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