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
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1$\begingroup$ Sandor's proof is certainly correct. You can also see Lemma 5.17 in Kollar-Mori. $\endgroup$– Karl SchwedeCommented Dec 13, 2011 at 0:35
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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.
answered Dec 13, 2011 at 0:09
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$\begingroup$ 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? $\endgroup$ Commented Dec 13, 2011 at 10:08
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$\begingroup$ 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 $\Delta$. $\endgroup$ Commented Dec 13, 2011 at 11:05