My question is if [this](https://mathoverflow.net/questions/68973/strict-transform-under-blow-up-along-nonsingular-subvariety "this") can be generalized as well:
Let $\beta:\widetilde{X}\mathrel{\mathop:}=\mathop{\mathrm{Bl}}\_Z(X)\to X$ be the blow-up of a smooth algebraic variety $X$ along a irreducible generically smooth subscheme $Z$. 
Let $E\mathrel{\mathop:}=\beta^{-1}(Z)$ be the exceptional divisor. Now, let us assume I have a divisor $D$ on $X$, with $Z$ closed subscheme of $D$ ($D$ smooth and irreducible, if it helps). 
Then I ask if $\beta^\ast D \sim \widetilde{D} + \alpha E$, for some integer $\alpha$, where $\widetilde{D}$ denotes the strict transform of $D$ and "$\sim$" is linear equivalence.



Strict Transform under Blow-Up along singular Subscheme
I think the answer is, almost. 


$E$ might have more than one component. Indeed, since you only assume that $Z$ is generically smooth, $E$ might have many different components. $E = \sum E\_i$.


You can always write $\beta^\* D = \widetilde{D} + \sum\_i \alpha\_i E\_i$, but you can't use the same $\alpha$ for all of them.


Let me give an example. Suppose $X = \mathbb{A}^2$, $Z = V(x \cdot (x,y) \cdot (x,y^2)) = V(x^3, x^2y^2, x^2y, xy^3)$.  

Certainly $Z$ is generically smooth and irreducible (its just not reduced, in higher dimensions, I'm think I can rig reduced examples as well). Set $D = \text{Div}(x^3)$. The blow-up of $X$ has two exceptional divisors.  

Note $Z$ is a closed subscheme of $D$ as well. By working in higher dimensions, one can rig examples where the support of $Z$ and $D$ are different.


Anyway, one chart of the blow-up looks like $k[x/y, y^2/x]$ with the obvious map to $\text{Spec} k[x,y]$. On that chart, the two exceptional divisors are $E\_1 = \text{Div}(x/y)$ and $E\_2 =\text{Div}(y^2/x)$. 


On this chart, we compute $\beta^\*D$. Note
$$x^3 = ((x/y)^2 \cdot (y^2/x))^3$$ and so $$\beta^\*D = \widetilde{D} + 6 E\_1 + 3 E\_2.$$


