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?**



Is a complex manifold projective just because its blow-up at a point is ?
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. Kollár--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 Kollár--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 $X$.


