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Background: By Chow's theorem, if a complex manifold can be embedded holomorphically into complex projective space, then this complex manifold must be algebraic.

Question: Suppose X is a compact complex manifold (not necessarily algebraic). Let $f:X--> {\mathbb{CP}}^n$ be a meromorphic map that is injective on its domain. Does this imply that X is algebraic?

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Such a manifold is a Moishezon manifold: it is not quite algebraic, but almost! Searching for Moishezon will give you some precise results. – damiano Apr 19 '10 at 12:00
thank you damiano – user2529 Apr 19 '10 at 12:04
The answer via alg. spaces is satisfying because GAGA holds for proper alg. spaces. Existence of such an $f$ is equivalent to existence of an $f$ with 0-dimensional fibers on the complement of a nowhere-dense analytic set in $X$; could be easier to "check" in practice. Also equiv. that for nowhere dense analytic set $Z$ in $X$ big enough so $X - Z$ contained in domain of $f$ and $f(X - Z)$ loc. analytic in $\mathbf{CP}^n$, $\dim f(X_i - Z_i) = \dim X_i$ for conn'd comps $X_i$ of $X$. Can formulate w/o smoothness (using compact Hausdorff analytic space in role of $X$), as in Chow's theorem. – BCnrd Apr 19 '10 at 13:43
So, your question is: "is a (compact) holomorphic manifold, bimeromorphic to a projective veriety, necessarily algebraic?" – Qfwfq Apr 19 '10 at 14:06
@Colin: everyone's answers and comments are assuming you know enough to see that your hypotheses do immediately imply bimeromorphicity; so you really are assuming it. If that implication is not clear then you need to learn more about complex geometry in higher dimensions (and double check you fully understand the meaning of "meromorphic map" in higher dimensions). – BCnrd Apr 20 '10 at 5:03
up vote 8 down vote accepted

If $X$ is not compact there are loads of problems so I follow you in adding compactness as a condition. Under the assumption of the question $X$ is bimeromorphic to a projective variety and hence by a result of Artin it is an algebraic space. Hence if you accept algebraic spaces as being algebraic the answer is yes. As there are smooth proper algebraic spaces that are not schemes if you don't the answer is no.

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