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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).

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Surely you want the alg. structure on $Y$ to be compatible with that on $X$ via the analytic $f$. And do you want that in such cases the alg. structure is also unique? (Recall that algebraizations, when they exist, are generally not unique in the non-compact case.) How much of the theory of coherent analytic sheaves are you willing to use? (For example, you must intend $\overline{f}$ to be a finite analytic morphism: proper with finite fibers. And finite analytic morphisms are "classified" by coherent sheaves of algebras...) Anyway, the equality $\pi_1(X) = \pi_1(X^{\rm{an}})$ is very deep. –  BCnrd Dec 25 '10 at 21:25
    
Hi BCnrd, <<And do you want that in such cases the alg. structure is also unique?>> Yes but I think that this follows from my setup. Once you know that $\overline{Y}$ is projective and by this I mean of course that this projective structure is compatible with the analyitc structure which is given on $\overline{Y}$ then we know that this projective structure is unique, this follows from the corollary on p. 30 of Serre's GAGA paper. <<you must intend to be a finite analytic morphism: proper with finite fibers.>> Yes of course I want that! I will add it –  Hugo Chapdelaine Dec 25 '10 at 22:30
    
<<Anyway, the equality $\pi_1(X) = \pi_1(X^{\rm{an}})$ is very deep>> Well you see here I take for granted the existence of the map $\overline{f}$ so there is no need to use Grauert and Remmert constructions which I think guarantee the existence of such a map $\overline{f}$ compatible with $f$. Personally I think that the existence of the map $\overline{f}$ is deeper than what I'm asking for. So here the whole point is I take the existence of this map as granted! I think that from there the argument should be "clever homological algebra" but I cannot figure it out by myself! –  Hugo Chapdelaine Dec 25 '10 at 22:32
    
It would be nice to have some kind of "hands on" description of this coherent sheaf $\mathcal{F}^{an}$ since after all it comes from the push forward of a map which is finite unramified outside some analytic divisor. –  Hugo Chapdelaine Dec 25 '10 at 22:32
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Dear Hugo: My comment about compatibility of alg. structures is indeed a triviality; I just said it to make precise the statement you really want. Anyway, indeed existence of $\overline{f}$ is where all difficulties lie. Granting that, the argument is quite simple (assuming you admit a good general theory of coherent analytic sheaves). Namely, as I said above, finite analytic maps correspond to coherent sheaves of algebras. So apply GAGA to get coherent sheaf of algebras on $\overline{X}$, then finite $\overline{X}$-scheme which must recover $\overline{Y}$. Finite over proj. is proj. QED –  BCnrd Dec 25 '10 at 22:51
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2 Answers

up vote 3 down vote accepted

(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 $\overline Y\simeq Z^{\rm an}$.

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For $\mathbf{C}$-scheme $S$ loc. of f.type & coh. $O_S$-algebra $A$, $q:S' = {\rm{Spec}}_S(A) \rightarrow S$ as loc. ringed spaces over $\mathbf{C}$ and canonical $A \simeq q_{\ast}(O_{S'})$ is final among pairs $(h:T \rightarrow S, \phi:A \rightarrow h_{\ast}(O_T))$ with $\mathbf{C}$-morphism $h$ & $O_S$-alg. map $\phi$. Use $T = {\rm{Spec}}_{S^{\rm{an}}}(A^{\rm{an}})$ & canonical $h$ and $\phi$ to get $T \rightarrow S'$ as loc. ringed spaces over $S$. Univ. property of analytification gives $f:T \rightarrow {S'}^{\rm{an}}$ over $S^{\rm{an}}$. This $f$ is isom, via completed stalk argument. –  BCnrd Dec 26 '10 at 1:12
    
Hi BCnrd, thanks a lot for Houzel's references. So in total Houzel's 4 papers "Geometrie analytique locale" consists of $12+22+25+15=74$ pages which is quite a lot of pages to read. So you made a very good remark by emphasizing the fact that $\mathcal{F}$ is not only a coherent sheaf of modules but of algebras. This remark is quite important. Nevertheless I would like to see at what places do we use the exact sequence which appears in (2) of my question. –  Hugo Chapdelaine Dec 26 '10 at 2:00
    
Are we somehow also using in the course of the argument the fact that on a Stein manifold $W$ a coherent sheaf of $O_W$-module is (1) generated by global sections and (2) has trivial cohomology groups in degrees larger than $0$. –  Hugo Chapdelaine Dec 26 '10 at 2:05
    
So you see the thing which I find fascinating about the algebraicity of $Y$ is the following. So we use the same notation as in the question. So from the existence of this analytic map $f:Y\rightarrow X$ we get that $Y$ is quasi-projective which implies that $Y$ has a lot of rational functions (so meromorphic)! So from the existence of only one analytic map we get the existence of many rational functions (even many regular functions if $X$ is affine for example)! So I'm trying to understand the heuristic behind that! –  Hugo Chapdelaine Dec 26 '10 at 2:06
    
And by many rational functions I mean enough functions so that you can separate points and tangents. This is kind of fascinating since a priori I don't quite see how to use the mere existence of $f$ to construct a meromorphic function $g:Y\rightarrow\mathbf{C}$ such that $g(P)=0$ and $g(Q)=1$ where $P$ and $Q$ are 2 points in the same fiber of the map $f:Y\rightarrow X$. –  Hugo Chapdelaine Dec 26 '10 at 2:26
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This is not an answer to my question. This is more like a comment, I want to see if I understand BCnrd's argument or if I miss the point completely. I still feel shaky with scheme theory. Feel free to make any comments.

So suppose that $X$ is as in the original question and assume furthermore (only for the sake of simplicity) that it is affine. Then we have that

$X=Spec(A)$ where $A=\mathbf{C}[x_1,\ldots,x_n]/(f_1,\ldots,f_r))$ where the $f_i$'s are polynomials in the $x_i$'s. Note that $A$ is a noetherian ring. Now let $\mathcal{F}$ be the $O_{\overline{X}}$-coherent sheaf of module on $\overline{X}$ which is obtained from GAGA. As BCnrd pointed out $\mathcal{F}$ is also simultaneously an $O_{\overline{X}}$-sheaf of algebra. In particular, it is a finite type $O_{\overline{X}}$-sheaf of algebra. From now on we will only use the fact that it is of finite type as a sheaf of algebra. Since $\mathcal{F}$ is a finite type sheaf of $O_{\overline{X}}$-algebra and $X$ is affine then $\mathcal{F}|X$ is generated by sections over $X$ (this is an easy exercise but important to point out). So in other words one has that

$B:=\mathcal{F}(X)=A[y_1,y_2,\ldots,y_m]$ for $y_1,\ldots,y_m\in \mathcal{F}(X)$. Now let $Y_1,\ldots, Y_m$ be formal variables and let $R:=A[Y_1,\ldots,Y_m]$. Note that $R$ is also an noetherian ring.

We have a natural map $\phi:R\rightarrow B$ which takes $Y_i\mapsto y_i$. Since $R$ is noetherian it follows that $ker(\phi)$ is finitely generated so let $ker(\phi)=(g_1,\ldots,g_t)$ where the $g_i$'s are polynomials in the $Y_i$'s and coefficients in $A$.

Now let $B:=A[Y_1,\ldots,Y_m]/(g_1,\ldots,g_t)$ so that $$ B=\mathbf{C}[x_1,\ldots,x_n,Y_1,\ldots,Y_m]/(f_1,\ldots,f_r,g_1,\ldots,g_t) $$ where we think now of the $g_i$'s as being polynomials in the $x_1,\ldots,x_n,Y_1,\ldots,Y_m$.

Now one has to verify that the "analytification" of the $\mathbf{C}$-scheme structure of $Spec(B)$ gives a space which is isomorphic (as an analytic variety) to $Y$ and compatible with the map $f$. This is easy to see. By construction, one has that $$ MaxSpec(B)\subseteq \mathbb{A}_{\mathbf{C}}^{n+m}\;\;\; (\star) $$

and that

$$ MaxSpec(A)\subseteq\mathbb{A}_{\mathbf{C}}^{n} $$ With respect to these embeddings the map $f$ is given by $$ (x_1,\ldots,x_n,Y_1,\ldots,Y_m)\mapsto (x_1,\ldots,x_n). $$ In particular, we see that the map $f$ is regular. Note that via the inclusion $(\star)$ the variety $Y=MaxSpec(B)$ inherits a complex structure which makes $f$ holomorphic. This implies that the original complex structure on $Y$ is compatible with the complex structure which comes from the inclusion $(\star)$.

Finally, because of the smoothness of $X$ and the fact that $f:Y\rightarrow X^{an}$ was analytically unramified it is easy to see that $Spec(B)$ is smooth. (the smoothness is equivalent to the regularity of the local ring which may be detected after completion of the local ring).

So it seems that the only place where the coherence was used was at the outset on the analytic sheaf $\mathcal{F}^{an}$ of $O_{\overline{X}}$-module on the $\mathbf{C}$-projective space $\overline{X}$. This allowed us to apply GAGA in order to get the existence of an algebraic sheaf of $O_{\overline{X}}$-algebra $\mathcal{F}$ on $\overline{X}$. And from there we only used the fact that $\mathcal{F}|X$ was a finite type sheaf of $O_{X}$-algebra.

Please let me know if I miss something important.

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