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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 $A[p^\infty]\sim X$, 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!

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Your argument is correct. –  anon Nov 6 '12 at 14:45

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