Let $C$ and $T$ be compact connected Riemann surfaces (or: smooth projective connected curves over $\mathbb{C}$) of genus at least two and let $X:=C\times T$. Let $(c,t)$ be a point of $X$, and let $X'\to X$ be the blow-up of $X$ in $(c,t)$. By Grauert's contraction theorem, we may contract the strict transform of $\{c\}\times T$ on $X'$ and obtain a normal complex-analytic surface $X'\to S$.



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Note that $S$ contains a unique rational curve (given by the image of the exceptional curve $E$ of $X'\to X$), and that $S$ has a unique singular point $\sigma$ in $S$.


My interest in this surface is related to Lang's conjectures, and I first learned about this surface from Frederic Campana. Indeed, the surface $S$ has the peculiar property that, for any point $s$ which does *not* lie on the rational curve and any pointed curve $(D,d)$, the set of pointed maps $(D,d)\to (S,s)$ is finite. However, for the pointed curve $(C,c)$ and the singular point $\sigma$, the space of pointed maps $(C,c)\to (S,\sigma)$ covers $S$.


I was not able to prove projectivity of $S$, not even whilst assuming it is proper (so that one might appeal to <https://arxiv.org/abs/1112.0975> )



Are any of these complex surfaces ever projective?
Here is a simple method for constructing projective examples:


Assume there exist maps $f:C \to \mathbb{P}^1$ and $g:T \to \mathbb{P}^1$ of the same degree which are totally ramified at $c$ and $t$. Let $X = C \times T$, $Y = \mathbb{P}^1 \times \mathbb{P}^1$, and consider the map $p:=(f,g): X \to Y$. Let $X'$ be the blowup of $X$ at $(c,t)$ and $Y'$ the blowup of $Y$ at $(f(c), g(t))$. An easy local computation shows that $p$ induces a morphism $p': X' \to Y'$.


Now let $Y\_1$ be the blow down of the strict transform of $\{f(c)\} \times \mathbb{P}^1 $ in $Y'$. The surface $Y\_1$ is projective, so its normalisation $X\_1$ in the function field of $X'$ is also projective. By using that the map $X' \to Y\_1$ factors through $X\_1$, it follows easily that $X\_1$ is equal to the surface $S$, so $S$ is projective.


