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I've been reading Kedlaya's paper http://arxiv.org/abs/math/0208027 on finiteness of rigid cohomology and there's something I can't quite resolve in my understanding of the topology on the Robba ring.

Let's take $K$ to be a complete $p$-adic field ($\mathbb{Q}_p$ if you want), with ring of integers $\mathcal{O}_K$ and uniformiser $\pi$. Then the Robba ring $\mathcal{R}_K$ over $K$ consists of those series $\sum_{i\in\mathbb{Z}} a_it^i$ with $a_i\in K$ such that for all $\eta<1$ $\vert a_i\vert\eta^i\rightarrow 0$ as $i\rightarrow \infty$, and there exists some $\xi<1$ such that $|\vert a_i\vert\xi^i\rightarrow 0$ as $i\rightarrow -\infty$. As far as I understand (although this may be one of the things I've got wrong), there are partially defined norms $\lVert\sum_i a_it^i \rVert_r=\sup_i\{\vert a_i \vert p^{-ir}\}$ for $r>0$, and these induces a topology on $\mathcal{R}_K$, for which a sequence $f_n\rightarrow 0$ iff for all sufficiently small $r>0$, the norms $\lVert f_n \rVert_r$ are all defined and tend towards zero. (This seems to be the definition of the topology on Robba rings given by Kedlaya.)

The Robba ring also contains an integral subring $\mathcal{R}^\mathrm{int}_K$ consisting of series $\sum_i a_it^i$ such that each $a_i\in \mathcal{O}_K$, and Definition 2.5.10 of Kedlaya's paper says that this subring can be topologically characterised as the set of elements $f$ such that for any $a\geq0$ and any $c\in K$ with $\vert c\vert<1$, the sequence $(cf^a)^n$ converges to $0$ in $\mathcal{R}_K$. Actually, it's not too difficult to see that $\mathcal{R}_K^\mathrm{int}$ has to be contained in the set of `topologically bounded' elements, and it seems plausible that the converse is also true. Either way, $\mathcal{R}^\mathrm{int}_K$ is a DVR, with maximal ideal generated by $\pi$, and the quotient $\mathcal{R}^\mathrm{int}_K/(\pi)$ is just the Laurent series field $k((t))$ over the residue field $k$ of $K$.

Again, looking at Definition 2.5.10 of Kedlaya's paper, there is a topological characterisation of the ideal $(\pi)$ inside $\mathcal{R}_K^\mathrm{int}$ - he says that the set of topologically nilpotent elements in $\mathcal{R}_K^\mathrm{int}$ forms an ideal and the quotient by this ideal is again isomorphic to $k((t))$. But now let's look at the element $t$ of $\mathcal{R}_K^\mathrm{int}$. Then for all $r>0$ we have $\lVert t^n \rVert_r = p^{-nr}\rightarrow 0$ as $n\rightarrow \infty$, which says that $t$ is topologically nilpotent. But $t$ is a unit in $\mathcal{R}_K^\mathrm{int}$, so something's gone wrong! Where have I made my mistake?

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  • $\begingroup$ I believe that your norms at the beginning are wrong and should be replaced by $\| \sum_i a_i t^i\|_r = \sup_i\{|a_i| p^{-ir}, |a_i|\}$. In geometric terms, you are considering the sup-norm on an annulus $C(0;p^{-r},1)$ and in your formula you only considered one of the two sides. $\endgroup$ Commented Oct 17, 2014 at 11:44
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    $\begingroup$ Yes, I wondered about that but I don't think so. There will be elements of $\mathcal{R}_K$ for which that norm is not defined for any $r$, since we require convergence on some semi-open annulus $p^{-r}\leq t <1$ rather than the closed annulus $p^{-r}\leq t\leq 1$. $\endgroup$
    – ChrisLazda
    Commented Oct 17, 2014 at 12:30
  • $\begingroup$ You are right of course. But still you should take into account what happens near radius 1. For $r>s>0$, let me denote $\|\sum_i a_i t^i\|_{r,s} = \sup_i\{|a_i|p^{-ir}, |a_i| p^{-is}\}$. Then I think you need convergence for $\|\cdot\|_{r,s}$, for some $r>0$ and all $s<r$. $\endgroup$ Commented Oct 17, 2014 at 13:20
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    $\begingroup$ But this does not answer your question. Thinking further, I am not sure that Kedlaya's statement is right. $\endgroup$ Commented Oct 17, 2014 at 13:29

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