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Let $R$ be a complete DVR with fraction field $K$, $X$ be a regular scheme flat over $R$. Let $L$ be a finite field extension of $K$ and $Q$ be the integral closure of $R$ in $L$. Denote by $Y:=X \times_R Q$ the base change of $X$. Is $Y$ a regular scheme? If not true in general, is there any additional assumption on $R$ for which this holds true?

PS. One can assume that the characteristic of $K$ is zero.

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closed as off-topic by Olivier, Karl Schwede, abx, Dima Pasechnik, Daniel Moskovich Apr 23 '15 at 13:58

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  • $\begingroup$ Check out formally smooth and/or geometrically regular. $\endgroup$ – Olivier Apr 22 '15 at 20:19
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No. Take $R=\mathbb{Z}_p$ and $X$ to be the affine scheme defined by $xy=p$. This is regular, but its base change to $\mathbb{Z}_p[\sqrt{p}]$ is not.

If $X$ is smooth over $R$, then $Y$ is smooth over $Q$ (and therefore regular in many useful situations). This is of course a much stronger assumption.

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  • $\begingroup$ And if you want an equal-characteristic example, take $R=\mathbf C[[t]]$ and $X$ be the closed subscheme of $\mathbf A^2_R$ defined by $xy=t$; it is regular, but its base change to $\mathbf C[[t^{1/2}]]$ is not. $\endgroup$ – ACL Apr 22 '15 at 18:56

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