pull-back of canonical divisor under blow-up of a singular point 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! 
 A: 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 $K_Y = f^{*}K_X$.
