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?


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 $uw = 0$).


Just to spell out what Sasha means, here is a way to see that the resolution in crepant. Once you believe in adjuction, $$ K_{X'} = (K_{\tilde{\mathbb{A}^3}} + X')_{\mid X'} = (\sigma^*K_{\mathbb{A}^3}+2E+\sigma^*Y - 2E)_{\mid X'} = \sigma^*(K_{\mathbb{A}^3}+Y)_{\mid X'} = \sigma^*K_Y. $$ First we use the adjunction formula on the blowup $\sigma:\tilde{\mathbb{A}^3}\to \mathbb{A}^3$, then explicit formulas for the blowup (recall that $Y$ has multiplicity 2), then adjunction on $\mathbb{A}^3$ itself.


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