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In Mumford's first paper on Surfaces in char $p$ [1], part 2 Step (II), he wants to show that, given an indecomposable curve of canonical type $D$ on a smooth projective surface $F$ with $p_g(F)=0, K_F^2=0, K_F$ nef, $|nD|$ is composite with a pencil of curves of canonical type for some $n$.

He goes on to show the existance of a disjoint indecomposable curve of canonical type $D'$, and that $|2K+2D+2D'|$ is composite with a pencil of curves of canonical type.

My question arises here: In the next sentence he says that since $D.(2K+2D+2D')=0$, $D$ must be a fiber of said pencil. This seems to make sense only if the map given by $|2K+2D+2D'|$ is in fact a morphism / defined everywhere, which does not seem obvious.

In [2, p. 96], Badescu follows the same argument, but adds some detail. He shows that $\phi_{|2K+2D+2D'|}(F)=B$ is of dimension $1$. He then concludes that

Thus $|2K+2D+2D'|$ is indeed composed with a pencil, and $\phi_{|2K+2D+2D'|}$ is a morphism.

Let $\Delta \in |2K+2D+2D'|$. As far as I know, since in this case $\kappa(\Delta) = \nu(\Delta)=1$ (and $R(L)$ is finitely generated; $\nu = $ numerical Kodaira dimension), wherefore $\Delta$ is semi-ample. So some multiple of $\Delta$ would give a morphism, and it seems the rest of the argument in the paper would still work.

However, afaik the reference I could find for this fact is much more recent than Mumford's paper (which is from 1977). So I assume there must be a better, more elementary way to see that $\phi_{|\Delta|}$ is a morphism?

EDIT The references I mentioned are [3] and [4].

[1] Mumford, D. - Enriques' classification of surfaces in char $p$: I, from "Global Analysis, Papers in Honor of K. Kodaira"

[2] Badescu, L. - Algebraic Surfaces

[3] Mourougane; Russo - Some Remarks on Nef and Good Divisors on an Algebraic Variety (link)

[4] Kawamata - Pluricanonical Systems on Minimal Algebraic Varieties (link)

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  • $\begingroup$ What precisely is your question? Are you asking for an earlier reference? $\endgroup$ Nov 28, 2017 at 10:39
  • $\begingroup$ @JasonStarr Sorry if I was not clear. Yes, either an earlier reference, or a more elementary way to see why $\phi$ is a morphism. $\endgroup$
    – numberjedi
    Nov 28, 2017 at 10:43

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