\documentstyle[mssymb,/usr2/tex/macros/old/cyr]{article}
%
\def\MW{Mordell-\mbox{\kern-.16em}Weil}
\newcommand\TS{Tate-\v Safarevi\v c}
\newcommand\BW{Barnes-\mbox{\kern-.16em}Wall}
\newcommand{\be}{\begin{equation}}
\newcommand{\ee}{\end{equation}}
\newcommand{\ra}{\rightarrow}
\newcommand{\teq}{\!=\!}
\newcommand{\C}{{\bf C}}
\newcommand{\Q}{{\bf Q}}
\newcommand{\R}{{\bf R}}
\newcommand{\Z}{{\bf Z}}
\newcommand{\Nmin}{{N_{\rm min}}}
\newcommand{\Ht}{\hat{h}}
\newcommand{\Sha}{{\cyr Sh}}
\newcommand{\End}{\mathop{\rm End}\nolimits}
\newcommand{\Hom}{\mathop{\rm Hom}\nolimits}
\newcommand{\Jac}{\mathop{\rm Jac}\nolimits}
\newcommand{\Leech}{{\Lambda_{24}}}
\newcommand{\Fbar}{{\bar{\bf F}_2}}
\newcommand{\0}{^{\phantom0}}
%
\begin{document}

\begin{center}
\begin{Large}
\MW\ lattices in characteristic~2 \\
II: The Leech lattice as a \MW\ lattice

\vspace*{2ex}
\end{Large}

Noam D. Elkies
\end{center}

\vspace*{4ex}

{\bf Introduction.}  In the first paper~\cite{E1} of this series
we constructed a family of lattices in dimensions $2^{n+1}$ for
positive integers~$n$.  In this paper we show that a modification
of the construction also recovers the Leech lattice $\Leech$
in dimension~24.  This yields a new description of $\Leech$:
it consists in effect of ${\bf 0}$ together with the solutions of the
equation $y^2+y=x^3+t^{13}$ in rational functions $x(t),y(t)$ defined
over $\Fbar$, the algebraic closure of the two-element field.
Call this set $L$ as long as we have not yet identified it
with~$\Leech$.  The addition law on an elliptic curve makes $L$
a commutative group, on which we define an integer-valued
function~$\Ht$ by the simple formula~(\ref{Hdef}) below.  

It is known that $\Leech$ is the only even unimodular lattice
of rank~24 with no vectors of norm~2.  We show using the methods
of~\cite{E1} that $L$ is a free abelian group of rank~24,
that $\Ht$ is a positive-definite quadratic form on that group,
that the pairing $\langle P,Q\rangle = (\Ht(P+Q)-\Ht(P)-\Ht(Q))/4$
on~$L$ associated to the quadratic form~$\Ht/2$ has determinant~1,
and that $\Ht/2$ does not attain the value~2 on~$L$.
This is enough to identify $L$ with~$\Leech$ once we prove that
$\Ht(P)/2$, already shown integral in~\cite{E1},
is even for all $P\in L$.  We do this by proving a more general
``lemma on linear descents''$\!$,
which will also be useful in a subsequent paper analyzing
the \MW\ lattices of rank $2^{n+1}$ for $n\leq4$.
We conclude with a description of the $(x,y)$
corresponding to the 196560 minimal vectors of~$\Leech$.

As in~\cite{E1} we can also interpret $L\cong\Leech$ as the
\MW\ lattice over~$\Fbar$ of a constant curve~$E_0$ over the
function field $\Fbar(C)$, where $C$\/ is a certain hyperelliptic curve
(see~(\ref{C13}) below) and $E_0$ is a supersingular elliptic curve.
Of the isometry group~$Co_0\0$ of~$\Leech$ our construction exhibits
only a very small subgroup of size $3744=24\cdot13\cdot12$,
generated by the 24 automorphisms of~$E_0$, the maps
$t\mapsto\beta t$\/ for each 13-th root of unity~$\beta$,
and the field automorphisms of~$\Fbar$ over~${\bf F}_2$.
Gross showed~\cite{GROSS} that if one uses instead
of~$C$\/ the Fermat quintic curve then one again obtains $\Leech$ as
a \MW\ lattice, this time with many more isometries (though still
few compared with the size of $Co_0\0$\,\ldots) due to the large
automorphism group PGU$_3({\bf F}_4)$ of the Fermat quintic in
characteristic~2; indeed it is thanks to those automorphisms
that Gross can prove that $\Ht(P)\equiv0\bmod4$ for all $P$\/ in his
\MW\ lattice, and thus identify that lattice with~$\Leech$.
In a later paper we shall obtain the same isometries from
our construction of~$\Leech$ by relating it with the \BW\ lattices.

{\bf Acknowledgements.}  As detailed in~\cite{E1}, I am grateful
to Conway, Sloane, Tsfasman and Gross for inspiration; to Gross, Serre
and Shioda for technical conversations and correspondence; and to
the Harvard Society of Fellows, the National Science Foundation,
and the Packard Foundation for funding.  I also acknowledge the
assistance of {\sc macsyma} with symbolic computations.

{\bf A \MW\ lattice.}  Let $k$\/ be the finite field of
$2^{12}\teq4096$ elements, and $K=k(t)$.  Fix $a_6\in k$\/ of trace
$\sum_{m=0}^{11} a_6^{2^m}=1$, and let $E/K$\/ be the potentially
constant supersingular curve
\be
y^2 + y = x^{3} + t^{13} + a_6,
\label{Edef}
\ee
and $L$ its \MW\ lattice; recall from~\cite{E1} that this is
the group $E(K)$ whose nonzero elements are solutions of~(\ref{Edef})
in rational functions $x(t),y(t)\in K$, a group endowed with the
positive-definite form
\be
\Ht: L = E(K) \ra \Z, \quad
{\bf 0} \mapsto 0,\ (x,y) \mapsto \deg(x).
\label{Hdef}
\ee

Our main goal is:

{\bf Theorem.}  {\em The curve $E/K$ has trivial \TS\ group.
Its \MW\ lattice~$L$\/ has rank\/~$24$, discriminant $2^{24}$,
minimal norm~$8$ and normalized center density~$\delta(L)=1$; it is
homothetic with the Leech lattice~$\Leech$.}

In this section we use the techniques of~\cite{E1} to prove
the claims concerning the rank, \TS\ group, and discriminant
of~$L$, and show that the minimal norm and normalized center
density are at least~$8$ and~$1$ respectively.

Let $E_0$ be the supersingular curve
\be
y^2 + y = x^3 + a_6
\label{E0}
\ee
over~$K$, and $C$\/ the hyperelliptic curve
\be
u^2 + u = t^{13}
\label{C13}
\ee
of genus~6.  As in~\cite{E1} we identify $L$\/ with the group of
$k$\/-rational maps: $C\ra E_0$ sending the point at infinity
to~$\bf0$, and thus with $\Hom_k(\Jac(C),E_0)$.
Clearly $C$\/ is dominated by the curve $u^2 + u = t^{65}$,
so $\Jac(C)$ is contained in the Jacobian of that curve.
Since that Jacobian is isogenous with a power of~$E_0$ (this
is the case $n\teq6$ of~\cite[Prop.1]{E1}), the same is true
of $\Jac(C)$.  Since $\Jac(C)$ is isogenous with $E_0^6$, the
rank of $\Hom_k(\Jac(C),E_0)$ is the same as that of
$\Hom_k(E_0^6,E_0)=(\End_k(E))^6$, that is, $6\cdot 4=24$.

Still proceeding as in~\cite{E1}, we find that $E/K$\/ has
good reduction at all finite primes, and additive bad reduction
of Kodaira type\footnote{The Kodaira type IV$^*$ did not arise
in~\cite{E1} because there the exponent $2^n+1$ was always congruent
to either~3 or~5 mod~6, whereas here we have $13\equiv1\bmod6$.}
IV$^*$ at $t\teq\infty$ with discriminant $t^{-36}$;
that its L-function is $L(E/k,s) = (|k|^{-s} - |k|^{-1})^{24}$,
so the analytic rank of~$E/K$\/ equals its arithmetic rank~24;
and thus that by Milne's theorem~\cite{MILNE} the conjecture of
Birch and Swinnerton-Dyer (formulated in this geometric setting by
Artin and Tate~\cite{TATEBSD}) holds for $E/K$, producing the formula
\be
\Delta(L)\, |\Sha(E/K)| = |k|^{2(\frac{36}{12}-1)} = 2^{24}.
\label{Disc}
\ee

Since $L$\/ is an even integral lattice, its discriminant
$\Delta=\Delta(L)$ is an integer.  Of course $|\Sha(E/K)|$
is an integer as well, so it follows from~(\ref{Disc}) that
both these integers are powers of~2.  In particular,
to demonstrate the triviality of $\Sha(E/K)$ it suffices
as in~\cite{E1} to prove that the rank of the Selmer group
for the 2-isogeny
\be
\phi: E \ra E,\quad (x,y) \mapsto (x^2,y^2+t^{13}+a_6)
\label{phi}
\ee
(with $\phi^2=-2$) is at most half the rank of~$L$, that is,
at most~12.  But as in \cite[Thm.2]{E1} we can use the ``descent map''
\be
\epsilon: E(K) / \phi E(K) \hookrightarrow K / K^2,\quad
{\bf 0} \mapsto 0,\ (x,y) \mapsto x \bmod K^2
\label{phi_descent}
\ee
to identify this Selmer group with a group of odd polynomials
of~$t$, the degree of each of whose nonzero elements is $<13/3$
and congruent to~13 mod~4.  Thus the Selmer group is contained
in $\{at|a\in k\} \cong (\Z/2)^{12}$, whence $\Sha(E/K)$ is
indeed trivial and $\Delta=2^{24}$.

If $P=(x,y)$ is a nonzero point in $E(K)$ then $x$ must have
a pole of even order $\geq 13/3$, i.e.\ at least~6, at $t\teq\infty$
for~(\ref{Edef}) to hold there.  Thus $\Ht(P)\geq6$ with equality
if and only if $x(t)$ is a polynomial of degree~6.  But a computation
reveals that this is not possible.  Indeed assume that
$x=\sum_{j=0}^6 x_j t^j$ with $x_j\in k$\/; since $\epsilon(P)$
is a linear polynomial we have $x_5=x_3=0$.  Let
\be
\eta = y^2+y = x^3+t^{13}+a_6 = \sum_{j \leq 18} \eta_j t^j.
\label{eta}
\ee
Then (cf.\ the Lemma in~\cite[Prop.3]{E1}) we must have
\be
\eta_{13} = \eta_9^2+\eta_{18}\0 = \eta_7^2+\eta_{14}\0
= \eta_5^2 + \eta_{10}\0 = 0.
\label{eta_cond}
\ee
Writing the $\eta_j$ in terms of the $x_j$ (this is easy because
\be
x^3 = x \cdot x^2 =
\left( \sum_{j=0}^6 x_j t^j \right)
\left( \sum_{j=0}^6 x_j^2 t^{2j} \right)
\label{x^3}
\ee
in characteristic~2) we thus obtain $x_1\0 x_6^2 = 1$,
$(x_1\0 x_4^2)^2 = x_6^3$, and
\be
x_6 (x_2 x_6 + x_4^2) = x_1^2 x_2^4 + x_2 (x_2 x_6 + x_4^2) = 0.
\label{x_cond}
\ee
But the first two equations imply that $x_1,x_4,x_6$ are nonzero;
then the vanishing of $x_6 (x_2 x_6 + x_4^2)$ means that
$x_2 x_6 = x_4^2$ and $x_2 \neq 0$, but then
\be
x_1^2 x_2^4 + x_2 (x_2 x_6 + x_4^2) = x_1^2 x_2^4
\label{x_contr}
\ee
cannot vanish, and we reach a contradiction.  Thus $L$ has
minimal norm at least~8, the smallest even integer $>6$, and
normalized center density
\be
\delta(L) = \Delta^{-1/2}(\Nmin/4)^{24/2} \geq 2^{-12} (8/4)^{12} = 1.
\label{delta_ineq}
\ee

\pagebreak

{\bf A lemma on linear descents.}
It is believed (see e.g.~\cite[p.13]{C-S}) that the maximal normalized
packing density for a 24-dimensional lattice is 1, and $\Leech$
is, up to scaling, the unique lattice attaining this maximum.  Thus
the estimate~(\ref{delta_ineq}) strongly suggests that our \MW\ lattice
is homothetic with~$\Leech$.  Since $\Leech$ is a lattice of
discriminant~1 all of whose points have even norm, this would
mean that every point $P\in L$ has $\Ht(P)$ divisible by~4.
Conversely, to identify $L$ with $2^{1/2}\Leech$ it is enough to
prove that $4|\Ht(P)$ for all $P\in L$, because it is known that
$\Leech$ is the unique unimodular lattice in dimension~24 with no
vectors of norm~2 \cite{CONWAY,NIEMEIER,VENKOV}.

Let $P\in L$, then, be a nonzero point, with coordinates $(x,y)$.
We have already seen by local methods that $\Ht(P)\teq\deg(x)$ is even.
Those methods do not suffice to determine $\Ht(P)\bmod 4$, but since
$\epsilon(P)$ is known to be a linear polynomial we can apply
the following global result:

{\bf Lemma.} {\em Let $k$\/ be a field of characteristic~2 and let
$K$\/ be the rational function field $K=k(t)$.  Fix a polynomial
$A_6(t)$ of odd degree~$d$, and let $E/K$\/ be the elliptic curve
with Weierstrass model
\be
y^2+y=x^3+A_6.
\label{Egen}
\ee
Let $P\in E(K)$ be a nonzero point with coordinates $(x,y)$, and
assume that $x(t)$ is of the form $\xi(t^2)+at$ for some rational
function~$\xi$ and nonzero $a \in k$\/ (equivalently, that $dx/dt=a$).
Then $x$ has a pole of order $(d-1)/2$ at $t=\infty$, and all its
other poles have degree~2.  Furthermore, let
\be
x = \sum_{j=-\infty}^{(d-1)/2} x_j t^j,
\label{x_Laurent}
\ee
be the Laurent expansion of~$x$ about $t\teq\infty$,
with $x_1=a$ and $x_j=0$ for odd $j\neq1$.  Then
\be
x_{-2}\0 = a^{-2}\ {\rm or}\ 0,
\label{x-2}
\ee
and
\be
\frac12 \left( \deg(x) - \frac{d-1}{2} \right)
\equiv a^2 x_{-2}\0 \bmod 2.
\label{lemma}
\ee
Thus $d$\/ and $x_{-2}\0$ determine $\Ht(P) \bmod 4$.
}

(Note that necessarily $d\equiv1\bmod4$ or $d=3$, else
$x=\xi(t^2)+at$ cannot have a pole of order $(d-1)/2$ at infinity.)

{\em Proof}\/: Without loss of generality we may assume that
$k$ is algebraically closed.  That $x$ has a pole at infinity
of degree $(d-1)/2$ we deduce as usual by differentiating~(\ref{Egen})
to obtain
\be
\frac{dy}{dt} = ax^2 + \frac{dA_6}{dt}
\label{diff}
\ee
and comparing valuations at $t\teq\infty$.  Now let $t_0\in k$
be a finite pole of~$x$ and~$y$, of order $2z,3z$ respectively.
Then $ax^2+dA_6/dt$ has a pole of order $4z$ at $t\teq t_0$,
while $dy/dt$ has a pole of order at most $3z+1$.  Thus $z=1$
and $t_0$ is a double pole of~$x$.  Moreover, if the leading
terms of $x,y$ at this pole are $\alpha^2 (t-t_0)^{-2}$ and
$\alpha^3 (t-t_0)^{-3}$ then by equating leading terms in~(\ref{diff})
we find $\alpha^3 = a \alpha^4$, whence $\alpha = a^{-1}$ (since
by assumption $\alpha\neq0$).

We can now prove (\ref{x-2},\ref{lemma}) by decomposing $x$ into partial
fractions.  Let $S\subset k$\/ be the set of finite poles of~$x$;
this is a finite set whose cardinality~$|S|$ equals the left-hand side
of~(\ref{lemma}).  Then
\be
x = \sum_{j\geq0} x_j t^j  +  a^{-2} \sum_{t_0\in S} \frac1{(t-t_0)^2}.
\label{x_pf}
\ee
Thus for each $j>0$ the $t^{-2j}$ coefficient in~(\ref{x_Laurent})
is given by
\be
x_{-2j} = a^{-2} \sum_{t_0 \in S} t_0^{2j-2}.
\label{x-2j}
\ee
In particular $x_{-2}\0 = a^{-2} |S|$, which proves both~(\ref{x-2})
and~(\ref{lemma}).~~$\Box$

In our case $A_6=t^{13}+a_6$ so $d=13$.  If $P\in\phi L$ then $\Ht(P)$
must be doubly even because $L$ is an even lattice and $\phi: L \ra L$
is a norm-doubling map.  Otherwise $P$\/ satisfies the hypotheses
of our Lemma.  Expand $x$ in the Laurent series~(\ref{x_Laurent})
and define $\eta_j$ again by~(\ref{eta}).  Then the
conditions~(\ref{eta_cond}) yield the same equations on
$x_1,x_2,x_4,x_6$, except that $\eta_{10}$ now has an additional
term $x_{-2}\0 x_6^2$.  Solving the first three equations
\be
x_1\0 x_6^2 = 1,\quad
(x_1\0 x_4^2)^2 = x_6^3,\quad
x_6 (x_2 x_6 + x_4^2) = 0
\label{x_cond'}
\ee
for $x_2,x_4,x_6$ in terms of $x_1\teq a$ yields
\be
(x_2,x_4,x_6) = (a^{-5/4},a^{-7/8},a^{-1/2}).
\label{x_solve}
\ee
The last equation $\eta_5^2=\eta_{10}\0$ then becomes
\be
0 = x_1^2 x_2^4 + x_2 (x_2 x_6 + x_4^2) + x_{-2}\0 x_6^2
= a^{-3} + 0 + a^{-1} x_{-2}\0 .
\label{x_finis}
\ee
Thus $x_{-2}\0 = a^{-2}$, and we conclude by~(\ref{lemma})
that $\deg(x) \equiv 0 \bmod 4$.  This completes the proof
of our Theorem.

{\bf Minimal vectors.}  The Leech lattice $\Leech$ has 196560
vectors of minimal norm~4.  Thus $L$ has as many points of
minimal height 8, each with $x$-coordinate
\be
x = a^{-1/2} t^6 + a^{-7/8} t^4 + a^{-5/4} t^2 + a t
  + x_0 + \frac{a^{-2}}{(t-t_0)^2}
\label{x_min}
\ee
for some $a\in k^*$ and $x_0,t_0\in k$.  Consider two such points
$P,Q$\/ with the same value of~$a$.  Then $P\pm Q \in \phi L$,
so either $P = \pm Q$\/ or $\Ht(P\pm Q) \geq 16$.  By the
parallelogram law
\be
\Ht(P+Q) + \Ht(P-Q) = 2\big(\Ht(P)+\Ht(Q)\big) = 2(8+8) = 32;
\label{parallelogram}
\ee
thus if $P \neq \pm Q$\/ then $\Ht(P+Q)=\Ht(P-Q)=16$ and
$P,Q$\/ are orthogonal in~$L$.  Now a set of pairwise orthogonal
nonzero vectors in~$L$ (or even in $L\otimes\R\cong\R^{24}$) can
have at most 24 elements; thus each of the $2^{12}-1=4095$
possible $a$ accounts for at most $48$ minimal points.
But $48\cdot 4095 = 196560$, the kissing number of~$L$!
Thus each $a$ must be represented by 24 pairs of minimal
vectors, forming an orthogonal frame.\footnote{All we really
used is that $\phi:L\ra L$ is a norm-doubling map; this argument
shows that any norm-doubling map of~$\Leech$ yields a partition
of its minimal vectors into 4095 orthogonal frames.}

To find for each $a\in k^*$ the 24 values of $(x_0,t_0)$ that make
(\ref{x_min}) the \hbox{$x$-coordinate} of a minimal point we solve
the two equations
\be
\eta_3^4 + \eta_6^2 + \eta_{12}\0 =
\eta_1^{16} + \eta_2^8 + \eta_{4}^4 + \eta_{8}^2 + \eta_{16}\0 = 0.
\label{more_conds}
\ee
Note that if $(x_0,t_0)$ works then so does $(x_0+\mu,t_0)$
for each cube root of unity~$\mu$, so (\ref{more_conds})
must give rise to equations in~$t_0$ and
\be
X_0 := x_0^4 + x_0\0.
\label{X0_def}
\ee
The equation $\eta_3^4 + \eta_6^2 + \eta_{12}\0 = 0$
yields (after multiplying by~$a$)
\be
X_0 = a^{13} + a^{13/4} + a^{-13/8} + a^{-5} t_0^8;
\label{X0}
\ee
substituting this into the second equation and taking fourth roots
we find that $t_0$ is one of the solutions of the sextic
\be
% a^{9/8} t_0^6 + a^{3/4} t_0^4 + a^{3/8} t_0^2 + a^{21/8} t_0\0
% = a^{273/8} + a^{39/4} + a^{39/8} + a^{117/32} + 1.
t_0^6 + a^{-3/8} t_0^4 + a^{-3/4} t_0^2 + a^{3/2} t_0\0
= a^{33} + a^{69/8} + a^{15/2} + a^{81/32} + a^{-9/8}.
\label{sextic}
\ee
(These solutions are necessarily distinct because the derivative
of~(\ref{sextic}) with respect to~$t_0$ is the nonzero constant
% $a^{21/8}$.) From each of these 6 solutions we obtain 4 possible~$x_0$
$a^{3/2}$.)  From each of these 6 solutions we obtain 4 possible~$x_0$
by solving (\ref{X0}) and~(\ref{X0_def}) for a total of 24 $(x_0,t_0)$
pairs; thus each of these must make~(\ref{x_min}) the $x$-coordinate
of a point of~$E(K)$ of minimal height~8.

In fact we could, and at first did, use these 196560 minimal points
to give an alternative, albeit unpleasant, proof that $L$ is
isometric with $2^{1/2}\Leech$, invoking the fact~(\cite[Ch.14]{C-S})
that the minimal vectors of~$\Leech$ comprise the unique solution
of the kissing number problem in dimension~24.  (Note that this also
independently confirms that $\Delta(L)=2^{24}$, assuming only that
$L$ has rank~24.)  To do this we must show that (\ref{x_min})
yields a pair of solutions of~(\ref{Edef}) in~$K$\/
for each choice of $a\in k^*$ and $x_0,t_0$ satisfying
(\ref{X0_def},\ref{X0},\ref{sextic}), which so far
we have only done with the foreknowledge that 196560 such
solutions exist.  It might seem that we would first have to prove
that $x_0,t_0\in k$\/ and the constant coefficient of $x^3+t^{13}+a_6$
is of the form $y_0^2+y_0\0$ for some $y_0\0\in k$\/ (how to prove
this directly?), but in fact it suffices to construct 196560 minimal
solutions in $\bar k(t)$: since $L$ already has maximal rank 24,
Gal$(\bar k/k)$ acts trivially on $L\otimes\R$, and thus also on~$L$,
so any solution in $\bar k(t)$ is automatically defined over~$k$.

Since the coefficients~$\eta_j$ of the Laurent expansion
of $\eta=x^3+t^{13}+a_6$ satisfy (\ref{eta_cond},\ref{more_conds}),
$\eta$ is locally of the form $y^2+y=0$ at $t\teq\infty$.  This
leaves only the pole at $t\teq t_0$, which imposes two further
conditions on $\eta$, one of which is already satisfied by the
choice of leading term $a^{-2}/(t-t_0)^2$ of~$x$.  To state the
remaining condition, let $c_0$ be the constant term in the expansion
of~$x$ in powers of $t-t_0$, which is
\be
c_0= a^{-1/2} t_0^6 + a^{-7/8} t_0^4 + a^{-5/4} t_0^2 + a t_0\0 + x_0\0,
\label{c0_def}
\ee
and let $c_2$ be the $(t-t_0)^2$ coefficient
$a^{-1/2} t_0^4 + a^{-5/4}$.  Then
$\eta=y^2+y$ locally at~$t\teq t_0$ if and only if
$a^{-4}(c_0^4 + c_0\0) = a^{-8} c_2^2 $, that is, if and only if
\be
c_0^4 + c_0\0 = a^{-5} t_0^8 + a^{-13/2} = b^4 + b
\label{c0}
\ee
by~(\ref{X0}), where
\be
b = x_0 + a^{13/4} + a^{-13/8}.
\label{b_def}
\ee
Thus $(c+b)^4 = c+b$.  But by~(\ref{sextic}) we see that $c+b$
is simply
\be
a^{65/2} + a^{65/8} + a^{65/32},
\label{b+c}
\ee
so the condition $(c+b)^4=c+b$ becomes $a^{130}=a^{65/32}$ or
\be
a^{64} (a^{4096}-a)=0,
\label{miracle}
\ee
which (since $a\neq0$) is precisely the condition $a\in k^*$!
Thus each such $a$ indeed gives rise to $2\cdot4\cdot6=48$
minimal points in~$L$ and we are done.

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{\bf 5} (1973), 142--178.
\bibitem[7]{TATEBSD} Tate, J.: On the conjecture of Birch and 
Swinnerton-Dyer and a geometric analog.
{\em S\'em.\ Bourbaki} {\bf 306} (1965--66).
\bibitem[8]{VENKOV} Venkov, B.B.: Even Unimodular 24-Dimensional
Lattices (trans.\ G.A.\ Kendall), Chapter~18 of~\cite{C-S}.
\end{thebibliography}

\vspace*{5ex}
\begin{small}
\noindent 
Dept.\ of Mathematics\\
Harvard University\\
Cambridge, MA 02138 USA\\ \\
November, 1993
\end{small}


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