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I'm not saying nothing really new wrt the previous answers,but anyway.

First, there is no contradiction. If you take another representative of the same linear system, this will need to have some negative coefficients. You can compute intersections counting points only when the intersection is transverse (or at least proper if you count multiplicities) and certainly $E$ is not transverse to itself.

From another point of view, the tangent space to the space deformations of $E$ in $S$ in the point $[E]$ is $T_{[E]} Def = H^0 ( N_{E/S})$, and the latter is zero. Indeed by adjunction $N_{E/S} = \mathcal{O}_S(E) |_E = \mathcal{O}_E(-1)$. So not only your linear system only contains the point $[E]$; there is no way to deform $E$ at all (even in a nonlinear way).

As for the computation, say $S$ is the blowup of $T$ in the point $p$, let $f \colon S \to T$ be the blowup. Take any curve $C$ passing through $p$ with multiplicity $1$; then $f^{*}(C) = \widetilde{C} + E$, where $\widetilde{C}$ is the strict transform.

By the push-pull formula $E \cdot f^{*}(C) = f_{*}(E) \cdot C = 0$, hence $E \cdot \widetilde{C} = - E^2$. But $E \cdot \widetilde{C} = 1$ because they intersect transversely in one point and you're done!

1

I'm not saying nothing really new wrt the previous answers,but anyway.

First, there is no contradiction. If you take another representative of the same linear system, this will need to have some negative coefficients. You can compute intersections counting points only when the intersection is transverse (or at least proper if you count multiplicities) and certainly $E$ is not transverse to itself.

From another point of view, the tangent space to the space deformations of $E$ in $S$ in the point $[E]$ is $T_{[E]} Def = H^0 ( N_{E/S})$, and the latter is zero. Indeed by adjunction $N_{E/S} = \mathcal{O}_S(E) |_E = \mathcal{O}_E(-1)$. So not only your linear system only contains the point $[E]$; there is no way to deform $E$ at all (even in a nonlinear way).

As for the computation, say $S$ is the blowup of $T$ in the point $p$, let $f \colon S \to T$ be the blowup. Take any curve $C$ passing through $p$ with multiplicity $1$; then $f^{*}(C) = \widetilde{C} + E$, where $\widetilde{C}$ is the strict transform.

By the push-pull formula $E \cdot f^{*}(C) = f_{*}(E) \cdot C = 0$, hence $E \cdot \widetilde{C} = - E^2$. But $E \cdot \widetilde{C} = 1$ because they intersect transversely in one point and you're done!