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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.)

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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.


Addition, to answer the extra question in the comments.

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.

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  • $\begingroup$ Thank you, sir! This all seems right to me, and your hunch about affine-ness matches mine. $\endgroup$
    – user47305
    Jan 5 '15 at 7:03
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    $\begingroup$ Thanks, but you don't need to call me "sir". Hopefully, I'm not that old yet... :) . $\endgroup$ Jan 5 '15 at 7:07
  • $\begingroup$ @ 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. $\endgroup$
    – GDR
    Jul 26 at 20:45
  • $\begingroup$ @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 $A=D+E$. $\endgroup$ Jul 27 at 0:57
  • $\begingroup$ @ 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? $\endgroup$
    – GDR
    Jul 27 at 8:40

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