(Note here that in order to answer Q2 you need to explain how we may think of $Y$ as the being locally the zero locus of a bunch of polynomials) So first my guess was that in order to get the existence of these polynomials one would have to use GAGA so this was the original set-up of my question. I was hoping that by using the definition of coherence one could try to define locally on analytic open sets of $\overline{Y}$ "enough meromorphic functions" on $\overline{Y}$ which then could be used to construct an embedding in a complex projective space of a suitable dimension. However in the answer that was suggested I don't see how this notion of coherence is used, it is kind of hidden and I just don't like that. At the end of the day one has to show that $Y$ may be viewed locally as the zero locus of polynomials).
(Q3) (less interesting) Now that $Y$ is a $\mathbf{C}$-scheme, \mathbf{C}$-scheme by (Q2), explain why the analytic map$f:Y\rightarrow X^{an}$induces a map of$\mathbf{C}$-scheme$f:Y\rightarrow X$. 6 added 1676 characters in body Added So I'll try to rephrase the problem a little bit in order to focus on the part that I'm really interested in. So let us assume that$X$is a smooth affine variety over$\mathbf{C}$. So concretely one may think of$X=Spec(\mathbf{C}[x_1,\ldots,x_n]/(f_1,\ldots,f_r))$where the$f_i$'s are polynomials in$n$variables which satisfy a suitable Jacobian condition which expresses the fact that$X$is smooth. So now suppose that$Y$is a smooth connected analytic varietyand that$f:Y\rightarrow X^{an}$is a surjective finite unramified analytic cover of$X^{an}$. (Q2) Is there a simple way to put a$\mathbf{C}$-scheme structure on$Y$which is compatible with its analytic structure? (Note here that in order to answer Q2 you need to explain how we may think of$Y$as the zero locus of a bunch of polynomials) (Q3) (less interesting) Now that$Y$is a$\mathbf{C}$-scheme, explain why the analytic map$f:Y\rightarrow X^{an}$induces a map of$\mathbf{C}$-scheme$f:Y\rightarrow X$. (Q4) (this might be very easy to answer) The map of$\mathbf{C}$-scheme$f:Y\rightarrow X$is quasi-finite. Is it necessarily finite, i.e., is$Y$necessarily an affine$\mathbf{C}$-scheme? Say that we solve (Q2) and that (Q4) is answered positively then we may think of$Y=Spec(\mathbf{C}[y_1,\ldots,y_m]/(g_1,\ldots,g_s)$and from this description it is easy to see that you have "enough meromorphic functions on$Y$". For example take twodistinct points$P,Q\in Y$then we may always find a linear polynomial$l(y_1,\ldots,y_m)$such that$l(P)=0$and$l(Q)=1$. Note that I mainly care about (Q2). 5 added 7 characters in body Let$X$be a smooth quasi-projective variety (so irreducible) over$\mathbf{C}$. We may think of$X$as a complex manifold which we denote by$X^{an}$. Of course the topology on$X^{an}$is finer than the Zarisiki topology on$X$. Now let us suppose that we have a surjective finite unramified analytic cover$f:Y\rightarrow X^{an}$. Now for the sake of simplicity (I'm quite sure that one may relax considerably these assumptions) we will assume that there exists a normal projective variety$\overline{X}\supseteq X$(as an open subset in the Z-topology) and that there exists a normal compact analytic variety$\overline{Y}\supseteq Y$( as an open subset in the analytic topology) and a finite ramified analytic covering map$\overline{f}:\overline{Y}\rightarrow\overline{X}^{an}$which extends the map$f$. Then one may look at the analytic coherent sheaf$O_{\overline{Y}}$push it forward by$f_{*}$and obtain the following analytic coherent sheaf on$\overline{X}^{an}$:$\mathcal{F}^{an}:=f_{*}{\mathcal{O}}_{\overline{Y}}$. Now by GAGA we know that there exists a unique algebraic coherent sheaf$\mathcal{F}$on$\overline{X}$such that the (1) The "analytification" of$\mathcal{F}$is equal to$\mathcal{F}^{an}$. By definition of coherence of$\mathcal{F}$we know that (2) For evey$x\in\overline{X}$there exists a Zariski open set$U$of$x$such that the sequence of algebraic sheaves$({O_{\overline{X}}|U})^n\ \rightarrow ({O_{\overline{X}}|U})^m\rightarrow\mathcal{F}|U\rightarrow 0$is exact for some integers$m,n\in\mathbf{Z}_{\geq 0}$(which may depend on$x$). Q: Now using$(1)$and$(2)$is there a simple way to deduce that$\overline{Y}$is projective. ? Note that once we know that$\overline{Y}$is projective then$\overline{Y}\backslash Y$is analytically closed and therefore Zariski closed which implies that$Y$is quasi-projective. The conclusion that I was interested in was$Y$is quasi-projective. So it seems that one may find a proof that$\overline{Y}$is projective in Chap 12 of SGA1, but I'm sure that there must be a direct and easy way to deduce the algebraicity of$\overline{Y}$using$(1)$and$(2)\$.