# Strict Transform under Blow-Up along nonsingular Subvariety

Let $\beta:\widetilde{X}\mathrel{\mathop:}=\mathop{\mathrm{Bl}}_Z(X)\to X$ be the blow-up of a nonsingular algebraic variety $X$ along a nonsingular subvariety $Z$. Let $E\mathrel{\mathop:}=\beta^{-1}(Z)$ be the exceptional divisor. Now, let us assume I have a divisor $D$ on $X$. Then I was told that $\beta^\ast D \sim \widetilde{D} + \alpha E$, where $\widetilde{D}$ denotes the strict transform of $D$ and "$\sim$" is linear equivalence. I was also told that $\alpha$ is the "multiplicity of $D$ along $Z$". First, what does multiplicity mean here and second, does anyone know (possibly by reference) a proof?

Note: This bears some relation to Hartshorne Exercise II.8.5. For fibred surfaces, there is also Proposition 9.2.23 in Liu's Book. However, it seemed very specific to the twodimensional case.

First of all, let $X$ be a smooth variety and $D$ an effective divisor on $X$. Denote $\operatorname{mult}_x(D)$ the multiplicity of $D$ at a point $x\in X$. The function $x\mapsto\operatorname{mult}_x(D)$ is known to be upper-semicontinuous on $X$. Therefore, if $Z\subset X$ is any irreducible subvariety, one can define the multiplicity of $D$ along $Z$, denoted $\operatorname{mult}_Z(D)$, to be the multiplicity $\operatorname{mult}_x(D)$ at a general point $x\in Z$.
Now, the second part of your question is just a matter of local computation. I will sketch it in the case where $X$ is a smooth surface and $Z$ is a point $x_0\in X$, leaving to you to work out the details in more general situations.
So, fix local coordinates $(x,y)$ centered at $x_0$ and consider the blow-up map $\mu\colon\widetilde X\to X$ given by $\mu(u,v)=(uv,v)$. In this chart, the exceptional divisor $E$ is given by the single equation $\{v=0\}$. Now, take a divisor $D\subset X$ whose local equation near $x_0$ is given by $\{f(x,y)=0\}$. Let $$f(x,y)=\sum_{j,k\ge 0}a_{jk}x^jy^k.$$ Then, $\operatorname{mult}_{x_0}(D)=m$, where $m=\inf\{j+k\mid a_{jk}\ne 0\}$. Finally, a local equation for $\mu^*D$ is given by $$f\circ\mu=f(uv,v)=\sum_{j,k\ge 0}a_{jk}u^jv^{j+k}=v^m\underbrace{\sum_{j,k\ge 0}a_{jk}u^jv^{j+k-m}}_{\text{holomorphic and does not vanish along E}}.$$ Thus, you see that $\mu^*D$ consist of the sum of one irreducible component given by $mE$ and the remaining part is just the proper transform of $D$.
• I'm sorry for being so dull, but I do not really understand your proof, especially not how to generalize it. I think my idea of the blow-up is very algebraic, and you seem to be strongly thinking of complex manifolds. For instance, how is $\mu(u,v)=(uv,v)$ a blow-up? I am usually thinking $\mathrm{Proj}\bigoplus_{d\ge 0} I(Z)^d$. How does this generalize to blowing up along arbitrary subvarieties? Also, can you explain why $m$ is equal to $\inf\left\{\,j+k\,\vert\,a_{jk}\ne 0\,\right\}$? Jun 28 '11 at 16:34
• Being algebraic-minded is very good, but I would suggest to pay attention also to more concrete ways to look at things, if you permit. For example, the map $\mu$ is the very definition of blow-up, as well as the multiplicity at one point of a regular function. You can find all this basic stuff for example on "Principle of algebraic geometry" by Griffiths and Harris. If you are still in trouble after you take a look at that, I would be glad to help you more! All the best! Jun 28 '11 at 22:36