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As an example, I want to find resolution of singularity of Segre variety in $M_{2,2}$, let this variety $V=V(x_1x_4-x_2x_3)$. Then blow-up of $V$ at $O$ is a closed subset of $\mathbb{A}^4 \times \mathbb{P}^3$ defined by the following equations; \begin{eqnarray} x_1y_2-x_2y_1=0, \nonumber\\ x_1y_3-x_3y_1=0, \nonumber\\ x_1y_4-x_4y_1=0, \nonumber\\ x_2y_3-x_3y_2=0, \nonumber\\ x_2y_4-x_4y_2=0, \nonumber\\ x_3y_4-x_4y_3=0, \nonumber\\ x_1x_4-x_2x_3=0. \nonumber \end{eqnarray} On some coordinate $y_4=1$, those relations can be expressed as follows; \begin{eqnarray} x_1=x_4y_1, \nonumber \\ x_2=x_4y_2, \nonumber\\ x_3=x_4y_3, \nonumber \\ x_1x_4=x_2x_3. \nonumber \end{eqnarray} Then it becomes $ \{ (x_4, y_1, y_2,y_3); y_1=y_2y_3 \}$. However, my question is if we let $g:U=\{(y_1,y_2,y_3,x_4)\} \mapsto \{(x_1,x_2,x_3,x_4)\}$, the Jacobian of it should be invertible, but my computation is zero. What is problem on my argument? Did I misunderstand the notion of blow-up?

Thanks for reading my question, please give me advice.

ddd
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  • The equation $y_1 = y_2y_3$ define a smooth quadric surface and similarly with the other charts, so it seems you did resolved the singularity. The exceptional divisor in this chart is given by $x_4 = y_1 - y_2y_3 = 0$. On the other hand, it's not surprising that $g$ is not invertible (if $g$ is the restriction of the blow-up to this chart) since it will contract the intersection of the exceptional divisor with the chart. The blow-up is an isomorphism only on the complement of the exceptional divisor. – Nicolas Hemelsoet Nov 09 '17 at 07:39

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