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Before stating my question, let me recall (part of) the classical result on the adjunction map for complex projective surfaces, due in this modern form to Beltrametti and Sommese:

Adjunction Theorem. Let $X \subset \mathbb{P}^n$ be smooth surface and $D$ the hyperplane class. If $(K_X+D)^2 >0$ then (apart from a finite list of exceptional cases) the adjunction map $$\varphi_{|K_X+D|} \colon X \longrightarrow X_1 \subset \mathbb{P}^N$$ defined by the complete linear system $|K_X +D|$ is birational onto a smooth surface $X_1$ and blows down precisely the $(-1)$-curves $E$ on $X$ with $DE=1.$

Now, I have a rational surface $X$ with an effective, ample and basepoint-free (but not very ample) divisor $D$ such that $h^0(X, \, D)=3$ and $D^2=3$, inducing a triple cover $f \colon X \longrightarrow \mathbb{P}^2$.

I have many computational evidences showing that the first adjunction map associated with the complete linear system $|K_X + D|$ should be birational and contract precisely the $(-1)$-curves $E$ on $X$ such that $DE=1$. Unfortunately, in order to rigorously prove this I cannot use the Adjunction Theorem in the form stated above, since $D$ is not very ample. So my question is:

Question. Is there any version of the Adjunction Theorem that holds for $D$ ample (maybe with some additional assumption) and that I can use in my situation?

Any answer, suggestion or reference to the existing literature will be greatly appreciated.

Edit. Let me add the description of one of the simplest examples I have (the others are more complicated, and I'm not yet able to understand them so explicitly as this one).

Let $X$ be the blow up of $\mathbb{P}^2$ at $13$ points that impose only $12$ independent conditions to plane quartics (we can construct such a set of points by a Cayley-Bacharach argument). Let $L$ be the strict transform of a line in $X$ and $E_i$ the excepitional divisors. Setting $$D = 4L - \sum_{i=1}^{13}E_i$$ we have $$D^2=16-13=3, \quad h^0(X, \, D)=15-12=3.$$ Notice that $K_X+D=L$, hence $h^0(X, \, K_X+D)=3$ and the adjunction map $\varphi_{|K_X+D|}$ is birational in this case, in fact it is precisely the blow-down morphism $X \longrightarrow \mathbb{P}^2$.

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  • $\begingroup$ I think the essence of the theorem is the fact that the map is well defined and is birational. If you know that, what is contracted is obvious: everything orthogonal to the new hyperplane section! $\endgroup$ Feb 23, 2015 at 10:53
  • $\begingroup$ Yes, I agree with you. Unfortunately, showing that the adjoint system $|K_X+D|$ is base-point free and that the associated map is birational is precisely what I'm not able to prove. $\endgroup$ Feb 23, 2015 at 10:56
  • $\begingroup$ Dear Francesco, I am curious about the example you describe. If I understand well, it requires that $K_X+D$ has at least as many sections as $D$, which seems strange for $X$ rational. Is there a special property of $X$ or $D$ or both that explains why this is the case? $\endgroup$
    – user5117
    Feb 23, 2015 at 13:31
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    $\begingroup$ Dear @ArtiePrendergast-Smith, I added one of the examples. The only special property that I can see is that in any case the morphism associated with $D$ is a triple cover of the plane. $\endgroup$ Feb 23, 2015 at 14:13

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