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.
 A: 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 $D$.
