics 228. Berlin: Springer (ISBN 0-387-23229-X/hbk). xv, 436 p. (2005). ZBL1062.11022.

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This is a subtle issue (which has come up before on this site several times, see e.g. is the modular curve X(N) defined over Q? for a related question).

Your $S(N)$ is naturally a scheme over $\mathbb{Z}[1/N, \zeta_N]$. Your $X_{\widetilde{\Gamma}(N)}$ is, as you have defined it, a scheme over $\mathbb{Z}[1/N]$. But there is actually a natural map $\mathbb{Z}[1/N, \zeta_N] \hookrightarrow \mathcal{O}(X_{\widetilde{\Gamma}(N)})$ which sends $\zeta_N$ to $e_N(P, Q)$; and if you regard $X_{\widetilde{\Gamma}(N)}$ as a $\mathbb{Z}[1/N, \zeta_N]$-scheme via this morphism, it coincides with $S(N)$.

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  • $\begingroup$ When you think of $e_N(P,Q)\in \mathcal{O}(X_{\widetilde{\Gamma}(N)})$, what is $P$ and $Q$? Are these some fixed generators of the $N$-torsion points of the universal elliptic curve? $\endgroup$
    – Coherent Sheaf
    Commented May 9, 2023 at 21:26
  • $\begingroup$ They are the canonical $N$-torsion points of the universal elliptic curve (which are part of what it means for the space to represent a functor). $\endgroup$
    – David Loeffler
    Commented May 10, 2023 at 0:05
  • $\begingroup$ I had two follow-up question. Firstly, to summarise what you have said, would it be correct to say that over $\mathbb{Z}[1/N,\zeta_N]$, $X_{\widetilde{\Gamma}(N)}(\mathbb{C})=\Gamma(N)\backslash \mathcal{H}$, but over $\mathbb{Z}[1/N]$ this is not true anymore? $\endgroup$
    – Coherent Sheaf
    Commented May 10, 2023 at 7:18
  • $\begingroup$ Secondly, over $\mathbb{Z}[1/N$], will we get $X_{\widetilde{\Gamma}(N)}(\mathbb{C})=\sqcup_{(\mathbb{Z}/N\mathbb{Z})^{\times}}\Gamma(N)\backslash \mathcal{H}$ $\endgroup$
    – Coherent Sheaf
    Commented May 10, 2023 at 7:21
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    $\begingroup$ Perhaps the same thing in better language: this is about whether we study maximal ideals of $O(X) \otimes_{\mathbb{Z}[1/N, \zeta_N]}\mathbb{C}$ or of $O(X) \otimes_{\mathbb{Z}[1/N]} \mathbb{C}$. The $\mathbb{C}$-points of the first are $\Gamma(N) \backslash \mathcal{H}$. The $\mathbb{C}$-points of the second are $\varphi(N)$ copies of that. ($X = X_{\tilde{\Gamma}(N)}$) $\endgroup$
    – David Loeffler
    Commented May 10, 2023 at 14:23

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