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\begin{document}
\title{Group structures of elementary supersingular abelian varieties over 
finite fields}
\author{Hui Zhu\\
\em MSRI, 1000 Centennial Drive, Berkeley, CA 94720, USA\\
{\tt zhu@msri.org}} 
\date{Version Nov 29, 1998}
\maketitle
\begin{abstract} 
Let $A$ be a supersingular abelian variety over a finite field ${\bf k}$ 
which is isogenous to a power of a simple abelian variety over
${\bf k}$.  Write the characteristic polynomial of the Frobenius
endomorphism of $A$ relative to ${\bf k}$ as $f=g^e$ for a monic
irreducible polynomial $g$ and a positive integer $e$, we show that
the group of ${\bf k}$-rational points $A({\bf k})$ on $A$ is
isomorphic to $(\zz/g(1)\zz)^e$ unless $A$'s simple component is 
of dimension $1$ or $2$, in which case
we prove that $A({\bf k})$ is isomorphic to
$(\zz/g(1)\zz)^a\times(\zz/{\frac{g(1)}{2}}\zz\times\zz/2\zz)^b$
for some non-negative integers $a,b$ with $a+b=e$. In particular, if
the characteristic of ${\bf k}$ is $2$ or $A$ is simple of dimension
greater than $2$, then $A({\bf k})\cong (\zz/g(1)\zz)^e$. 

AMS classification: 14K22, 06B15.
\end{abstract}

\section{Introduction}\label{S1} 

We list some notation and terminology for this paper as follows: ${\bf k}$
is a finite field of characteristic $p$ with $q$ elements. Let
$\bar{\bf k}$ be an algebraic closure of ${\bf k}$. 
Let $A$ be an abelian variety of dimension $d$ defined over
${\bf k}$. Let $\pi$ be the Frobenius endomorphism of $A$ relative to ${\bf k}$
and $f$ its characteristic polynomial.

An abelian variety over ${\bf k}$ is {\it elementary} if it is ${\bf k}$-isogenous
to a power of a simple abelian variety over ${\bf k}$. This definition is
different from that 
of~\cite{Waterhouse:69} (see~\cite[page~54]{Waterhouse-Milne:71}). 
An abelian variety $A$ is
elementary if and only if $f=g^e$ for some monic irreducible
polynomial $g$ over $\qq$ and some positive integer $e$. An arbitrary
abelian variety is ${\bf k}$-isogenous to a product of elementary abelian
varieties, and $f=\prod_{i=1}^{t}g_i^{e_i}$ for distinct monic
irreducible polynomials $g_i$ over $\qq$ and positive integers $e_i$.
An abelian variety $A$ over ${\bf k}$ is {\it supersingular} if each complex
root of $f$ can be written in the form $\zeta\sqrt{q}$,
the product of
some root of unity $\zeta$ and the positive square root $\sqrt{q}$. 

\begin{theorem}\label{Telementary} %% theorem 1.1
Let $A$ be an elementary supersingular abelian variety over ${\bf k}$ and
$f=g^e$ as above. Then $A({\bf k})$ is isomorphic as an abelian group to
$(\zz/g(1)\zz)^e$ except in the following cases:
\begin{enumerate} 
\item[(1)] $p\equiv 3\bmod
4$, $q$ is not a square, and $A$ is ${\bf k}$-isogenous to a power of a supersingular
elliptic curve with $g=X^2+q$,
\item[(2)] $p\equiv 1\bmod 4$, $q$ is not a square,
and $A$ is ${\bf k}$-isogenous to a power of a two dimensional abelian variety
with $g=X^2-q$.
\end{enumerate} 
In these two exceptional cases, there are
non-negative integers $a,b$ with $a+b=e$ such that
$$A({\bf k})\cong (\zz/g(1)\zz)^a\times
\left(\zz/{\tfrac{g(1)}{2}}\zz\times\zz/2\zz\right)^b.$$
\end{theorem}

This result is particularly striking when $p=2$ or $A$
is simple with $d > 2$ for then
$A({\bf k})\cong_{\zz}(\zz/g(1)\zz)^e$. In the latter case $A({\bf k})$ will be either
cyclic or a product of two cyclic groups, since $e=1$ or $2$. (See
Proposition~\ref{Pe12}).

We call an elementary supersingular abelian variety $A$ {\it
exceptional} if it belongs to either of the two isogeny classes stated in
Theorem~\ref{Telementary} (1) and (2). We will show 
(see Proposition~\ref{Paprime}) that if $A$ is exceptional, 
then for every pair of
non-negative integers $a',b'$ with $a'+b'=e$, there exists an
abelian variety $A'$ which is ${\bf k}$-isogenous to $A$ 
and $$A'({\bf k})\cong (\zz/g(1)\zz)^{a'}\times
\left(\zz/{\tfrac{g(1)}{2}}\zz\times\zz/2\zz\right)^{b'}.$$

In this paper $\End_{\bf k}(A)$ denotes the ring of ${\bf k}$-endomorphisms of
$A$.  Write $\End_{\bf k}^0(A)=\End_{\bf
k}(A)\otimes_{\zz}\qq$. Let $\qq[\pi]$ be the $\qq$-subalgebra of
$\End_{\bf k}^0(A)$ generated by $\pi$, let $\O$ be its maximal order,
and $\zz[\pi]$ its $\zz$-subalgebra generated by $\pi$.  The group  
$A(\bar{\bf k})$  is naturally an $\End_{\bf k}(A)$-module.  
Our results describe $A(\bar{\bf k})$ 
as a module over any subring of $\End_{\bf k}(A)\cap\qq[\pi]$
that contains $\pi$.  For any prime number $l$ we write $R_{(l)}$ for
the localization of a commutative ring $R$ at $l$, this notation
should not be confused with that for the $l$-adic completion $R_l$.

\begin{theorem}\label{Tmodule-elementary} % Theorem 1.2
Let $A$ be an elementary supersingular abelian variety over ${\bf k}$
of dimension $d$. Let $R$ be a ring with 
$\zz[\pi]\subseteq R\subseteq \End_{\bf k}(A)\cap \qq[\pi]$. 
Then there is a surjective $R$-module homomorphism 
$$\varphi: A(\bar{\bf k})\rightarrow (R_{(p)}/R)^e$$
such that the cardinality of the kernel of $\varphi$ divides $2^d$. 
Furthermore, $\varphi$ is an isomorphism when $p=2$.
\end{theorem}

Suppose $A$ is a simple supersingular abelian variety over ${\bf k}$
and $R$ the endomorphism ring $\End_{\bf k}(A)\cap\qq[\pi]$: if   
$d\neq 2$, then $A(\bar{\bf k})\cong_R
(R_{(p)}/R)^e$; if $d=2$, then 
$A(\bar{\bf k})\cong_R (R_{(p)}/R)^a\times (\O_{(p)}/\O)^b$ for some
non-negative integers $a,b$ with $a+b=e$. (See Proposition~\ref{Psimple2}.)  

The group structure of the ${\bf k}$-rational points and $\bar{\bf k}$-rational points 
on an elliptic curve were studied by~\cite{Deuring:41} 
(see also~\cite[Chapter~V]{Silverman:86}
and~\cite{HWL:96}). The group structure of the ${\bf k}$-rational points on a
supersingular elliptic curve can be found
in~\cite[Chapter~4]{Schoof:87}.  Descriptions of that nature for
higher dimensional abelian varieties remain open.  (Recently,
independent of our work, the group structure of dimensional two
supersingular abelian varieties was studied in~\cite{Xing:96}.)  

We develop the following idea for studying a supersingular abelian
variety $A$ over ${\bf k}$: we show that the ring $\zz[\zeta\sqrt{q}]$
is locally almost a discrete valuation ring at every prime; in fact it
is a {\it Bass order} over some suitable subring (see
section~\ref{S5}). Next we describe the Tate modules of $A$ over
$\zz[\pi]$. Finally the group structure of $A({\bf k})$ follows by
viewing $A({\bf k})$ as the kernel of the isogeny $\pi-1$ on $A(\bar{\bf k})$ 
(see section~3).  The group and module structures
on an arbitrary supersingular abelian varieties over finite fields are
studied in~\cite{paper2}.

This paper is based on a portion of the author's Berkeley Ph.D thesis.
The author is deeply grateful to her advisor Professor Hendrik
Lenstra for inspiration and guidance.  The author also wish to
thank Bjorn Poonen and Phil Ryan for their comments on an earlier
version of this paper. The author was supported as a MSRI 
postdoctoral fellow while preparing this paper. 

%section 2
\section{Torsion-free modules over Bass orders}\label{S5}

\subsection{Orders}\label{S51}

We begin this section by introducing some notions from commutative algebra, 
then we prove several auxiliary results from algebraic number theory. 
The material largely follows~\cite[Introduction and Chapter
3]{Curtis-Reiner:90}.  Here we assume all rings are commutative and
modules are {\em finitely generated}.  Let $K$ be a local or global field of
zero characteristic and $\O_K$ its ring of integers.

Suppose $L$ is a finite dimensional separable $K$-algebra.  An
$\O_K$-algebra $\Lambda$ is called an {\it $\O_K$-order} (in $L$) if
it is a finitely generated projective $\O_K$-module (and
$\Lambda\otimes_{\O_K}K=L$).  A $\zz$-order is simply called an {\it
order}.  Let $\Lambda$ be an $\O_K$-order in $L$. We denote the unique
maximal $\O_K$-order in $L$ by $\O_L$.  If $M$ is a $\Lambda$-module
which is projective over $\O_K$, then $M$ is called a {\it
$\Lambda$-lattice}.

For any prime $\wp$ of $\O_K$, we denote by $(\O_K)_\wp, \Lambda_\wp,
M_\wp$ their $\wp$-adic completions, respectively. If $K=\qq$ and
$\wp = l$ for some prime number $l$, then we write $(\O_K)_l, \Lambda_l,
M_l$ for their $l$-adic completions.

A $\Lambda$-module $M$ is called {\it torsion-free} if $\alpha m\neq 0$ for
any non-zerodivisor $\alpha\in \Lambda-\{0\}$ and $m\in
M-\{0\}$. In particular, when $\Lambda$ is a domain 
then this is equivalent to the standard definition.
A $\O_K$-module is projective if and only if it is 
torsion-free~\cite[II.4 (4.1)]{Frohlich-Taylor:91}.  
So $M$ is a torsion-free $\Lambda$-module 
if and only if it is a $\Lambda$-lattice.  If $M$ is a torsion-free
$\Lambda$-module, then it is torsion-free over $\O_K$, hence there is
a natural embedding $M\hookrightarrow M\otimes_{\O_K}K$, where
$M\otimes_{\O_K}K$ has a natural $L$-module structure. 
If $M\otimes_{\O_K}K$ is free of rank $e$ over $L$ for some integer $e$,
then $M$ is said of {\it rank} $e$.
We shall note here that 
$e$ is used to denote an arbitrary positive integer in this section.  

Suppose $M'$ and $M''$ are any torsion-free $\O_K$-modules with 
{$M'\otimes_{\O_K}K = M''\otimes_{\O_K}K$},
then the {\it ideal index} $[M'':M']$
is defined to be the fractional $\O_K$-ideal generated by all
determinants of the $K$-linear maps
$M''\otimes_{\O_K}K\stackrel{\eta}{\rightarrow}M'\otimes_{\O_K}K$ such
that $\eta(M'')\subseteq M'$.  
It is an integral $\O_K$-ideal, that is,
$[M'':M']\subseteq \O_K$, if and only if $M'\subseteq M''$.  Let
$\Tr_{L/K}$ denote the trace map from $L$ to $K$.  Denote the trace
dual (also known as the complementary module in \cite{Lang:86}) of
$\Lambda$ by $\Lambda^{\dagger}=\{ a\in
L\;|\;\Tr_{L/K}(a\Lambda)\subseteq \O_K\}.$ It is easy to see that
$\Lambda^{\dagger}$ is a fractional ideal of $\Lambda$, and
$\Lambda\subseteq \Lambda^\dagger$.  Define the discriminant (ideal)
of $\Lambda$ over $\O_K$ by $\Delta_{\Lambda/\O_K}=[\Lambda^\dagger:
\Lambda]$. Note that $\Lambda\subseteq \O_L\subseteq\O_L^\dagger
\subseteq \Lambda^\dagger$ and
$[\Lambda^\dagger:\O_L^\dagger]=[\O_L:\Lambda]$.  The discriminant of
$L$ over $K$ is defined by
$\Delta_{L/K}=\Delta_{\O_L/\O_K}=[\O_L^\dagger:\O_L]$ and so
$[\O_L:\Lambda]^2\,\Delta_{L/K}=\Delta_{\Lambda/\O_K}$.  Therefore,
the relationship between the ideal index of $\Lambda$ in $\O_L$ and
the discriminant of $\Lambda$ is
$[\O_L:\Lambda]^2\mid\Delta_{\Lambda/\O_K}.$ 

Suppose $L$ is a finite field extension of $K$.  
Let $\alpha$ be an integral element in $L$ and $h\in
\O_K[X]$ be its (monic) minimal polynomial.  Let
$\Lambda=\O_K[\alpha]$. Then $\Delta_{\Lambda/\O_K}=\O_K\,\Delta(h)$.
Let $\wp$ be any non-zero prime ideal of $\O_K$.  
Then $\Lambda_\wp$ is semisimple and $\Lambda_\wp\cong \prod_{Q\mid \wp}\Lambda_\wp$ where 
the product ranges over all prime ideals $Q$ of $\Lambda$ lying over $\wp$.
There is a bijective correspondence between these $Q$'s of $\Lambda$ 
and the set of (monic) irreducible factors $\bar{h_0}$ of
$\bar{h}=(h\bmod \wp)\in ({\O_K}/\wp)[X]$.  (See~\cite[Chapter I,
Proposition~25, page 27]{Lang:86}.)  If $Q$ corresponds to $\bar{h_0}$
in this bijection, then $Q=(\wp,h_0(\alpha))$ in $\Lambda$. 
We use the following notation throughout this paper: for any prime ideal $v$ of $\O_L$
lying over a prime $\wp$ of $\O_K$, let $\gamma(v/\wp)$,
$\kappa(v/\wp)$ and $\varrho(v/\wp)$ denote the {\em ramification
index}, {\em residue field degree} and {\em decomposition degree},
respectively. In particular, when $\Lambda = \O_L$ then
$\kappa(Q/\wp)=\dim_{{\O_K}/\wp}\Lambda/Q=\deg(\bar{h_0})$ and
$\gamma(Q/\wp)$ equals the multiplicity of $\bar{h_0}$ as a factor of
$\bar{h}$.  We have the following lemma whose proof is due to Hendrik
Lenstra.

\begin{lemma}\label{Llenstra} 
Let the notation be as above. Then $Q$ is not invertible if and only
if $\bar{h_0}^2\mid\bar{h}$ and all coefficients of the remainder of
$h$ upon division by $h_0$ are in $\wp^2$.  The $(\O_K)_\wp$-order
$\Lambda_\wp$ is not the maximal order if and only if there is a monic
irreducible factor $\bar{h_0}$ of $\bar{h}$ such that
$\bar{h_0}^2\mid\bar{h}$, and all coefficients of the remainder of $h$
upon division by $h_0$ are in $\wp^2$.
\end{lemma}
\begin{proof} 
Write $J: = (\wp,h_0(X))$ in $\O_K[X]$, it is a prime ideal. 
Then the natural surjective map $\O_K[X]\rightarrow
\Lambda$ induces a surjective map $\theta: J/J^2\rightarrow Q/Q^2$ with
$\kernel(\theta)$ generated by $h$. Write $h$ in base $h_0$ and
obtain $h=r_2h_0^2+r_1h_0+r_0$ for some $r_2,r_1,r_0\in {\O_K}[X]$ with
$\deg(r_1), \deg(r_0)<\deg(\bar{h_0})$. Then $h\in J$ if and only if $r_0\in
\wp[X]$, while $h\in J^2$ if and only if $r_1\in \wp[X]$ and $r_0\in
\wp^2[X]$. So $\dim_{\Lambda/Q}J/J^2=1+\dim_{{\O_K}/\wp}\wp/\wp^2=2$ and hence  
$\dim_{\Lambda/Q}Q/Q^2=\dim_{\Lambda/Q}J/J^2-\dim_{\Lambda/Q}\kernel(\theta)
=2-\dim_{\Lambda/Q}\kernel(\theta)$, 
where
$\dim_{\Lambda/Q}\kernel(\theta)$ is $0$ or $1$. 
Therefore, $\dim_{\Lambda/Q}Q/Q^2\neq 1$ if and only if 
$h\in J^2$. We conclude that $Q$ is not invertible if and
only if $h\in J^2$, which is equivalent to $\bar{h_0}^2\mid \bar{h}$ and
all coefficients of the remainder of $h$ upon division by $h_0$ are in
$\wp^2$.
Thus the semilocal ring $\Lambda_\wp$ is maximal if and only if 
$\Lambda_Q$ is maximal for each prime ideal $Q$ over $\wp$, 
which is equivalent to $Q$ is invertible, and so follows our assertion. 
\end{proof}

\begin{corollary}\label{Clenstra}
Let the notation be as in Lemma~\ref{Llenstra}.
If $h_0=X-\beta$ with $\beta\in \O_K$, then
$Q$ is not invertible if and only if $h(\beta)\equiv 0\bmod \wp^2$ 
and $h'(\beta)\equiv 0\bmod \wp$, where $h'$ denotes
the derivative of $h$.
\end{corollary}
\begin{proof}
The condition $\bar{h_0}^2\mid \bar{h}$ is equivalent to that 
$h(\beta)\equiv 0\bmod \wp$ and 
$h'(\beta)\equiv 0\bmod \wp$. 
The condition that all coefficients of the remainder of 
$h$ upon division by $h_0$ are in
$\wp^2$ is equivalent to $h(\beta)\equiv 0\bmod \wp^2$.
\end{proof}

The following is an easy exercise in algebraic number theory and so
we leave its proof to the reader. In this paper $(\frac{\cdot}{p})$
denotes the Jacobi symbol.

\begin{lemma}\label{Lpage12}
Let $p$ be a prime number in $\zz$ and $m$ a positive integer.
Then \begin{enumerate}
\item[(1)] $\Delta_{\qq(\sqrt{p})/\qq}=p$ if $p\equiv 1\bmod 4$
while $\Delta_{\qq(\sqrt{p})/\qq}=4p$ if $p\not\equiv 1\bmod 4$;  
\item[(2)] $\sqrt{p}\in\qq(\zeta_m)$ if
and only if $\Delta_{\qq(\sqrt{p})/\qq}\mid m$;  
\item[(3)] let $p\neq 2$, if $p\mid m$ then  
$\qq(\sqrt{(\frac{-1}{p})p})\in\qq(\zeta_m)$.
\end{enumerate}
\end{lemma}

\subsection{Bass orders}
A reference for concepts in this subsection is~\cite[Chapter~4]{Curtis-Reiner:90}.
We call an $\O_K$-order $\Lambda$ a {\it Gorenstein order} if every exact
sequence of $\Lambda$-modules
$0\rightarrow \Lambda\rightarrow M\rightarrow N\rightarrow 0,$ in
which $M$ and $N$ are $\Lambda$-lattices is split over
$\Lambda$. If $\Lambda$ has the additional property
that every $\O_K$-order in $L$ containing $\Lambda$ is also a Gorenstein
order, then we call $\Lambda$ a {\it Bass order}. Note that 
being a Bass order is a local property, in other
words, $\Lambda$ is a Bass $\O_K$-order if and only if
$\Lambda_\wp$ is a Bass $(\O_K)_\wp$-order for every prime $\wp$ in
$\O_K$. 

\begin{prop}\label{Pequivalent}
The following are equivalent:
\begin{enumerate} 
\item[(1)] $\Lambda$ is a Bass $\O_K$-order;
\item[(2)] $\O_L/\Lambda$ is a cyclic $\Lambda$-module;
\item[(3)] every ideal of $\Lambda$ can be generated by two elements;
\item[(4)] for every maximal ideal $Q$ of $\Lambda$ we have 
$\dim_{\Lambda_Q/Q\Lambda_Q}(\O_L)_Q/Q(\O_L)_Q \leq 2$;
\item[(5)] the multiplicity of $\Lambda$ at each maximal ideal $Q$ is $\leq 2$.
\end{enumerate}
\end{prop}
\begin{proof}
The first three parts are equivalent according 
to~\cite[Theorem~2.1]{Levy:85}. 
The last two parts are equivalent to (1) by~\cite[Theorem~2.1]{Greither:82}.
\end{proof}

\begin{remark}\label{Xbass}
Here are some examples of Bass orders of interest.
\begin{enumerate}
\item[(i)] 
If $L$ is a quadratic field extension over $K$, then 
$(\O_L)_\wp/\Lambda_\wp$ is cyclic over $\Lambda_\wp$ for 
every prime $\wp$ of $\O_K$ and thus $\Lambda$ is a Bass order over $\O_K$.
\item[(ii)] All maximal orders in number fields are Bass orders.
\end{enumerate}
\end{remark}

We are interested in describing torsion-free modules $M$ over $\Lambda_\wp$
of rank $e$. Recall that $\Lambda_\wp$ is a semilocal ring whose
maximal ideals are those prime ideals $Q$ lying over $\wp$, so there is
a corresponding decomposition of $M$ as $M\cong \prod_{Q\mid \wp} M_Q$.  

If $\Lambda_\wp=(\O_L)_\wp$, then $M_Q$ is
torsion-free over the principal ideal domain $(\O_L)_Q$ of rank $e$,
so $M_Q\cong\Lambda_Q^e$ for all $Q$. Thus $M\cong\Lambda_\wp^e$.
If $\Lambda_\wp\neq (\O_L)_\wp$, then it is generally hard to 
classify such modules $M$ in terms of orders in $L_\wp$   
(see~\cite[Chapter~3]{Curtis-Reiner:90}). 
However, torsion-free modules over Bass orders can be described as follows. 

\begin{theorem}[Bass]\label{Tbass}
Let $K$ be a local field, $\O_K$ its ring of integers, 
and $\Lambda$ a Bass $\O_K$-order in a finite field extension $L$ over $K$. 
Then every indecomposable
torsion-free $\Lambda$-module is a projective $\Lambda'$-module for some
$\O_K$-order $\Lambda'$ in $L$ containing $\Lambda$.
\end{theorem}
\begin{proof} 
Follows from~\cite[Theorem (37.13)]{Curtis-Reiner:90}
and the definition of Bass orders.
\end{proof}

%%%%%%
\subsection{Supersingular $q$-numbers}\label{S3}

In this subsection there are the definition of $\E$ and the proof of
Lemma~\ref{Lindex}, the latter of which is crucial for the technical
part of this paper.

For the rest of the paper $l$ is a prime number different from $p$.
Let $\zeta_m=e^{\frac{2{\rm\pi}\sqrt{-1}}{m}}$ for any positive
integer $m$. The primitive $m$-th roots of unity are the $\zeta_m^\nu$
with positive integers $\nu$ that are coprime to $m$.  We write
$\Phi_m(\cdot)$ for the $m$-th cyclotomic polynomial, i.e., the
minimal polynomial of $\zeta_m$.

An algebraic number $\alpha\in \cc$ is called a {\it supersingular
$q$-number} if it is of the form $\zeta\sqrt{q}$.  
Let $\pi=\zeta_m^{\nu}\sqrt{q}$.  Let $K=\qq(\pi^2)$ and let $\O, \O_K$ be
the ring of integers of $\qq(\pi),K$, respectively.  
Obviously $K=\qq(\zeta_{m/(2,m)})$ and $[\qq(\pi):K]=1$ or $2$.
(In this paper $(m_1,m_2):=gcd(m_1,m_2)$ for any integer $m_1, m_2$.)
 
\begin{lemma}\label{LRF}
Suppose $q$ is a non-square. Then  $\qq(\pi)=K$ if and only if
\begin{eqnarray*}
(1)\ \Delta_{\qq(\sqrt{p})/\qq}\mid m,\quad\mbox{and}\quad 
(2)\ \Delta_{\qq(\sqrt{p})/\qq}\nmid m/(2,m)\mbox{ if }4\mid m.
\end{eqnarray*}
In this case $2\mid \gamma(v/p)$ for   
any prime $v$ of $\O$ lying over $p$.
\end{lemma}
\begin{proof}
We note $\qq(\pi)=\qq\left(\zeta_{m/(2,m)},\sqrt{p\zeta_{m/(2,m)}}\right)
=K\left(\sqrt{p\zeta_{m/(2,m)}}\right).$ 
Thus $\qq(\pi)=K$ if and only if 
$\sqrt{p\zeta_{m/(2,m)}}\in K$, and if and only if 
$$K\left(\sqrt{p}\right)=K\left(\sqrt{\zeta_{m/(2,m)}}\right),$$ 
that is,
$\qq\left(\zeta_{m/(2,m)},\sqrt{p}\right)=\qq(\zeta_m)$. This is equivalent to 
$$(1a)\ \sqrt{p}\in\qq(\zeta_m),\quad\mbox{and}\quad (2a)\
\sqrt{p}\notin\qq(\zeta_{m/(2,m)}) \mbox{ if } 4\mid m,$$ which is
equivalent to (1) and (2) respectively by Lemma~\ref{Lpage12}~(2).

To show the second assertion it is enough to prove it for just one prime 
$v$ over $p$ since all primes lying over $p$ are conjugate as 
$\qq(\pi)$ is the cyclotomic field $\qq(\zeta_{m/(2,m)})$.
We claim $(2,p)p\mid\frac{m}{(2,m)}$. In fact, if $p=2$ then (1) implies $8\mid
m$ by Lemma~\ref{Lpage12}~(1) 
so our claim follows; if $p\neq 2$ then (1) 
implies $p \mid m$. But since $p\neq 2$, 
we have $(2,p)p\mid \frac{m}{(2,m)}$.
By Lemma~\ref{Lpage12} (3), we thus see that  
$\qq(\zeta_{m/(2,m)})$ contains a quadratic field 
$\qq(\sqrt{(\frac{-1}{p})p})$ over $\qq$ where $p$
is totally ramified. Hence $2\mid \gamma(v/p)$.
\end{proof}

For any prime number $l$, let $m_l$ and $m_{(l)}$
denote, respectively, the $l$-part and the non-$l$-part of a positive
integer $m$.
Let $\E$ be the set of supersingular $q$-numbers 
$\zeta_m^{\nu}\sqrt{q}$ which satisfy the following conditions:
$p\neq 2$, $q$ is not a square, $p\nmid m$, and 
\begin{eqnarray*}
\mbox{(1) $4\nmid m$ when $p\equiv
1\bmod 4$;  and  (2) $4||m$ when $p\equiv 3\bmod 4$.}
\end{eqnarray*}

We remark here that $\alpha\in \Lambda_\wp$ 
is a unit if and only if $\alpha$ is coprime to $\wp$.

\begin{lemma}\label{Lindex} 
Let the notation be as above. 
If $(l,\pi)\notin \{2\}\times\E$ then
$\zz[\pi]_l=\O_l$. If $(l,\pi)\in
\{2\}\times\E$
then $\zz[\pi]_2\subsetneq \O_2$; 
let $\wp$ be any prime in $\O_K$ lying over $2$,
then 
\begin{enumerate} 
\item[(1)] $\qq(\pi)$ is a quadratic extension over $K$ where $\wp$ is split 
if $p\equiv \pm 1\bmod 8$, and $\wp$ is inert if $p\equiv \pm 3\bmod 8$.
\item[(2)] \begin{sloppypar}
$\zz[\pi]_\wp$ is a local ring and a Bass $(\O_K)_\wp$-order such that
$\O_\wp$ is the only $(\O_K)_\wp$-order in $\qq(\pi)_\wp$ that
properly contains $\zz[\pi]_\wp$. Moreover, 
$\O_\wp/\zz[\pi]_\wp\cong_{(\O_K)_\wp}(\O_K)_\wp/\wp$.\end{sloppypar}
\end{enumerate}
\end{lemma}
\begin{proof} 
If $q$ is a square, then $\pi\notin \E$ and 
$\zz[\pi]_l=\zz[\zeta_m]_l=\O_l$. For the rest of the proof, we
assume $q$ is not a square. We consider the following two cases.

\noindent{\fbox{Case 1. $l\neq 2$ or $p\mid m$.}}

We claim $\zz[\pi]_l=\O_l$. 
Since $l\neq p$, we note that $\zz[\pi^2]_l=\zz[\zeta_{m/(2,m)}]_l=(\O_K)_l$. 
Suppose $l\neq 2$, obviously 
$\zz[\pi^2]_l=(\O_K)_l\subseteq \zz[\pi]_l\subseteq \O_l$. 
If $\qq(\pi)=K$ then $\O_l=(\O_K)_l$ and so $\zz[\pi]_l=\O_l$.
If $\qq(\pi)\neq K$ then $[\O:\zz[\pi]]_l^2\mid
\Delta_{\zz[\pi]/(\O_K)};$ 
but since $\zz[\pi]\cong \O_K[X]/(X^2-q\zeta_{m/(2,m)})$, 
we have $\Delta_{\zz[\pi]/\O_K}=\O_K\Delta(X^2-q\zeta^\nu_{m/(2,m)})
=4q\O_K.$ So 
$(\Delta_{\zz[\pi]/\O_K})_l=(\O_K)_l$ since $4q$ is coprime to $l$.
Therefore $[\O:\zz[\pi]]_l$ is the unit ideal and so $\zz[\pi]_l=\O_l$. 

Now let $l=2$ and $p\mid m$.
By Lemma~\ref{Lpage12} (3),  
$\sqrt{(\frac{-1}{p})p}\in\zz[\zeta_{m/(2,m)}]_2=\zz[\pi^2]_2
\subseteq \zz[\pi]_2$. 
Moreover, the norm of $\sqrt{(\frac{-1}{p})p}$ over $\qq$ 
is $\pm p$ which is coprime to $2$ so $\sqrt{(\frac{-1}{p})p}$ 
is a unit in $\zz[\pi]_2$. 
Therefore,
$\zz[\pi]_2=\zz[\pi\sqrt{(\frac{-1}{p})p}]_2
=\zz[\zeta^\nu_m\sqrt{(\frac{-1}{p})}]_2$.  This proves our claim.

\noindent {\fbox{Case 2. $l=2$ and $p\nmid m$.}}  

Write $m=2^jm_{(2)}$.
It is easy to verify
that $\qq(\pi)=\qq(\zeta_{m_{(2)}},\alpha)$ where $\alpha=\zeta^\mu_{2^j}\sqrt{p}$ for some 
$2^j$-th primitive root of unity $\zeta^\mu_{2^j}$.
We note that $\qq(\zeta_{m_{(2)}})$ and $\qq(\alpha)$ are linearly
disjoint and that the minimal polynomial of $\alpha$ over
$\qq(\zeta_{m_{(2)}})$ is $h=X^{2^{j-1}}+p^{2^{j-2}}$ if $j\geq 2$,
and is $h=X^2-p$ if $j<2$.

Let $\wp'$ be any prime ideal in the ring of integers of 
$\qq(\zeta_{m_{(2)}})$ lying over $2$.
We show $\zz[\pi]_{\wp'}=\zz[\zeta_{m_{(2)}},\alpha]_{\wp'}$. 
The inclusion $\zz[\pi]_{\wp'}\subseteq \zz[\zeta_{m_{(2)}},
\alpha]_{\wp'}$ is trivial.
Conversely, since $\pi^{2^j}=\zeta_{m_{(2)}}^{2^j}q^{2^{j-1}}$ and
$\alpha=\pi\zeta_{m_{(2)}}^{-\mu}$, we have $\zeta_{m_{(2)}}, \alpha\in
\zz[\pi]_{\wp'}$. Thus $\zz[\zeta_{m_{(2)}},\alpha]_{\wp'}\subseteq 
\zz[\pi]_{\wp'}$. 
That is, $\zz[\pi]_{\wp'}=\zz[\zeta_{m_{(2)}},\alpha]_{\wp'}$. 
Hence, $\zz[\pi]_{\wp'}=(\zz[\zeta_{m_{(2)}}]_{\wp'})[\alpha]$. 

If $j\geq 2$, then $h\equiv (X-1)^{2^{j-1}} \bmod \wp'$.  Note that
$\zz[\zeta_{m_{(2)}}]_{\wp'}$ is a discrete valuation ring, so we have
by Corollary~\ref{Clenstra} that $\zz[\pi]_{\wp'}$ is not maximal if
and only if $h(1)=1+q^{2^{j-2}}\equiv 0\bmod {\wp'}^2$, that is,
$j=2$ and $p\equiv 3\bmod 4$. Similarly, if $j<2$ then $h\equiv
(X-1)^2\bmod \wp'$ and so $\zz[\pi]_{\wp'}$ is not maximal if and only
if $p\equiv 1\bmod 4$.  Note $\zz[\pi]_2=\prod_{\wp'\mid
2}\zz[\pi]_{\wp'}$.  By Lemma~\ref{Llenstra} and the above argument,
$\zz[\pi]_2$ is not maximal if and only if $\pi\in\E$.

In the special case $\pi\in\E$, we have 
$K=\qq(\zeta_{m_{(2)}})$ and
$\qq(\pi)=K(\sqrt{(\frac{-1}{p})p})$ is quadratic over $K$.  
Moreover, $\zz[\pi]_\wp = (\O_K)_\wp[\sqrt{(\frac{-1}{p})p}]$ and 
$\wp$ is totally ramified in $\zz[\pi]_\wp$. This proves 
that $\zz[\pi]_\wp$ is a local ring. 
The decomposition of $\wp$ in the quadratic extension $\qq(\pi)$ over $K$
corresponds to that of $2$ in $\qq(\sqrt{(\frac{-1}{p})p})$ over
$\qq$, which is as in our assertion.  Since $\zz[\pi]_\wp$ is a
quadratic order over the complete discrete valuation ring
$(\O_K)_\wp$, it is a Bass order by Remark~\ref{Xbass}~(1).  As
$(\O_K)_\wp$-orders, $\zz[\pi]_\wp
\subset\O_\wp\cong (\O_K)^2_\wp$. 
There is an injection 
$\zz[\pi]_\wp/(\O_K)_\wp\hookrightarrow \O_\wp/(\O_K)_\wp\cong
(\O_K)_\wp$, under which $\zz[\pi]_\wp/(\O_K)_\wp\cong
\wp^i (\O_K)_\wp$ for some positive integer $i$.
In other words, $\zz[\pi]_\wp=(\O_K)_\wp+\wp^i\O_\wp$ and so
$\O_\wp/\zz[\pi]_\wp\cong (\O_K)_\wp/\wp^i$.  But 
$\Delta_{\zz[\pi]_\wp/(\O_K)_\wp}
=(\O_K)_\wp\Delta(X^2-(\frac{-1}{p})p)=4(\O_K)_\wp$ and hence
$[\O_\wp:\zz[\pi]_\wp]^2=[(\O_K)_\wp:\wp^i]^2=2^{2i}(\O_K)_\wp\mid
4(\O_K)_\wp$. Thus $i=1$, that is,
$\O_\wp/\zz[\pi]_\wp\cong(\O_K)_\wp/\wp$ as $(\O_K)_\wp$-modules.  
Hence $\O_\wp$ is
the only $(\O_K)_\wp$-order in $\qq(\pi)_\wp$ that properly contains
$\zz[\pi]_\wp$.
\end{proof}

\subsection{Torsion-free modules over Bass orders}\label{S54}

Let the notation be as in section~\ref{S3}. For any ring $R$ we use
$R^*$ to denote its group of units.  
Henceforth in this section we
assume that $R$ is an order in $\qq(\pi)$ containing $\zz[\pi]$.  Let $M$
be a torsion-free $R_l$-modules (as defined in section~\ref{S51}) of
rank $e$, our goal here is to describe
all such modules. We recall that all modules are
assumed to be finitely generated. 

\begin{lemma}\label{Lindex2} 
Let $\wp$ be any prime ideal in $\O_K$ lying over $2$.  Let $N$ be an
indecomposable torsion-free $\zz[\pi]_\wp$-module. Suppose $(l,\pi)\in
\{2\}\times\E$.  If $\wp$ is inert in $\qq(\pi)$, then $N\cong
\zz[\pi]_\wp$ or $\O_\wp$. If $\wp$ is split, i.e., $\wp=\wp_1\wp_2$ for some prime 
ideals $\wp_1$, $\wp_2$ in $\qq(\pi)$,
then $N\cong \zz[\pi]_\wp$, $\O_{\wp_1}$, or $\O_{\wp_2}$.
\end{lemma}
\begin{proof}
By Lemma~\ref{Lindex}, the $\zz[\pi]_\wp$ is a local ring and an $(\O_K)_\wp$-order,
so we invoke Theorem~\ref{Tbass}. 
If $N$ is projective over the local ring $\zz[\pi]_\wp$ 
then $N\cong \zz[\pi]_\wp$. Otherwise, 
$N$ is projective over
$\O_\wp$. 
Suppose $\wp$ is inert in $\qq(\pi)$, that is, $\O_\wp$ is a
discrete valuation ring then $N\cong_{\O_\wp}
\O_\wp$. If $\wp$ splits into $\wp_1$ and $\wp_2$ in $\qq(\pi)$, that
is, if $\O_\wp\cong\O_{\wp_1}\times
\O_{\wp_2}$, then $N\cong_{\O_\wp} \O_{\wp_1}$ or 
$\O_{\wp_2}$.  Therefore
$$N\cong_{\zz[\pi]_\wp}\zz[\pi]_\wp,
\O_{\wp_1},\mbox{  or}\quad\O_{\wp_2}.$$This finishes the proof. \end{proof}

\begin{prop}\label{Pindecomposable} 
There is the following isomorphism of $R_l$-modules:
$$M\cong_{R_l}
\left\{
\begin{array}{ll}
R_l^e&\mbox{if $(l,\pi)\notin \{2\}\times\E$},\\
\prod_{\wp\mid l} (R_\wp^{a_\wp} \times
\O_\wp^{b_\wp})&\mbox{if $(l,\pi)\in \{2\}\times\E$}
\end{array}\right.$$
where $\wp$ ranges over all prime ideals in $\O_K$ lying over $2$, and  
$a_\wp$, $b_\wp$ are non-negative integers such that $a_\wp+b_\wp=e$.
\end{prop}
\begin{proof} 
Suppose $(l,\pi)\notin \{2\}\times\E$.  By Lemma~\ref{Lindex}, the
$\zz_l$-order $R_l$ is maximal and our assertion follows from 
the argument preceding Theorem~\ref{Tbass}.

Suppose $(l,\pi)\in \{2\}\times\E$.  Since $M_\wp$ is a torsion-free
$R_\wp$-module of rank $e$, by the Krull-Schmidt-Azumaya
theorem~\cite[Theorem (30.6)]{Curtis-Reiner:90}, $M_\wp$ can be
expressed as a finite direct sum of indecomposables with the summands
unique up to isomorphism and order of occurrence.  If $\wp$ is inert
in $\qq(\pi)$, then by Lemma \ref{Lindex2} there are non-negative
integers $a_\wp, b_\wp$ with $a_\wp+b_\wp=e$ such that $M_\wp\cong
R_\wp^{a_\wp}\times \O_\wp^{b_\wp}$. Now suppose $\wp$ is split in
$\qq(\pi)$. Then $M_\wp\cong R_\wp^{a_{\wp}}\times
\O_{\wp_1}^{b_\wp}\times \O_{\wp_2}^{c_\wp}$ for some non-negative
integers $a_\wp, b_\wp, c_\wp$; by comparing ranks in
$\qq(\pi)_\wp^e\cong_{\qq(\pi)_\wp} \qq(\pi)_\wp^{a_\wp}\times
\qq(\pi)_{\wp_1}^{b_\wp}\times \qq(\pi)_{\wp_2}^{c_\wp}$, we are
forced to have $b_\wp=c_\wp$.  Thus, $M_\wp\cong R^{a_\wp}_\wp \times
(\O_{\wp_1}\times \O_{\wp_2})^{b_\wp}\cong R_\wp^{a_\wp}\times
\O_\wp^{b_\wp}$ for $a_\wp, b_\wp$ with $a_\wp+b_\wp=e$. Therefore
$$M\cong\prod_{\wp\mid 2} M_\wp \cong_{R_2}\prod_{\wp\mid 2}
(R_\wp^{a_\wp} \times \O_\wp^{b_\wp}).$$ This finishes our
proof. \end{proof}

The following corollary is prepared for the next section.

\begin{corollary}\label{C58}
If $M$ is a torsion-free $R_l$-module of rank $e$ then we have 
$M/(\pi-1)M\cong_{R_l} (R_l/(\pi-1))^e$
unless $l=2$, $q$ is not a square, and
$\pi=\pm\sqrt{(\frac{-1}{p})q}$, in which case 
there are non-negative integers $a,b$ with $a+b=e$ such that
$$M/(\pi-1)M \cong_{R_2} (R_2/(\pi-1))^a \times (\O_2/(\pi-1))^b.$$
\end{corollary}
\begin{proof} 
First of all we show that $m\notin 2^\zz$ if and only if
$R_2/(\pi-1)=0$, that is, $\pi-1\in R_2^*$. Indeed, $m\notin 2^\zz$
implies $\zeta_m^\nu-1\in \zz[\pi]_2^*\subseteq R_2^*$.  Write
$\pi-1=(\zeta_m^\nu-1)\sqrt{q} + (\sqrt{q}-1)$.  If $p=2$ then
$(\zeta_m^\nu -1)\sqrt{q}$ lies in a prime over $2$ while
$\sqrt{q}-1\in R_2^*$ so their sum lies in $R_2^*$; if $p\neq 2$, then
$R_2^*\sqrt{q}=R_2^*$ and $\sqrt{q}-1$ lies in a prime over $2$ thus
their sum also lies in $R_2^*$. This proves our claim.  Consequently,
if $m\in 2^\zz$ then $M/(\pi-1)M\cong (R_2/(\pi-1))^e$ since they are
both trivial.  By Proposition~\ref{Pindecomposable}, we have
$M/(\pi-1)M\cong_{R_l}(R_l/(\pi-1))^e$ unless $l=2, \pi\in\E$ and
$m\in 2^\zz$.  By the definition of $\E$, we have
$\pi=\zeta_m^\nu\sqrt{q}\in\E$ if and only if $q$ is not a square and
$m=1$ or $2$ if $p\equiv 1\bmod 4$; while $m=4$ if $p\equiv 3\bmod 4$.
That is, we have $l=2$, $q$ is not a square and
$\pi=\pm\sqrt{(\frac{-1}{p})q}$.
\end{proof}

%section 3
\section{Elementary supersingular abelian varieties}\label{S7}

\subsection{Preliminaries}\label{S2}

This section provides some auxiliary results on abelian varieties over
finite fields. We shall quote from~\cite{Milne}
and~\cite{Mumford:74} without comment.  

Recall that $l$ is any prime different from $p$.  If $G$ is an abelian
group we denote by $G[l^\infty]$ the subgroup of all elements in $G$
whose order is a $l$-power. For every ${\bf k}$-isogeny
$r:A\rightarrow A$, we denote by $A[r]$ the kernel of the induced map
on $A(\bar{\bf k})$ as abelian groups.  The $l$-adic Tate module
$T_l(A)=\projlim_n A[l^n]$ is free of rank $2d$ over $\zz_l$. Since
the Frobenius endomorphism $\pi$ acts faithfully on it, $T_l(A)$ is
torsion-free $\zz[\pi]_l$-module.  $V_l(A)=T_l(A)\otimes_{\zz_l}\qq_l$
is a $\qq[\pi]_l$-module.  We also know that $\qq[\pi]$ is a
semisimple $\qq$-algebra. If the characteristic polynomial of the
Frobenius is $f=\prod_{i=1}^{t}g_i^{e_i}$ as in section~1, then
\begin{eqnarray}
\qq[\pi]_l  \cong \prod_{i=1}^{t} \qq[\pi]_l/(g_i(\pi)),\quad 
V_l(A) \cong \prod_{i=1}^{t}
(\qq[\pi]_l/(g_i(\pi)))^{e_i}.\label{E1}
\end{eqnarray}
In particular, if $A$ is elementary so $\qq[\pi]$ is a field, then $V_l(A)\cong
\qq(\pi)_l^e$. Thus $T_l(A)$ is a torsion-free module of rank $e$ over
any $\zz_l$-order of $\qq(\pi)_l$ containing $\zz[\pi]_l$.

$T_l$ defines a (covariant) functor from the category of abelian
varieties $A'$ over ${\bf k}$ with a ${\bf k}$-isogeny $r:
A\rightarrow A'$ to the category of $\zz[\pi]_l$-modules $T_l(A')$
with an injective $\zz[\pi]_l$-module homomorphism $T_l(A)\rightarrow
T_l(A')$. For each positive integer $n$ we have
$l^{-n}T_l(A)/T_l(A)\cong A[l^n]$ and so, taking direct limits,
$V_l(A)/T_l(A)\cong A[l^\infty]$.  Mapping the exact sequence
$0\rightarrow T_l(A)\rightarrow V_l(A)\rightarrow
A[l^\infty]\rightarrow 0$ to that of $A'$ by $r$ induces an injective
$\zz[\pi]_l$-module homomorphism $r: T_l(A)\rightarrow T_l(A')$ with cokernel
$T_l(A')/r T_l(A)$ and an isomorphism $r: V_l(A)\rightarrow
V_l(A')$. Let $r^{-1}T_l(A')$ be the pullback of $T_l(A')\subset
V_l(A')$ under this isomorphism, there is an isomorphism $T_l(A')/r
T_l(A)\cong r^{-1}T_l(A')/T_l(A)$.  By applying the Snake Lemma to the
above resulting diagram, we have $\kernel(r)[l^\infty]\cong
r^{-1}T_l(A')/T_l(A)$.  In fact, every $\zz[\pi]_l$-lattice (as $\zz_l$-order) of
$V_l(A)$ containing $T_l(A)$ arises in this way:

\begin{prop}\label{P21} 
Let $\theta: V_l(A)/T_l(A)\stackrel{\sim}{\rightarrow} A[l^\infty]$ be
the isomorphism as above.  For every $\zz[\pi]_l$-lattice $M$
containing $T_l(A)$, there is an abelian variety $A'$
with a ${\bf k}$-isogeny $r:A\rightarrow A'$ such that
$M=r^{-1}T_l(A')$ in $V_l(A)$ and $\theta(M/T_l(A))=\kernel(r)(\bar{\bf k})$.
\end{prop} 
\begin{proof}
Write $G=\theta(M/T_l(A))$, it is a finite subgroup of $A[l^\infty]$. 
By our assumption, there is an \'etale subgroup scheme ${\cal G}$ 
of $A$ over ${\bf k}$ with ${\cal G}(\bar{\bf k})=G$. 
Take $A'= A/{\cal G}$ and the ${\bf k}$-isogeny
$r:A\rightarrow A'$ with $\kernel(r)={\cal G}$. The argument preceding the
proposition indicates $\theta(r^{-1}T_l(A')/T_l(A))={\cal G}[l^\infty]=G$.
\end{proof}

Define $T_p(A)=\projlim_nA[p^n]$ in an analogous manner.
It is free $\zz_p$-module of rank between $0$ and $d$ (inclusive).

\begin{prop}\label{PAr1}
For any ${\bf k}$-isogeny $r:A\rightarrow A$, 
there is an isomorphism of abelian groups:
$$A[r]\cong\prod_{l} T_l(A)/r T_l(A)$$  
where $l$ ranges over all prime numbers.
\end{prop} 
\begin{proof}
The finite abelian group $A[r]$ 
has the decomposition
$A[r]\cong \prod_{l}(A[r])[l^\infty]$, 
where each component is isomorphic to $T_l(A)/r T_l(A)$. 
\end{proof}

Note that $A({\bf k})=A[\pi-1]$.

\subsection{Elementary supersingular abelian varieties}

An abelian variety over ${\bf k}$ is supersingular if and only if the
eigenvalues of its Frobenius $\pi$ are supersingular $q$-numbers. This
is equivalent to that the Newton polygon of $A$ is a straight line of
slope $1/2$, which is equivalent to that $A$ is $\bar{\bf
k}$-isogenous to a power of a supersingular elliptic curve (as a
special case of~\cite[Theorem~4.2]{Oort:74}).  A supersingular abelian
variety $A$ over ${\bf k}$ has $A[p] = 0$. 
We remark in brief that the converse is not true
when $d>2$. In fact, an abelian variety has $A[p]=0$ if and only if
its Newton polygon has no $0$-slope segment, which does not imply
being a straight line of slope $1/2$ when $d>2$.

For the rest of this section we assume that $A$ is an elementary
supersingular abelian variety over ${\bf k}$ whose Frobenius relative
to ${\bf k}$ is $f=g^e$ for some monic irreducible polynomial $g$ over 
$\qq$ and a positive integer $e$. 

Since $\qq[\pi]=\qq(\pi)$ is a field, we fix an embedding of
$\qq(\pi)$ in $\cc$ and identifying $\pi$ with its image, we write
$\pi=\zeta_m^{\nu}\sqrt{q}$ for some primitive $m$-th root of unity
$\zeta_m^\nu$.  We resume the notation from section~\ref{S3}:
$K=\qq(\pi^2)=\qq(\zeta_{m/(2,m)})$, its ring of integers
$\O_K=\zz[\zeta_{m/(2,m)}]$, and $\O$ the ring of integers of
$\qq(\pi)$.

Let $\Q$ be the set of all supersingular $q$-numbers
$\zeta_m^\nu\sqrt{q}$ for some primitive root of unity $\zeta_m^\nu$
such that either of the following two conditions is satisfied:
(1) $m=1$ or $2$; (2) $q$ is a square, $(2,p)p\nmid m$
and $p$ is of odd order in the group $(\zz/m_{(p)}\zz)^*$.

\begin{prop}\label{Pe12}
Suppose $A$ is simple supersingular over ${\bf k}$.
\begin{enumerate}
\item[(1)] When $\pi\in\Q$ then $e=2$ and $\End_{\bf k}^0(A)$ 
is a quaternion algebra 
over $\qq(\pi)$; 
\item[(2)] 
When $\pi\not\in\Q$ then $e=1$ and $\End_{\bf k}^0(A)$ 
is commutative and equal to $\qq(\pi)$.
\end{enumerate}
\end{prop}
\begin{proof} Let $v$ be any place of $\qq(\pi)$ (including
both finite and infinite primes).  Let $e_v$ denote the denominator of
the Hasse invariant, $\inv_v(\End_{\bf k}^0(A))$, of $\End_{\bf
k}^0(A)$ at $v$.  By~\cite[Th\'eor\`eme 1]{Tate1:68} we have
$$\inv_v(\End_{\bf k}^0(A))=\frac{\ord_v(\pi)[\qq(\pi)_v:
\qq_p]}{\ord_v(q)}=
\frac{[\qq(\pi)_v:\qq_p]}{2}=\frac{\gamma(v/p)\kappa(v/p)}{2} \bmod
1,$$ for all primes $v$ lying over $p$, so $e_v=1$ or $2$.  Now
$e_v=1$ for all complex $v$ and also for all finite primes $v$ not
lying over $p$, while $e_v=2$ for all real $v$. We have $e=\lcm_v(e_v)
= 2$ if either (1)' $v$ is real or (2)' $\gamma(v/\wp)\kappa(v/\wp)$
is odd; and $e=1$ otherwise.  It is obvious that (1)' is equivalent to
(1).  We show below that if $v$ is not real prime then (2)' is
equivalent to (2):

Suppose $q$ is not a square: we claim that $e_v=1$ for all finite
primes $v$ over $p$.  Now $[\qq(\pi):K]=1$ or $2$. The former implies
$2\mid \gamma(v/p)$. Consider the latter case, if $\sqrt{p}\in\qq(\pi)$,
then $2\mid \gamma(v/p)$ and so $e_v=1$; otherwise, we would have quadratic
extensions $\qq(\zeta_m,\sqrt{p}) \supset \qq(\pi)\supset K$.  But if $p$
was unramified in $\qq(\pi)/K$, then it would be unramified in 
$\qq(\zeta_m,\sqrt{p})/\qq(\zeta_m)$, which is
absurd; so we must conclude that $p$ is totally ramified in
$\qq(\pi)/K$ and hence $2\mid \gamma(v/p)$ and so $e_v=1$.

Suppose $q$ is a square: so that
$\qq(\pi)=\qq(\zeta_m)$. Then for any finite prime $v$ over $p$, we have
that $\kappa(v/p)$ equals the order of $p$ in 
$(\zz/m_{(p)}\zz)^*$; Let $\phi(\cdot)$
denote the Euler phi-function here, then $\gamma(v/p)=\phi(m_p)$, 
which is odd if and only if $(2,p)p\nmid m$.
This finishes our proof.
\end{proof}

\begin{remark}\label{Re12}
Suppose $A$ is simple supersingular over ${\bf k}$. 
If $\pi\in \E$, then $\pi\in \Q$ if and only if $d = 2$.
This follows from the above proposition and the definitions of $\E$ and $\Q$.
\end{remark}

\begin{remark}\label{R1}
Let $A$ be a simple supersingular abelian variety with
odd dimension $d>2$, then $e=1$ and $\End_{\bf k}^0(A)$ must be commutative.
Indeed, recall~\cite[Th\'eor\`eme 1]{Tate1:68} that $2d=e[\qq(\pi):\qq]$
and so it suffices to show $2\mid [\qq(\pi):K][\qq(\zeta_{m/(2,m)}):\qq]$. 
Either $[\qq(\pi):K]=1$ or $2$, in
the former case $[\qq(\zeta_{m/(2,m)}):\qq]=\phi(m/(2,m))>1$ and so is even.
\end{remark}

\subsection{Module structures}

Let $R$ be a subring in $\qq(\pi)$ with $\zz[\pi]\subseteq R\subseteq
\End_{\bf k}(A)\cap \qq(\pi)$. For any finite group $G$, we write 
$\#G$ for its order.

\begin{lemma}\label{L52} 
Let $M'\subseteq M''$ be modules over any ring $R$. Let $r\in R$ be
such that $r R$ is of finite index in $R$ and $r$ acts faithfully on
$M', M''$.
\begin{enumerate}
\item[(1)] If $M'$ contains a free $R$-module of rank
$s$ as a submodule of finite index, then $$\#M'/r M'=(\#(R/r R))^s.$$
\item[(2)] If $M', M''$ contain a free $R$-module of rank $s$ 
as a submodule of finite index in $M',M''$, respectively,
then there are  homomorphisms 
$\rho': M'/r M'\rightarrow M''/r M''$ 
and  {$\rho'': M''/r M''\rightarrow M'/r M'$} 
with $$\#\kernel(\rho')=\#\cokernel(\rho')
=\#\kernel(\rho'')=\#\cokernel(\rho'')
\;\mid\;\# M''/M'.$$
\end{enumerate}
\end{lemma} 
\begin{proof} 
(1) Since $r$ acts faithfully on $M'$ and $R^s$, the injective map  
$r: M'\rightarrow M'$ induces an injective map 
$r: R^s\hookrightarrow R^s$. 
On the other hand,  the given injection $R^s\hookrightarrow M'$
is of finite index, we thus have
$$\#(M'/r M')\cdot\#(M'/R^s)=\#(R^s/r
R^s)\cdot\#(M'/R^s).$$ Therefore, $\#M'/r M'=\#(R/r
R)^s$.

(2) Let $r$ act on the short exact sequence of $R$-modules 
$$0\rightarrow M'\rightarrow
M''\rightarrow M''/M'\rightarrow 0,$$ and apply the Snake lemma. 
We then get the desired map $\rho'$ with $\#\cokernel(\rho')$ 
dividing $\#M''/M'$. 
By part (1), we have $\#M'/r M'=\#M''/r M''$ as they both equal
$\#(R/r R)^s$. Thus $\#\kernel(\rho')=\#\cokernel(\rho')$.
Any finite $R/r R$-module $N$ has an isomorphic dual 
$\Hom_\zz(N,\qq/\zz)$, our assertion on $\rho''$ follows by 
taking the dual of $\rho'$.
\end{proof}

\begin{prop}\label{PAr} % 3.7
Let $r$ be an isogeny in $R$. Then there
is an $R$-module homomorphism 
$$\varphi_r: A[r]\longrightarrow \prod_{l\neq p}(R_l/r R_l)^e$$ which
is an isomorphism except when $\pi\in\E$ in which case
$\#\kernel(\varphi_r)$ and $\#\cokernel(\varphi_r)$ are equal and divide 
$2^d_{(p)}.$
\end{prop}
\begin{proof}
By Proposition~\ref{PAr1} and the fact
$A[p]=0$, we have $A[r]\cong\prod_{l\neq p}T_l/rT_l$.  
Recall that $T_l$ is a torsion-free $R_l$-module of rank $e$, so we
invoke Proposition~\ref{Pindecomposable}. 
If $\pi\not\in\E$ or $p=2$, then $T_l/rT_l\stackrel{\sim}{\rightarrow} 
(R_l/r R_l)^e$ for each $l\neq p$, and we obtain the desired 
isomophism $\varphi_r$. Now suppose $\pi\in\E$. 
Lemma~\ref{Lindex}~(2) implies
$\#\O_\wp/R_\wp\;\mid\;\#\O_\wp/\zz[\pi]_\wp =\#(\O_K)_\wp/\wp =
2^{\kappa(\wp/2)}.$ Clearly $\kappa(\wp/2)\varrho(\wp/2)\mid [K:\qq]$
and $[K:\qq]=[\qq(\pi):\qq]/2$ by Lemma~\ref{Lindex}~(1). 
For each $l$, we have a map
$T_l/rT_l{\rightarrow}(R_l/rR_l)^e$ which is an isomorphism if $l\neq 2$. 
When $l=2$, Lemma~\ref{L52} indicates the size of its kernel and cokernel 
are equal and divide $(\# T_2/R_2^e)_{(p)}$. Taking product over all $l\neq p$ 
we obtain the desired map $\varphi_r$ with $\#\kernel(\varphi_r)=\#\cokernel(\varphi_r)$ 
and divides
$$(\#\O_2/R_2)_{(p)}^e
\;\mid\; 2^{\kappa(\wp/2) \varrho(\wp/2)e}_{(p)}
\;\mid\; 2^{e[K:\qq]}_{(p)} 
\;\mid\; 2^{e[\qq(\pi):\qq]/2}_{(p)}$$
where the last number equals $2^d_{(p)}$. 
\end{proof}

\noindent {\bf Proof of Theorem~1.2.} Let $S=\zz-p\zz$. By
Proposition~\ref{PAr}, there is an $R$-module homomorphism $\varphi_n:
A[n]\rightarrow ((\frac{1}{n}R)/R)^e$ for every $n\in S$. Let $W_n$ be
the set of such homomorphisms.  If {$m\mid n$}, then by passing to the
largest submodule annihilated by $m$ we see that any $R$-module homomorphism
$\varphi_n$ maps the submodule $A[m]$ of $A[n]$ to
$((\frac{1}{m})R/R)^e$, so there is a restriction map $W_n\rightarrow
W_m$. Since the projective limit of a system of non-empty finite sets
is non-empty, the projective limit of the sets $W_n$ is
non-empty. Therefore we can make a simultaneous choice of
$R$-module homomorphisms $\varphi_n$ that commute with the inclusions
$A[m]\subseteq A[n]$ and $((\frac{1}{m}R)/R)^e\subseteq
((\frac{1}{n}R)/R)^e$. Taking the injective limit over $n\in S$, we
get an $R$-module homomorphism $\injlim A[n]
\rightarrow\injlim((\tfrac{1}{n}R)/R)^e$, that is $\varphi:
A(\bar{\bf k})\stackrel{\sim}{\rightarrow} (R_{(p)}/R)^e.$ Since
$A(\bar{\bf k})$ and $(R_{(p)}/R)^e$ are both divisible as abelian groups,
the cokernel of $\varphi$ is also divisible, but it is finite and hence
trivial. So $\cokernel(\varphi)\cong\injlim\cokernel(\varphi_n)$ is
trivial and $\varphi$ is surjective.  In $A(\bar{\bf k})$ we have
$\kernel(\varphi)\cong\injlim\kernel(\varphi_n)$.  Thus $\varphi$ is
an isomorphism except when $\pi\in\E$, in which case
$\#\kernel(\varphi)$ divides $2^d_{(p)}$ since $\#\kernel(\varphi_n)$
divides $2^d_{(p)}$ for each $n$.  \qed


\begin{prop}\label{Psimple2}
Let $A$ be a simple supersingular abelian variety 
over ${\bf k}$ with $f=g^e$. Let $R=\End_{\bf k}(A)\cap\qq(\pi)$. 
If $p=2$ or $d\neq 2$, then 
$A(\bar{\bf k})\cong_R (R_{(p)}/R)^e$. 
If $p\neq 2$ and $d=2$, then there are
non-negative integers $a,b$ with $a+b=e$ and
$$A(\bar{\bf k})\cong_R (R_{(p)}/R)^a \times (\O_{(p)}/\O)^b.$$
\end{prop}
\begin{proof} 
Let $\varphi: A(\bar{\bf k})\rightarrow (R_{(p)}/R)^e$ be defined as in
Theorem~1.2, which is an isomorphism unless $\pi\in\E$. Suppose
$\pi\in\E$. Then $\#\kernel(\varphi) = \#\cokernel(\varphi)$ is a
$2$-power.  Suppose $d\neq 2$.  Then $e=1$ by Remark~\ref{Re12} and so
$T_2$ is a torsion-free $R_2$-module of rank 1.  Recall that $K$ is a
cyclotomic field.  For any prime $\wp$ in $\O_K$ lying over $2$, write
$T_\wp$ for the $\wp$-adic completion of $T_2$ and $T_\wp:T_\wp=\{r\in
R_\wp\mid r T_\wp\subseteq T_\wp\}$.  Then $T_\wp:T_\wp = R_\wp$, so
$T_\wp$ is a fractional ideal of $R_\wp$.  Recall from
Lemma~\ref{Lindex} that $R_\wp$ is a Bass $(\O_K)_\wp$-order and thus
$T_\wp\cong_{R_\wp}R_\wp$ by~\cite[\S~2.6]{Buch-HWL:94}.  So
$T_2\cong_{R_2}R_2$ and this induces isomorphism 
$T_2/2T_2\stackrel{\sim}{\rightarrow}R_2/2R_2$ by Lemma~\ref{L52}.  
Thus $\varphi$ is an isomorphism.
Suppose $d=2$. Then $\pi\in\E$ implies $\pi=\pm\sqrt{(\frac{-1}{p})q}$
and $e=2$ by Remark~\ref{Re12}. In this case, $\wp=2$, so by
Proposition~\ref{Pindecomposable}, we have $A[n]\cong \prod_{l\neq p}T_l/nT_l
\cong (\frac{1}{n}R/R)^a\times
(\frac{1}{n}\O/O)^b$ for all $n\in S=\zz-p\zz$.  
Take injective limit both sides over $n\in S$, we have
$$A(\bar{\bf k}) \cong \injlim((\tfrac{1}{n}R/R)^a\times
(\tfrac{1}{n}\O/\O)^b)\cong_R (R_{(p)}/R)^a\times
(\O_{(p)}/\O)^b.$$This finishes our proof.\end{proof}

\subsection{Group structures}

In this subsection we shall apply the results of the previous
subsection to our study of the group structure of $A({\bf k})$. 

If $A$ is exceptional, $\qq(\pi)=\qq(\sqrt{(\tfrac{-1}{p})q})
=\qq(\sqrt{(\tfrac{-1}{p})p})$, so 
$\O=\zz[\frac{1+\sqrt{(\frac{-1}{p})p}}{2}].$    
By Lemma~\ref{Lindex}~(2) we notice 
$\O_2/\zz[\pi]_2\cong \zz_2/2\zz_2\cong \zz/2\zz.$

\vspace{3mm}
\noindent {\bf Proof of Theorem~1.1.}  
Apply Corollary~\ref{C58} to $M=T_l(A)$ and $R=\zz[\pi]$.
Now 
$$A({\bf k})\cong_{\zz[\pi]}(\zz[\pi]/(\pi-1))^e
\cong_\zz (\zz/g(1)\zz)^e$$ 
unless $A$ is exceptional, in which case
the argument preceding the proof 
implies that $(\pi-1)/2\in \O_2$ while $(\pi-1)/4\notin \O_2$. Since 
$\#\O_2/(\pi-1)=\#\zz[\pi]_2/(\pi-1)=|g(1)|_2$, we have
$$\O_2/(\pi-1)\cong_{\zz_2}\zz_2/2\zz_2\times
\zz_2/\tfrac{g(1)}{2}\zz_2.$$ Hence there are non-negative integers
$a, b$ with $a+b=e$ such that
$$\begin{array}{lll}
A(k)\!\!&\cong_{\zz[\pi]}&\!\!\!(\zz[\pi]_2/(\pi-1))^a\times(\O_2/(\pi-1))^b
\times \prod_{l\neq 2}(\zz[\pi]_l/(\pi-1))^e\\
\!\!&\cong_{\zz} &\!\!\!
    ((\zz_2/g(1)\zz_2)^a \times
    (\zz_2/\tfrac{g(1)}{2}\zz_2\times\zz_2/2\zz_2)^b)
   \times\prod_{l\neq 2}(\zz_l/g(1)\zz_l)^{e}\\ 
\!\!&\cong_{\zz}
    &\!\!\!(\zz/g(1)\zz)^{a}\times
    (\zz/\tfrac{g(1)}{2}\zz\times \zz/2\zz)^{b}.
\end{array}$$
This proves our theorem.
\qed

\begin{prop}\label{Paprime}
Let the notation be as in Theorem~\ref{Telementary}. If $A$ is
exceptional, then for every pair of non-negative integers $a', b'$ with
$a'+b'=e$ there exists an abelian variety $A'$ isogenous over ${\bf k}$
to $A$ such that
$$A'({\bf k})\cong_\zz (\zz/g(1)\zz)^{a'}\times
(\zz/\tfrac{g(1)}{2}\zz \times \zz/2\zz)^{b'}.$$ 
\end{prop}
\begin{proof} 
Let $A$ be exceptional. By Theorem~\ref{Telementary}, 
there are non-negative integers $a,b$ with
$a+b=e$ such that $T_2\cong_{\zz[\pi]_2}\zz[\pi]_2^{a}\times
\O_2^{b}$ and $$A({\bf k})\cong_\zz (\zz/g(1)\zz)^{a}\times
(\zz/\tfrac{g(1)}{2}\zz\times \zz/2\zz)^{b}.$$ 
If ${b'}=b$, then we are done.
If ${b'}<b$, let 
$$M=\zz[\pi]_2^{a}\times\O_2^{b'}\times 
(\tfrac{1}{2}\zz[\pi]_2)^{b-{b'}};$$ 
if $b'>b$, let 
$$M=\zz[\pi]_2^{a'}\times\O_2^{b'},$$ 
in either case 
$M\cong_{\zz[\pi]_2}\zz[\pi]_2^{a'}\times\O_2^{b'}$. 
By the argument preceding the proof of Theorem~1.1,
we know that 
$\O_2\subseteq \frac{1}{2}\zz[\pi]_2\subset \qq(\pi)_2$.
By Proposition~\ref{P21}, there exits an abelian variety $A'$ over ${\bf k}$ with
$T_2(A')=M$ and a ${\bf k}$-isogeny $A\stackrel{r}{\rightarrow}A'$
with $\kernel(r)(\bar{\bf k})\cong T_2(A')/T_2(A)$ while
$T_l(A')=T_l(A)$ for all $l\neq 2$. Thus
$$\begin{array}{lll}
A'({\bf k})&\cong&\prod_{l\neq p}T_l(A')/(\pi-1)T_l(A')\\
&\cong_{\zz[\pi]}&(\zz_2[\pi]/(\pi-1))^{a'}\times(\O_2/(\pi-1))^{b'}
\times \prod_{l\neq 2}(\zz[\pi]_l/(\pi-1))^e\\
&\cong_\zz& (\zz/g(1)\zz)^{a'}\times
(\zz/\tfrac{g(1)}{2}\zz \times \zz/2\zz)^{b'}.
\end{array}$$
This finishes the proof. 
\end{proof}

\begin{corollary}\label{Csimple}
Suppose $A$ is a simple supersingular abelian variety over ${\bf k}$
of dimension $d>2$ with $f=g^e$, 
then $A({\bf k})\cong_\zz
(\zz/g(1)\zz)^e$ with $e=1$ or $2$. 
If $d =1$ or $2$ then $A({\bf k})\cong_\zz (\zz/g(1)\zz)^e$ or 
$A({\bf k})\cong_\zz \zz/\tfrac{q+1}{2}\zz \times \zz/2\zz$. 
\end{corollary}
\begin{proof} 
If $A$ is simple over ${\bf k}$ of dimension $d>2$, then
$A$ is never exceptional, so $A({\bf k})\cong_\zz (\zz/g(1)\zz)^e$,
where $e=1$ or $2$ as we have seen in Proposition~\ref{Pe12}. 

If $A$ is an elliptic curve, then 
$A({\bf k})\cong_\zz (\zz/g(1)\zz)^e$ unless $A$ is exceptional in which case
$A({\bf k})\cong_\zz (\zz/g(1)\zz)^e$ or 
$A({\bf k})\cong_\zz \zz/\tfrac{q+1}{2}\zz \times \zz/2\zz$. 
Both cases may occur because of Proposition~\ref{Paprime}.
(This result can be found in~\cite[Chapter~4]{Schoof:87}.) 

If $A$ is of dimension $2$, then 
$A({\bf k})\cong_\zz(\zz/g(1)\zz)^2$ 
unless $A$ is exceptional in which case 
$A({\bf k})\cong_\zz(\zz/g(1)\zz)^2$ or 
$A({\bf k})\cong_\zz \zz/\frac{q-1}{2}\zz \times \zz/2\zz$. 
Both cases may occur because of Proposition~\ref{Paprime}.
\end{proof}

In particular, by Remark~\ref{R1}, if $A$ is simple supersingular of odd 
dimension $d>2$, then $A({\bf k})\cong \zz/g(1)\zz$ is cyclic.

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