MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose $(X,\Delta)$ is a log canonical pair and $A$ is an ample divisor on $X$. Could you give me an esay proof of the fact that there exists $A'$ that is $\mathbb{Q}$-linearly equivalent to $A$ such that $(X,\Delta+A')$ is again log canonical? Thanks a lot

share|cite|improve this question
Sandor's proof is certainly correct. You can also see Lemma 5.17 in Kollar-Mori. – Karl Schwede Dec 13 '11 at 0:35
Thank you for the reference! – Gianni Bello Dec 13 '11 at 10:08
up vote 5 down vote accepted

If $B$ is a general member of a basepoint-free linear system (say $|mA|$ for $m\gg 0$), then a log resolution $f:Y\to X$ of $(X,\Delta)$ is also a log resolution of $(X,\Delta+\frac 1m B)$ because $B$ will be transversal to all the strata related to the resolution. It follows that $f^*B=f^{-1}_*B$ so the discrepancies don't change either and hence what has been (log) terminal/canonical, remains that.

share|cite|improve this answer
Thank you very much! Just to br sure: what do you exactly mean by "the strata related to the resolution"? Are they something like the varieties that need to be blown-up in order to log-resolve the pair? – Gianni Bello Dec 13 '11 at 10:08
The preimage of $\Delta$ on a log resolution is a simple normal crossing divisor. Take the intersection of an arbitrary subset of the irreducible components. The image of this on $X$ is a stratum. The point is, that if $B$ is general, then it will be transversal to any centers of any exceptional divisors and to $\Delta$. Hence its pull-back is the same as its strict transform and the same map is still a log resolution if you add $B$ to $\Delta$. – Sándor Kovács Dec 13 '11 at 11:05

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.