Deformation of ordinary p-divisible groups via Grothendieck-Messing - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T11:42:57Z http://mathoverflow.net/feeds/question/84273 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/84273/deformation-of-ordinary-p-divisible-groups-via-grothendieck-messing Deformation of ordinary p-divisible groups via Grothendieck-Messing B. Cais 2011-12-25T16:12:56Z 2011-12-25T16:12:56Z <p>I am hoping that someone can point out the error in the "proof" of the following "theorem":</p> <blockquote> Theorem: Let $k$ be a perfect field of characteristic $p>2$ and let $G$ be an ordinary $p$-divisible group over $W(k)$. Then the connected-etale sequence of G is split: $G\simeq G^m \times G^{et}$. </blockquote> <p>The "Theorem" is decidedly false. For example, extensions of $\mathbf{Q}_p/\mathbf{Z}_p$ by $\mathbf{G}_m[p^{\infty}]$ over $W(k)$ are classified by the abelian group $1+pW(k)$, so any nonzero element of this group gives an ordinary $p$-divisible group with non-split connected-etale sequence.</p> <blockquote> Proof: Let $G_0:=G\times_{W(k)} k$ be the special fiber of $G$. Since $k$ is perfect, $G_0= G_0^{m}\times G_0^{et}$. By Messing (LNM 264, Chap V. Theorem 1.6), there is an equivalence of categories between deformations of $G_0$ to $W(k)$ and free $W(k)$-submodules $L$ of $D(G_0)(W(k))$ lifting $\omega_{G_0}\hookrightarrow D(G_0)(k)$. This equivalence is induced by sending a lift $G'$ of $G_0$ to $\omega_{G'}$. Now $G' = G^{m}\times G^{et}$ and $G$ both lift $G_0$, and these lifts correspond to the submodules $\omega_{G'}$ and $\omega_G$ of $D(G_0)(W(k))$, respectively. But since $G$ is ordinary, so $G^0=G^{m}$, the pullback map $\omega_{G}\rightarrow \omega_{G^m}$ is an isomorphism. Via this isomorphism, the map $\omega_{G'}\rightarrow D(G_0)(W(k))$ coincides with the composite $$\omega_{G'}\simeq \omega_{G^m}\times \omega_{G^{et}} = \omega_{G^m}\simeq \omega_G\rightarrow D(G_0)(W(k))$$ and we conclude that $\omega_{G'}=\omega_G$ as submodules of $D(G_0)(W(k))$ lifting $\omega_{G_0}$. It then follows from Messing's Theorem above that $G\simeq G'$, as claimed. </blockquote> <p>I must be making some silly and obvious mistake...can you find it?</p>