 **Note:**  By an "analytic non-algebraic" surface below I mean a two dimensional compact analytic variety $X$ (over $\mathbb{C}$) which is not an algebraic variety.


A property of Nagata's example (see the end of the post for the construction) of a non-algebraic normal analytic surface $X$ is the following: 


($\star$) $\quad$ There is a point $P$ on $X$ such that every (compact) algebraic curve $C$ on $X$ passes through $P$. 


In a paper I am writing I also constructed (to my surprise) some examples of non-algebraic normal analytic surfaces which have this peculiar property. 


 **Questions:** Is this sort of behaviour "normal" for such surfaces? Or, more precisely, if an analytic surface does not satisfy ($\star$), is it necessarily algebraic? How about for higher dimensions?


 **Nagata's Construction (following Bădescu's book on surfaces):** Start with a smooth plane cubic $C$ and a point $P$ on $\mathbb{P}^2$ such that $P - O$ is not a torsion point (where $O$ is any of the inflection points of $C$) on $C$. Let $X\_1$ be the blow up of $\mathbb{P}^2$ at $P$, and for each $i \geq 1$, let $X\_{i+1}$ be the blow up of $X\_i$ at the point of intersection of the strict transform of $C$ and the exceptional divisor on $X\_i$. Each blow up decreases the self-intersection number of the strict transform $C\_i$ of $C$ by $1$, so that on $X\_{10}$ the self-intersection number of $C\_{10}$ is $-1$. $X$ is the blow down of $X\_{10}$ along $C\_{10}$. By some theorems of Grauer and Artin, $X$ is a normal analytic surface.



Pathologies of analytic (non-algebraic) varieties.
The answer is **no**.


A counterexample is provided by the so-called *Hopf surfaces* (they were actually constructed by Kodaira, see Donu Arapura's comment). 


A Hopf surface of type $\alpha=(\alpha\_1, \, \alpha\_2)$, where $0 < |\alpha\_1| \leq |\alpha\_2| < 1$, is the compact complex surface $H\_{\alpha}$ obtained as the quotient of $\mathbb{C}^2 \setminus (0,0)$ by the infinite cyclic group generated by the automorphism $$ (z\_1, \, z\_2) \to (\alpha\_1 z\_1, \, \alpha\_2 z\_2).$$


One can prove that $H\_{\alpha}$ is a compact complex surface diffeomorphic to $S^1 \times S^3$, so it admits no Kähler metric. In particular, it is *not* algebraic.


However, $H\_{\alpha}$ does *not* satisfy your property $(\star)$. In fact, there is the following result:



> 
> The Hopf surface $H\_{\alpha}$ is an elliptic fibre space over $\mathbb{P}^1$ if and only if $\alpha\_1^l=\alpha\_2^k$ for some $l, \, k \in \mathbb{Z}$. Otherwise $H\_{\alpha}$ contains exactly two compact curves, which are disjoint (they are the images of the $z\_1$-axis and the $z\_2$-axis). 
> 
> 
> 


So $H\_{\alpha}$ contains either infinitely many or exactly two *disjoint* elliptic curves. 


For more details, see [Barth-Peters-Van de Ven, *Compact Complex Surfaces*, Chapter V]. 


