Let $f:X\rightarrow Y$ be a proper and surjective morphism of projective $k$-varieties, where $k$ is a field of characteristic $0$. Let $L$ be a line bundle over $X$. Assume that $L$ is nef and generically ample ( here generically ample means $L$ is ample on the generic fiber of $f$). Does there exists an line bundle $A$ on $Y$ such that $L\otimes f^*A$ is big?

## 1 Answer

This holds for *any* ample line bundle $A$ on $Y$. Indeed, then $f^*A$ is nef [Laz, Ex. 1.4.4(i)], hence so is $D= L + f^*A$. Setting $n = \dim X$ and $m = \dim Y$, we see that $D^n \geq 0$ [Laz, Thm. 1.4.9] and $D$ is big if and only if $D^n > 0$ [Laz, Thm. 2.2.16].

For positivity of $D^n$, we may replace $(L,A)$ by $(dL,dA)$ for $d \gg 0$ to assume $A$ is very ample. There is a dense open $U \subseteq Y$ with preimage $V = f^{-1}(U)$ such that $f \colon V \to U$ is flat (and proper) of relative dimension $r = n-m$ and $L|_V$ is relatively $f$-ample [Laz, Thm. 1.2.17]. General hyperplanes $H_1,\ldots,H_m \in \lvert A\rvert$ intersect transversally in a finite (reduced) scheme $S \subseteq U$. We find $$L^r \cdot (f^*A)^m = L^r \cdot f^*S = \sum_{s \in S} \big(L\big|_{X_s}\big)^r > 0.$$ Since $f^*A$ and $L$ are nef, we get $L^i \cdot (f^*A)^{n-i} \geq 0$ for all $i$ [Laz, Ex. 1.4.16], hence $$D^n = \big(L + f^*A\big)^n = \sum_{i=0}^n {n \choose i}L^i \cdot \big(f^*A\big)^{n-i} \geq {n \choose r} L^r \cdot \big(f^* A\big)^m > 0.\tag*{$\square$}$$

**References.**

[Laz] R. Lazarsfeld, *Positivity in algebraic geometry I*. Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge **48**, Springer, Berlin (2004). ZBL1093.14501.