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
1 Answer
If $B$ is a general member of a basepointfree 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.

$\begingroup$ 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 blownup in order to logresolve the pair? $\endgroup$ Dec 13, 2011 at 10:08

$\begingroup$ 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 pullback is the same as its strict transform and the same map is still a log resolution if you add $B$ to $\Delta$. $\endgroup$ Dec 13, 2011 at 11:05