Recall that a discrete valuation ring $R$ is excellent if the extension $\widehat{K}/K$ is separable, where $\widehat{R}$ is the completion of $R$ (with respect to the maximal ideal), $K = \mathrm{Frac}(R)$, and $\widehat{K} = \mathrm{Frac}(\widehat{R})$. Suppose therefore that $R$ is excellent and consider a local injection $R \hookrightarrow R^{\prime}$ of discrete valuation rings such that the induced residue field extension is separable and such that a uniformizer of $R$ is also a uniformizer of $R^{\prime}$. Then, is $R^{\prime}$ necessarily excellent?


No. Let $R'$ be any non-excellent dvr whatsoever, hence of equicharacteristic $p > 0$, and let $t \in R'$ be a uniformizer. Let $R = \mathbf{F}_p[t]_{(t)}$. The local inclusion $R \hookrightarrow R'$ has induced residue field extension that is separable since $\mathbf{F}_p$ is perfect. And of course $R$ is excellent for any of a million reasons. So this seems to be a counterexample.

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    $\begingroup$ Thank you. Then this is a counterexample to the second claim of Lemma 2 of section 3.6 of Bosch, Lutkebohmert, Raynaud "Neron models." (The elementary question math.stackexchange.com/questions/1003548/… was also about this.) $\endgroup$ – Question Mark Nov 3 '14 at 5:16
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    $\begingroup$ Ah, their Lemma just has a (bad) typo in its statement. The entire content of the proof of the Lemma is its second sentence (which never uses the excellence of $R$, and whose proof is correct), and the third has the typo: the implication they meant is that $R$ is excellent if $R'$ is excellent (not the other way around, which is what is written), and that is an immediate consequence of what they prove. So in the statement of their Lemma 2, replace $R$ with $R'$ in the first sentence and replace the final $R'$ with $R$ in the last sentence. (That circumvents your other question linked above.) $\endgroup$ – user27920 Nov 3 '14 at 5:57
  • $\begingroup$ I agree that the proof of the second sentence is correct, but it seems to me that excellence of $R$ is used (to get that $\widetilde{R}$ is a finite $R$-module). I also don't see how to descend excellence from $R'$ to $R$, if the claim was intended as you suggest. Could you clarify? $\endgroup$ – Question Mark Nov 3 '14 at 6:18
  • $\begingroup$ @QuestionMark: My recollection is that one can get the conclusions about $\widetilde{R} \otimes_R R'$ being a dvr (with such-and-such uniformizer) without excellence, by using other results from general valuation theory and commutative algebra. I'll try to come back to this later. But once that is shown then ${\rm{Frac}}(\widehat{R'})$ is separable over ${\rm{Frac}}(R)$ by transitivity through separability of ${\rm{Frac}}(R')$ over ${\rm{Frac}}(R)$, so the intermediate field ${\rm{Frac}}(\widehat{R})$ is also separable over ${\rm{Frac}}(R)$ (the "opposite" of the other question you asked). $\endgroup$ – user27920 Nov 3 '14 at 16:05
  • $\begingroup$ Thanks. It would be helpful if you could clarify about $\widetilde{R} \otimes_R R'$. Using Bourbaki "Algebre commutative" VI, no. 1 Cor. 3 and no. 3 Thm. 1 and the uniqueness of an extension of a valuation to a purely inseparable extension of degree $p$, I can see that $\widetilde{R}$ is a DVR even when $R$ is not excellent, but I am not sure how to conclude that $\widetilde{R} \otimes_R R'$ is, e.g., Noetherian; presumably one should use excellence of $R'$? Also, there seems to be one more case to consider: when both the ramification index and the residual degree of $\widetilde{R}/R$ are 1. $\endgroup$ – Question Mark Nov 3 '14 at 17:35

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