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?

  • 1
    $\begingroup$ An ample divisor $D$ does not always guarantee $\phi$ to be an embedding, while a very ample divisor does. $\endgroup$
    – Yuchen Liu
    May 15, 2013 at 12:31
  • $\begingroup$ oops! thanks let me fix that $\endgroup$ May 15, 2013 at 12:44
  • 1
    $\begingroup$ I don't really have time to write out the details, but the answer (in general, not just for surfaces) is that $\operatorname{dim} \varphi(S) = \operatorname{max} \left\{k \, | \, D^k >0 \right\}$. I think that's not too hard to show, by thinking about what the fibres of $\varphi$ really are. $\endgroup$
    – user64687
    May 15, 2013 at 14:20
  • $\begingroup$ @AsalBeagDubh. Sorry I don't get this. What do you mean with $D^k$ when $k>2$? $\endgroup$ May 20, 2013 at 23:22
  • 1
    $\begingroup$ @Heitor: $D^k$ means the intersection of $D$ with itself $k$ times; this is a cycle of codimension $k$ on $S$. (Remember I was talking about the general case, not just the surface case.) When I say $D^k>0$ in my comment, I mean that this intersection is a nonzero effective cycle. $\endgroup$
    – user64687
    May 21, 2013 at 16:04

1 Answer 1


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .