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Is there some condition on a complex algebraic surface that implies it has a smooth canonical divisor? I am searching for the sharpest possible condition, but sufficient criteria would be nice as well.

For example, the question is quite simple for surfaces that can be embedded in $P^3$. Roughly, just notice that in this case the linear system of the canonical line bundle is base point free and hence, by Bertini's theorem, we can find a smooth canonical divisor.

When we cannot avoid having a singular canonical divisor, then we are left with some singular curves. I am interested in computing the Euler characteristic of their symmetric product. I will post a question about this in a follow up enquire.

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Is it implicit in the question that $K_X$ admits an effective representative (i.e. that $H^0(X,\Omega^2)$ admits a non-zero section)? –  Emerton Jan 17 '10 at 3:17
That is the case I am most interested on, thank you for your comment. –  James O Jan 17 '10 at 3:35

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up vote 7 down vote accepted

Any smooth projective surface with nonempty $K_X$ is obtained by blowing up finitely many points on its unique minimal model. From the formula $K_X=f^*K_Y+E$ for the blowup, you see that the exceptional divisors of the blowup are always in the base locus of $|K_X|$. Thus, the problem is reduced to the minimal model.

Then you have to go through the classification of the minimal models of surfaces, that's been known for a hundred years now. (Using a book such as van de Ven "Complex surfaces", or Shafarevich et al, or Beauville...)

For a minimal surface $Y$ Kodaira dimension 0 for example, $12K_Y=0$. So either $K_Y\ne 0$ and then $|K_X|=\emptyset$, or $K_Y=0$ and then any divisor in $|K_X|$ is $\sum a_i E_i$, where $E_i$ are the exceptional divisors of the blowups.

For a minimal surface $Y$ of Kodaira dimension 2, the question is still somewhat tricky. If looking at higher multiples $|mK_Y|$ suffices, then by a well known theorem (Bombierri? certainly I. Reider gave a very nice proof), $|5K_Y|$ is free, so a general element is smooth (in characteristic 0). For $|K_Y|$ I don't think the answer is known but why not search mathscinet.

Finally, for Kodaira dimension 1, an elliptic surface $\pi:Y\to C$, there is a well-known Kodaira's formula for the canonical class $K_Y=\pi^*K_C + R$ with explicit rational coefficients in $R$. I'd play with that. Again, for higher multiples I think $|12K_Y|$ works.

Of course, to your example of a hypersurface in $\mathbb P^3$ you can add the case of complete intersections, and other surfaces for which $K_X$ is either zero or $\pm K_X$ is very ample.

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A small quibble: a min surface $Y$ of Kodaira dim zero has $K_Y$ torsion but not necessarily zero (think Enriques surface). But it matters little, since if $K_Y$ is torsion but non zero there can be no effective canonical divisor. –  Tim Perutz Jan 17 '10 at 15:29
You are right, of course. –  VA. Jan 17 '10 at 16:04
Thanks a lot again, VA and Tim Perutz! –  James O Jan 19 '10 at 0:19

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