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Assume that $X$ is a smooth 3-fold (over $\mathbb C$). Let $V$ be a smooth divisor on $X$ and let $S_1,S_2$ be prime divisors on $X$. Assume that given a curve $C$ on $X$ not contained in $V$, then $a:=C\cdot V>0$ (intersection product of C and V) and assume that the multiplicity of $S_i$ along the generic point of C is $b>0$.

My question is: how to bound (from below) the intersection multiplicity of the free components of $S_{1|V}$ and $S_{2|V}$ at a given point of $C\cap V$ (here $S_{i|V}$ is the restriction of $S_i$ on $V$)?

Since it is a local problem, we may assume that $C\cap V$ is a single point $p\in V$. If we assume that $S_1,S_2$ and $T$ meet exactly at $p$, then I believe that the answer is easy, as I would expect that $$I_p(V;S_{1|V},S_{2|V})=I_p(X;S_1,S_2,V)\ge b^2 C\cdot V=a\cdot b^2$$

Assuming that this is correct, my question is, is there a way to get a similar bound assuming that $S_1$ and $S_2$ meet in a curve contained in $V$?

In other words, I am assuming that $S_{i|V}=B_i\cup B$ where $B$,$B_1$ and $B_2$ are curves on $V$ (not necessarily reduced or irreducible) such that $B_1\cap B_2$ is a finite set of points (we may assume $B_1\cap B_2=p$).

Besides the obvious lower bounds, I am expecting that $$I_p(V;B_1,B_2)\ge a\cdot (b-c)^2$$ where $c$ is the multiplicity of the curve $B$ at the point $p$. This is the answer we get assuming that $S_1$ and $S_2$ have a common component $T$ such that $T_{|V}=B$. So I guess my question is: is the inequality true in general?


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What's the relationship of $C$ to the other objects? Also, when you first use $T$ it is not yet defined. Are we supposed to assume the later definition for $T$? – Sándor Kovács Mar 16 '11 at 23:32
Both $B$ and $C$ are curves contained in the intersection of $S_1$ and $S_2$, but $B$ is contained in $V$ and $C$ is not. Moreover, if $T$ is a (not nec. reduced) surface contained in the intersection of $S_1$ and $S_2$ and such that $T_{|V}=B$ then I believe that the inequality I wrote is correct. In general, $T$ might not exist (as $S_1$ and $S_2$ might be irreducible). I was wondering if the inequality still holds. I am not sure if this answer your questions. – user8229 Mar 17 '11 at 8:15

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