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Let $X\subset \mathbb{A}^4_\mathbb{C}$ be a three fold defined by the equation $xy-zw=0$
This variety has a singularity at origin of $\mathbb{A}^4_\mathbb{C}$
If we blow up this three fold in two ways i.e by the ideals $(x,z)$ and $(y,z)$, we can get two resolutions $Y_0\rightarrow X,Y_1\rightarrow X$ with exceptional curves $\mathbb{P}^1_\mathbb{C}$

How can I prove $Y_0$ and $Y_1$ are not isomorphic over $X$.

I would like to prove this claim in order to prove $X$ has no minimal resolution.

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The strict transform of the plane $x = z = 0$ in one of the resolutions is still a plane, and in the other it is isomorphic to the blowup of the plane.

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  • $\begingroup$ Thanks. I would like to ask you a one more question. $\endgroup$
    – George
    Commented Apr 17 at 0:20
  • $\begingroup$ Why are these blow up resolutions? Even if we restrict these maps to the complement of the fiber of singular locus , we cannot get bijection on to it's image $\endgroup$
    – George
    Commented Apr 17 at 0:26
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    $\begingroup$ This is a computation. Over the complement of the singularity, the blowup center is a Cartier divisor, so the blowup is the identity. $\endgroup$
    – Sasha
    Commented Apr 17 at 4:43
  • $\begingroup$ Thanks. I understand why these blowup are resolutions. How to show this three fold has no minimal resolution by using these fact. $\endgroup$
    – George
    Commented Apr 17 at 11:24
  • $\begingroup$ @George: for $i=0,1$, if $Y_i \rightarrow Z \rightarrow X$ is a factorisation with $Z$ smooth then $Y_i \rightarrow Z$ must be an isomorphism over $X$. Since $Y_0$ and $Y_1$ are not isomorphic over $X$, this means that the two morphisms $Y_i \rightarrow X$ cannot both factor through a morphism $Z \rightarrow X$ for some smooth $Z$. $\endgroup$
    – Lazzaro Campeotti
    Commented Apr 18 at 15:13

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