Suppose $\mu:X\rightarrow Y$ is a blow up of a smooth irreducible subvariety $Z$ of a smooth projective variety $Y$. Let $L$ be an ample Cartier divisor on $Y$. Let $E$ be the exceptional divisor of $\mu$. Is it true that there exists $0<\epsilon\ll 1$ such that $\mu^*L - \epsilon E$ is $\mathbb{Q}$-ample?

It works for surfaces and blow up of a point. I don't see how to prove this for general blow up.

  • $\begingroup$ Is $Y$ itself also smooth (then the answer is yes, see for instance Hartshorne's section on blowups)? If not, are you defining $E$ to the subscheme defined by $I_Z \cdot O_X$? Or are you defining it to the the divisorial part of the exceptional set (with what scheme structure)? Depending on your answer, the answer is also yes... $\endgroup$ Jul 22, 2015 at 0:28
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    $\begingroup$ Thanks Prof. Schwede. I did need it for the nicest scenario. I made the necessary changes. Now I'm curious how it works in case of smooth blow up in a possibly non smooth variety, if I consider E to be divisorial part of the exceptional locus. $\endgroup$
    – yagna
    Jul 22, 2015 at 18:21

1 Answer 1


Let me start with a little nitpicking:

  1. What on Earth do you mean by $\mu^*L-\epsilon E$ when you said that $L$ was a line bundle? You cannot add a line bundle and a divisor! So, let's assume that you said that $L$ was a Cartier divisor (or at least $\mathbb Q$-Cartier).
  2. It looks like you are assuming that Y is smooth, or at least that it is smooth along Z, but you didn't say so. If $Y$ is not necessarily smooth, then the exceptional set is not necessarily a divisor or irreducible, so your description of $E$ as the exceptional divisor is totally unsubstantiated. So let's assume that $Y$ is $\mathbb Q$-factorial and that you want an exceptional effective $\mathbb Q$-divisor $E$ such that $\mu^*L-E$ is $\mathbb Q$-ample.

With that, you can do this:
Let $H$ be a very ample Cartier divisor on $X$ (Homework: why is there such a divisor?). Then $\mu_*H$ is a Weil divisor on $Y$ and if $Y$ is $\mathbb Q$-factorial, then it is $\mathbb Q$-Cartier and hence $\mu^*\mu_*H$ makes sense and is a $\mathbb Q$-Cartier divisor.

Then it is easy to see that $H-\mu^*\mu_*H$ is a $\mu$-very ample $\mathbb Q$-Cartier divisor and also that it is negative effective $\mu$-exceptional. Finally, since $L$ is ample it follows that $a\mu^*L+H-\mu^*\mu_*H$ is ample for $a\gg 0$. (One could for example argue that $aL-\mu_*H$ is ample (actually nef is enough) for $a\gg0$ and hence $a\mu^*L+H-\mu^*\mu_*H$ is also ample).


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