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Let $(X,0)$ be a normal surface singularity. Suppose that it does not admit a smoothing.

Is it possible that there exists an isolated surface singularity $(Y,0)$ reduced near $0$ which is not irreducible and $(X,0)$ is one of its irreducible components and such that $(Y,0)$ does admit a smoothing? Is there some obstruction for this to happen?

Are these kind of questions treated somewhere?

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  • $\begingroup$ Are you assuming that $Y$ is reduced near $0$? $\endgroup$
    – Jason Starr
    Commented Nov 15, 2018 at 13:38
  • $\begingroup$ Yes, I assume that. I edited the equestion. Although, I would be happy with $Y$ being reduced along a component isomorphic to $(X,0)$. $\endgroup$
    – user131261
    Commented Nov 15, 2018 at 14:53
  • $\begingroup$ If $Y$ is reduced near $0$, then you should be able to rule out deformations of singularities of $(Y,0)$. $\endgroup$
    – Jason Starr
    Commented Nov 15, 2018 at 15:00
  • $\begingroup$ I am not sure I understand your comment. Do you mean that $(Y,0)$ does not admit a smoothing because one of its components does not? $\endgroup$
    – user131261
    Commented Nov 15, 2018 at 16:07
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    $\begingroup$ In your case $X$ is a connected component of the normalisation $Z$ of $Y$. On p. 255 of dam.brown.edu/people/mumford/alg_geom/papers/… Mumford attributes to Rim the result that $Z$ is local (so that $Z=X$) if $Y$ is smoothable. This seems to resolve your issue. $\endgroup$
    – inkspot
    Commented Nov 15, 2018 at 21:20

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I will adhere to Mumford's notation as much as possible. Let $y$ denote the closed point of $Y$, $\mathcal O=\Gamma(Y,\mathcal O_Y)$, $\widetilde{\mathcal O}=\Gamma(Y-\{y\},\mathcal O_Y)$ and $Z=\operatorname{Spec}\widetilde{\mathcal O}$, which is the normalization of $Y$ in this situation. Then (Rim) $\widetilde{\mathcal O}$ is local if $Y$ is smoothable, so $Z$ is connected. Then $X=Z$, and then $X=Y$.

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