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Actually I have two related questions.

Here is the first...

Suppose $X$ is a, possibly singular, complex projective variety. Let $D$ be an effective Cartier divisor on $X$ and $x\in X$ a closed point such that the multiplicity of $D$ at $x$, usually denoted by $mult_x(D)$, is $k>0$.

Let $\mu:X'\rightarrow X$ be the blowing up at $x$, and let $E$ be the exceptional divisor.

Is it true that the order of $\mu^*(D)$ along $E$ is $k$?

I'm sure about it in the smooth case and I suppose it is true also in the singular case but I don't have a reference.

The second question is the following:

Suppose $(X,\Delta)$ is a KLT pair, and $f:Y\rightarrow X$ is a log resolution of the pair $(X,\Delta)$.

Let $E$ be an exceptional divisor on $Y$ mapping to a point on $X$ and let $a(E,X,\Delta)$ be its the discrepancy. In other words $a(E,X,\Delta)$ is the order along $E$ of the divisor $K_Y-\mu^*(K_X+\Delta)$.

I know $a(E,X,\Delta)\leq 1$ if $X$ is smooth.

It really seems to me this is also true in the singular case. Do I wrong? In any case do tou have a proof or a reference for this?

Thanks a lot!

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1 Answer 1

For the first question, I think the answer is no. Consider the following example:

$X = Spec k[x,y,z]/(xy - z^2)$ a quadric cone. Consider the Cartier divisor $D = V(z)$. It has two irreducible components corresponding to the ideals $(x,z)$ and $(y,z)$ respectively (these are non-Cartier, Q-Cartier divisors). Both components smooth and they meet at the origin and so the multiplicity of $D$ (of this nodal singularity) is $2$. By multiplicity, I assume you mean the multiplicity of the scheme $D$ at a point.

On the other hand, if you blow up the origin $(x,y,z)$ you get a chart $$k[x/z, y/z, z]/( (x/z)(y/z) - 1).$$ The pull back of $D$ on this chart is just $z = 0$ (one copy of the exceptional divisor) so the order of $\mu^*(D)$ along $E$ is equal to 1. (The order of the components along the exceptional divisor is $1/2$ in each case, but they are $\mathbb{Q}$-Cartier)

There's a deeper problem in your first question though. If I recall correctly, in general, when you blow-up a point $x \in X$ on a singular variety, there isn't a unique prime exceptional divisor lying over $x$. There are probably multiple such divisors. To make matters worse, the pull back of your given Cartier divisor can have different multiplicities along these different exceptional divisors.

For the second question:

You assume that the pair $(X, \Delta)$ is klt, and you define the discrepancy at $E$ to be the order along $E$ of $K_Y - \mu^*(K_X + \Delta)$. Then you say that you know that $a(E, X, \Delta) \leq 1$ if $X$ is smooth. This isn't true.

I assume you know that the definition of klt implies that these discrepancies are all $> -1$. However, consider the following example.

$X = Spec k[x,y,z]$ and $\Delta = 0$. This pair is certainly klt. When you blow up the origin though, the relative canonical divisor $K_{Y/X} = 2E$, two copies of the exceptional (if you blow up the origin in $\mathbb{A}^n$, you get $n-1$ copies of the exceptional divisor). If you blow up points on that exceptional divisor (and repeat), you get further exceptional divisors with greater and greater discrepancy.

Hopefully I didn't misunderstand the question.

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Thank you very much for your answer, it has been really helpful to clarify my ideas and understand my mistakes. I must admit my question was really badly written! However I want to abuse of your patience and try with a different question (that I hope to formulate in the right way). Let $(X,\Delta)$ be a KLT pair as above and let $D$ be a Cartier divisor on $X$ with multiplicity arbitrarily large at a point $x \in X$. Can I conclude that there exists a prime divisor $E$ over $X$ (i.e. lying on a variety birational to $X$) such that the discrepancy $a(E,X,\Delta+D)\leq -1$? –  Gianni Bello Jun 8 '10 at 8:43
1  
I don't know an answer off the top of my head but perhaps someone else does (I'm sure it's easier than what I'm about to suggest). In particular, one approach would be to try to bound the multiplicity of a klt pair in general. This was done by Helmke in the case that the ambient space is smooth. One approach would then be to make the ambient smooth. If $X$ is an lci use inversion of adjunction (see a paper of Ein-Mustata-Yasuda) or if the normalization of $D$ is smooth (see Kawakita, Inventiones ...). In either case, you should be able to reduce to the case of a smooth ambient variety. –  Karl Schwede Jun 8 '10 at 13:07
    
Ok. Thank you again. I'll try... –  Gianni Bello Jun 8 '10 at 17:03

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