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This question already has an answer here:

Let $k$ be a field, let $X/k$ be a stable curve. Is it always possible to find a deformation $\mathcal{X}/k[[t]]$ such that $\mathcal{X}$ is regular?

(Sorry for the confusion, this is a duplication of one of my previous post....The answer to this question is yes, by Theorem B.2 in Brian Conrad’s Appendix to “Specialization of linear systems from curves to graphs” by Matthew Baker.)

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marked as duplicate by Qixiao, user44191, Dima Pasechnik, LSpice, Denis Nardin Jul 15 at 18:02

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ It looks like you ask a question and answer it at the same time … what is the question, then? $\endgroup$ – LSpice Jul 14 at 13:23
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    $\begingroup$ @LSpice Sorry for the confusion, I asked the question and realized it was a duplicate, but I couldn’t close the question. $\endgroup$ – Qixiao Jul 14 at 21:49
  • $\begingroup$ In that situation, you can flag your post for moderator attention. $\endgroup$ – LSpice Jul 15 at 1:51
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Yes for $k$ algebraically closed. First of all, for each singularity, we can choose a local deformation over $k[[t]]$ such that the total space around this singularity is regular. In some local coordinates it can be written as $k[[x, y, t]] / (xy-t)$. To get a global deformation, one uses the local-global principle (1.5) of Deligne--Mumford.

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