Strict transform under resolution of singularity along a singular $\mathbb{Q}$-Cartier divisor

Question

$\DeclareMathOperator\Bl{Bl}$Let $f: Y=\Bl_0^\omega(\mathbb{C}^3)\to \mathbb{C}^3$ be a weighted blow up of $\mathbb{C}^3$ with weights $w(x,y,z)=(1,1,2)$. Then $Y$ and the exceptional divisor $E\cong \mathbb{P}(1,1,2)$ are singular at the same point. Locally, the germ of singularity in $Y$ is isomorphic to a cone over Veronese surface, living in $\mathbb{C}^6$, and $E$, as a Weil divisor, can be thought of as a cone over the image of a quadratic curve mapped in $\mathbb{P}^5$ by the Veronese embedding $v: \mathbb{P}^2\to\mathbb{P}^5$.

Blow up $\mathbb{C}^6$ at the vertex of the cone, $g:Bl_0\mathbb{C}^6\to \mathbb{C}^6$ singularities in $Y$ and $E$ should both be resolved. And we should have some $\mathbb{Q}$-linear equivalence: $$g^*E\sim_{\mathbb{Q}} \tilde{E}+dE'$$ where $E'$ is the exceptional divisor of $g$ and $\tilde{E}$ is the strict transform of $E$ under $g$.

My questions is: How to compute $d$?

What I have tried is to write, locally, $Y$ as $\mathbb{C}^3/\mathbb{Z}_2$ and realized as a cone over Veronese surface, so can be $\operatorname{Spec} \mathbb{C}[x^2, y^2, z^2, xy, yz, xz]$, and $E$ can be cut by $z=0$.

But I am confused about how to define multiplicity of $E$ at the vertex, as a divisor in $Y$. I don't know how to get $d$ by similar argument as the Proposition 3.6, Chapter V in Hartshorne's book.

Answer 1

The cone $Y$ is the weighted projective space $\mathbb{P}(2,1,1,1)$. The vertex is the point $[1:0:0:0]$. Call $[x:y:z:w]$ the coordinates on $\mathbb{P}(2,1,1,1)$.

Now, a cone in $\mathbb{P}(2,1,1,1)$ over a conic is given by an equation of the form $f(y,z,w)=0$ where $f$ is an homogeneous polynomial of degree two in $y,z,w$.

Now, call $u_0,u_1,u_2,u_3,e$ the coordinates on the blow-up $Z$ of $\mathbb{P}(2,1,1,1)$ at $[1:0:0:0]$, where $e$ corresponds to the exceptional divisor $E'$. The blow-up morphism is given by $\pi:Z\rightarrow\mathbb{P}(2,1,1,1)$, $\pi(u_0,u_1,u_2,u_3,e) = (u_0,u_1e,u_2e,u_3e)$, and hence $\pi^{*}(f(y,z,w)=0) = {e^2f(u_1,u_2,u_3)=0}$. Since the variable $x$ has weight two in your notation you get $d=1$.