# Relative form of Kodaira's lemma?

If $X$ is a smooth projective variety, Kodaira's lemma states that a big line bundle $D$ can be decomposed (as $\mathbb Q$-divisors) as $A+E$, with $A$ ample and $E$ effective.

I am wondering what the correct form of this lemma in the relative setting is, and where I can read about it. Suppose that $f : X \to Y$ is a projective morphism, and that $D$ is an $f$-big divisor. Is it true that $D$ can be written as the sum of an $f$-ample divisor $A$ and an effective divisor $E$? (Or maybe an "$f$-effective divisor" $E$, meaning $f_*(\mathcal O_X(E)) \neq 0$?) Remember that $f$-big means $D$ is big on the generic fiber, whereas $f$-ample means that $A$ is big on every fiber.

Sometimes people want to assume that $Y$ is affine, though I am not sure where this comes in. Feel free to assume it. (I suppose this means $f$-effective implies effective, and $f$-ample implies ample.)

The usual proof of Kodaira's lemma should work:

Let $$D$$ be $$f$$-big and let $$A$$ and arbitrary relatively ample and effective divisor. Then consider the short exact sequence:

$$0\to \mathscr O_X(mD-A) \to \mathscr O_X(mD) \to \mathscr O_X(mD)\left|_A\right. \to 0$$

In the absolute case, one observes that as $$m\to\infty$$, the dimension of $$H^0(X, \mathscr O_X(mD))$$ grows as $$m^n$$ where $$n=\dim X$$ while the dimension of $$H^0(A, \mathscr O_X(mD)\left|_A\right.)$$ can't grow faster than $$m^{n-1}$$, so for $$m\gg 0$$ the induced map cannot be injective and hence $$H^0(X,\mathscr O_X(mD-A))\neq 0$$ and we're done.

For the relative case one can do the same thing, just take $$f_*$$ instead of $$H^0$$. We have the exact $$0\to f_*\mathscr O_X(mD-A) \to f_*\mathscr O_X(mD) \to f_*\mathscr O_X(mD)\left|_A\right.$$ and we may observe that the rank of $$f_*\mathscr O_X(mD)$$ and that of $$f_*\mathscr O_X(mD)\left|_A\right.$$ can be computed as $$H^0$$ on the general fiber. We get the same conclusion, that is, that the map $$f_*\mathscr O_X(mD) \to f_*\mathscr O_X(mD)\left|_A\right.$$ can't be injective and hence $$f_*\mathscr O_X(mD-A)\neq 0$$.

Remark: I suppose one reason why one might want to take $$Y$$ to be affine is that in that case $$f_*$$ is given by an $$H^0$$ and hence $$f$$-effective actually implies effective.

In the situation as described in the question, assume that $$Y$$ is quasi-projective and let $$H$$ be a very ample effective divisor on $$Y$$. Further assume that $$H$$ is chosen so that $$A=D+f^*H$$ is big on $$X$$ (i.e., not just $$f$$-big). In particular, we may assume that $$A$$ is effective. Letting $$E=f^*H$$ shows that the required decomposition is possible. If $$H$$ is chosen generally in its linear system, then one may even assume that $$E$$ does not have any exceptional components.

• Thank you, sir! This all seems right to me, and your hunch about affine-ness matches mine.
– user47305
Jan 5 '15 at 7:03
• Thanks, but you don't need to call me "sir". Hopefully, I'm not that old yet... :) . Jan 5 '15 at 7:07
• @ Sándor Kovács: I know it has been a long time. Anyway, I wanted do ask you: does your argument imply that an $f$-big divisor $D$ can be written as $D = A - E$ where $A,E$ are effective divisors on $X$ and $f(E)$ is a divisor in $Y$? Thanks a lot.
– GDR
Jul 26 at 20:45
• @GDR: Assuming you mean equality as Q-divisors, I suspect that you meant to ask something slightly different. What you asked holds trivially: Let $E$ be an effective divisor as you want and let $A=D+E$. Jul 27 at 0:57
• @ Sándor Kovács: Are you saying that if $D\subset X$ is a divisor whose restriction to the generic fiber of $f:X\rightarrow Y$ is big the $D$ is effective?
– GDR
Jul 27 at 8:40