p-divisible group over an algebraically closed field of characteristic p arises from abelian variety - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T16:58:01Z http://mathoverflow.net/feeds/question/111620 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/111620/p-divisible-group-over-an-algebraically-closed-field-of-characteristic-p-arises-f p-divisible group over an algebraically closed field of characteristic p arises from abelian variety Taisong Jing 2012-11-06T05:38:21Z 2012-11-06T14:25:14Z <p>It may be trivially true or trivially false, just a quick ask, if $k=\overline{k}$ and char k = p>0, X is a p-divisible group over $k$, suppose the Newton polygon of $X$ is symmetric, then there always exists an abelian variety $A$ over $k$ such that $A[p^\infty]\cong X$. Is this statement true?</p> <p>Y. Manin made his conjecture that if $X$ has a symmetric Newton polygon then it is isogeneous to some $A[p^\infty]$, it has been proved, however nobody pointed out that whether this isogeny is necessary. A straightforward argument is like this: if $A[p^\infty]\sim X$, then there exists a finite subgroup scheme $G$ of $A[p^\infty]$ such that $A[p^\infty]/G\cong X$. But then $G$ is also a subgroup of $A$, so $A/G$ is an abelian variety over $k$ with p-divisible group isomorphic to $X$.</p> <p>I am just confused by no such comments showed up in the literature, and I am not sure whether I have made some stupid mistakes in this argument, so maybe someone can help me assure or deny it. Thank you!</p>