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?
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No. Let $R'$ be any nonexcellent 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.

2$\begingroup$ 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

1$\begingroup$ 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 suchandsuch 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