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In a book I am reading, "Travaux de Gabber sur l'uniformisation locale et la cohomologie etale des schemas quasi-excellents" by Luc Illusie, Yves Laszlo, Fabrice Orgogozo (https://arxiv.org/abs/1207.3648), there is the following assertion whose proof I would like to understand.

Let $e_1, ..., e_n$ be natural numbers, not all zero. For any trait $S$ with uniformizer $\pi$, define $$ V(S,\pi,e_1,...,e_n) = Spec(O_S[T_1,\ldots,T_n]/(T_1^{e_1}...T_n^{e_n}-\pi)). $$ It is claimed that this scheme is regular in Exposé XVI, Lemma 3.5.13. However the proof is only briefly outlined. If someone can provide details, I would appreciate it.

The first case is if at least one of $e_i$ are "invertible in $\eta$" (generic point of $S$), which, I don't even see why that would be relevant, but this case is supposed to be an easy exercise. Is there a regularity criterion which would be easy to apply here?

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    $\begingroup$ Maybe I'm mistaken, but this has dimension $n$ and $T_1, \ldots, T_n$ is a regular sequence. After dividing by the ideal $(T_1, \ldots, T_n)$ you obtain $\mathcal{O}_S/(\pi)$, which is a field. The underlying criterion is that if $f$ is a nonzerodivisor in a local ring $R$ such that $R/(f)$ is regular, then $R$ is regular. $\endgroup$ – Piotr Achinger Jun 4 at 7:03
  • $\begingroup$ For intuition, let's consider $\mathcal{O}_S = k[\pi]$, then your ring is $k[\pi][T_1, \ldots, T_n]/(T_1^{e_1} \ldots T_n^{e_n} - \pi) = k[T_1, \ldots, T_n]$ i.e. just a polynomial ring. (And of course it is irrelevant for regularity that we have a monomial in the $T_i$.) $\endgroup$ – Piotr Achinger Jun 4 at 7:16
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    $\begingroup$ @Piotr Achinger: the ring is not local. $\endgroup$ – abx Jun 4 at 7:33
  • $\begingroup$ @abx Oops, of course you are right! Part of my brain was thinking of $\mathcal{O}_S[[T_1, \ldots, T_n]]/(\ldots)$... Thinking of an easy fix that doesn't need $\mathcal{O}_S$ quasi-excellent. $\endgroup$ – Piotr Achinger Jun 4 at 7:41
  • $\begingroup$ That was exactly my first thought as well, but indeed the ring doesn't seem to be local as far as I could see. $\endgroup$ – VeridisQuo Jun 4 at 10:07
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Here is another argument to prove the desired regularity, along the lines of Piotr Achinger's comment.

As explained in the linked seminar notes, the essential case to consider is $V(S, \pi, e_1, \dots, e_n)$ (for which we write $V$), where for some $i$, the exponent $e_i$ is invertible in $\eta$.

From the hypothesis that $e_i$ is invertible we may conclude that $V_\eta$ is smooth over $\eta$. Therefore, all points of $V$ in the fiber over $\eta$ are regular. To show that $V$ is regular, it thus remains to show that the points in the special fiber are regular, i.e. it remains to show that any prime ideal of $\mathcal{O}_S[T_1, \dots, T_n]/(T_1^{e_1} \dots T_n^{e_n} - \pi)$ (for which we write $R$) containing $\pi$ is generated by a regular sequence.

Consider a prime ideal $P$ of $R$ containing $\pi$. Since we have $\pi = T_1^{e_1} \dots T_n^{e_n}$ in $R$, we have $T_j$ in $P$ for some $j$ such that $e_j$ is non-zero. The quotient ring $R/T_j R$ is isomorphic to $(\mathcal{O}_S/\pi \mathcal{O}_S)[T_1, \dots, T_{j-1}, T_{j+1}, \dots, T_n]$. (Here we use the hypothesis that $e_j$ is non-zero: the ideal of $\mathcal{O}_S[T_1, \dots, T_n]$ generated by $T_1^{e_1} \dots T_n^{e_n} - \pi$ and $T_j$ is the same as the ideal generated by $\pi$ and $T_j$.) Since $R/T_j R$ is a polynomial ring over a field, any prime ideal of $R/ T_j R$ is generated by a regular sequence. In particular, the prime ideal $P(R/T_j R)$ is generated by a regular sequence. Let $g_1, \dots, g_k$ be elements of $R$ mapping to such a regular sequence. We then have that the sequence $T_j, g_1, \dots, g_k$ generates $P$. Furthermore, $T_j$ is regular in $R$, since $R/T_j R$ is a domain, and so, by the definition of ``regular sequence'', we have that $T_j, g_1, \dots, g_k$ is a regular sequence in $R$.

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There is an effective action of the torus $G=\mathbb G_{m,S}^{n-1}=\operatorname{Spec} O_S[s_1,...,s_n]/(s_1...s_n-1)$ on $V$ (the co-action is given by $T_i\mapsto s_i\otimes T_i$) such that the closed point $0$ defined by the ideal $(T_1,...,T_n,\pi)$ lies in the closure of every $G$-orbit in the closed fiber of $V\to S$. Since the non-regular locus of $V$ is closed, it is enough to verify that $V$ is regular at $0$, for which Piotr Achinger's first comment suffices.

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