I was checking an example of canonical singularities from surface.


We consider the surface $X:(xz=y^2)\subset \mathbb A^3$. The only singular point is the origin. We write down one affine piece of the blow-up of $\mathbb A^3$ at the origin, which is the map $\sigma:\mathbb A^3\to \mathbb A^3$ given by 
$$x=u,y=uv,z=uw.$$


Note that $\omega\_X=\frac{1}{z}dy\wedge dz=\frac{1}{2y}dx\wedge dz=\frac{1}{x}dx\wedge dy$ is a rational canonical differntial of $X$ which is regular outside the origin. Note that $dx\wedge dy=udu\wedge dv$ and $Y=\sigma^\*X:(w=v^2)\subset \mathbb A^3$. It seems to me that we have $K\_Y=\sigma^\* K\_X+E$. But we have already known that $K\_Y=\sigma^\* K\_X$, I must make some mistakes. Could anyone explain this for me ? Thanks! 



pull-back of canonical divisor under blow-up of a singular point
Here is the computation:


let us consider the action:
$$
\begin{array}{ccc}
\mu\_{2}\times\mathbb{A}^{2} & \longrightarrow & \mathbb{A}^{2}\\
(\epsilon,x\_{0}, x\_{1}) & \longmapsto & (\epsilon x\_{0},\epsilon x\_{1})
\end{array}
$$
The ring of invariants is given by:
$$k[x\_0^2,x\_0x\_1,x\_1^2]\cong k[y\_0,y\_1,y\_2]/(y\_0y\_2-y\_1^2)$$
and we see that the singularity $X=\mathbb{A}^{2}/\mu\_{2}$ corresponds to the vertex $v$ of the affine cone $$X=Spec(k[x\_0^2,x\_0x\_1,x\_1^2])\cong k[y\_0,y\_1,y\_2]/(y\_0y\_2-y\_1^2)$$
that is the vertex of a quadric cone $Q\subset\mathbb{P}^2$ or equivalently the singularity $\frac{1}{2}(1,1)$ of the weighted projective plane $\mathbb{P}(1,1,2)$. Now, $dx\_0\wedge dx\_1$ is a basis of $\bigwedge^2\Omega\_{\mathbb{A}^2}$, and $(dx\_0\wedge dx\_1)^{\otimes 2}$ is invariant under the action. The form
$$\omega = \frac{(dy\_0\wedge dy\_1)^{\otimes 2}}{y\_0^2}\in (\bigwedge^2\Omega\_{k(X)})^{\otimes 2}$$
is a basis of $(\bigwedge^2\Omega\_{X})^{\otimes 2}$ because the quotient map $\pi:\mathbb{A}^2\rightarrow X$ is \'etale on $X\setminus\{v\}$, and $\pi^{\*}\omega = 4(dx\_0\wedge dx\_1)^{\otimes 2}$.
Blowing-up the vertex $v$ we get a resolution $f:Y\rightarrow X$. If $[\lambda\_0:\lambda\_1:\lambda\_2]$ are homogeneous coordinates on $\mathbb{P}^2$ then the equations of $Y$ in $\mathbb{A}^3\times\mathbb{P}^2$ are:
$$
\left\lbrace\begin{array}{l}
y\_0\lambda\_1-y\_1\lambda\_0=0, \\ 
y\_0\lambda\_2-y\_2\lambda\_0=0, \\ 
y\_1\lambda\_2-y\_2\lambda\_1=0, \\
y\_0y\_2-y\_1^2. 
\end{array}\right. 
$$
Therefore, $y\_1 = \frac{\lambda\_1}{\lambda\_0}y\_0$, and $\frac{\lambda\_2}{\lambda\_1}=\frac{\lambda\_1}{\lambda\_0}$ yields $y\_2 = \frac{\lambda\_1}{\lambda\_0}y\_1 = (\frac{\lambda\_1}{\lambda\_0})^2y\_0$. Then, in $Y$ we have an affine chart isomorphic to $\mathbb{A}^2$ with coordinates $(y\_0,t)$ where the resolution is given by $(y\_0,t)\mapsto (y\_0,y\_0t,y\_0t^2)$, with $t = \frac{\lambda\_1}{\lambda\_0}$, and the exceptional divisor $E$ over $v$ is given by $\{y\_0=0\}$. We have
$$f^{\*}\omega = (dy\_0\wedge dt)^{\otimes 2}.$$
Therefore, $f^{\*}\omega$ has neither a pole nor a zero along $E$, and we may write $K\_Y = f^{\*}K\_X$.


