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.



Strict Transform under Blow-Up along nonsingular Subvariety
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 $D$.


