Let $X$ be a smooth complex projective variety and $p:Y\to X$ be a locally trivial in analytic topology $\mathbb CP^k$-bundle. Suppose we have a line bundle $L$ on $Y$, restricting to $\mathcal O(1)$ on $\mathbb CP^k$-fibres.


**Question.** Is it true that there is a line bundle $L'$ on $X$ such that $p^\*L'\otimes L$ is very ample on $Y$?




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I think one could be able to prove this using Serre's (+ Kodaira?) vanishing, but I can't prove it so far.


**Idea.** Here is an idea of how one could try to solve this. So, first all, one can take $L''$ such that $p^\*L''\otimes L$ is ample on $Y$. Next, one can try to use Kodaira vanishing, it says that $K\_Y\otimes p^\*L''\otimes L$ has zero higher cohomology. Now, we could take the line bundle $K\_Y\otimes (p^\*L''\otimes L)^{k+2}$, and this bundle will restrict to each fibre as $\mathcal O(1)$. By taking $L''$ positive enough, it should be possible to make $K\_Y\otimes (p^\*L''\otimes L)^{k+2}$ ample. I think that since this bundle is ample and its higher cohomology vanish, by Grothendiek-Riemann-Roch it will have a lot of sections (especially if $L''$ is very ample and has a lot of sections). This seems to be not far from proving that $K\_Y\otimes (p^\*L''\otimes L)^{k+2}$ is very ample... And I guess this bundle is $p^\*L'\otimes L$ for some $L'$?



Constructing a very ample line bundle on a projective bundle
Since $R^ip\_\*L=0$ for $i>0$, by semi-continuity theorem, you see that $p\_\*L$ is a vector bundle of rank $k+1$. You can twist by a sufficiently ample bundle $L'$ on $X$ to make it globally generated. Thus, you have $O\_X^m\to p\_\*L\otimes L'$ surjective and thus you get an embedding $Y\subset X\times \mathbb{P}^{m-1}$. Further, the $O(1)$ of the second factor restricts to $L\otimes p^\*L'$ on $Y$. Now, taking a very ample line bundle $M$ on $X$, $M\otimes O(1)$ is very ample on $X\times\mathbb{P}^{m-1}$ and then its restriction $L\otimes p^\*(L'\otimes M)$ is very ample on $Y$.


