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I think that the statement is false for all ruled surfaces over a curve of strictly positive genus.

In fact, let $C$ be a curve of genus $\geq 1$ and let us consider two distinct points $x, \, y \in C$. Take now the two 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 as follows. By contradiction, assume that they are isomorphic. Then there exists 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.

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.

I think that the answer is no, already for ruled surfaces over an elliptic curve.

In fact, let $C$ be a curve of genus $\geq 1$ and let us consider two distinct points $x, \, y \in C$. Take now the two 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 $C \times \mathbb{P}^1$ has countably many distinct minimal models.

The argument is as follows. By contradiction, assume that they are isomorphic. Then there exists 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 has positive genus, and the latter by degree reasons.

I think that the statement is false for all ruled surfaces over a curve of strictly positive genus.

In fact, let $C$ be a curve of genus $\geq 1$ and let us consider two distinct points $x, \, y \in C$. Take now the two 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 as follows. By contradiction, assume that they are isomorphic. Then there exists 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.

I think that the answer is no, already for ruled surfaces over an elliptic curve.

In fact, let $C$ be a curve of genus $\geq 1$ and let us consider two distinct points $x, \, y \in C$. Take now the two rank $2$ vector bundles on $C$ given 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 $\mathbb{P}(\mathscr{E}_x)$ and $\mathbb{P}(\mathscr{E}_y)$ are not isomorphic.

By contradiction, assume that they are isomorphic. Then there 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 $\mathscr{L}=\mathscr{O}_C(y).$

Bot cases are impossible: the former since since $x \neq y$ and the curve has positive genus, and the latter because it implies $\mathscr{O}_C(y) \oplus \mathscr{O}_C(x+y)= \mathscr{O}_C \oplus \mathscr{O}_C(y).$

I think that the answer is no, already for ruled surfaces over an elliptic curve.

In fact, let $C$ be a curve of genus $\geq 1$ and let us consider two distinct points $x, \, y \in C$. Take now the two 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 $C \times \mathbb{P}^1$ has countably many distinct minimal models.

The argument is as follows. By contradiction, assume that they are isomorphic. Then there exists 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 has positive genus, and the latter by degree reasons.