Let $n:\widetilde X\rightarrow X$ be the normalization of a complex (quasi-projective) variety $X$. Assume $\widetilde X$ is smooth, that $n$ is an isomorphism outside a smooth connected subvariety $Y\subset \widetilde X$ of codimension $>1$ and that $n_{|Y}:Y\rightarrow n(Y)$ has degree $2$ (the quotient of $Y$ by an involution).
Are the blow-ups $Bl_Y(\widetilde X)$ and $Bl_{n(Y)}(X)$ necessarily isomorphic?
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1$\begingroup$ That fails if $X$ is a tacnodal curve. $\endgroup$– Jason StarrCommented May 4, 2021 at 17:33
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1$\begingroup$ Oops! A tacnodal curve fails the "codimension $>1$" hypothesis for $Y$. Instead, consider a "tacnodal surface", e.g., the zero scheme in affine $4$-space $\text{Spec}\ \mathbb{C}[s,t,u,v]$ of the ideal $\langle u,v\rangle \cap \langle u-s^2,v-t^2 \rangle$. $\endgroup$– Jason StarrCommented May 4, 2021 at 18:15
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$\begingroup$ Thank you very much for your counter-example. I edited the question to be closer to the situation I am interested in. $\endgroup$– pi_1Commented May 4, 2021 at 18:16
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$\begingroup$ Thanks. The codimension $>1$ does not add anything: it does not guarantee that the projectivized normal is disconnected. $\endgroup$– pi_1Commented May 4, 2021 at 18:32
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$\begingroup$ You can easily modify my example to be connected: simply take the product with some smooth variety $Y$ that admits a finite, unbranched, 2-to-1 cover, $Y'\to Y$. My example variety $X$ admits an involution that permutes the two branches at the tacnode, namely $(s,t,0,0) \leftrightarrow (s,t,s^2,t^2)$. Form the quotient of $X\times Y'$ by the "diagonal" involution. Then the singular is connected, isomorphic to $Y$, and the normalization is isomorphic to the quotient of $\widetilde{X}\times Y'$ by the diagonal involution. So the inverse image of the singular locus is isomorphic to $Y'$. $\endgroup$– Jason StarrCommented May 20, 2021 at 19:21
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