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One can get modular curves by the following procedures: first take the uper half plane and the rationan numbers on the x-axis, then we consider the quotient by a congruence subgroup. Now we get a compact Riemann surface, and by Chow's results, it is algebraic--an algebraic variety. Here we start with an analytic object and finally we get an algebraic one. But can we use algebraic methords only (e.g. by quotients of group schemes) to get modular curves? Or, can we find a meanful moduli problem solved by a modular curve?

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BTW the fact that compact Riemann surfaces are algebraic is due to Riemann, not to Chow. Chow's result is that compact complex subvarieties of $\mathbb{P}^n$ are algebraic. – Felipe Voloch Apr 18 '10 at 18:04

Modular curves are moduli spaces of elliptic curves with additional (torsion) structure and can be constructed purely algebraically. If you start with an elliptic curve $E$ with transcendental $j$-invariant $j$, over an arbitrary algebraically closed field $k$, and adjoin to $k(j)$ the $x$-coordinates of the $N$-torsion points of $E$ you get a function field over $k$ which is the function field of $X(N)$. See Rohrlich's paper in the Cornell-Silverman-Stevens volume on Fermat's Last Theorem or Katz-Mazur.

Edit: Note, however, Brian's warning in the comments.

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Felipe, sorry to complain, but can anything nontrivial and arithmetically interesting be rigorously proved with the "function field" defn (even if shrink k to be suitable number field)? For arithmetic aspects of modular forms (let alone FLT), that perspective seems inadequate. (I struggled with non-scheme approach early in grad school before concluding after total failure that such efforts were doomed. Avoiding serious alg. geom. for this stuff is like avoiding number fields in study of Diophantine problems over Q: why try?) Serre's "Lectures on MW" punts to DR (as does KM for Tate curve!). – BCnrd Apr 18 '10 at 18:52
No, I'm not advocating doing without schemes or serious algebraic geometry. I'm just pointing out you don't need the upper half plane to define modular curves. You'll still need to do some stuff with the cusps (via the Tate curve, which can be done algebraically or rigid analytically) and, by hand, above $j=0,1728$. I just went through this exercise in a course I am teaching. Eventually you want all points of view, as they complement each other. – Felipe Voloch Apr 18 '10 at 19:31
Felipe, the point of my "complaint" is you mention Rohrlich's article as reference, but that article suggests fn field viewpoint is adequate to develop useful foundations (which is false). It is frustrating for grad students to struggle with that perspective which is too weak, so I have a bit of a pet peeve about ever encouraging the idea (even implicitly) that it is a good starting point for the theory. I agree that the description of the fn field is good to know early on, but just as a description. Analogy: we don't suggest to define or prove assoc. of gp law on ell. curve by brute force! – BCnrd Apr 18 '10 at 19:42
I'm not sure I understand Brian's objections --- Shimura worked a lot with function fields and really did prove things. However, I would agree that this is completely the wrong way to do things. – JS Milne Apr 18 '10 at 20:59
Jim, my memory is that Shimura worked "birationally over $\mathbf{Z}$" (i.e., inverted some unknown finite set of places) but otherwise did make extensive use of actual varieties (going rather beyond function field techniques) following Weil or using his own theory of "reduction mod $p$" to handle intersection theory, using CM points and so forth to pin down the actual variety and fields of defn. That goes far beyond the fn field. Rohrlich's article follows the Shimura viewpoint, but someone not raised on Weil Foundations can't convert it into a useful method if not a master like Shimura. – BCnrd Apr 18 '10 at 21:33

It is not possible to get the modular curves via a quotient construction purely in the category of schemes. On the one hand one needs to start with the upper halfplane, which is simply not an algebraic object, and on the other hand one needs to take the quotient by an infinite discrete group, which is not an algebraic kind of quotient.

On the other hand, resoundingly yes, the modular curves corresponding to congruence subgroups of $\operatorname{PSL}_2(\mathbb{Z})$ are the (at least) coarse moduli schemes attached to natural moduli problems involving moduli of elliptic curves with level $N$-structure.

For a very brief introduction, see p. 12 of

For a serious study of these moduli problems, see e.g. Arithmetic Moduli of Elliptic Curves by Katz and Mazur.

Addendum: In the notes linked to above, I briefly describe 4 different methods of constructing the canonical model of a modular curve $X(\Gamma)$ over an appropriate number field (for instance, if $\Gamma(N) \subset \Gamma$, then the "appropriate field" is contained in $\mathbb{Q}(\zeta_N)$). Briefly: #1) rationality at the cusps; #2) modular polynomials; #3) generic elliptic curves as in Rohrlich's paper referred to in Felipe's answer; #4) moduli problem. All but the first are independent of the description of $X(\Gamma)$ as a uniformized Riemann surface.

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There's a dirty little secret in Katz-Mazur: when it comes time to invoke the analytic theory (i.e., analytification of algebraic moduli scheme over $\mathbf{C}$ gives the expected quotient of upper half-plane analytically and not just set-theoretically, so it's connected, for example), they pass over that issue in silence. Rigorous proof requires real work, not just defns (esp. to use method which works for Siegel half-spaces and Mumford's modular varieties). I do not know anywhere in literature (apart from sketches in a couple of Deligne papers) which faces this issue head-on. – BCnrd Apr 18 '10 at 18:09
@Brian: I'm sorry to hear that. Do you have any ambitions of writing this up yourself? – Pete L. Clark Apr 18 '10 at 18:15
@Brian: to be clear, they pass over which issue in silence? The claim that $X^\textrm{an}$ is complex-analytically isomorphic to $\mathfrak{h}^*/\Gamma$? – Jay Apr 18 '10 at 18:25
@Pete: I have written it up for myself, embedded in you-know-what which I have to finish some day. @Jay: Yes. Justify the analyticity of the natural set-theoretic map from Siegel half-space to analytification of one of Mumford's fine moduli schemes (over C); pfs by elliptic functions are illegal. :) Get a bijection mod discrete group by GAGA, but that doesn't prove bijection is analytic. Half-space qts are moduli spaces in category of complex analytic spaces (once one defines moduli problem), but even over complex manifolds this requires a lot of work to do rigorously (as far as I know!). – BCnrd Apr 18 '10 at 18:38
"I agree PVHS is the way to prove analyticity (but not evident to beginners" well, they could try reading [the next version] of my Introduction to Shimura Varieties. More seriously, this argument in the one-dimensional case is pretty trivial, and Katz and Mazur were surely aware of it, so I find the statement in your first comment a bit strong. Perhaps they considered it too obvious to write out (or just forgot). – JS Milne Apr 18 '10 at 21:15

Passing comment: Mumford's GIT constructs modular curves as quotients---not of the upper half plane, but of some parameter space of subspaces of projective space, by an algebraic group. As for the last part of your question: sure there is a moduli problem solved by a modular curve! Isomorphism classes of elliptic curves plus level structure (e.g. point of order N) is represented by a modular curve (Y_1(N) in this case, as long as N>=4).

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Kevin, in the spirit of what I was trying to say above, to prove such a universal property on the analytic side one needs to do something beyond the set-theoretic description: prove that a family of elliptic curves over an analytic base (maybe you want to assume smooth, though not necessary) is the analytic quotient of a line bundle modulo a "fiberwise co-compact" rank-2 local system of free $\mathbf{Z}$-modules. So requires some analytic theory, on par with early stuff in Ch. 2 of KM (after defining what is meant by "family of elliptic curves" over an analytic base). – BCnrd Apr 19 '10 at 13:50
To prove it---sure! But all I'm asserting is that it's true ;-) – Kevin Buzzard Apr 19 '10 at 17:41

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