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This is a light question about notation, but I received no answer in Stackexchange.

Let $k$ be an algebraically closed field of characteristic $p>0$ and let $A=A_{/k}$ be an ordinary abelian variety of dimension $g\geq1$ which we assume principally polarized for the sake of simplicity. One knows that the $p$-torsion of $A$ is a product: $$A[p]=\hat A[p]\times T_p(A)\otimes(\Bbb Z_p/p\Bbb Z_p).$$ Here $\hat A[p]$ is the maximal connected subgroup, $T_p(A)\otimes(\Bbb Z_p/p\Bbb Z_p)$ is etale, both factors are subgroups of rank $p^g$ and they're Cartier dual of each other.

The completely inseparable isogeny obtained by taking the quotient by $\hat A[p]$ is the relative Frobenius $F_k:A\rightarrow A^{(p)}$ where $A^{(p)}$ is the abelian variety obtained by twisting the $k$-structure of $A$ by the geometric Frobenius i.e. the automorphism $x\mapsto x^{1/p}$ of $k$.

The isogeny $A\rightarrow A^\prime$ obtained by taking the quotient by $T_p(A)\otimes(\Bbb Z_p/p\Bbb Z_p)$ is etale and after an identification $A\simeq(A^\prime)^{(p)}$ is the Verschiebung $V_k:(A^\prime)^{(p)}\rightarrow A^\prime$ associated with $A^\prime$.

Is there a standard notation for the abelian variety that I denoted $A^\prime$? I have checked a few textbooks and lecture notes about abelian varieties and/or group schemes, but I seemed not to be able to find any.

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I guess you could call it $A^{(1/p)}$? Of course $A^{(p)}$ makes sense over any scheme where $p=0$ but $A^{(1/p)}$ relies on your base being e.g. a perfect field, so it won't come up as much I guess. –  Kevin Buzzard Apr 29 '12 at 13:06
    
Or perhaps you could call it $A^\sigma$ where $\sigma$ is the $p$th root map on the base. –  Kevin Buzzard Apr 29 '12 at 13:07
    
@Kevin Buzzard: I actually like the idea of calling it $A^{(1/p)}$! I'm sort of surprised, though, that there is no standard notation for it, like if this object had not been given much consideration. –  Andrea Mori Apr 29 '12 at 15:24
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I’ve never had occasion to do so in a publication, but what I use in my own notes is $A^{(p^{-1})}$. –  Lubin May 24 '12 at 20:11
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