MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).

## Return to Question

2 added 4 characters in body

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?

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 $A[p^\infty]~X$, 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$. I am just confused by no such comments showed up in the literature, and I am not sure whether I have made some stupid mistakes in this argument, so maybe someone can help me assure or deny it. Thank you! 1 # p-divisible group over an algebraically closed field of characteristic p arises from abelian variety It may be trivially true or trivially false, just a quick ask, if$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$. Is this statement true? Y. Manin made his conjecture that if$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]~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\$.

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!