Proposition 11.14.2FD54983C4E79

Proof. Put $D=K_ X+f=-2S-(e+1)f$. The four line bundles $L_ i^*\otimes L_ j(D)$ have classes

Each has a negative coefficient in the basis $S,f$, and hence no nonzero section. Applying $\operatorname {Hom}_ X(L_ i,\cdot )$ to the extension twisted by $D$ gives $\operatorname {Hom}_ X(L_ i,V(D))=0$ for $i=1,2$. Applying $\operatorname {Hom}_ X(\cdot ,V(D))$ to the original extension then gives $\operatorname {Hom}_ X(V,V(D))=0$. By Serre duality this is the prioritary condition $\operatorname {Ext}^2_ X(V,V(-f))=0$.

Restriction to a ruling fiber is exact, because the quotient is locally free. The restricted line bundles have degrees five and minus one, and the restricted extension splits, since $\operatorname {Ext}^1_{\mathbb P^1}(\mathcal O(-1),\mathcal O(5)) =H^1(\mathbb P^1,\mathcal O(6))=0$. The quotient of negative degree on this curve excludes ampleness. $\square$