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So the fact that you need characteristic 0 means that it does not follow formally from the universal property of the Neron model. So how does the characteristic 0 assumption intervene? Is related somehow to the fact that in characteristic zero finite flat group schemes are etale?
Well in general we have the following result: Thm (a) Let $Y$ be a quasi-projective variety defined over $\overline{\mathbf{Q}}$ and let $f:X\rightarrow Y$ be an finite etale morphism defined over $\mathbf{C}$ then $X$ can be defined over $\overline{\mathbf{Q}}$. One direction of Belyi's theorem is a special case of Thm (a) above when you take $Y=P^1-\{0,1,\infty\}$. I guess that the "tentative theorem" of the question readily implies Thm (a).
Yes there might be some algebraic relations but I don't think it is a problem. I should probably rephrase the question in terms of derivations of $\mathbb{C}$ over $\overline{\mathbb{Q}}$. thanks for your comment.
If $Y$ is a variety over a field $K$ of characteristic $0$ then by definition it is of finite type over $Spec(K)$. So any such variety may be viewed a scheme over $Spec(\mathbb{Q}[t_1,\ldots,t_n])$ where the $t_i$'s are the various coefficients which appear in a choice of a set of defining equations of $Y$.
By a model of a variety $X$ over a field $K$ I simply mean that $X$ may be defined as the zero locus of a bunch of polynomials with coefficients in $K$.
You wrote <<Once you learn finite analytic maps better, you'll see removing branch locus is unnecessary to make various arguments (such as algebraicity constructions) and it simplifies things to keep branch pts in the picture.>> I'm willing to buy that and I would really like to learn more about finite analytic maps. It is still a bit annoying (at least to myself) not being able to see the heuristic about how this key notion of COHERENCE of this sheaf of algebras $\mathcal{F}$ can be used to construct enough meromorphic functions on $Y$. There must be some yoga behind that!
Hi BCnrd, <<I meant just that coherence is as ubiquitous in C-analytic geom. as in alg. geom.>>. I certainly agree with this statement of yours. Also you wrote <<fn on compact Riemann surface requires serious analysis with any approach.>> I completely agree with that! In fact I never looked at all the fine details of this proof. For sure it only works in complex dimension $1$.
Second approach: Prove the existence of a single meromorphic non-constant function $f:X\rightarrow P^1(C)$ where $X$ is your Riemann surface. Let $X'$ be $X$ minus the ramification locus. Then $f$ gives you a finite unramified analytic cover of $P^1(C)$ minus the branch points. Then you conclude that $X'$ is an an algebraic curve which implies that $X$ is an algebraic curve over $C$ I find the second approach fascinating since no harmonic analysis is involved but we still need the analytic map $f$ as an input which makes it weaker than approach 1.
