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 $(2)$.


**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 variety
and 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 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 by (Q2), 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 two
distinct 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).



Finite unramified analytic coverings vs finite etale coverings
(Using the notation from the question) $\mathcal F$ is a coherent sheaf of $\mathcal O\_{\overline X}$-algebras. Then $Z={\rm Spec}\_{\overline X}\, \mathcal F\to \overline{X}$ is a finite morphism between projective schemes. Looking at the construction of ${\rm Spec}\_{\overline X} \mathcal F$ should tell you that $\overline Y\simeq Z^{\rm an}$.


