# How to tell if it's a Moishezon morphism

Suppose that $f \colon X\rightarrow S$ is a proper morphism of reduced and irreducible complex spaces and $f$ is a smooth deformation in the sense of Kodaira and Spencer. If we know each fiber $X_s$, $s\in S$ is a Moishezon manifold, can we conclude $f \colon X\rightarrow S$ is a Moishezon morphism? Here Moishezon morphism is defined as following: $f$ is called Moishezon if $f$ is bimeromorphic to a locally projective morphism $g\colon Y\rightarrow S$.

• The morphism $f$ need not be Moishezon. One example is when $X$ is a Hopf surface, $(\mathbb{C}^2 \setminus\{(0,0)\})/\sim$ with $(x,y)\sim(qx,qy)$ for a nonzero complex number $q$ of modulus $\neq 1$. The target $S$ is $\mathbb{CP}^1 = (\mathbb{C}^2\setminus\{(0,0)\})/\mathbb{C}^\times$, and every fiber of $f$ is isomorphic to the elliptic curve $E=\mathbb{C}^\times/q^{\mathbb{Z}}$. All of the fibers are Moishezon, but the morphism is not Moishezon. – Jason Starr May 23 '16 at 14:50
• @JS: how does one prove that $f$ is not Moishezon? – Qfwfq May 23 '16 at 19:54
• @JS: Also, is the conclusion not compatible with HassanJolany's answer below or am I missing something? – Qfwfq May 23 '16 at 19:55
• You are missing something. Hassan Jolany tells you that $f$ is locally Moishezon, over any small neighborhood of a point in $S$. This is much weaker than being globally Moishezon. In the example of Jason Starr the fibration is even locally trivial. If this $f$ were Moishezon, the Hopf surface would be Moishezon, which it is not. – abx May 24 '16 at 5:21

A Moishezon manifold $M$ is a compact complex manifold such that the field of meromorphic functions on each component $M$ has transcendence degree equal the complex dimension of the component: $\dim_\mathbf{C}M=a(M)=\operatorname{tr.deg.}_\mathbf{C}\mathbf{C}(M).$
Let $\pi: X\to S$ be a proper morphism of complex spaces. If $\pi$ is a Moishezon morphism, i.e., bimeromorphic over $S$ to a projective morphism then each fiber of $\pi$ is a Moishezon space. Conversely, if $\pi$ is smooth and each fiber is Moishezon, then for every $s \in S$ there exists a neighborhood set $U$ such that $\pi|_U$ is Moishezon
• Daniel Barlet recently showed(there is anoher proof also by Dan Popovici mat.univie.ac.at/~esiprpr/esi2238.pdf) that Let $\pi : X \to S$ a proper surjective map between irreducible complex spaces. Let $S^∗$ be a dense Zariski open set in $S$. Assume that for each $s ∈ S^∗$ the fiber $X_s$of $π$ at $s$ are Moishezon, then the central fibre $X_0$ is Moishezon arxiv.org/pdf/1705.01743.pdf – user21574 Dec 6 '17 at 3:49