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

$$\DeclareMathOperator\Bl{Bl}$$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.

• Note that the spelling is 'Cartier' in your header. Also, please set your question in a shaded area by starting a paragraph with > followed by space. Commented Oct 13, 2017 at 15:12
• @JimHumphreys Thank you! I just edited it as you advised. Commented Oct 13, 2017 at 19:51

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 $d=1$.
• I think $d$ should be $\frac{1}{2}$. Otherwise, assume $d=1$, then $K_{Z/X}=3\tilde{E}_1+\frac{7}{2}E_2$ and it is not Cartier. But $Z$ and $X$ are both smooth. Commented Oct 27, 2017 at 1:23