Suppose that we have a compact Kaehler manifold $X$ with big and nef canonical class $c_1(K_{X})$, does it imply that $X$ is projective? By the base point free theorem, big and nef implies semi ample but it is for projective algebraic manifolds. So it seems to suggest that big and nef does not necessarily imply projectivity. But I have seen in literature that people claim that big and nef does imply projectivity.
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7$\begingroup$ I guess if $X$ has a big line bundle then it is bimeromorphic to a projective variety, hence it is Moishezon. But Moishezon plus Kaehler equals projective. $\endgroup$– PopCommented Jul 8, 2020 at 22:03
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$\begingroup$ @Pop: IMHO, this comment should be expanded into an answer $\endgroup$– Francesco PolizziCommented Jul 9, 2020 at 15:59
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If $X$ has a big line bundle $L$ then for an appropriate natural number $m$, sections of $L^m$ define a meromorphic map $\varphi: X \dashrightarrow \mathbf P^N$ which is bimeromorphic onto its image. Therefore $X$ is bimeromorphic to the projective variety $\overline{\varphi(X)}$, hence it is Moishezon. But Moishezon plus Kähler equals projective.
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$\begingroup$ It seems to me that the OP was only assuming that $c_1(K_X)$ is big (and nef), in the sense that it contains a Kahler current, not that $K_X$ itself is big. So you need to first use a result of Ji-Shiffman in doi.org/10.1007/BF02921329 to get that $K_X$ is a big line bundle. $\endgroup$ Commented Jul 12, 2020 at 18:16
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$\begingroup$ Another comment: by another result of Demailly-Paun (Thm 2.12 in arxiv.org/pdf/math/0105176.pdf) it suffices to assume that $c_1(K_X)$ is nef and has strictly positive self-intersection, since this implies that $c_1(K_X)$ contains a Kahler current. Nefness of course here is in the analytic sense, i.e. it lies in the closure of the Kahler cone. $\endgroup$ Commented Jul 12, 2020 at 18:18