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Consider a compact connected complex manifold $X$ of dimension $n$. Siegel proved in 1955 that its field of meromorphic functions $\mathcal M (X)$ has transcendence degree over $\mathbb C$ at most $n$. Moishezon studied those complex manifolds for which the degree is $n$, and consequently these manifolds are now called Moishezon manifolds.

In dimension 2, every Moishezon surface is projective algebraic according to a theorem of Chow-Kodaira proved in 1952, long before Moishezon formally introduced his concept. However in dimensions 3 and more, there exist nonprojective Moishezon manifolds (you can see an example in Shafarevich's book Basic Algebraic geometry).

Nevertheless a Moishezon manifold $X$ is close to projective: Moishezon's main result is that after a finite number of blow-ups with smooth centers, $X$ becomes algebraic projective. So if a blow-up of $X$ is projective, you can't deduce that $X$ was projective. However this main result says nothing about the dimensions of the manifolds you blow up. I've heard it claimed that if only one point is blown-up, you can't get from a non-projective to a projective manifold, but I could obtain neither precise reference nor proof. Hence my question :

If a compact complex manifold becomes projective algebraic after blowing-up a point, was it already projective algebraic?

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A reference which indicates that the answer is "yes" is the thesis of Laurent Bonavero, arXiv:alg-geom/9512013. There he says (p.13, after "Remarque") "Or, de façon générale, si X est une variété compacts et si X^~ est la variété éclatée au point x, alors X est projective si est seulement si X^~ l'est". –  Artie Prendergast-Smith Feb 10 '11 at 9:38
    
(I'm not sure what went wrong with those accents...) –  Artie Prendergast-Smith Feb 10 '11 at 9:40
    
Dear Artie, I knew about Bonavero's article and the sentence you quote:it was implicitly included in the category "I've heard it claimed.." . But the absence of proof or reference is pretty frustrating. Anyway, thanks a lot for taking the trouble to dig up the article and telling me about it. –  Georges Elencwajg Feb 10 '11 at 11:07
    
You're welcome! I guessed that you probably knew about the Bonavero reference, but I thought it was worth mentioning it here for other readers of the question. (To elaborate on Georges' last comment: after the sentence quoted above, Bonavero refers to Kleiman's famous paper "Towards a numerical theory of ampleness". But as far as I could tell, that paper doesn't say anything about the matter at hand.) –  Artie Prendergast-Smith Feb 10 '11 at 13:16
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3 Answers 3

up vote 14 down vote accepted

Dear Georges, maybe the following argument works. (It's quite possible a sign went wrong somewhere, though.)

Let $\pi: Y \rightarrow X$ be the blowup. By assumption Y is projective, so it carries an ample line bundle A say. Let E denote the exceptional divisor of the blowup, and consider line bundles of the form $A+nE$ (for positive integers n). If C is any curve in Y which is not contained in E, then $(A+nE).C = A.C + nE.C$ is positive, for any n. On the other hand, let L be a line in E: then $E.L = -1$. (Note that all other curves in E are numerical multiples of this one.) So if we set n=A.L (positive, by ampleness of A) then we have $(A+nE).L = 0$. So the line bundle A+nE is nef, and has degree 0 exactly on those curves which lie in E.

I claim that $A+nE$ is in fact basepoint-free. To see this, it suffices (by the Basepoint-free Theorem, see e.g. Koll\'ar--Mori Chapter 3) to show that the line bundle $m(A+nE)-K_Y$ is nef and big, for some positive integer m. Now A is ample and E is effective, so $A+nE$ is big for all positive n (ample+effective=big --- this is also in Koll\'ar--Mori). Moreover, bigness is an open condition, so for m sufficiently large, $m(A+nE)-K_Y$ is still big. So it remains to prove nefness.

Recall that we were free to choose A to be any ample line bundle, so choose it to satisfy the condition that $A-K_Y$ is itself ample (again, using the fact that ampleness is open). Then $m(A+nE)-K_Y = (mA-K_Y) +mnE$, so in particular it has positive degree on any curve C which is not contained in E. On the other hand, if L is a line in E, then $(m(A+nE)-K_Y).L = -K_Y.L$ (by the calculations in the first paragraph.). Moreover, $-K_Y=\pi^\ast(-K_X)-(dim X-1)E$, so $-K_Y.L=-(dim X-1)E.L>0$. So $m(A+nE)-K_Y$ has nonnegative degree on all curves, i.e. it is nef.

Putting all this together, we have that $m(A+nE)$ is basepoint-free for suitable positive integers m and n. So it defines a contraction morphism $p: Y \rightarrow Z$ to another projective variety Z. But the morphism p contracts exactly those curves on which A+nE has degree 0, which by construction are exactly the curves contained in E. Therefore Z is exactly the blow-down of E to a point, hence isomorphic to X. Since Z is projective, so is X.

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Of course, it would be more satisfying and plausible if I could give an example to show why the same idea need not work if we blow up something of higher dimension. I don't have one at the moment, but let me just say vaguely: if the exceptional divisor E of the blowup contains curves with distinct numerical classes in Y, things can clearly get more complicated.... –  Artie Prendergast-Smith Feb 10 '11 at 8:09
    
In particular, suppose the curves in E span a face of the cone of curves of Y which is not rational polyhedral: then it is not at all clear to me that we can choose n as in the first paragraph so that A+nE is nef with degree 0 exactly on the contracted curves. –  Artie Prendergast-Smith Feb 10 '11 at 8:10
    
Thanks a lot, Artie. Your mastery of concepts like "nef" or "big" is impressive. I'm happy to accept your answer. –  Georges Elencwajg Feb 10 '11 at 11:14
    
Georges: this seems to follow from the Moishezon criterion for ampleness. i.e. a divisor D on an n manifold is ample if it has self intersection D^n >0, and also D^r meets every r dim'l irreducible subvariety >0. (One needs the Moishezon, not the Nakai criterion, since one needs to allow that the variety is not projective a priori.) I.e. if D^r meets E on the blow up it seems to still do so on the original variety. Sorry if this is stupidly wrong. I admit I just thought about this, and did not check it on paper. So this may be equivalent to Artie's much better answer. –  roy smith Feb 12 '11 at 4:18
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I apologize for these half baked remarks. But if you blow up a curve in a 3 fold, and take an ample divisor in the blowup, then it is not obvious to me why the blow down of that divisor should have positive intersection with the curve that was blown up, because it will usually contain that curve. –  roy smith Feb 12 '11 at 4:31
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There is a sledgehammer available for this particular nut: Mori's results on extremal rays, in his paper (Annals 1979?) on smooth projective varieties $Y$ where $K_Y$ is not nef: if $Y=Bl_PX$, with exceptional divisor $E$, then any line in $E$ will span an extremal ray and then can be contracted in the category of projective varieties. Since all curves in $E$ lie in the same ray, $E$ must be contracted to a point, so the result of the contraction must be $Y$.

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Dear inkspot, Indeed, one of the key ingredients in the "sledgehammer" result you mention is the basepoint-free theorem (as invoked in my answer). So the two answers are very close in spirit. –  Artie Prendergast-Smith Mar 11 '11 at 11:19
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Can anyone please tell me if there's something wrong with the following reasoning?

Sticking to the notation in the answer above, let $A$ be an ample divisor on $Y$. Then we can write $A \sim \pi^*D - kE$ for some divisor $D$ on $X$. Since $A \cdot l>0$, $\pi^*D \cdot l =0$, and $E \cdot l =-1$ we must have $k>0$. In fact, we can take $k=1$ since for any curve $C$ on $Y$ not contained in $E$, $(\pi^*D-kE)\cdot C \leq (\pi^*D-E)\cdot C$ because $E\cdot C \geq 0$, and we still have $(\pi^*D-E)\cdot l =1$. So take $A \sim \pi^*D -E$ to be our ample divisor on $Y$. By Seshadri's criterion, there exists an $0< \epsilon <1$ such that for any curve $C'$ on $Y$, $A \cdot C' \geq \epsilon ~ m(C')$, where $m(C')=sup_{q \in C'} m_q(C')$ is the multiplicity of the curve.

Now let $C$ be any curve on $X$ and denote by $C'$ its strict transform under the blowup. If $C$ does not pass through the center $p \in X$, then $C' \cong C$ and $C'$ does not meet $E$, so $E \cdot C' =0$. Then $A \cdot C' = \pi^*D \cdot C' = D \cdot C \geq \epsilon ~ m(C') = \epsilon ~ m(C)$.

If $C$ does pass through $p$ with multiplicity $m$, then $E \cdot C' =m$ and we have $A \cdot C' = \pi^*D \cdot C' -E \cdot C' = D \cdot C - m \geq \epsilon ~ m(C')$. But this means that $D \cdot C \geq \epsilon ~ m(C') + m \geq \epsilon ~ m(C)$. Hence $D$ is ample on $X$ by Seshadri's criterion (which still holds on complete non-projective schemes) so $X$ is projective.

The problem is I'm not quite sure where I'm using the smoothness of $X$, since blowing up points on a (singular) non-projective surface will make the blowup projective. (Maybe in the first line where it's assumed that $Pic(Y) \cong Pic(X) \oplus \mathbb Z$.) In any case, the above answer (and this one if it's correct) gives a quick proof of the Chow-Kodaira theorem, which says that any smooth complete algebraic surface must be projective.

I expect the statement to hold for singular complete algebraic varieties in dimension $\geq 3$. That is, if $X$ is any complete algebraic variety of dimension $\geq 3$, then $X$ is projective if and only if $Bl_p(X)$ is projective.

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