Consider the action of $\mathbb{Z}\_3\subset SL\_2(k)$ on $\mathbb{A}^2$, we have the quotient $Y$ as in the title. According to the classification of Du Val singularity, we know that the crepant resolution $X$ of $Y$ has two exceptional curves with self-intersection -2, which intersects at a points. 


On the other hand, if one just blow up $(0,0,0)$ in $\mathbb{A}^3$ and consider the strict transformation $X'$ of $Y$. One has on each patch smooth surfaces defined by (suppose coordinate (x,y,z,u:v:w))


$k[x,v,w]/(w-xv^3)=k[x,v]$ $(u=1)$


$k[y,u,w]/(y-uw)=k[u,w]$ $(v=1)$


$k[z,u,v]/(u-zv^3)=k[z,v]$ $(w=1)$


hence $X'$ is also smooth.


It seems to me that $X'$ and $X$ are not isomorphic, and I feel $X'$ is more of a minimal resolution of $Y$.


My questions : Is $X'$ a crepant resolution?


How are $X'$ and $X$ related?



Crepant resolution of $Y=k[x,y,z]/(xz-y^3)$
It is easy to see that the canonical class of $X'$ is trivial (combine the formula for the canonical class of the blowup of $A^3$ and the adjunction formula). Hence $X'$ is crepant. But a crepant resolution of a surface is unique, so $X = X'$.


By the way, if you look at the exceptional divisor of $X'$, you will note that this is a union of two curves (in your second chart given by the equation $uw = 0$).


