I would like to know if the following statement is true or false:



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> Given a non-singular complex projective surface $S$, it has at most a countable number of minimal models (up to isomorphism).
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We know that it is true for non-ruled surfaces (here we have uniqueness of the minimal model) and for rational surfaces. But what about ruled irrational surfaces?



Number of minimal models of a surface
Your statement is actually **false** for all ruled surfaces over a curve of strictly positive genus.


In fact, let $C$ be a curve of genus $\geq 1$, consider two distinct points $x, \, y \in C$ and take the rank $2$ vector bundles on $C$ defined by $$\mathscr{E}\_x = \mathscr{O}\_C \oplus \mathscr{O}\_C(x), \quad \mathscr{E}\_y = \mathscr{O}\_C \oplus \mathscr{O}\_C(y).$$


I claim that the two projective bundles over $C$ given by $\mathbb{P}(\mathscr{E}\_x)$ and $\mathbb{P}(\mathscr{E}\_y)$ are *not* isomorphic, hence the birational class of $C \times \mathbb{P}^1$ contains *uncountably many* distinct minimal models.


The argument is the following. If the projective bundles were isomorphic, then there would exist a line bundle $\mathscr{L}$ on $C$ such that $\mathscr{E}\_x \otimes \mathscr{L}=\mathscr{E}\_y$, that is $$\mathscr{L} \oplus \mathscr{L}(x) = \mathscr{O}\_C \oplus \mathscr{O}\_C(y).$$


Since by the Krull-Schmidt theorem the decomposition of a vector bundle into indecomposable summands is unique up to permutations of the summands, we must have either $\mathscr{L} = \mathscr{O}\_C$ or $\mathscr{L}=\mathscr{O}\_C(y).$ 


Now, both cases are impossible: the former since $x \neq y$ and the curve $C$ has strictly positive genus, and the latter by degree reasons.


