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Let $\Gamma\subseteq \Gamma'\subset SL_2(\mathbb Z)$ be congruence subgroups, and $X(\Gamma)$, $X(\Gamma')$ be the associated smooth projective modular curves over $\mathbb C$. The inclusion $\Gamma\subseteq \Gamma'$ induces a (canonical) non-constant morphism $p:X(\Gamma)\to X(\Gamma')$ of curves over $\mathbb C$.

Question: 1) Suppose $N,M$ are positive integers with $N\mid M$. Then $\Gamma=\Gamma_0(M)\subseteq \Gamma'=\Gamma_0(N)$ and there exist canonical $\mathbb Q$-models $X$ and $X'$ of $X(\Gamma)$ and $X(\Gamma')$ respectively. Does there exist a $\mathbb Q$-morphism $p_{\mathbb Q}:X\to X'$ whose base change to $\mathbb C$ is $p$?

2) More generally: Let $F$ be a number field such that $X(\Gamma)$ and $X(\Gamma')$ have models which are defined over $F$ (one can always find such a number field). Does there exist $F$-schemes $X$ and $X'$ whose base change to $\mathbb C$ (with respect to an embedding $\sigma:F\hookrightarrow \mathbb C$) are $X(\Gamma)$ and $X(\Gamma')$ respectively, and an $F$-morphism $p_F:X\to X'$ such that the base change to $\mathbb C$ (with respect to $\sigma$) of $p_F$ is $p$?

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A variety over $\mathbf{C}$ that descends to a subfield can generally be descended in numerous non-isomorphic ways. Thus, saying a variety over a given field is "defined over" a subfield is too imprecise to do much. Without being more specific/clearer about the $F$-descents you are using, it seems impossible to answer the question. (If you follow recipes dictated by specific moduli problems on the category of $F$-schemes then it is possible to give an affirmative answer in some cases. But one can also make choices of moduli problems for which the answer is negative.) – user29283 May 21 '13 at 14:55
Dear Xuhan, thanks for explaining this. I will now edit the question with the hope to make it more reasonable. – Guillaume Pastorini May 21 '13 at 18:07
In general, for canonical models of Shimura varieties, all canonical maps are defined over the same fields as the varieties. – anon May 21 '13 at 21:41
The revised question has the same defects as the original one: depending on which $F$-descents you choose for the modular curves (you don't specify which descents!), the answer can be positive or negative. For example, for $\Gamma_0(N)$ I could choose the moduli problem based on embeddings of $\mu_N$ and for $\Gamma_0(M)$ I could choose the one based on embeddings of $\mu_M$ or of $\mathbf{Z}/M\mathbf{Z}$. These all give moduli schemes over $\mathbf{Q}$. For the first choice of the latter the answer is affirmative, and for the 2nd choice of the latter the answer is negative (when $N > 2$). – user29283 May 22 '13 at 1:43
Dear Xuhan, in 1) I mean the canonical model (which I believe is unique up to $\mathbb Q$-morphism); I now mention that I mean the canonical model. I think 2), which is formulated as an existence question, makes sense without specifying the $F$-descents. – Guillaume Pastorini May 23 '13 at 17:28
up vote 4 down vote accepted

Yes. Please see Theorem 7.1.3 of Katz-Mazur.

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For information about exactly which number field you can use, please see Chapter 9 of the same book. – stankewicz May 21 '13 at 18:56
@stankewicz: As I indicate in my comments, one cannot say that the answer is "yes" or "no" until the OP makes clearer exactly which $F$-descents (for specific $F$) are being considered. The theory of canonical models provides a systematic recipe giving affirmative answers, and the theory of moduli schemes does too, but in both cases one has to make clear exactly what recipes are being used in order to determine if the answer is affirmative or negative (can go either way, especially when taking the approach through moduli problems). – user29283 May 22 '13 at 1:45
Well perhaps I'm being stupid here, but it seems to me that Pastorini is asking an existence question and that the theory of moduli as given in Katz-Mazur is sufficient to say that yes, there exist schemes and a map defined over some number field which base extend to the canonical map. I know quite well that you could twist any part of this data and screw something up, but for the question as asked I don't see the objection. – stankewicz May 22 '13 at 2:36
@stankewicz: It seems plausible that the OP has specific $F$-descents in mind (generally needed if one is going to use an affirmative answer to do something) but has not said which ones. The questions as posed remain ambiguous about which such structures are desired (if any). It not clear if the OP has fully appreciated this concern about the formulation. That is why I don't think one can (yet) say "yes" as an answer. More context for the question from the OP should be helpful to determine the kind of answer that the OP would find useful. – user29283 May 22 '13 at 5:06
Dear Stankewicz, thanks for the references. They completely answer my (probably ambigous?) questions. – Guillaume Pastorini May 23 '13 at 17:30

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