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In the paper "Moduli of $p$-divisible groups" by Scholze and Weinstein (see http://math.bu.edu/people/jsweinst/Moduli/Moduli.pdf), one finds the following claim in Lemma 5.2.7:

Lemma: Let $K$ be a spherically complete nonarchimedean field. Let $\Lambda_0 \supset \Lambda_1 \supset ... \Lambda_n \supset ...$ be a descending sequence of principal ideals of $O_K$ such that $\cap_n \Lambda_n \neq 0$. Then $R^1 \lim_n \Lambda_n = 0$.

No proof is given in the paper, and I have failed at reconstructing the proof myself. Any help or pointers on why this is true would be highly appreciated.

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  • $\begingroup$ First you forgot a condition, second: there IS given a proof in that paper! $\endgroup$
    – user1688
    Commented Dec 31, 2015 at 7:24
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    $\begingroup$ @Corbennick: The OP is asking about the final assertion in the proof of Lemma 5.2.7 in that paper, a step which is left to the reader (and so not proved in the paper), upon imposing the omitted condition to which you allude (i.e., each nonzero ideal $\Lambda_i$ is finitely generated, or equivalently is a "closed disk"). Exercise 3.5.1 in Weibel's book on homological algebra is a good reference for Toby to see that the requested vanishing is literally a restatement of the spherical completeness of $K$ (using that the $\Lambda_i$'s are closed discs with radii bounded away from 0). $\endgroup$
    – nfdc23
    Commented Dec 31, 2015 at 9:48
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    $\begingroup$ Your use of spherical completeness is too limited (why apply to that single descending sequence of discs, which anyway all contain 0!); it really gives you much more. Please do the exercise in Weibel's book, and then you should see the rest. I shouldn't have made the statement about "bounded away from 0" for radii; it isn't used here (but highlights the relevance of the comment about possibly non-unique limits near the end of the exercise in Weibel's book). $\endgroup$
    – nfdc23
    Commented Dec 31, 2015 at 17:30
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    $\begingroup$ Sorry for being slow. My proof of the exercise in Weibel's book is to use the short exact sequence $0 \to \{A_i\} \to \{A\} \to \{A/A_i\} \to 0$, apply derived functors, observe that the middle system has a vanishing $\lim^1$ as its constant, and thus arrive at: $A \to \lim A/A_i$ is surjective (i.e., each Cauch sequence has a limit) iff $\lim^1 A_i = 0$. This doesn't seem to help shed light on the present situation though. $\endgroup$
    – Toby
    Commented Dec 31, 2015 at 18:55
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    $\begingroup$ It sheds a lot of light on the present situation (and in fact explains everything): think amount the geometric meaning of surjectivity of $K \rightarrow \lim K/\Lambda_i$ and then stare at the definition of spherical completeness. $\endgroup$
    – nfdc23
    Commented Dec 31, 2015 at 22:02

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