Let $f:X\to Y$ be a projective morphism between two normal quasi-projective varieties, and $L$ a $f$-ample line bundle on $Y$. Then the claim is: There is a compactication $\bar{f}:\overline{X}\to\overline{Y}$ and a globally ample line bundle $\overline{L}$ on $\overline{X}$, (i.e., $\overline{L}$ is ample on $\overline{X}$) such that $\overline{f}|_X=f$ and $\overline{L}|_{\overline{X}}\sim_f L$, where $\overline{X}$ and $\overline{Y}$ are normal projective varieties.

Note that if we forget about the line bundle $L$, then getting an $\overline{f}$ satisfying the given properties is not so hard. Indeed there exist (normal) projective varieties $X'$ and $\overline{Y}$ be such that $X\hookrightarrow X'$ and $Y\hookrightarrow\overline{Y}$ are open immersions. Then $f:X\to Y$ gives rise to a rational map $f:X'--->\overline{Y}$. Let $\overline{X}$ be normalization of the graph of this rational map with projection morphisms $g:\overline{X}\to X'$ and $\bar{f}:\overline{X}\to\overline{Y}$. Clearly $g$ is an isomorphism over the open set $X\subset X'$; so identifying $X$ with $g^{-1}X$ we see that $\bar{f}|_{\overline{X}}=f$.

I don't know how to find a globally ample line bundle on $\overline{X}$ which will restrict to a relatively ample line bundle on $X$. I realize that this is somewhat of an elementary question, so I apologies in advance for posting it here. But I need to know the proof of this result in order to use it confidently. So any help with references for these sort of ''elementary'' things, or hints on how to prove it will be greatly appreciated!

Quick note: $\overline{L}|_{\overline{X}}\sim_f L$ means $\overline{L}|_{\overline{X}}\cong L\otimes f^*M$ for some line bundle $M$ on $Y$.

  • $\begingroup$ From your argument you also get that you can extend $L$ to $\overline X$ as a relatively ample line bundle, say $\widehat L$. Then take a very ample line bundle $M$ on $Y$ and $\overline L:=\widehat L\otimes f^*M$ will do the trick. $\endgroup$ Mar 27, 2018 at 4:20
  • $\begingroup$ @SándorKovács, I am sure I am being very stupid here, but how do I extend L to a relatively ample line bundle $\hat{L}$ on $X$? Given that $\iota:X\hookrightarrow\overline{X}$, only thing I can think of is $\hat{L}:=\iota_*L$. $\endgroup$ Mar 27, 2018 at 4:29
  • $\begingroup$ I guess you might need to take a power of $L$, although I am not sure if it is really necessary. It definitely makes your life easier. In any case $\overline X$ is projective over $\overline Y$ so it comes with a natural relatively very ample line bundle. If $L$ is also relatively very ample then from the fact that you constructed $\overline X$ as the closure of a quasi-projective embedding of $X$ induced by $L$ tells you that that line bundle restricts to $L$ on $X$. The point is to think relatively over $Y$. $\endgroup$ Mar 27, 2018 at 6:22
  • $\begingroup$ Thank you @SándorKovács! I see what's going on there now. $\endgroup$ Mar 27, 2018 at 16:52


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