Consider the following setting: let $(R,\mathfrak{m})$ be a Noetherian complete discrete valuation ring (with maximal ideal $\mathfrak{m}$) and let $K$ be its field of fractions. Now take $L$ be the completion of an algebraic extension, even infinite, of $K$, and let $S$ be the integral closure of $R$ inside $L$. What I want to prove is that if I take an étale finite group scheme $G$, represented by the étale Hopf algebra $A$, I have a bijection of sets \begin{equation} G(S/\mathfrak{m}^{i+1}S):=\text{Hom}_{\text{RAlg}}(A,S/\mathfrak{m}^{i+1}S)\simeq \text{Hom}_{\text{RAlg}}(A,S/\mathfrak{m}^{i}S):=G(S/\mathfrak{m}^{i}S) \end{equation} given for every $i$ by the composition with the projection to the next quotient. I guess that the main ingredient to prove this could be Hensel Lemma, since $R$ is Henselian, but I really don't see the proof. Any hint will be very appreciated. Thanks!
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$\begingroup$ use the fact that etale implies formally etale. $\endgroup$– Will SawinCommented May 24, 2015 at 7:47

2$\begingroup$ Thank you! But I still don't see how this can help me. Maybe it's because I don't know well the notion of formally etale. Can you help me more? $\endgroup$– rimeCommented May 24, 2015 at 8:00

1$\begingroup$ Every object in your question matches up with something in the definition of formally etale. Using Wikipedia's notation, you want to take $C= S/m^{i+1}S$, $B=G$, $J= m^i S$. Then the statement it gives is precisely the statement you ask for. The key point is that $m^i$ is nilpotent in $S/m^{i+1}$. $\endgroup$– Will SawinCommented May 24, 2015 at 12:44
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