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During my reading of Peter Scholze and Jared Weinstein's paper ``Moduli of $p$-divisible groups'' I found this assertion in the proof of Proposition 6.1.3. Consider the following situation. Let $k$ be a perfect field of characteristic $p>0$, and let $W:=W(k)$ be its associated ring of Witt vectors, and consider a deformation problem for a given $p$-divisible group over $k$, called $H$, given by a functor called $\text{Def}^{\text{isog}}_{H}$ sending a $p$-adically complete flat $W$-algebra $R$ equipped with the $p$-adic topology to the set of deformations $(G,\rho)$ of $H$, modulo quasi-isogenies. Here a deformaation of $H$ to $R$ is a couple $(G,\rho)$, where $G$ is a $p$-divisible group over $R$, and $\rho: H\otimes_{k}R/p\rightarrow G\otimes_{R}R/p$. Notice that the paper specify explicitly that giving a quasi-isogeny over $R$ is a stricly stronger condition than giving a compatible system of quasi-isogenies over $R/p^{n}$. After this introduction of the notation, Scholze and Weinstein discuss some properties of a period map attached to the moduli problem of this kind of deformations. During the proof of the first proposition regarding the period map, I found this strange comment. Suppose we have two liftings $(G_{i},\rho_{i})$ with $i=1,2$ of $H$ to an open and bounded subring (still $p$-adically complete) $R_{0}$ of $R$. Then we get a quasi-isogeny $\sigma :G_{1}\otimes_{R_{0}}R_{0}/p\rightarrow G_{2}\otimes_{R_{0}}R/p$. What the paper says now is that, since we have by hypothesis, a compatibility with the Hodge filtration, we can lift this quasi-isogeny to a quasi isogeny over $R_{0}$. Now, I know that Grothendieck-Messing theory requires the hypothesis of $p$ at least locally nilpotent in $R_{0}$, which is surely not satisfied in this case. I agree that we can lift the map at each $R/p^{n}$ level, and I can figure that this quasi-isogeny remains a quasi-isogeny, simply because we can prove this for isogenies and it's quite obvious. But how is it possible to lift the system to $R_{0}$? I think we have an isomorphism:

$$\text{Hom}_{\text{Spf}(R_{0})}(G_{1},G_{2})\simeq \underset{n}{\underset{\rightarrow}{\lim}}\text{Hom}_{\text{Spec}(R_{0}/p^{n})}(G_{1}\otimes_{R_{0}}R_{0}/p^{n}, G_{2}\otimes_{R_{0}}R_{0}/p^{n})$$

Am I right? Moreover, how can this isomorphism preserve quasi-isogenies? If I am wrong, do you have any suggestion? And finally, doesn't this kind of argument contradict the fact that giving a quasi-isogeny over $R_{0}$ is in general a strictly stronger condition than giving a compatible system of quasi-isogenies on the reduction? Thanks in advance for any possible suggestion!

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    $\begingroup$ In the directed limit you've written down, the subsets of quasi-isogenies on the right-hand side are bijectively identified for all n (!) - check out Theorem 1.1.13 here: math.leidenuniv.nl/scripties/wang.pdf $\endgroup$ Commented Jun 9, 2016 at 9:18
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    $\begingroup$ Sorry, not "subsets"; the quasi-isogenies only give subsets of what you've written after applying $ \otimes \mathbb{Q}$ outside all the Hom groups. $\endgroup$ Commented Jun 9, 2016 at 9:20
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    $\begingroup$ As David points out, quasi-isogenies always lift over nilpotent thickenings, but S-W are actually applying Grothendieck-Messing directly to the formal divided power thickening $R_0\rightarrow R_0/p$. If $p=2$, there is an issue because the divided powers of $2$ are not nilpotent, but this is okay, since you can first lift to $R_0/4$ by the first remark. $\endgroup$ Commented Jun 9, 2016 at 17:01
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    $\begingroup$ The PD structure is given simply by the usual divided powers on multiples of $p$ (i.e. $p\mapsto p^n/n!$). When $p\neq 2$, these divided powers tend to $0$, i.e. they're topologically nilpotent. When $p=2$, then the same is true for divided powers of $4$. $\endgroup$ Commented Jun 10, 2016 at 15:14
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    $\begingroup$ As for applying G-M to such a thickening, one just applies it at all finite levels, and observes that giving an actual isogeny (not quasi, but an honest homomorphism) of $p$-divisible groups over $R_0$ is equivalent to giving a compatible family over all $R_0/p^nR_0$. $\endgroup$ Commented Jun 10, 2016 at 15:16

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