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Does there exist a closed Moishezon manifold with zero second Betti number?

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    $\begingroup$ $M = \mathrm{pt}$ works, though you're probably looking for a positive-dimensional example. $\endgroup$ – Arun Debray Dec 8 '18 at 20:19
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If $X$ has positive dimension, the answer is no. In fact, the following holds:

Proposition. Let $X$ be a compact manifold such that $a(X)= n >0$. Then $b_2(X)>0$.

The proof is essentially based on the well-known fact that the assumption $a(X)=n$ implies that $X$ is a bimeromorphic modification of a projective manifold. Look at Lemma 1.4 of the paper

Campana, Frédéric; Demailly, Jean-Pierre; Peternell, Thomas, The algebraic dimension of compact complex threefolds with vanishing second Betti number, Compos. Math. 112, No. 1, 77-91 (1998). ZBL0910.32032.

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    $\begingroup$ Note: $a(X)$ is the transcendence degree of the field of meromorphic functions of $X$ over $\mathbf{C}$ — and $X$ is implicitly assumed connected. $\endgroup$ – YCor Dec 9 '18 at 13:56

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