Dimension of the image of the morphism associated to a Divisor Let $S$ be an algebraic smooth surface over $\Bbb{C}$.
Let $D\in\mathrm{Div}(S)$ be such that the complete linear system $|D|$ is base-point free and suppose $h^0(D)=N+1$ with $N>0$. To $D$ is then associated a morphism $\varphi:S\rightarrow\Bbb{P}^N$.
By definition if $D$ is very ample then $\varphi$ is an embedding; in particular $\dim\varphi(S)=2$.
Question: what are some possible conditions on $D$ (e.g. self-intersection) to determine $\dim\varphi(S)$ ?
For example if $D^2=0$ can we say the image of $\varphi$ is a curve? Why?
 A: As AsalBeagDubh seems to be busy, let me try an answer. 

If $D^2>0$, then $\dim\phi(S)=2$. Otherwise $D^2=0$ and $\dim\phi(S)= 1$. 

Proof. Let $D'\in |D|$, let $H$ be a hyperplane of $\mathbb P^N$ not containing $\phi(S)$. As $\phi^*H \sim D$ by construction of $\phi$, we have by the projection formula (see Hartshorne, Appendix A), 
$$ \phi_*D'. H=D'.\phi^*H=D^2.$$ 
Therefore: 
(i) if $D^2>0$, then $\phi(D')$ is not finite (vary $H$, see the argument below with $H'$), hence $\phi(D')$ has dimension $1$;
(ii) if $D^2=0$, then $\phi(D')$ is finite because $H$ has positive intersection number with any curve in $\mathbb P^N$.  
(Here we can notice that $D^2=\phi_*D'.H$ is always non-negative because $H$ is ample.) 
Now suppose $\dim\phi(S)=2$. Then $S\to \phi(S)$ is a surjective morphism of irreducible varieties of the same dimension, so outside a codimension $1$ closed subset, $\phi : S\to \phi(S)$ is quasi-finite. This implies that $\dim\phi(C)=1$ for all but finitely many curves $C$ in $S$. This excludes the case (ii) because the $D'$'s cover $S$, thus $D^2>0$. 
If $\dim\phi(S)=1$,  then $\phi(D')$ is finite. Let $H'$ be a hyperplane (hypersurface if the ground field is finite, but we can indeed extend the ground field to an algebraic closure) not meeting $\phi(D')$. Then 
$D^2=\phi_*D'.H'=0$.  
Finally $\dim\phi(S)=0$ can't happen because otherwise $\phi(S)$ is a single point, and then $D\sim 0$ which would mean that $|D|$ is empty. 
