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Let $X$ be an regular integral scheme, flat and projective over $S$, where $S$ is a smooth projective curve over an algebraically closed field $k$

Then, $X$ is a regular projective surface over $k$. Thus, it has a very ample divisor $A$. I would like to understand what such a divisor would look like.

There's a reasonable intersection theory on $X$ viewed as a fibered surface, where the pairing is defined on pairs $(D,E)\in\text{Div}(X)\times\text{Div}(X)$ with at least one of $D,E$ contained in a fiber $X_s$ of $X\rightarrow S$. Then, this pairing is bilinear and insensitive to addition by principal divisors. Furthermore, if $E\subset X_s$, and $i : X_{\mathcal{O}_{S,s}}\hookrightarrow X$ is the inclusion corresponding to the pullback of $X/S$ to $\text{Spec }\mathcal{O}_{S,s}\rightarrow S$, then we have an equality: $$D\cdot E = i^*D\cdot i^*E$$ This implies that for every closed point $s\in S$, $X_s\cdot X_s = 0$, and hence the intersection pairing restricted to divisors supported on $X_s$ is negative semidefinite. Furthermore, if $\Gamma$ is a vertical prime divisor, then $D\cdot \Gamma = \text{deg }\mathcal{O}_X(D)|_\Gamma$.

Now suppose we have a very ample effective divisor $A$, then we must have $\text{deg }\mathcal{O}_X(A)|_\Gamma > 0$ for every vertical prime divisor $\Gamma$, but this means that $A$ cannot itself be vertical, since then if $A = \sum_i n_i\Gamma_i$ with $n_i\ge 1$ and $\Gamma_i$ prime divisors, then $A\cdot A = \sum_i n_i(A\cdot\Gamma_i)$ which must be $\le 0$ since self-intersections of vertical divisors is $\le 0$, but on the other hand each term in the sum is positive since $A$ is ample, which is a contradiction.

Thus, am I right to say that $A$ cannot be vertical? In fact, I think this means $A$ cannot be contained in any fiber of any flat projective map from $X$ to any smooth projective curve $S'$.

How might one "construct" an ample divisor on $X$?

I apologize if this question seems trivial, but I'm trying to exercise my newly acquired "understanding" of intersection theory. In particular, if possible I'd like to see a method/"narrative" of constructing an ample divisor by throwing in more and more prime divisors to ensure its intersection with any prime divisor is positive.

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  • $\begingroup$ Yes, you're right to say $A$ can't be vertical. You need to do something like taking the class of a fibre and adding (some multiple of) the class of a section. $\endgroup$ – bertram Apr 24 '17 at 19:28
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Let $\pi:X\to S$ be your fibration. By replacing $S$ with its normalization in $k(X)$ (called Stein factorization) you may assume that general fiber is irreducible, which we call $f$ (note that they are numerically equivalent). Now, let $C$ be a multisection. That is, $C\subset X$ is an irreducible curve and $\pi:C\to S$ is finite (Do you know why such a $C$ exist?). Now, consider $D=C+nf$ for $n>>0$. You easily check that $D\cdot E\geq 0$ for every irreducible curve and $D^2>0$. (This is what is termed as nef and big). The only curves with $D\cdot E\leq 0$ are in some finitely many fibers and for all of them we have $D\cdot E_i=0$. so call these $E_1,\ldots, E_n$. Then the intersection form restricted to these $E_i$s is negative definite and thus you can find a divisor $G=\sum a_iE_i, a_i>0$ such that $G\cdot E_i<0$ for all $i$. Now, for $N>>0$, one checks that $ND-G$ has intersection with all curves positive and its self intersection is positive. Thus, $ND-G$ is ample.

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  • $\begingroup$ by $f$, do you mean any irreducible closed fiber? (Is irreducibility important here?) I can see that $D\cdot E\ge 0$ for any irreducible curve $E$ (pick $n$ large enough so that $nf\cdot C > -C^2$), but I don't see why $D^2 > 0$. We have $D^2 = C^2 + nf\cdot(2C+nf) = C^2$, but why must $C^2 > 0$? $\endgroup$ – user355183 Apr 24 '17 at 21:37
  • $\begingroup$ Yes, irreducibility is important, otherwise there could be infinitely many curves with $D\cdot E=0$, then I am not sure how to argue. $D^2=(C+nf)^2=C^2+2n C\cdot f$, since $f^2=0$. $C\cdot f>0$, so if $n>>0$, we have $C^2+2n C\cdot f>0$. Your calculation, $nf\cdot (2C+nf)=0$ is wrong, since $C\cdot f>0$. $\endgroup$ – Mohan Apr 24 '17 at 21:57
  • $\begingroup$ Okay so these $E_i$'s are just the components of special fibers which don't intersect $C$? How do you find this $G$? It's clear if the matrix of the intersection form is diagonal, but how do you do this in general? $\endgroup$ – user355183 Apr 24 '17 at 22:49
  • $\begingroup$ May be you should red carefully the proof of Hodge index theorem. $\endgroup$ – Mohan Apr 24 '17 at 23:51
  • $\begingroup$ The proof of the Hodge index theorem wasn't very helpful, but in the end I guess the answer to my question follows from the elementary fact that negative definite matrices are invertible. $\endgroup$ – user355183 Apr 25 '17 at 18:37

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