# Finite quadratic modules and lattices

Xiao-Jie Zhu

###### Abstract

We give a new proof of the fact that any finite quadratic module can be
decomposed into indecomposable ones. For any indecomposable finite quadratic
module, we construct a lattice, and a positive definite lattice, both of which
are of the least rank, whose discriminant module is the given one. The
resulting lattices are given by their Gram matrices explicitly.

††footnotetext: 2010 Mathematics Subject Classification. 11E12 (primary),
16D70 (secondary).††footnotetext: Key words and phrases. Finite quadratic
modules, lattices, discriminant modules.††footnotetext: This work is supported
by the Fundamental Research Funds for the Central Universities of China (Grant
No. 22120180508).††footnotetext: Address. School of Mathematical Sciences,
Tongji University, 1239 Siping Road, Shanghai, P.R. China††footnotetext:
E-mail<EMAIL_ADDRESS>
###### Contents

  1. 1 Introduction
  2. 2 Bilinear map modules and quadratic map modules
  3. 3 Lattices and finite quadratic modules
  4. 4 Discriminant modules of even lattices
  5. 5 Discriminant modules of positive definite even lattices
  6. 6 Some Remarks

##  1 Introduction

A finite quadratic module is a finite abelian group equipped with a
nondegenerate quadratic map assuming values in the quotient
$\mathbb{Q}/\mathbb{Z}$, and a lattice is a free $\mathbb{Z}$-module of finite
rank equipped with a nondegenerate symmetric bilinear form taking values in
the reals. The results of this paper concern these two concepts. Our first
result is a new proof of the structure theorem of finite quadratic modules,
which says any such module can be decomposed as an orthogonal direct sum of
three kinds of indecomposable modules. Our proof involves only some theory of
finite abelian groups, not using methods of $p$-adic fields and $p$-adic
lattices. To our best knowledge, the first proof of this theorem stated in
such form appears in [13], where the author used classical results from the
theory of quadratic forms over $p$-adic numbers. We must mention that, Wall
has actually proved this in [14, §6], but he did not state it explicitly. Our
second result is that, we find an even lattice of the least rank whose
discriminant module is any given indecomposable finite quadratic module. These
lattices are presented by their Gram matrices. In this direction, Wall is the
first person to prove any finite quadratic module is the discriminant module
of some even lattice. See [14]. Comparing to Wall’s result, our result has the
advantage of giving the lattice explicitly for any indecomposable module. Our
third result is analogous to the second result, which gives, for any
indecomposable module, a positive definite even lattice instead of a general
even lattice, still of the least rank. The motivation of such constructions is
the need of providing concrete examples of Jacobi forms of lattice index.

We now briefly review relevant results not mentioned above in the literature.
The importance of the concept and theory of finite quadratic modules, to our
best knowledge, lies in two facts. One fact is that the concept of finite
quadratic modules is a useful tool to study non-unimodular integral even
lattices. This was shown by Nikulin in [7]. More specifically, Nikulin gives a
necessary and sufficient condition for the existence of even lattices with
given inertia and discriminant module, which generalizes a theorem of John
Milnor dealing with unimodular lattices. See [7, Theorem 1.1.1] and [7,
Theorem 1.10.1]. The other fact is that, to any finite quadratic module, one
can associate a Weil representation of $\mathrm{Mp}_{2}(\mathbb{Z})$, the
metaplectic cover of the modular group. (For background materials on Weil
representations, see [5], or the original paper of Weil [15].) The Weil
representations of this kind are very interesting objects in Representation
Theory, since they include all representations of the modular group (or its
double cover) whose kernels are congruence subgroups. See [12] for a thorough
discussion and a proof. Also see [11]. They are also interesting in the area
of modular forms, since any modular form of level $N\in\mathbb{Z}_{\geq 1}$
with the multiplier system given by a Dirichlet character can be lifted to a
modular form for a Weil representation associated to some finite quadratic
module. One can find an explicit formula of this lift in [9, Theorem 5.7].
Also, this kind of Weil representations have important applications in the
theory of automorphic forms on orthogonal groups. The key concept of this
application is that of a Bocherds’ singular theta lift, which maps a modular
form for some Weil representation to an automorphic form on some orthogonal
group. See [1] for this theory. This concludes the discussion of the
importance of the concept and theory of finite quadratic modules. Besides the
above discussion, we suggest another good refenrence on finite quadratic
modules and relevant topics, the book [3] (more exactly, §14.5).

We list the notational convention. As usual, the symbol $\mathbb{Z}$,
$\mathbb{Q}$, $\mathbb{R}$, and $\mathbb{C}$ denote the ring of integers, of
rationals, of reals, and of complex numbers respectively. We use
$\mathbb{Z}_{\geq n}$ to denote the set of integers greater than or equal to
$n$. For two integers $a$ and $b$, the notation $(a,b)$ denotes the greatest
common divisor of these two numbers, and $\genfrac{(}{)}{}{}{a}{b}$ denotes
the Kronecker’s extension of the Legendre symbol. For a definition of
Kronecker symbol $\genfrac{(}{)}{}{}{a}{b}$, see [3, §3.4]. For a ring $R$, we
use $R^{\times}$ to denote its group of units, and use $\mathrm{GL}_{n}(R)$ to
denote the group of invertible $n\times n$ matrices with entries in $R$, and
$\mathrm{SL}_{n}(R)$ the subgroup of matrices whose determinant equals 1 in
$\mathrm{GL}_{n}(R)$. For an $n\times n$ matrix $A$, the symbol
$A^{\mathrm{T}}$ denotes its transpose, and $A^{-1}$ its inverse. The
indentity matrix of size $n$ would be denoted by $\mathrm{I}_{n}$. The symbol
$\mathop{\mathrm{diag}}\left(a_{1},a_{2},\dots,a_{n}\right)$ denotes the
$n\times n$ diagonal matrix whose entries on the main diagonal are $a_{1}$,
$a_{2}$, …, and $a_{n}$. The symbol $\delta_{ij}$ denotes Kronecker’s
$\delta$, i.e., the function assumes the value $1$ if $i=j$ and the value $0$
if $i\neq j$. For a set $S$, by $|S|$ we mean its cardinality. For a finite
abelian group $G$, by $\widehat{G}$ we mean its dual group, i.e., the group of
all complex linear characters on $G$ with the usual multiplication of
functions. If $\sigma$ is a homomorphism of two algebraic structures with an
underlying abelian group structure (written additively), then $\ker(\sigma)$
denotes the inverse image of the zero element. Other notations can be found
where we give definitions.

The structure of this paper is as follows. In Section 2, we review some basic
facts concerning modules, bilinear maps, quadratic maps, orthogonal sums,
matrices of bilinear forms, how to take quotients, and the concept of
nondegeneracy. This is necessary for our purpose: we shall use these basic
facts in our proof of the structure theorem of finite quadratic modules and
also in our construction of a lattice whose discriminant module is any given
finite quadratic module. In Section 3, after a review of basic facts on finite
quadratic modules and lattices, a proof of the structure theorem of finite
quadratic modules is given. The proof is separated into five parts: Lemma
3.15, Lemma 3.16, Lemma 3.17, Lemma 3.18, and Theorem 3.19. We make some
comments on this proof. Lemma 3.16 deals with the primary decomposition of a
finite quadratic module (which is unique). The proof of this lemma is
according to [2, Proposition 1.6], where Boylan considers finite quadratic
modules over number fields. Lemma 3.17 and Lemma 3.18 deal with decompositions
of $p$-primary modules. The former assumes $p>2$, and the latter $p=2$. The
proofs of these two lemmas are original. There is a technically difficult
point in the proof of Lemma 3.18, for which we state a new lemma, Lemma 3.15,
to solve this difficulty. Finally, we combine these lemmas, to obtain Theorem
3.19. In Section 4, we turn to our second result—constructions of
nondegenerate even lattices with given indecomposable finite quadratic
modules. The resulting lattices are given by their Gram matrices in Theorem
4.2, Theorem 4.5, and Theorem 4.7. Section 5 has a similar structure,
comparing to Section 4, in which we construct positive definite even lattices
with given indecomposable finite quadratic modules as their discriminant
modules. The resulting lattices are given by their Gram matrices in Theorem
5.6, Theorem 5.7, Theorem 5.8, Theorem 5.9, and Theorem 5.10.

##  2 Bilinear map modules and quadratic map modules

######  Convention.

In this section, we fix a commutative ring $R\neq\\{0\\}$ with an identity
$1=1_{R}$.

The constructions presented here are common to both finite quadratic modules
and $\mathbb{Z}$-lattices.

Let $M$, $M_{1}$, $M_{2}$ and $N$ be $R$-modules. We use
$\mathrm{Hom}_{R}(M;N)$ to denote the set of linear maps from $M$ to $N$, and
$\mathrm{Hom}_{R}(M_{1},M_{2};N)$ the set of bilinear maps from $M_{1}\times
M_{2}$ to $N$. They both become $R$-modules under ordinary addition of
functions and scalar multiplication of functions by elements of $R$. One can
verify straightforwardly that the map

$\displaystyle\mathrm{Hom}_{R}(M_{1},M_{2};N)$
$\displaystyle\longrightarrow\mathrm{Hom}_{R}(M_{1};\mathrm{Hom}_{R}(M_{2};N))$
$\displaystyle B$ $\displaystyle\longmapsto(m_{1}\mapsto(m_{2}\mapsto
B(m_{1},m_{2}))$ (2.1)

is an isomorphism of $R$-modules. Similarly the map

$\displaystyle\mathrm{Hom}_{R}(M_{1},M_{2};N)$
$\displaystyle\longrightarrow\mathrm{Hom}_{R}(M_{2};\mathrm{Hom}_{R}(M_{1};N))$
$\displaystyle B$ $\displaystyle\longmapsto(m_{2}\mapsto(m_{1}\mapsto
B(m_{1},m_{2}))$ (2.2)

is also an isomorphism. Some authors say that
$B\in\mathrm{Hom}_{R}(M_{1},M_{2};N)$ is nondegenerate (or nonsingular) if
both the values of the map (2) and (2) at $B$ are $R$-linear isomorphisms from
$M_{1}$ to $\mathrm{Hom}_{R}(M_{2};N)$ and from $M_{2}$ to
$\mathrm{Hom}_{R}(M_{1};N)$ respectively. In some situations the term
“nondegenerate” refers to a different meaning; it means that both the values
of the map (2) and (2) at $B$ are monomorphisms. To distinguish the two
meanings we shall call $B$ nondegenerate (strongly nondegenerate respectively)
if the values at $B$ are monomorphims (isomorphisms respectively) in this
paper.

We are particularly interested in the case $M_{1}=M_{2}=M$. In this case we
write $\underline{M}=(M,B)$ and call it a bilinear map module. If $N=R$ then
$B$ is called a bilinear form and $\underline{M}=(M,B)$ a bilinear form
module. We say that $\underline{M}$ is nondegenerate (strongly nondegenerate
respectively) if B is nondegenerate (strongly nondegenerate respectively).

Among all the bilinear maps there are three important classes: the symmetric
ones, the skew-symmetric (or antisymmetric) ones, and the alternating (or
symplectic) ones. Assume that $M_{1}=M_{2}=M$. We say $B$ is symmetric (or
$\underline{M}$ is a symmetric bilinear map module) if $B(x,y)=B(y,x)$ for any
$x,y\in M$. We say that $B$ is skew-symmetric if $B(x,y)=-B(y,x)$ for any
$x,y\in M$ and that $B$ is alternating if $B(x,x)=0$ for any $x\in M$. We
concern mainly the case that $B$ is symmetric.

There is an associated orthogonality relation on any bilinear map module
$\underline{M}=(M,B)$. For $m_{1},m_{2}\in M$, we say that $m_{1}$ is
orthogonal to $m_{2}$ (or $m_{1}\perp m_{2}$ symbolically) if
$B(m_{1},m_{2})=0$. One can verify that if $B$ is symmetric, skew-symmetric,
or alternating, then the orthogonality relation is a symmetric relation, that
is, $m_{1}\perp m_{2}$ implies $m_{2}\perp m_{1}$. When $R$ is an integral
domain and $N=R$ one can also prove the converse: if orthogonality is a
symmetric relation, then $B$ is symmetric or alternating, but this requires
some effort. A proof can be found in [8, p. 266] for $R$ being a field.
However, that proof is also valid for any integral domain $R$.

From now on, we always assume that $B$ is symmetric, skew-symmetric, or
alternating, so that orthogonality is a symmetric relation. For a subset $S$
of $M$, denote by $S^{\perp}$ the set of all elements in $M$ which is
orthogonal to elements in $S$. The subset $S^{\perp}$ must be a submodule.
Here is a useful and basic property concerning orthogonality, which we omit
the direct proof.

######  Lemma 2.1.

Let $M$, $N$ be $R$-modules and $M_{0}$ a submodule of $M$. Let
$B\in\mathrm{Hom}_{R}(M,M;N)$ be symmetric, skew-symmetric, or alternating.
Denote by $\phi\colon M\rightarrow\mathrm{Hom}_{R}(M_{0};N)$ the composite of
the value of the map (2) at $B$ and the canonical map
$\mathrm{Hom}_{R}(M;N)\rightarrow\mathrm{Hom}_{R}(M_{0};N)$ that maps $f$ to
the restriction of $f$ to $M_{0}$. Then $M_{0}^{\perp}=\ker(\phi)$.
Consequently, if $\phi$ is surjection, then we have
$M/M_{0}^{\perp}\cong\mathrm{Hom}_{R}(M_{0};N)$.

Note that Lemma 2.1 remains true if we substitute the map (2) for (2), since
$B$ is symmetric, skew-symmetric, or alternating. Also note that we need
$\phi$ to be surjection for the isomorphism relation holding. Hence, we give a
useful condition for $\phi$ being surjection as follows:

######  Lemma 2.2.

Use notations and conditions of Lemma 2.1. Let $M_{1}$ be a submodule of $M$
containing $M_{0}$. If the restriction of $B$ to $M_{1}\times M_{1}$ is
strongly nondegenerate, and each map in $\mathrm{Hom}_{R}(M_{0};N)$ extends to
a map in $\mathrm{Hom}_{R}(M_{1};N)$, then $\phi$ is a surjection.

The proof is also direct, so we omit it. To use this lemma in practice, the
key point is to choose a suitable $M_{1}$ satisfying the two conditions. One
way of doing this in the case $N=R$ is to use Lemma 2.5.

The following lemma is well-known and useful, so we give a detailed proof.
(See, for instance, Milnor and Husemoller [6] and Wall [14].)

######  Lemma 2.3.

Use notations and conditions of Lemma 2.1. Suppose that the restriction of $B$
to $M_{0}\times M_{0}$ is strongly nondegenerate. Then we have

  1. 1.

$M=M_{0}\oplus M_{0}^{\perp}$.

  2. 2.

If $B$ is also strongly nondegenerate, then so is the restriction of $B$ to
$M_{0}^{\perp}\times M_{0}^{\perp}$.

###### Proof.

We prove the first part. Need to prove that $M_{0}\cap M_{0}^{\perp}=\\{0\\}$
and $M=M_{0}+M_{0}^{\perp}$. To prove the former assertion, let $x\in
M_{0}\cap M_{0}^{\perp}$. Then for any $y\in M_{0}$ we have $B(x,y)=0$. It
follows that $x=0$ since $x\in M_{0}$ and the restriction of $B$ to
$M_{0}\times M_{0}$ is strongly nondegenerate. Next we prove the latter
assertion. Let $x\in M$ be arbitary. Consider the map $y\mapsto B(x,y)$ on
$M_{0}$. It is a linear map on $M_{0}$, as easily verified. Hence by strong
nondegeneracy of the restriction of $B$ to $M_{0}\times M_{0}$, there exists a
unique $x_{0}\in M_{0}$ such that $B(x,y)=B(x_{0},y)$ for $y\in M_{0}$. It
follows that $x-x_{0}\in M_{0}^{\perp}$, and $x=x_{0}+(x-x_{0})\in
M_{0}+M_{0}^{\perp}$. This concludes the proof of the first part.

We now prove the second part. It suffices to prove that the map $\pi\colon
M_{0}^{\perp}\rightarrow\mathrm{Hom}_{R}(M_{0}^{\perp};N)$, $x\mapsto(y\mapsto
B(x,y))$ is an isomorphism of $R$-modules. By the first part of this lemma,
any map in $\mathrm{Hom}_{R}(M_{0}^{\perp};N)$ extends to a map in
$\mathrm{Hom}_{R}(M;N)$. Since $B$ is strongly nondegenerate, for any
$f\in\mathrm{Hom}_{R}(M_{0}^{\perp};N)$, there exists an $x\in M$ such that
$B(x,y)=f(y)$ for $y\in M_{0}^{\perp}$. We write $x=x_{1}+x_{2}$ with
$x_{1}\in M_{0}$ and $x_{2}\in M_{0}^{\perp}$ by the first part of this lemma.
One then verifies immediately thay $\pi(x_{2})=f$, hence $\pi$ is surjective.
Since $\pi$ is obviously $R$-linear, it remains to prove that $\pi$ is
injective, i.e., that $\ker(\pi)=0$. Let $x\in\ker(\pi)$, that is, $B(x,y)=0$
for any $y\in M_{0}^{\perp}$. It follows that $x\perp M$. Consequently $x=0$
by strong nondegeneracy of $B$. This concludes the proof of the second part. ∎

We now turn to the subject of quadratic maps. Let $M,N$ be $R$-modules and let
$Q\colon M\rightarrow N$. We call $Q$ a quadratic map if the following
conditions hold:

  1. 1.

$Q(rx)=r^{2}Q(x)$ for $r\in R$, $x\in M$.

  2. 2.

The associated map $B_{Q}(x,y):=Q(x+y)-Q(x)-Q(y)$ is bilinear.

Define $\underline{M}=(M,Q)$ and call it a quadratic map module or quadratic
module (over the ring $R$). The bilinear map module $(M,B_{Q})$ is called the
associated bilinear map module of $\underline{M}$. Note that the bilinear map
$B_{Q}$ associated with $Q$ is necessarily symmetric. In fact, when the module
$N$ satisfies that for any $v\in N$, there exists a unique $w\in N$ such that
$v=w+w$ (denote w by $2^{-1}v$ even if 2 is not a unit in $R$) there is a one-
to-one correspondence between quadratic maps and symmetric bilinear maps. The
correspondence is given by the following two maps: the map $Q\mapsto B_{Q}$
from quadratic maps onto symmetric bilinear maps, and the map
$B\mapsto(x\mapsto 2^{-1}B(x,x))$ from symmetric bilinear maps onto quadratic
maps. They are inverse to each other, as straightforwardly verified. Thus,
under this circumstance, the theory of symmetric bilinear maps and quadratic
maps are indistinguishable. However, when $N$ does not satisfy that condition,
the two theories are different. This happens, for instance, when $R$ is a
filed of characteristic 2 and $N\neq\\{0\\}$, or when $N$ has torsion elements
of order $2$. The relation of $Q$ and $B_{Q}$ can be easily generalized to the
case of more than two summands by induction:

$Q\left(\sum_{1\leq i\leq n}x_{i}\right)=\sum_{1\leq i\leq
n}Q(x_{i})+\sum_{1\leq i<j\leq n}B_{Q}(x_{i},x_{j}).$ (2.3)

We can form orthogonal sums, quotients, and tensor products of bilinear map
modules or quadratic modules. The construction of tensor products need $N$ to
be an associative algebra over $R$ and plays an important role in the theory
of bilinear form module: It is used to define the multiplication on the Witt
ring of $R$. We shall not need this construction. For details, see Milnor and
Husemoller [6, p. 14]. Nevertheless, the constructions of orthogonal sums and
quotients are important in subsequent sections, so we review these
constructions here.

Let $\underline{M}_{i}=(M_{i},B_{i}\colon M_{i}\times M_{i}\rightarrow N)$,
$i\in I$ be a family of bilinear map modules. Recall that we always assume
that $B$ is symmetric, skew-symmetric, or alternating. Define its orthogonal
(externalaaaSimilarly one can consider orthogonal internal direct sum which is
isometrically isomorphic to orthogonal external direct sum. By abuse of
language, we do not distinguish these two concepts.) direct sum as
$\bigoplus_{i\in I}M_{i}$ equipped with the bilinear map defined as follows:

$B\left((x_{i})_{i\in I},(y_{i})_{i\in I}\right)=\sum_{i\in
I}B_{i}(x_{i},y_{i}).$

Note that external direct sum $\bigoplus_{i\in I}M_{i}$ is the set of elements
in Cartesian product $\prod_{i\in I}M_{i}$ with a finite support; hence the
sum in the above formula makes sense. It can be deduced that if all $B_{i}$’s
are nondegenerate, then $B$ is nondegenerate. Moreover, assume that $I$ is
finite. Then the condition that all $B_{i}$’s are strongly nondegenerate
implies that $B$ is strongly nondegenerate. In addition, if all $B_{i}$ are
symmetric (skew-symmetric, alternating respectively), then $B$ is symmetric
(skew-symmetric, alternating respectively).

The construction of orthogonal sums also applies to quadratic modules. Let
$\underline{M}_{i}=(M_{i},Q_{i}\colon M_{i}\rightarrow N)$, $i\in I$ be a
family of quadratic modules. Then define its orthogonal sum as
$\bigoplus_{i\in I}M_{i}$ equipped with the quadratic map defined as follows:

$Q\left((x_{i})_{i\in I}\right)=\sum_{i\in I}Q_{i}(x_{i}).$

It follows that $B_{Q}\left((x_{i})_{i\in I},(y_{i})_{i\in
I}\right)=\sum_{i\in I}B_{Q,i}(x_{i},y_{i})$, where $B_{Q}$ is the associated
bilinear map of $Q$, and $B_{Q,i}$ the associated bilinear map of $Q_{i}$.
Consequenty, if all the summands are nondegenerate quadratic modules, then
their orthogonal sum is nondegenerate. When $I$ is finite this fact remains
true if we substitute strong nondegeneracy for nondegeneracy.

We also need the concept of quotients of bilinear map modules or quadratic
modules. Let $M$, $N$ be $R$-modules and let $B\in\mathrm{Hom}_{R}(M,M;N)$. We
write $\underline{M}=(M,B)$. Let $M_{0}$ and $N_{0}$ be submodules of $M$ and
$N$ respectively. Suppose that $B(x,y)\in N_{0}$ for any $(x,y)\in M\times
M_{0}$ or $M_{0}\times M$. Then we define the quotient of $\underline{M}$ with
respect to $(M_{0},N_{0})$ as the $R$-module $M/M_{0}$ equipped with the
bilinear map

$\displaystyle M/M_{0}\times M/M_{0}$ $\displaystyle\longrightarrow N/N_{0}$
$\displaystyle(x+M_{0},y+M_{0})$ $\displaystyle\longmapsto B(x,y)+N_{0}.$

One can immediately verify that this map is well-defined and is indeed
$R$-bilinear, hence gives $M/M_{0}$ a bilinear map module structure. In
addition, if $B$ is symmetric (skew-symmetric, or alternating respectively),
then the map on quotient module is alse symmetric (skew-symmetric, or
alternating respectively).

In the quadratic module case, $B$ is replaced by a quadratic map $Q\colon
M\rightarrow N$ and $\underline{M}=(M,Q)$. The condition for $B$ is replaced
by a condition for $Q$: $Q(x)\in N_{0}$ for any $x\in M_{0}$ and
$B_{Q}(x,y)\in N_{0}$ for any $(x,y)\in M\times M_{0}$. Then we can define the
quotient of $\underline{M}$ with respect to $(M_{0},N_{0})$ the $R$-module
$M/M_{0}$ equipped with the quadratic map

$\displaystyle M/M_{0}$ $\displaystyle\longrightarrow N/N_{0}$ $\displaystyle
x+M_{0}$ $\displaystyle\longmapsto Q(x)+N_{0}$

which gives $M/M_{0}$ a quadratic module structure.

Like all other mathematical structures, there are morphisms between bilinear
map modules which make them a category. A morphism from a bilinear map module
$\underline{M}_{1}=(M_{1},B_{1}\colon M_{1}\times M_{1}\rightarrow N)$ to
another $\underline{M}_{2}=(M_{2},B_{2}\colon M_{2}\times M_{2}\rightarrow N)$
is a $R$-linear map $\pi$ from $M_{1}$ into $M_{2}$ preserving bilinear maps,
i.e. satisfying $B_{1}(x,y)=B_{2}(\pi(x),\pi(y))$. If, in addition, $\pi$ is a
bijection, we call $\pi$ an isometric isomorphism. For quadratic modules, we
can define morphisms and isometric isomorphisms similarly: just replace the
condition $B_{1}(x,y)=B_{2}(\pi(x),\pi(y))$ with $Q_{1}(x)=Q_{2}(\pi(x))$.

We conclude this section by introducing methods of matrix theory into the
theory of bilinear map modules. To this end, we need $N$ to be an unitary
associative $R$-algebra, that is, $N$ is both an unitary ring and $R$-module
such that the ring multiplication are $R$-bilinear. Thus we have a category of
matrices whose objects are positive integers and morphisms from $n$ to $m$ are
matrices of $m$ rows and $n$ columns with entries in $N$. The composition of
this category is the usual matrix multiplication. A direct verification (take
care that we do not assume that $N$ is commutative) shows these data indeed
make up a category. In another words, the matrix multiplication over the ring
$N$ is associative and for each positive integer $n$ there is a unique
$n\times n$ identity matrix even if $N$ is non-commutative.

Recall that we let $M_{1}$ and $M_{2}$ be $R$-modules and
$B\in\mathrm{Hom}_{R}(M_{1},M_{2};N)$. Let
$\mathbf{x}=(x_{1},x_{2},\ldots,x_{s})$ and
$\mathbf{y}=(y_{1},y_{2},\ldots,y_{t})$ be finite sequences in $M_{1}$ and
$M_{2}$ respectively. Define the Gram matrix of $B$ with respect to the pair
$(\mathbf{x},\mathbf{y})$ (or simply, with respect to $\mathbf{x}$ if
$\mathbf{x}=\mathbf{y}$) to be the matrix$\left(B(x_{i},y_{j})\right)_{1\leq
i\leq s,\,1\leq j\leq t}$. Denote it by $G_{B}$ if the sequences $\mathbf{x}$
and $\mathbf{y}$ are clear. A fundamental method to calculate the value of a
bilinear map is as follows. Let $\alpha_{1},\alpha_{2},\ldots,\alpha_{s}$ and
$\beta_{1},\beta_{2},\ldots,\beta_{t}$ be in $R$ and denote the multiplication
identity of $N$ by $1_{N}$. Then we have

$B\left(\sum_{i=1}^{s}\alpha_{i}x_{i},\sum_{j=1}^{t}\beta_{j}y_{j}\right)=\left(\alpha_{1}1_{N},\ldots,\alpha_{s}1_{N}\right)\cdot
G_{B}\cdot\left(\beta_{1}1_{N},\ldots,\beta_{t}1_{N}\right)^{\mathrm{T}}.$
(2.4)

Recall that if $A$ is a matrix with $m$ rows and $n$ columns, then by a left
inverse of $A$ we mean a matrix $B$ with $n$ rows and $m$ columns such that
$B\cdot A$ is the $n\times n$ identity matrix. Similarly we can define the
notion of right inverses of $A$.

######  Lemma 2.4.

Assume that the map $R\rightarrow N,\,r\mapsto r\cdot 1_{N}$ is injective. If
$G_{B}$ has a right inverse, then $(x_{1},x_{2},\ldots,x_{s})$ is $R$-linearly
independent. Likewise, if $G_{B}$ has a left inverse, then
$(y_{1},y_{2},\ldots,y_{t})$ is $R$-linearly independent.

###### Proof.

We only prove for the case that $G_{B}$ has a right inverse. Let
$\alpha_{1},\alpha_{2},\ldots,\alpha_{s}$ be in $R$ such that
$\sum_{i=1}^{s}\alpha_{i}x_{i}=0$. It follows from (2.4) that

$\left(B\left(\sum_{i=1}^{s}\alpha_{i}x_{i},y_{1}\right),\dots,B\left(\sum_{i=1}^{s}\alpha_{i}x_{i},y_{t}\right)\right)=\left(\alpha_{1}1_{N},\ldots,\alpha_{s}1_{N}\right)\cdot
G_{B}\cdot\mathrm{I}_{t}$

where $\mathrm{I}_{t}$ is the $t\times t$ identity matrix. Multiplying a right
inverse of $G_{B}$ on both sides and using the fact that
$\sum_{i=1}^{s}\alpha_{i}x_{i}=0$ we obtain that
$\alpha_{1}1_{N}=\ldots=\alpha_{s}1_{N}=0_{N}$. Hence
$\alpha_{1}=\ldots=\alpha_{s}=0_{N}$ since the map $r\mapsto r\cdot 1_{N}$ is
injective. This proves that $(x_{1},x_{2},\ldots,x_{s})$ is linearly
independent. ∎

In the case $M_{1}=M_{2}=M$ and $N=R$ one can prove more.

######  Lemma 2.5.

Let $M_{1}=M_{2}=M$ and $N=R$. Let Let $\mathbf{x}=(x_{1},x_{2},\ldots,x_{s})$
be a sequence in $M$ and $G_{B}$ the Gram matrix with respect to $\mathbf{x}$.
Define $M^{\prime}=\sum_{1\leq i\leq s}Rx_{i}$. Then $G_{B}$ is invertible if
and only if the sum defining $M^{\prime}$ is a direct sum and the restriction
of $B$ to $M^{\prime}\times M^{\prime}$ is strongly nondegenerate.

##  3 Lattices and finite quadratic modules

In this section, we recall some basic definitions and properties on lattices
over the integers, and on finite quadratic modules. Then we give a complete
proof of the Jordan decomposition theorem of finite quadratic modules.

######  Definition 3.1.

Let $L$ be a $\mathbb{Z}$-module, and $B$ a symmetric bilinear form from
$L\times L$ into the real numbers $\mathbb{R}$. Set $\underline{L}=(L,B)$. We
say that $\underline{L}$ is a lattice (over the integers) if $L$ is free of
finite rank (i.e. has a finite $\mathbb{Z}$-basis).

The quadratic form $Q\colon L\rightarrow\mathbb{R},\,x\mapsto 2^{-1}B(x,x)$ is
called the quadratic form of the lattice $\underline{L}$. If the image of $B$
is contained in $\mathbb{Z}$, then we call $\underline{L}$ integral; moreover,
if the image of $Q$ is contained in $\mathbb{Z}$ we call $\underline{L}$ even
integral (or simply call it even). It follows that $\underline{L}$ is even if
and only if $B(x,x)\in 2\mathbb{Z}$ for any $x\in L$. We call an integral
lattice which is not even an odd integral lattice, or simply an odd lattice.
If for any nonzero $x\in L$ we have $Q(x)>0$, then we call $\underline{L}$ a
positive definite lattice.

######  Remark 3.2.

If we have an ambient real sysmetric bilinear form vector space
$\underline{V}=(V,B\colon V\times V\rightarrow\mathbb{R})$), then we say that
$L$ is a $\mathbb{Z}$-lattice in $\underline{V}$ in a more strict sense: It
means not only that $L$ is a free $Z$-module of rank $\dim_{\mathbb{R}}V$
contained in $V$, but also that $L$ has a $\mathbb{Z}$-basis which is also a
$\mathbb{R}$-basis of $V$ additionally.

Let $R$ be a commutative extension ring of $\mathbb{Z}$, we shall use the
tensor product $R\otimes_{\mathbb{Z}}L$ which is a free $R$-module. Note that
the bilinear form $B$ extends to a bilinear form on $R\otimes_{\mathbb{Z}}L$.
We are interested mainly in the case that $R$ is the field of rationals
$\mathbb{Q}$, of reals $\mathbb{R}$, or of complex numbers $\mathbb{C}$. We
choose realizations of tensor products such that $L\subseteq\mathbb{Q}\otimes
L\subseteq\mathbb{R}\otimes L\subseteq\mathbb{C}\otimes L$.

######  Convention.

We are only interested in nondegeneracy case. In the followings, we always
assume that the bilinear extension $B^{\prime}$ of $B$ to $\mathbb{R}\otimes
L$ is strongly nondegenerate. Note that $B^{\prime}$ is strongly nondegenerate
if and only if it is nondegenerate, since $\mathbb{R}$ is a field and
$\dim\mathbb{R}\otimes L<\infty$. The following example shows that saying that
$B$ is nondegenerate is not enough.

######  Example 3.3.

Let $L=\mathbb{Z}^{2}$ and $(e_{1},e_{2})$ be the standard basis. Define a
bilinear form $B$ as
$B(e_{1},e_{1})=1,\,B(e_{1},e_{2})=B(e_{2},e_{1})=\sqrt{2},\,B(e_{2},e_{2})=2$.
Then the map
$L\rightarrow\mathrm{Hom}_{\mathbb{Z}}(L;\mathbb{R}),\,v\mapsto(w\mapsto
B(v,w))$ is a $\mathbb{Z}$-monomorphism, thus $B$ is nondegenerate in the
sense in Section 2. However, since the Gram matrix with respect to this basis
is not invertible even over the field $\mathbb{R}$, we have that the bilinear
extension of $B$ to $\mathbb{R}\otimes L$ is not nondegenerate.

We define the dual lattice of $\underline{L}$ which is important in the theory
of lattices as follows.

######  Definition 3.4.

The dual lattice of $\underline{L}$ is the lattice contained in the
$\mathbb{R}$-space $\mathbb{R}\otimes L$ whose element $v$ satisfies
$B(v,w)\in\mathbb{Z}$ for any $w\in L$. The dual lattice is denoted by
$L^{\sharp}$ which inherits the bilinear form $B$ defined on
$\mathbb{R}\otimes L$. When $L$ is a lattice in an ambient vector space
$\underline{V}=(V,B\colon V\times V\rightarrow\mathbb{R})$ with a
nondegenerate sysmetric bilinear form $B$, we can also define the dual lattice
of $L$ in $\underline{V}$ in a similar manner.

Let $(e_{1},\dots,e_{n})$ be a $\mathbb{Z}$-basis of $\underline{L}$, namely,
$L=\bigoplus_{1\leq i\leq n}\mathbb{Z}e_{i}$. Let $G_{B}$ be the Gram matrix
with respect to this basis. Then $\mathop{\mathrm{det}}G_{B}\neq 0$. Thus, for
each $\mathbb{R}$-linear form $\mathbb{R}\otimes
L\rightarrow\mathbb{R},\,e_{j}\mapsto\delta_{ij}$, we can find an
$e_{i}^{\ast}\in\mathbb{R}\otimes L$ such that
$B(e_{i}^{\ast},e_{j})=\delta_{ij}$. One verifies immediately that
$L^{\sharp}=\bigoplus_{i}\mathbb{Z}e_{i}^{\ast}$. The basis
$(e_{1}^{\ast},\dots,e_{n}^{\ast})$ of $L^{\sharp}$ is called the dual basis
of $(e_{1},\dots,e_{n})$ with respect to $B$ in $\mathbb{R}\otimes L$.

######  Proposition 3.5.

Let $\underline{L}=(L,B)$ be a lattice of rank $n\in\mathbb{Z}_{\geq 1}$ and
let $(e_{1},\dots,e_{n})$ be a $\mathbb{Z}$-basis of $\underline{L}$. Let
$G_{B}$ be the Gram matrix of $B$ with respect to this basis. Then

  1. 1.

The dual basis can be expressed as
$(e_{1}^{\ast},\dots,e_{n}^{\ast})^{\mathrm{T}}$ =
$G_{B}^{-1}\cdot(e_{1},\dots,e_{n})^{\mathrm{T}}$.

  2. 2.

The Gram matrix of $\underline{L}^{\sharp}$ with respect to
$(e_{1}^{\ast},\dots,e_{n}^{\ast})$ is $G_{B}^{-1}$.

  3. 3.

The double dual lattice $(L^{\sharp})^{\sharp}$ as a lattice in
$\mathbb{R}\otimes L$ is equal to $L$.

  4. 4.

The lattice $\underline{L}$ is integral if and only if $L\subseteq
L^{\sharp}$, if and only if entries of $G_{B}$ are all integers.

###### Proof.

Write $(e_{1}^{\ast},\dots,e_{n}^{\ast})^{\mathrm{T}}$ =
$A\cdot(e_{1},\dots,e_{n})^{\mathrm{T}}$, where $A=(a_{ij})$ is an $n\times n$
matrix. It follows from the fact $B(e_{i}^{\ast},e_{j})=\delta_{ij}$ and (2.4)
that $\mathrm{I}_{n}=AG_{B}\mathrm{I}_{n}$. The first part of the theorem then
follows. The second part follows from the first part, (2.4) and the fact that
$G_{B}^{-1}$ is symmetric. To prove the third part, note that
$e_{i}^{\ast\ast}=e_{i}$. The last part of the theorem is obvious by
definition. ∎

Let $\underline{L}=(L,B)$ be an integral lattice. Then by Proposition 3.5 we
have $L\subseteq L^{\sharp}$. We now use the construction of quotients in
Section 2. Since $B(L,L^{\sharp})=B(L^{\sharp},L)\subseteq\mathbb{Z}$ and
$B(L^{\sharp},L^{\sharp})\subseteq\mathbb{Q}$, the map $L^{\sharp}/L\times
L^{\sharp}/L\rightarrow\mathbb{Q}/\mathbb{Z},\,(v+L,w+L)\mapsto
B(v,w)+\mathbb{Z}$ is well-defined and $\mathbb{Z}$-bilinear where
$\mathbb{Q}/\mathbb{Z}$ is regarded as a quotient of $\mathbb{Z}$-modules and
hence is itself a $\mathbb{Z}$-module. It can be verified that the bilinear
map module $(L^{\sharp}/L,(v+L,w+L)\mapsto B(v,w)+\mathbb{Z})$ is
nondegenerate by Proposition 3.5(3).

If, moreover, $\underline{L}=(L,B)$ is an even integral lattice, we can regard
$\underline{L}$ as a quadratic map module (In this situation, the theory of
quadratic maps and symmetric bilinear maps are indistinguishable, as explained
in Section 2.). Now we have $Q(L)\subseteq\mathbb{Z}$ and
$B(L,L^{\sharp})\subseteq\mathbb{Z}$. Thus the map
$L^{\sharp}/L\rightarrow\mathbb{Q}/\mathbb{Z},\,v+L\mapsto Q(v)+\mathbb{Z}$ is
well-defined and quadratic, as explained in Section 2. This defines a
quadratic module $(L^{\sharp}/L,\,v+L\mapsto Q(v)+\mathbb{Z})$ whose
corresponding bilinear map is $L^{\sharp}/L\times
L^{\sharp}/L\rightarrow\mathbb{Q}/\mathbb{Z},\,(v+L,w+L)\mapsto
B(v,w)+\mathbb{Z}$. It follows that the quadratic module just defined is
nondegenerate. We will give a proof that it is actually strongly nondegenerate
later! See Corollary 3.9.

No matter $L$ is even or old, the quotient $L^{\sharp}/L$ must be a finite
abelian group since elements in $L^{\sharp}/L$ are all torsion elements and
$L^{\sharp}/L$ is finitely generated. A well-known argument shows that
$|L^{\sharp}/L|=|\mathop{\mathrm{det}}G_{B}|$ where $G_{B}$ is the Gram matrix
with respect to any $\mathbb{Z}$-basis of $L$.

The finite abelian group (i.e. finite $\mathbb{Z}$-module) $L^{\sharp}/L$
together with the quadratic map $v+L\mapsto Q(v)+\mathbb{Z}$ where $L$ is even
is a primary example of the following concept:

######  Definition 3.6.

Let $M$ be a finite abellian group, and $Q:M\rightarrow\mathbb{Q}/\mathbb{Z}$
be a quadratic map. If the corresponding bilinear map $(x,y)\mapsto
Q(x+y)-Q(x)-Q(y)$ is nondegenerate in the sense defined in Section 2, then the
pair $\underline{M}=(M,Q)$ is called a finite quadratic module.

Note that comparing to the general definition of quadratic map modules in
Section 2, here we add an additional condition of nondegeneracy. In the
followings, when we speak of a finite quadratic module, we always mean a
nondegenerate one. As a convention, we also treat the zero module with the
zero map as a finite quadratic module (in fact, this convention coincides with
the concepts in Section 2).

######  Example 3.7.

Let $\underline{L}=(L,B)$ be an even lattice whose quadratic form is $Q$, then
$(L^{\sharp}/L,\,v+L\mapsto Q(v)+\mathbb{Z})$ is a finite quadratic module, as
discussed before Definition 3.6. This is called the discriminant module
(group) of $\underline{L}$ and is denoted by $D_{\underline{L}}$. Note that
the discriminant module of the zero lattice is the zero module.

The constructions discussed in Section 2, such as orthogonal sums, isometric
isomorphisms on general quadratic map modules or bilinear map modules carry
over to finite quadratic modules.

The following lemma is useful and basic in the theory of finite quadratic
modules.

######  Lemma 3.8.

Denote by $\mathbb{C}^{\times}$ the multiplicative group of nonzero complex
numbers (regarding as a $\mathbb{Z}$-module). Then the map
$\mathbb{Q}/\mathbb{Z}\rightarrow\mathbb{C}^{\times},\,a+\mathbb{Z}\mapsto\mathrm{e}^{2\uppi\mathrm{i}a}$
is a $\mathbb{Z}$-module monomorphism whose image is the torsion elements in
$\mathbb{C}^{\times}$. Moreover, let $M$ be a finite $\mathbb{Z}$-module and
denote by $\widehat{M}$ the dual group of $M$, i.e., the group of all linear
characters of $M$. Then $\mathrm{Hom}_{\mathbb{Z}}(M;\mathbb{Q}/\mathbb{Z})$
is isomorphic to $\widehat{M}$ as $\mathbb{Z}$-modules.

###### Proof.

It is obviously that $a+\mathbb{Z}\mapsto\mathrm{e}^{2\uppi\mathrm{i}a}$ is
monomorphism and the image is the torsion elements in $\mathbb{C}^{\times}$.
It remains to prove that $\mathrm{Hom}_{\mathbb{Z}}(M;\mathbb{Q}/\mathbb{Z})$
is isomorphic to $\widehat{M}$. Define a map $\pi$ from
$\mathrm{Hom}_{\mathbb{Z}}(M;\mathbb{Q}/\mathbb{Z})$ to $\widehat{M}$ as
follows: for $f\in\mathrm{Hom}_{\mathbb{Z}}(M;\mathbb{Q}/\mathbb{Z})$, let
$\pi(f)$ be the map whose value at $x\in M$ is
$\mathrm{e}^{2\uppi\mathrm{i}f(x)}$. One can verify that $\pi$ is well-defined
and gives the desired isomorphism. ∎

######  Corollary 3.9.

Let $\underline{M}=(M,Q)$ be a finite quadratic module (or more generally a
finite bilinear map module with the bilinear map nondegenerate). Then
$\underline{M}$ is strongly nondegenerate.

###### Proof.

By definition of strong nondegeneracy in Section 2, we need to prove that the
map
$M\rightarrow\mathrm{Hom}_{\mathbb{Z}}(M;\mathbb{Q}/\mathbb{Z}),\,x\mapsto(y\mapsto
B(x,y))$ is an isomorphism where $B$ is the bilinear map associated with $Q$
(i.e., $B(x,y)=Q(x+y)-Q(x)-Q(y)$). It is already a monomorphism since $B$ is
nondegenerate by assumption. It is indeed an isomorphism since
$|\mathrm{Hom}_{\mathbb{Z}}(M;\mathbb{Q}/\mathbb{Z})|=|\widehat{M}|=|M|<\infty$
by Lemma 3.8 and elementary theory of finite abelian groups. ∎

The next is another important property of finite quadratic modules.

######  Proposition 3.10.

Let $\underline{M}=(M,Q)$ be a finite quadratic module and $M_{0}$ be a
submodule. Then the map $\phi$ in Lemma 2.1 (with $R=\mathbb{Z}$ and
$N=\mathbb{Q}/\mathbb{Z}$) is actually a surjection. Hence
$M/M_{0}^{\perp}\cong\mathrm{Hom}_{\mathbb{Z}}(M_{0};\mathbb{Q}/\mathbb{Z})$
(the canonical map is an isomorphism). Moreover, we have
$|M_{0}|\cdot|M_{0}^{\perp}|=|M|$ and $(M_{0}^{\perp})^{\perp}=M_{0}$.

A proof can be found in [2, Proposition 1.7]. For the reader’s convenience, we
also give a proof.

###### Proof.

Let $B$ be the associated bilinear map (i.e., $B(x,y)=Q(x+y)-Q(x)-Q(y)$).
Firstly, we prove that the map $\phi$ is a surjection. By Lemma 2.2 it
suffices to prove that $B$ is strongly nondenegerate, and each map in
$\mathrm{Hom}_{\mathbb{Z}}(M_{0};\mathbb{Q}/\mathbb{Z})$ extends to a map in
$\mathrm{Hom}_{\mathbb{Z}}(M;\mathbb{Q}/\mathbb{Z})$. Corollary 3.9 implies
immediately the former assertion. To prove the latter, note that it is
equivalent to saying that each linear character of $M_{0}$ extends to a linear
character of $M$ by Lemma 3.8. This is well-known. See, for example, [10,
Chap. VI, Sect. 1, Proposition 1]. Secondly, it follows from Lemma 2.1 and the
fact just proved immediately that
$M/M_{0}^{\perp}\cong\mathrm{Hom}_{\mathbb{Z}}(M_{0};\mathbb{Q}/\mathbb{Z})$.
Finally, we prove that $|M_{0}|\cdot|M_{0}^{\perp}|=|M|$ and
$(M_{0}^{\perp})^{\perp}=M_{0}$. By the second part of this proposition and
Lemma 3.8 we have
$|M/M_{0}^{\perp}|=|\mathrm{Hom}_{\mathbb{Z}}(M_{0};\mathbb{Q}/\mathbb{Z})|=|M_{0}|$.
Hence $|M_{0}|\cdot|M_{0}^{\perp}|=|M|$. Substitute $M_{0}^{\perp}$ for
$M_{0}$; we obtain $|M_{0}^{\perp}|\cdot|(M_{0}^{\perp})^{\perp}|=|M|$, and
hence $|M_{0}|=|(M_{0}^{\perp})^{\perp}|$. It follows immediately that
$M_{0}=(M_{0}^{\perp})^{\perp}$ since $M_{0}\subseteq(M_{0}^{\perp})^{\perp}$
and $M_{0}$ is finite. ∎

We now turn to give concrete examples of finite quadratic modules. The modules
given in the following definition turn out to be cornerstones of all finite
quadratic modules—any finite quadratic module can be decomposed as an
orthogonal sum of these modules (up to isomorphism).

######  Definition 3.11.

Let $p$ be a prime, $r$ a positive integer, $a$ an integer not divisible by
$p$. Define the following finite quadratic modules:

$\displaystyle\underline{A}_{p^{r}}^{a}$
$\displaystyle:=\left(\mathbb{Z}/p^{r}\mathbb{Z},\,x+p^{r}\mathbb{Z}\mapsto\frac{a}{p^{r}}x^{2}+\mathbb{Z}\right)\,\text{for
}p>2,$ (3.5) $\displaystyle\underline{A}_{2^{r}}^{a}$
$\displaystyle:=\left(\mathbb{Z}/2^{r}\mathbb{Z},\,x+2^{r}\mathbb{Z}\mapsto\frac{a}{2^{r+1}}x^{2}+\mathbb{Z}\right)\,\text{for
}p=2,$ (3.6) $\displaystyle\underline{B}_{2^{r}}$
$\displaystyle:=\left(\mathbb{Z}/2^{r}\mathbb{Z}\times\mathbb{Z}/2^{r}\mathbb{Z},\,(x+2^{r}\mathbb{Z},y+2^{r}\mathbb{Z})\mapsto\frac{x^{2}+xy+y^{2}}{2^{r}}+\mathbb{Z}\right),$
(3.7) $\displaystyle\underline{C}_{2^{r}}$
$\displaystyle:=\left(\mathbb{Z}/2^{r}\mathbb{Z}\times\mathbb{Z}/2^{r}\mathbb{Z},\,(x+2^{r}\mathbb{Z},y+2^{r}\mathbb{Z})\mapsto\frac{xy}{2^{r}}+\mathbb{Z}\right).$
(3.8)

######  Remark 3.12.

One can verify that the quadratic maps of these modules are well-defined and
the associated bilinear maps are nondegenerate. Thus they are indeed finite
quadratic modules. Moreover, the modules $\underline{A}_{p^{r}}^{a}$ are
indecomposable as abelian groups since they are primary cyclic. On the other
hand, although the modules $\underline{B}_{2^{r}}$ and $\underline{C}_{2^{r}}$
are decomposable as abelian groups, they can not be written as orthogonal sums
of proper submodules.

######  Remark 3.13.

It can be verified that modules of different types described in Definition
3.11 are not isomorphic. Maybe the only case which needs explaining is that
$\underline{B}_{2^{r}}\not\cong\underline{C}_{2^{r}}$. This can be verified by
observing that there are exactly $4^{r}-4^{r-1}$ elements in
$\underline{B}_{2^{r}}$ such that the value of the quadratic map is an odd
number divided by $2^{r}$ mod $\mathbb{Z}$, while there are only $4^{r-1}$
elements in $\underline{C}_{2^{r}}$ satisfying this condition. Now consider
modules of the same type. For an odd prime $p$ and $a,b$ not divisible by $p$,
we can prove that $\underline{A}_{p^{r}}^{a}$ is isomorphic to
$\underline{A}_{p^{r}}^{b}$ if and only if the Legendre symbol
$\genfrac{(}{)}{}{}{a}{p}=\genfrac{(}{)}{}{}{b}{p}$. Thus, for a fixed prime
$p>2$ and fixed $r\in\mathbb{Z}_{\geq 1}$, the modules
$\underline{A}_{p^{r}}^{a}$ give exactly two isomorphism classes when $a$ runs
through the integers not divisible by $p$. Now for $a,b$ not divisible by $2$,
we have $\underline{A}_{2^{r}}^{a}$ is isomorphic to
$\underline{A}_{2^{r}}^{b}$ if and only if $ab^{-1}$ is a quadratic residue
modulo $2^{r+1}$. Thus, for $p=2$ and $r=1$, the modules
$\underline{A}_{2}^{a}$ give exactly two isomorphism classes, and for $p=2$
and $r>1$, the modules $\underline{A}_{2^{r}}^{a}$ give exactly four
isomorphism classes when $a$ runs through odd integers.

######  Example 3.14.

Besides modules in Definition 3.11, there are still other modules which can be
defined in a natural way, although they may be decomposable, or isomorphic to
a module given in Definition 3.11. We define

$\displaystyle\underline{A}_{m}$
$\displaystyle:=\left(\mathbb{Z}/m\mathbb{Z},\,x+m\mathbb{Z}\mapsto\frac{1}{m}x^{2}+\mathbb{Z}\right)\,\text{for
}2\nmid m,$ (3.9) $\displaystyle\underline{A}_{m}$
$\displaystyle:=\left(\mathbb{Z}/m\mathbb{Z},\,x+m\mathbb{Z}\mapsto\frac{1}{2m}x^{2}+\mathbb{Z}\right)\,\text{for
}2\mid m,$ (3.10) $\displaystyle\underline{D}_{2^{r}}^{a,b,c}$
$\displaystyle:=\left(\mathbb{Z}/2^{r}\mathbb{Z}\times\mathbb{Z}/2^{r}\mathbb{Z},\,(x+2^{r}\mathbb{Z},y+2^{r}\mathbb{Z})\mapsto\frac{ax^{2}+bxy+cy^{2}}{2^{r}}+\mathbb{Z}\right).$
(3.11)

Note that in above definitions, we require that $m\in\mathbb{Z}_{\geq 1}$,
$r\in\mathbb{Z}_{\geq 1}$ and $a,b,c\in\mathbb{Z}$ with $b$ odd. One can
verify immediately that these are indeed finite quadratic modules. Also note
that when $b$ is even in (3.11), then $\underline{D}_{2^{r}}^{a,b,c}$ is
degenerate.

######  Lemma 3.15.

Let $b$ be an odd integer and $r$ be a positive integer. If $a$ and $c$ are
both odd integers, then $\underline{D}_{2^{r}}^{a,b,c}$ defined in (3.11) is
isometrically isomorphic to $\underline{B}_{2^{r}}$; otherwise it is
isomorphic to $\underline{C}_{2^{r}}$.

###### Proof.

Note that isomorphisms of
$\mathbb{Z}/2^{r}\mathbb{Z}\times\mathbb{Z}/2^{r}\mathbb{Z}$ as an abelian
group are exactly isomorphisms of that as a
$\mathbb{Z}/2^{r}\mathbb{Z}$-module, and that
$\mathbb{Z}/2^{r}\mathbb{Z}\times\mathbb{Z}/2^{r}\mathbb{Z}$ is a free
$\mathbb{Z}/2^{r}\mathbb{Z}$-module. Hence we use invertible matrices with
entries in $\mathbb{Z}/2^{r}\mathbb{Z}$ to represent isomorphisms of
$\mathbb{Z}/2^{r}\mathbb{Z}\times\mathbb{Z}/2^{r}\mathbb{Z}$: the $2\times 2$
invertible matrix $A$ represents the isomorphism $(a,b)^{\mathrm{T}}\mapsto
A\cdot(a,b)^{\mathrm{T}}$, where $a,\,b\in\mathbb{Z}/2^{r}\mathbb{Z}$.

Suppose at least one of $a$ and $c$ are even. We shall give an isometric
isomorphism of $\underline{D}_{2^{r}}^{a,b,c}$ and $\underline{C}_{2^{r}}$. To
this end, we define a “valuation” $f$ on the ring $\mathbb{Z}/2^{r}\mathbb{Z}$
as follows. For nonzero $x\in\mathbb{Z}/2^{r}\mathbb{Z}$, if $x\in
2^{j}\mathbb{Z}/2^{r}\mathbb{Z}$ but $x\notin
2^{j+1}\mathbb{Z}/2^{r}\mathbb{Z}$, then $f(x)=j$; moreover, $f(0)=r$. We have
$f(xy)=\min\\{f(x)+f(y),\,r\\}$. We also need the map
$\sigma_{a}\colon(\mathbb{Z}/2^{r}\mathbb{Z})^{\times}\rightarrow(\mathbb{Z}/2^{r}\mathbb{Z})^{\times}$,
$t\mapsto at+t^{-1}$, where $(\mathbb{Z}/2^{r}\mathbb{Z})^{\times}$ denotes
the group of units, and $a$ satisfies $f(a)>0$. It is not hard to see that
$\sigma_{a}$ is well-defined. We assert that this map is an injection, hence a
bijection since $(\mathbb{Z}/2^{r}\mathbb{Z})^{\times}$ has finite
cardinality. To prove this, let $t_{1}$ and $t_{2}$ be in
$(\mathbb{Z}/2^{r}\mathbb{Z})^{\times}$ such that
$at_{1}+t_{1}^{-1}=at_{2}+t_{2}^{-1}$. Then
$at_{1}t_{2}(t_{1}-t_{2})=t_{1}-t_{2}$. Apply the function $f$ to both sides;
we obtain $t_{1}=t_{2}$. This concludes the proof of injectivity. Denoted by
$\tau_{a}$ the inverse function of $\sigma_{a}$. Now one can verify that the
isomophism given by the matrix ($a$, $b$, $1$ denote their image under the
canonical map from $\mathbb{Z}$ onto $\mathbb{Z}/2^{r}\mathbb{Z}$)

$\begin{pmatrix}a&(\tau_{ac}(b))^{-1}\\\ 1&c\cdot\tau_{ac}(b)\end{pmatrix}$

is indeed an isometric isomorphism from $\underline{D}_{2^{r}}^{a,b,c}$ onto
$\underline{C}_{2^{r}}$.

On the other hand, suppose both of $a$ and $c$ are odd. We use induction on
$r$ to show that $\underline{D}_{2^{r}}^{a,b,c}$ is isometrically isomorphic
to $\underline{B}_{2^{r}}$ with an isomorphism given by a matrix whose entries
on the main diagonal are non-units in the ring $\mathbb{Z}/2^{r}\mathbb{Z}$,
and entries off the main diagonal are units. For $r=1$, the desired
isomorphism is given by the matrix whose entries on the main diagonal are $0$,
and entries off the main diagonal are $1$. For the induction step, assume that
we have found an isometric isomorphism from $\underline{D}_{2^{r}}^{a,b,c}$
onto $\underline{B}_{2^{r}}$, which is given by the matrix

$\begin{pmatrix}u&v\\\ w&z\end{pmatrix}$

where $u$ and $z$ are non-units in $\mathbb{Z}/2^{r}\mathbb{Z}$, and $v$ and
$w$ are units in $\mathbb{Z}/2^{r}\mathbb{Z}$. By abuse of language, we still
use $u$ ($v,\,w,\,z$ respectively) to denote an element of the inverse image
of $u$ ($v,\,w,\,z$ respectively) under the map
$\mathbb{Z}/2^{r+1}\mathbb{Z}\rightarrow\mathbb{Z}/2^{r}\mathbb{Z}$,
$t+2^{r+1}\mathbb{Z}\mapsto t+2^{r}\mathbb{Z}$. (In the followings, we also
use these letters to represent elements of the inverse image of corresponding
values under the canonical map
$\mathbb{Z}\rightarrow\mathbb{Z}/2^{r+1}\mathbb{Z}$.) We want to find the
desired isomorphism from $\underline{D}_{2^{r+1}}^{a,b,c}$ onto
$\underline{B}_{2^{r+1}}$ among the following matrices:

$\begin{pmatrix}u+2^{r}j_{0}&v+2^{r}j_{1}\\\
w+2^{r}j_{2}&z+2^{r}j_{3}\end{pmatrix}$

with
$j_{0},\,j_{1},\,j_{2},\,j_{3}\in\\{0+2^{r+1}\mathbb{Z},\,1+2^{r+1}\mathbb{Z}\\}$.
For any choice of $j_{0},\,j_{1},\,j_{2},\,j_{3}$, the entries on the main
diagonal of the above matrix are non-units and those off the main diagonal are
units in $\mathbb{Z}/2^{r+1}\mathbb{Z}$. Particularly, this matrix is
invertible, hence gives an isomorphism from $\underline{D}_{2^{r+1}}^{a,b,c}$
onto $\underline{B}_{2^{r+1}}$ as abelian groups. So we shall choose
$j_{0},\,j_{1},\,j_{2},\,j_{3}$ such that the quadratic maps of
$\underline{D}_{2^{r+1}}^{a,b,c}$ and $\underline{B}_{2^{r+1}}$ are preserved
by the corresponding map, that is,

$\displaystyle(u+2^{r}j_{0})^{2}+(u+2^{r}j_{0})(w+2^{r}j_{2})+(w+2^{r}j_{2})^{2}\equiv
a\mod 2^{r+1},$ $\displaystyle
2(u+2^{r}j_{0})(v+2^{r}j_{1})+(u+2^{r}j_{0})(z+2^{r}j_{3})$
$\displaystyle+(v+2^{r}j_{1})(w+2^{r}j_{2})+2(w+2^{r}j_{2})(z+2^{r}j_{3})\equiv
b\mod 2^{r+1},$
$\displaystyle(v+2^{r}j_{1})^{2}+(v+2^{r}j_{1})(z+2^{r}j_{3})+(z+2^{r}j_{3})^{2}\equiv
c\mod 2^{r+1},$

which are equivalent to

$\displaystyle 2^{r}(j_{0}w+j_{2}u)$ $\displaystyle\equiv
a-(u^{2}+uw+w^{2})\mod 2^{r+1},$ (3.12) $\displaystyle
2^{r}(j_{0}z+j_{1}w+j_{2}v+j_{3}u)$ $\displaystyle\equiv b-(2uv+uz+vw+2wz)\mod
2^{r+1},$ (3.13) $\displaystyle 2^{r}(j_{1}z+j_{3}v)$ $\displaystyle\equiv
c-(v^{2}+vz+z^{2})\mod 2^{r+1}.$ (3.14)

Since the right-hand sides of the above three congruences are both congruent
to $0$ or $2^{r}$ by induction hypothesis, $u$, $z$ (considered as in
$\mathbb{Z}$)are divisible by $2$, and $v$, $w$ (considered as in
$\mathbb{Z}$)are not divisible by $2$, we can sovle for $j_{0}$ and $j_{3}$
from (3.12) and (3.14). Then we can solve for $j_{1}$ and $j_{2}$ from (3.13).
Namely, the desired $j_{0},\,j_{1},\,j_{2},\,j_{3}$ exist, which completes the
induction step, hence the whole proof. ∎

We now proceed to show that modules given in Definition 3.11 actually serve as
cornerstones of all finite quadratic modules. This result can be found in, for
example, [13, Proposition 2.11]. We give a new proof here.

First of all, we consider the primary decomposition.

######  Lemma 3.16.

Let $\underline{M}=(M,Q)$ be a finite quadratic module. For a prime $p$,
denote by $M(p)$ the set of all elements in $M$ whose order is a power of $p$.
Then, $M(p)$, with the quadratic form inherited from $\underline{M}$, is
itself a finite quadratic module. Moreover, we have an orthogonal
decomposition $M=\bigoplus_{p}M(p)$ where all but finitely many $M(p)$’s are
zero. This decomposition is unique: if $M=\bigoplus_{p}N_{p}$ where $N_{p}$
contains only elements whose order are powers of $p$, then $N_{p}=M(p)$.

###### Proof.

Denote $B(x,y)=Q(x+y)-Q(x)-Q(y)$. It is obvious that $M(p)$ is a
$\mathbb{Z}$-submodule of $M$, and by the primary decomposition theorem of
finite abelian groups we have $M=\bigoplus_{p}M(p)$. For primes $p\neq q$, and
$x\in M(p)$, $y\in M(q)$, suppose that $p^{\alpha}x=0$ and $q^{\beta}y=0$.
Since $p^{\alpha}$ and $q^{\beta}$ are co-prime, there exist
$a,b\in\mathbb{Z}$ such that $ap^{\alpha}+bq^{\beta}=1$. Hence
$B(x,y)=(ap^{\alpha}+bq^{\beta})B(x,y)=B(ap^{\alpha}x,y)+B(x,bq^{\beta}y)=0$.
That is, the direct sum $\bigoplus_{p}M(p)$ is orthogonal. The uniqueness also
follows from the primary decomposition theorem of finite abelian groups. It
remains to prove that all $M(p)$ are finite quadratic modules, i.e., the
bilinear map restricted on each $M(p)$ is nondegenerate. This follows from the
fact that $\underline{M}$ is nondegenerate and that the sum
$\bigoplus_{p}M(p)$ is orthogonal. ∎

Secondly, we deal with $p$-primary finite quadratic modules, i.e., finite
quadratic modules whose order is a positive power of $p$. The case $p=2$ and
$p>2$ are very different, so we treat them separately in the following two
lemmas.

######  Lemma 3.17.

Any $p$-primary finite quadratic module where $p$ is an odd prime is
isometrically isomorphic to an orthogonal direct sum of modules of the form
(3.5).

###### Proof.

Let $\underline{M}=(M,Q)$ be a $p$-primary finite quadratic module and $B$ the
associated bilinear map $B(x,y)=Q(x+y)-Q(x)-Q(y)$. By the theorem of cyclic
decompositions of $p$-primary finite abelian groups, without loss of
generality, we can assume that
$M=\mathbb{Z}/p^{r_{1}}\mathbb{Z}\times\dots\times\mathbb{Z}/p^{r_{t}}\mathbb{Z}$
with $r_{1}\geq r_{2}\geq\dots r_{t}\geq 1$. For $1\leq j\leq t$, set
$\mathbf{1}_{j}=(\dots,1+p^{r_{j}}\mathbb{Z},\dots)$, where the
$j^{\prime}$-th coordinate equals $0+p^{r_{j^{\prime}}}\mathbb{Z}$ if $1\leq
j^{\prime}\leq t$ and $j^{\prime}\neq j$. It can be verified that
$Q(\mathbf{1}_{j})=\frac{a_{j,j}}{p^{\beta_{j,j}}}+\mathbb{Z}$ and
$Q(\mathbf{1}_{i}+\mathbf{1}_{j})=\frac{a_{i,j}}{p^{\beta_{i,j}}}+\mathbb{Z}$
with $a_{i,j}$ integers not divisible by $p$ and $\beta_{i,j}$ nonnegative
integers not exceeding $\max(r_{i},r_{j})$.

We now proceed by induction on $t$. The base case $t=1$ is clear: since
$\underline{M}$ is nondegenerate we have $\beta_{1,1}=r_{1}$; hence
$\underline{M}$ is isomorphic to $\underline{A}_{p^{r_{1}}}^{a_{1,1}}$. Assume
that the assertion holds for the case $t=t_{0}$; we shall prove it for the
case $t=t_{0}+1$. Choose a pair $(i_{0},j_{0})$ such that the quantity
$\max(r_{i_{0}},r_{j_{0}})-\beta_{i_{0},j_{0}}$ is the least among all
quantities $\max(r_{i},r_{j})-\beta_{i,j}$. Then we have
$\max(r_{i_{0}},r_{j_{0}})-\beta_{i_{0},j_{0}}=0$ since the negation
$\max(r_{i_{0}},r_{j_{0}})-\beta_{i_{0},j_{0}}>0$ contradicts the assumption
that $\underline{M}$ is nondegenerate (for example, the element
$p^{r_{1}-1}\mathbf{1}_{1}$ is orthogonal to the whole $M$ in this case.).
Denote by $G$ the subgroup generated by $\mathbf{1}_{j_{0}}$ (in the case
$i_{0}=j_{0}$) or by $\mathbf{1}_{i_{0}}+\mathbf{1}_{j_{0}}$ (in the case
$i_{0}>j_{0}$), both of which are isometrically isomorphic to
$\underline{A}_{p^{r_{j_{0}}}}^{a_{i_{0},j_{0}}}$. Apply Lemma 2.3(1) and
Corollary 3.9; we have an orthogonal direct sum $M=G\oplus G^{\perp}$. The
$\mathbb{Z}$-module $G^{\perp}$ is again a finite quadratic module by Lemma
2.3(2). Moreover any decomposition of $G^{\perp}$ into cyclic groups has
$t_{0}$ summands by elementary theory of finite abelian groups. Hence we
conclude the proof by applying the induction hypothesis to $G^{\perp}$. ∎

######  Lemma 3.18.

Any $2$-primary finite quadratic module is isometrically isomorphic to an
orthogonal direct sum of modules of the form (3.6), (3.7), or (3.8).

###### Proof.

Similarly as in the proof of Lemma 3.17, let $\underline{M}=(M,Q)$ be a
$2$-primary finite quadratic module and $B$ the associated bilinear map.
Assume that
$M=\mathbb{Z}/2^{r_{1}}\mathbb{Z}\times\dots\times\mathbb{Z}/2^{r_{t}}\mathbb{Z}$
with $r_{1}\geq r_{2}\geq\dots r_{t}\geq 1$ without loss of generality. Again,
for $1\leq j\leq t$, set $\mathbf{1}_{j}=(\dots,1+2^{r_{j}}\mathbb{Z},\dots)$,
where the $j^{\prime}$-th coordinate equals $0+2^{r_{j^{\prime}}}\mathbb{Z}$
if $1\leq j^{\prime}\leq t$ and $j^{\prime}\neq j$. At this point, something
different from the case $p>2$ happens: we can verify that
$Q(\mathbf{1}_{j})=\frac{a_{j,j}}{2^{\beta_{j,j}}}+\mathbb{Z}$ and
$Q(\mathbf{1}_{i}+\mathbf{1}_{j})=\frac{a_{i,j}}{2^{\beta_{i,j}}}+\mathbb{Z}$
with $a_{i,j}$ integers not divisible by $2$ and $\beta_{i,j}$ nonnegative
integers not exceeding $\max(r_{i},r_{j})+1$ instead of the quantity
$\max(r_{i},r_{j})$ in the case $p>2$.

Proceed by strong induction on $t$. The base case $t=1$ is clear as in the
proof of Lemma 3.17: $\underline{M}$ must be isometrically isomorphic to
$\underline{A}_{2^{r_{1}}}^{a_{1,1}}$. Assume that the statement is true for
the case $t<t_{0}$; we aim to prove it for the case $t=t_{0}$. Choose a pair
$(i_{0},j_{0})$ such that the quantity
$\max(r_{i_{0}},r_{j_{0}})+1-\beta_{i_{0},j_{0}}$ is the least among all
quantities $\max(r_{i},r_{j})+1-\beta_{i,j}$. There are exactly two cases
$\max(r_{i_{0}},r_{j_{0}})+1-\beta_{i_{0},j_{0}}=0$ or $1$ to consider, since
other cases contradict the assumption that $\underline{M}$ is nondegenerate
(for example, the element $2^{r_{1}-1}\mathbf{1}_{1}$ is orthogonal to the
whole $M$ in these cases.). We will prove in the next paragraph that in both
cases we can find a submodule $G$ isometrically isomorphic to one of the
modules $\underline{A}_{2^{r}}^{a}$, or $\underline{B}_{2^{r}}$, or
$\underline{C}_{2^{r}}$. The rest of the proof is almost the same as the last
few sentences of the proof of Lemma 3.17.

For the case $\max(r_{i_{0}},r_{j_{0}})+1-\beta_{i_{0},j_{0}}=0$, we proceed
similarly as in the proof of Lemma 3.17: define $G$ as the subgroup generated
by $\mathbf{1}_{j_{0}}$ (in the case $i_{0}=j_{0}$) or by by
$\mathbf{1}_{i_{0}}+\mathbf{1}_{j_{0}}$ (in the case $i_{0}>j_{0}$), each of
which is isometrically isomorphic to
$\underline{A}_{2^{r_{j_{0}}}}^{a_{i_{0},j_{0}}}$. For another case
$\max(r_{i_{0}},r_{j_{0}})+1-\beta_{i_{0},j_{0}}=1$, we have
$\beta_{i,j}\leq\max(r_{i},r_{j})$ for any $i,j$. Thus
$B(2^{r_{1}-1}\mathbf{1}_{1},\mathbf{1}_{1})=0+\mathbb{Z}$. Howerver, since
$B$ is nondegenerate, there exists a $j_{1}\neq 1$ such that
$B(2^{r_{1}-1}\mathbf{1}_{1},\mathbf{1}_{j_{1}})\neq 0+\mathbb{Z}$. Define $G$
as the subgroup of $M$ generated by $\mathbf{1}_{1}$ and $\mathbf{1}_{j_{1}}$.
We assert that $G$ is isometrically isomorphic to $\underline{B}_{2^{r_{1}}}$
or $\underline{C}_{2^{r_{1}}}$. First of all, we show that $r_{j_{1}}=r_{1}$.
Since
$B(\mathbf{1}_{1},\mathbf{1}_{j_{1}})=Q(\mathbf{1}_{1}+\mathbf{1}_{j_{1}})-Q(\mathbf{1}_{1})-Q(\mathbf{1}_{j_{1}})$,
we have $B(\mathbf{1}_{1},\mathbf{1}_{j_{1}})=\frac{b}{2^{r}}+\mathbb{Z}$ for
some $r\leq r_{1}$ and some odd integer $b$. However,
$B(2^{r_{1}-1}\mathbf{1}_{1},\mathbf{1}_{j_{1}})\neq 0+\mathbb{Z}$ implies
that $r=r_{1}$. Moreover, we have
$Q(\mathbf{1}_{1})=\frac{a}{2^{r_{1}^{\prime}}}+\mathbb{Z}$ and
$Q(\mathbf{1}_{j_{1}})=\frac{c}{2^{r_{j_{1}}^{\prime}}}+\mathbb{Z}$ for some
$r_{1}^{\prime}\leq r_{1}$, $r_{j_{1}}^{\prime}\leq r_{j_{1}}\leq r_{1}$ and
some odd integers $a$ and $c$. A direct calculation shows that

$B(x_{1}\mathbf{1}_{1}+y_{1}\mathbf{1}_{j_{1}},x_{2}\mathbf{1}_{1}+y_{2}\mathbf{1}_{j_{1}})\\\
=\frac{2^{r_{1}-r_{1}^{\prime}+1}ax_{1}x_{2}+b(x_{1}y_{2}+y_{1}x_{2})+2^{r_{1}-r_{j_{1}}^{\prime}+1}cy_{1}y_{2}}{2^{r_{1}}}+\mathbb{Z}$
(3.15)

for integers $x_{1}$, $y_{1}$, $x_{2}$, $y_{2}$. We now seek for integers
$x_{1}$ and $y_{1}$ such that
$x_{1}\mathbf{1}_{1}+y_{1}\mathbf{1}_{j_{1}}\perp G$, i.e.,
$B(x_{1}\mathbf{1}_{1}+y_{1}\mathbf{1}_{j_{1}},\mathbf{1}_{1})=0$ and
$B(x_{1}\mathbf{1}_{1}+y_{1}\mathbf{1}_{j_{1}},\mathbf{1}_{j_{1}})=0$. Let
$x_{2}=1$ and $y_{2}=0$ in (3.15); then
$B(x_{1}\mathbf{1}_{1}+y_{1}\mathbf{1}_{j_{1}},\mathbf{1}_{1})=0$ if and only
if $2^{r_{1}}\mid 2^{r_{1}-r_{1}^{\prime}+1}ax_{1}+by_{1}$. Similarly, let
$x_{2}=0$ and $y_{2}=1$; then
$B(x_{1}\mathbf{1}_{1}+y_{1}\mathbf{1}_{j_{1}},\mathbf{1}_{j_{1}})=0$ if and
only if $2^{r_{1}}\mid bx_{1}+2^{r_{1}-r_{j_{1}}^{\prime}+1}cy_{1}$. Hence,
$x_{1}\mathbf{1}_{1}+y_{1}\mathbf{1}_{j_{1}}\perp G$ if and only if
$2^{r_{1}}\mid\left(u2^{r_{1}-r_{1}^{\prime}+1}a+vb\right)x_{1}+\left(ub+v2^{r_{1}-r_{j_{1}}^{\prime}+1}c\right)y_{1}$
for any $u,\,v\in\mathbb{Z}$. Since the determinant of the matrix

$\begin{pmatrix}2^{r_{1}-r_{1}^{\prime}+1}a&b\\\
b&2^{r_{1}-r_{j_{1}}^{\prime}+1}c\end{pmatrix}$

is odd, we obtain that $x_{1}\mathbf{1}_{1}+y_{1}\mathbf{1}_{j_{1}}\perp G$ if
and only if $x_{1}$ and $y_{1}$ are both multiples of $2^{r_{1}}$.
Consequently, we have $r_{j_{1}}=r_{1}$. Secondly, we can prove that $G$, with
the quadratic map inherited from $\underline{M}$, is a finite quadratic
module. This also follows from the equivalence just proved. Finally, we have
$G$ is isometrically isomorphic to $\underline{B}_{2^{r_{1}}}$ or
$\underline{C}_{2^{r_{1}}}$, which follows from Lemma 3.15 and the fact that
$G$ is isometrically isomorphic to
$\underline{D}_{2^{r_{1}}}^{2^{r_{1}-r_{1}^{\prime}}a,b,2^{r_{1}-r_{j_{1}}^{\prime}}c}$.
∎

Now we can state the structure theorem of finite quadratic modules:

######  Theorem 3.19.

Every finite quadratic module can be decomposed as an orthogonal direct sum of
indecomposable modules defined in Definition 3.11.

###### Proof.

This follows immediately from Lemma 3.16 , Lemma 3.17, and Lemma 3.18. ∎

Note that the decomposition of the zero module is an empty sum by convention.

##  4 Discriminant modules of even lattices

In the last section, we illustrated that we have a finite quadratic module
associated to any even integral lattice—the discriminant module of the given
lattice. We now prove the converse: every finte quadratic module is
isometrically isomorphic to the discriminant module of an even integral
lattice, which is important for our later paper on Jacobi forms of lattice
index. This result was first discovered and proved in [14, Theorem 6]. We give
a proof here by presenting a concrete lattice represented by its Gram matrix
for any of the modules in Definition 3.11. In this section, we give an even
lattice of the least rank, while in the next section, we give a positive
definite even lattice of the least rank, for each of these finite quadratic
modules.

We begin with modules $\underline{A}_{p^{r}}^{a}$ (See Definition 3.11.).

######  Lemma 4.1.

Let $p$ be an odd prime, $r$ be a positive integer, and $a\in\mathbb{Z}$ with
$(p,a)=1$. Let $\underline{L}=(L,B)$ be an (nondegenerate) even lattice of
rank $n\in\mathbb{Z}_{\geq 1}$. Then the modules $D_{\underline{L}}$ (See
Example 3.7 for definition) and $\underline{A}_{p^{r}}^{a}$ are isometrically
isomorphic if and only if there exists a basis of $L$ such that the Gram
matrix of $\underline{L}$ with respect to this basis is of the form
$\mathop{\mathrm{diag}}\left(p^{r},1,\dots,1\right)\cdot S$, where
$S=(s_{ij})_{1\leq i,j\leq n}$ is an invertible integral matrix (i.e., $S\in
GL_{n}(\mathbb{Z})$) satisfying

  1. 1.

$s_{11},\dots,s_{nn}\in 2\mathbb{Z}$,

  2. 2.

The Legendre symbol
$\genfrac{(}{)}{}{}{s_{11}/2}{p}=\genfrac{(}{)}{}{}{a}{p}$,

  3. 3.

$p^{r}s_{1j}=s_{j1}$ for $2\leq j\leq n$ and $s_{ij}=s_{ji}$ for $2\leq
i,j\leq n$.

###### Proof.

We denote by $Q$ the quadratic map of $D_{\underline{L}}$, and $B_{Q}$ the
associated bilinear map.

Suppose that $D_{\underline{L}}$ and $\underline{A}_{p^{r}}^{a}$ are
isomorphic. Then there is a basis $(e_{1},e_{2},\dots,e_{n})$ of $L$ such that
$(\frac{1}{p^{r}}e_{1},e_{2},\dots,e_{n})$ is a basis of $L^{\sharp}$. Let $G$
be the Gram matrix of $\underline{L}$ with respect to
$(e_{1},e_{2},\dots,e_{n})$. Since
$G^{-1}(e_{1},e_{2},\dots,e_{n})^{\mathrm{T}}$ is also a basis of $L^{\sharp}$
by Proposition 3.5, there exists an $S\in GL_{n}(\mathbb{Z})$ such that
$S\cdot
G^{-1}(e_{1},e_{2},\dots,e_{n})^{\mathrm{T}}=(\frac{1}{p^{r}}e_{1},e_{2},\dots,e_{n})^{\mathrm{T}}$,
i.e., $G=\mathop{\mathrm{diag}}(p^{r},1,\dots,1)\cdot S$. We claim that this
$S=(s_{ij})_{1\leq i,j\leq n}$ satisfies the required three conditions. First
of all, $s_{ii}$ are all even integers since $\underline{L}$ is an even
lattice. Secondly, we have
$\genfrac{(}{)}{}{}{s_{11}/2}{p}=\genfrac{(}{)}{}{}{a}{p}$ by Remark 3.13 and
the fact that $Q(x_{1}\frac{1}{p^{r}}e_{1}+x_{2}e_{2}+\dots
x_{n}e_{n}+L)=\frac{s_{11}/2}{p^{r}}x_{1}^{2}+\mathbb{Z}$ with
$x_{1},\dots,x_{n}\in\mathbb{Z}$. Finally, the third condition on $S$ holds,
because $G$ is symmetric.

Conversely, suppose that there is a basis $(e_{1},e_{2},\dots,e_{n})$ of $L$
such that the Gram matrix $G$ of $\underline{L}$ with respect to this basis
can be decomposed as
$G=\mathop{\mathrm{diag}}\left(p^{r},1,\dots,1\right)\cdot S$, where
$S=(s_{ij})_{1\leq i,j\leq n}\in GL_{n}(\mathbb{Z})$ satisfies the three
conditions listed in the lemma. Then
$G^{-1}(e_{1},e_{2},\dots,e_{n})^{\mathrm{T}}$, hence $S\cdot
G^{-1}(e_{1},e_{2},\dots,e_{n})^{\mathrm{T}}$, is a basis of $L^{\sharp}$. It
follows that $Q(x_{1}\frac{1}{p^{r}}e_{1}+x_{2}e_{2}+\dots
x_{n}e_{n}+L)=\frac{s_{11}/2}{p^{r}}x_{1}^{2}+\mathbb{Z}$ for
$x_{1},\dots,x_{n}\in\mathbb{Z}$ using (2.3). So $D_{\underline{L}}$ is
isomorphic to $\underline{A}_{p^{r}}^{s_{11}/2}$, and hence isomorphic to
$\underline{A}_{p^{r}}^{a}$, since
$\genfrac{(}{)}{}{}{s_{11}/2}{p}=\genfrac{(}{)}{}{}{a}{p}$. ∎

######  Theorem 4.2.

Let $p$ be an odd prime, $r$ be a positive integer, and $a\in\mathbb{Z}$ with
$(p,a)=1$.

  1. 1.

If $\genfrac{(}{)}{}{}{a}{p}=1$, then the discriminant module of the even
nondegenerate lattice given by the Gram matrix

$\begin{pmatrix}2p^{r}&p^{r}\\\
p^{r}&\frac{1}{2}\left(p^{r}-\genfrac{(}{)}{}{}{-1}{p}^{r}\right)\end{pmatrix}$
(4.16)

is isometrically isomorphic to $\underline{A}_{p^{r}}^{a}$. Moreover, there is
no even nondegenerate lattice of a smaller rank with this property.

  2. 2.

If $\genfrac{(}{)}{}{}{a}{p}=-1$ and $r$ is even, we construct an even
nondegenerate lattice as follows: choose another odd prime $q\neq p$ such that
$\genfrac{(}{)}{}{}{q}{p}=-1$, and let $v$ be a solution of the system of
congruences

$p^{r}v^{2}\equiv 1\mod q,\qquad v^{2}\equiv 1\mod 4.$

Define the required lattice via the Gram matrix

$\begin{pmatrix}2qp^{r}&p^{r}v\\\
p^{r}v&\frac{p^{r}v^{2}-1}{2q}\end{pmatrix}.$ (4.17)

Then the discriminant module of this lattice is isometrically isomorphic to
$\underline{A}_{p^{r}}^{a}$. Moreover, there is no even nondegenerate lattice
of a smaller rank with this property.

  3. 3.

If $\genfrac{(}{)}{}{}{a}{p}=-1$, $r$ is odd, and $p\equiv 3\mod 4$, then the
discriminant module of the even nondegenerate lattice given by the Gram matrix

$\begin{pmatrix}-2p^{r}&p^{r}\\\
p^{r}&-\frac{1}{2}\left(p^{r}+1\right)\end{pmatrix}$ (4.18)

is isometrically isomorphic to $\underline{A}_{p^{r}}^{a}$. Moreover, there is
no even nondegenerate lattice of a smaller rank with this property.

  4. 4.

If $\genfrac{(}{)}{}{}{a}{p}=-1$, $r$ is odd, and $p\equiv 1\mod 4$, then
there exist no even nondegenerate lattices of rank $1$ or $2$ whose
discriminant module is isometrically isomorphic to
$\underline{A}_{p^{r}}^{a}$. Moreover, there is an even nondegenerate lattice
of rank $4$ whose discriminant module is isomorphic to
$\underline{A}_{p^{r}}^{a}$. To give such a lattice, we first choose an odd
prime $q$ such that $q\equiv 2\mod 3$ and $\genfrac{(}{)}{}{}{q}{p}=-1$. Then
we let $v$ be a solution of the system of congruences

$p^{r}v^{2}\equiv-\frac{4q+1}{3}\mod q,\qquad v^{2}\equiv-\frac{4q+1}{3}\mod
4,$

and set $x=\frac{1}{4q}(p^{r}v^{2}+\frac{4q+1}{3})$. The lattice is then given
by the Gram matrix

$\begin{pmatrix}2qp^{r}&p^{r}v&0&0\\\ p^{r}v&2x&1&0\\\ 0&1&2&1\\\
0&0&1&2\end{pmatrix}$ (4.19)

######  Remark 4.3.

In the fourth part we did not tell anything about whether there is a lattice
of rank $3$ satisfying the required condition. It turned out that such
lattices do not exist by Milgram’s formula (See, for instance, [6, Appendix
4]). We will get back to this matter later in Remark 5.3.

###### Proof of Theorem 4.2.

First of all, note that there is no even nendegenerate lattice of rank $1$
whose discriminant module is isomorphic to $\underline{A}_{p^{r}}^{a}$, since
such discriminant module must have an even cardinality. As a corollary, the
statements in Part 1 and Part 3 follows, using Lemma 4.1.

Secondly, we prove the statement in Part 2. Since $r$ is even, $v$ must exist,
and $\frac{p^{r}v^{2}-1}{2q}$ must be an even integer. Now the statement also
follows from Lemma 4.1.

Finally, we prove the statement in Part 4. There exists no even nondegenerate
lattice of rank $2$ whose discriminant module is $\underline{A}_{p^{r}}^{a}$.
This can be proved by contradicition. Assume that there exists such lattice
$\underline{L}$. By Lemma 4.1, the Gram matrix of $\underline{L}$ with respect
to some basis is $\mathop{\mathrm{diag}}(p^{r},1)\cdot S$, with S being

$\begin{pmatrix}2x&v\\\ p^{r}v&2y\end{pmatrix}$

for some $x,y,v\in\mathbb{Z}$ with $\genfrac{(}{)}{}{}{x}{p}=-1$ and
$4xy-p^{r}v^{2}=\pm 1$. By reduction mod $4$ in the latter formula, only the
minus symbol is possible. If there is an odd prime factor $q$ of $x$ such that
$\genfrac{(}{)}{}{}{q}{p}=-1$, then
$\genfrac{(}{)}{}{}{-1}{q}\cdot\genfrac{(}{)}{}{}{p^{r}}{q}\cdot\genfrac{(}{)}{}{}{v^{2}}{q}=\genfrac{(}{)}{}{}{-1}{q}$,
which contradicts the quadratic reciprocity law. Therefore, $2\mid x$, and
$\genfrac{(}{)}{}{}{2}{p}=-1$. It follows that $8\mid p^{r}v^{2}-1$ and
$p\equiv 5\mod 8$, which contradicts each other. It remains to prove that the
given lattice (4.19) actually produces the required module. The prime $q$ must
exist. Moreover, we have
$\genfrac{(}{)}{}{}{-3}{q}=\genfrac{(}{)}{}{}{-1}{q}\cdot\genfrac{(}{)}{}{}{3}{q}=(-1)^{\frac{q-1}{2}}\cdot\genfrac{(}{)}{}{}{q}{3}\cdot(-1)^{\frac{q-1}{2}}=-1$.
Let $m=-\frac{4q+1}{3}$, so
$\genfrac{(}{)}{}{}{-3}{q}\cdot\genfrac{(}{)}{}{}{m}{q}=\genfrac{(}{)}{}{}{4q+1}{q}$,
which implies that $\genfrac{(}{)}{}{}{m}{q}=-1$. It follows that $v$ exists
and $x$ is an integer. Now we can verify straightforwardly that the Gram
matrix (4.19) satisfies the condition in Lemma 4.1. ∎

Next we deal with the modules $\underline{A}_{2^{r}}^{a}$ (See Definition
3.11.).

######  Lemma 4.4.

Let $r$ be a positive integer, and $a$ be an odd integer. Let
$\underline{L}=(L,B)$ be an (nondegenerate) even lattice of rank
$n\in\mathbb{Z}_{\geq 1}$. Then the modules $D_{\underline{L}}$ (See Example
3.7 for definition) and $\underline{A}_{2^{r}}^{a}$ are isometrically
isomorphic if and only if there exists a basis of $L$ such that the Gram
matrix of $\underline{L}$ with respect to this basis is of the form
$\mathop{\mathrm{diag}}\left(2^{r},1,\dots,1\right)\cdot S$, where
$S=(s_{ij})_{1\leq i,j\leq n}$ is an invertible integral matrix (i.e., $S\in
GL_{n}(\mathbb{Z})$) satisfying

  1. 1.

$s_{22},\dots,s_{nn}\in 2\mathbb{Z}$,

  2. 2.

$s_{11}\notin 2\mathbb{Z}$ and $s_{11}a$ is a quadratic residue mod $2^{r+1}$,

  3. 3.

$2^{r}s_{1j}=s_{j1}$ for $2\leq j\leq n$ and $s_{ij}=s_{ji}$ for $2\leq
i,j\leq n$.

###### Proof.

The proof is almost the same as that of Lemma 4.1. Just replace $p$ by $2$. We
omit the details. ∎

To state the theorem for $\underline{A}_{2^{r}}^{a}$, we need first recall
some basic facts in elementary number theory. Denote by
$\left(\mathbb{Z}/2^{r+1}\mathbb{Z}\right)^{\times}$ the multiplicative group
of all units in the quotient ring $\mathbb{Z}/2^{r+1}\mathbb{Z}$, where $r\geq
1$. Then the map

$\displaystyle\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2^{r-1}\mathbb{Z}$
$\displaystyle\longrightarrow\left(\mathbb{Z}/2^{r+1}\mathbb{Z}\right)^{\times}$
$\displaystyle(n_{0}+2\mathbb{Z},n_{1}+2^{r-1}\mathbb{Z})$
$\displaystyle\longmapsto(-1)^{n_{0}}5^{n_{1}}+2^{r+1}\mathbb{Z}$

is a group isomorphism, where $\mathbb{Z}/2\mathbb{Z}$ and
$\mathbb{Z}/2^{r-1}\mathbb{Z}$ are regarded as the additive group of the
ambient rings. Moreover, let $U$ be the group of squares of elements in
$\left(\mathbb{Z}/2^{r+1}\mathbb{Z}\right)^{\times}$. Then by the above
isomorphism, one can verify that
$\left[\left(\mathbb{Z}/2^{r+1}\mathbb{Z}\right)^{\times}\mathbin{:}U\right]$
equals $2$ if $r=1$, and equals $4$ if $r>1$. The set $\\{\pm
1+2^{r+1}\mathbb{Z},\pm 5+2^{r+1}\mathbb{Z}\\}$ forms a complete system of
representatives of the quotient
$\left(\mathbb{Z}/2^{r+1}\mathbb{Z}\right)^{\times}/U$, when $r>1$. Moreover,
the subgroup $U$ can be described explicitly as the set of elements
$5^{2n}+2^{r+1}\mathbb{Z}$, where $n$ is nonnegative integers, which turns out
to be the set of elements $8n+1+2^{r+1}\mathbb{Z}$, where $n\in\mathbb{Z}$.
Thus, when we treat $\underline{A}_{2^{r}}^{a}$ for $r>1$, there are four
cases to consider, according to which element $\pm 1+2^{r+1}\mathbb{Z},\pm
5+2^{r+1}\mathbb{Z}$ that $a+2^{r+1}\mathbb{Z}$ is congruent to modulo $U$,
i.e., according to which number $\pm 1,\pm 5$ that $a$ is congruence to modulo
$8$.

######  Theorem 4.5.

Let $r$ be a positive integer greater than $1$, and $a$ be an odd integer.

  1. 1.

If $a$ is congruent to $1$ (or to $-1$, respectively) modulo $8$, then the
discriminant module of the even nondegenerate lattice given by the $1\times 1$
Gram matrix $(2^{r})$ (or the matrix $(-2^{r})$, respectively) is
isometrically isomorphic to $\underline{A}_{2^{r}}^{a}$.

  2. 2.

If $a$ is congruent to $\varepsilon\cdot 5$ modulo $8$, where $\varepsilon=\pm
1$, and $r$ is odd, we define an even nondegenerate lattice as follows: let
$v$ be a solution of the congruence $2^{r-1}v^{2}\equiv-\varepsilon\mod 5$,
and set $x=\frac{2^{r-1}v^{2}+\varepsilon}{5}$; then the required lattice is
given by the Gram matrix

$\begin{pmatrix}\varepsilon\cdot 5\cdot 2^{r}&2^{r}v&0\\\
2^{r}v&\varepsilon\cdot 2x&1\\\ 0&1&2\end{pmatrix}.$ (4.20)

The discriminant module of this lattice is isometrically isomorphic to
$\underline{A}_{2^{r}}^{a}$. Moreover, there is no even nondegenerate lattice
of a smaller rank with this property.

  3. 3.

If $a$ is congruent to $\varepsilon\cdot 5$ modulo $8$, where $\varepsilon=\pm
1$, and $r$ is even, we define an even nondegenerate lattice as follows: let
$v$ be a solution of the congruence $2^{r-1}v^{2}\equiv-1\mod 3$, and set
$x=\frac{2^{r-1}v^{2}+1}{3}$; then the required lattice is given by the Gram
matrix

$\begin{pmatrix}-\varepsilon\cdot 3\cdot 2^{r}&2^{r}v&0\\\
2^{r}v&-\varepsilon\cdot 2x&1\\\ 0&1&-\varepsilon\cdot 2\end{pmatrix}.$ (4.21)

The discriminant module of this lattice is isometrically isomorphic to
$\underline{A}_{2^{r}}^{a}$. Moreover, there is no even nondegenerate lattice
of a smaller rank with this property.

  4. 4.

If $a$ is congruent to $1$ (or to $-1$, respectively) modulo $4$, then the
discriminant module of the even nondegenerate lattice given by the $1\times 1$
Gram matrix $(2)$ (or the matrix $(-2)$, respectively) is isometrically
isomorphic to $\underline{A}_{2}^{a}$.

###### Proof.

The statements in Part 1 and 4 can be verified straightforwardly.

Now proceed to prove the statements in Part 2 and 3. First of all, one can
prove that there is no even nondegenerate lattice of rank $1$ whose
discriminant module is isomorphic to $\underline{A}_{2^{r}}^{a}$. Secondly, we
will prove that there is no such lattice of rank $2$ by contradiction. Assume
that there is a such lattice $\underline{L}$ of rank $2$. By Lemma 4.4,
$\underline{L}$ has a Gram matrix (with respect to some basis) of the form

$\begin{pmatrix}2^{r}a&2^{r}v\\\ 2^{r}v&2x\end{pmatrix}$

with $a,v,x\in\mathbb{Z}$ and $2ax-2^{r}v^{2}=\pm 1$. This is impossible,
since the determinant must be even. Finally, we prove that the lattices given
by (4.20) and (4.21) actually serve the purpose. Note that the congruence in
Part 2 and in Part 3 are both solvable, hence $x$’s are integers in both
cases. Then we can verify that (4.20) and (4.21) satisfy the conditions in
Lemma 4.4, which implies the desired isomorphisms. ∎

Finally, we deal with the modules $\underline{D}_{2^{r}}^{a,b,c}$ (For
definition, see (3.11)). Note that this includes the modules
$\underline{B}_{2^{r}}$ and $\underline{C}_{2^{r}}$ (defined in Definition
3.11).

######  Lemma 4.6.

Let $a,b,c$ be integers with $b$ odd, and let $r$ be a positive integer. Let
$\underline{L}=(L,B)$ be an even nondegenerate lattice of rank
$n\in\mathbb{Z}_{\geq 1}$. Then $D_{\underline{L}}$ and
$\underline{D}_{2^{r}}^{a,b,c}$ are isometrically isomorphic if and only if
$n\geq 2$, and there exists a basis of $L$ such that the Gram matrix of
$\underline{L}$ with respect to this basis is of the form
$\mathop{\mathrm{diag}}(2^{r},2^{r},1,\dots,1)\cdot S$, where
$S=(s_{ij})_{1\leq i,j\leq n}\in GL_{n}(\mathbb{Z})$ satisfies

  1. 1.

$s_{33},\dots,s_{nn}\in 2\mathbb{Z}$,

  2. 2.

$s_{11}$, $s_{22}$ are even with $s_{11}s_{22}\equiv 4ac\mod 8$, and
$s_{12}=s_{21}$ are odd,

  3. 3.

$2^{r}s_{1j}=s_{j1}$ and $2^{r}s_{2j}=s_{j2}$ for $3\leq j\leq n$, and
$s_{ij}=s_{ji}$ for $3\leq i,j\leq n$.

###### Proof.

Denote by $Q$ the quadratic map of $D_{\underline{L}}$, and $B_{Q}$ the
associated bilinear map.

Suppose that $D_{\underline{L}}$ and $\underline{D}_{2^{r}}^{a,b,c}$ are
isomorphic. Then there is a basis $(e_{1},e_{2},\dots,e_{n})$ of $L$ such that
$(\frac{1}{2^{r}}e_{1},\frac{1}{2^{r}}e_{2},e_{3},\dots,e_{n})$ is a basis of
$L^{\sharp}$. Hence $n\geq 2$. Let $G$ be the Gram matrix of $\underline{L}$
with respect to this basis. Then there is an $S\in GL_{n}(\mathbb{Z})$ such
that $S\cdot
G^{-1}=\mathop{\mathrm{diag}}(\frac{1}{2^{r}},\frac{1}{2^{r}},1,\dots,1)$,
i.e., $G=\mathop{\mathrm{diag}}(2^{r},2^{r},1,\dots,1)\cdot S$. We shall show
that this $S$ satisfies the desired conditions. Since $\underline{L}$ is even
we have $s_{33},\dots,e_{nn}\in 2\mathbb{Z}$. Since $B$ is symmetric, we have
$2^{r}s_{1j}=s_{j1}$ and $2^{r}s_{2j}=s_{j2}$ for $3\leq j\leq n$, and
$s_{ij}=s_{ji}$ for $3\leq i,j\leq n$. By (2.3), we have
$Q(x_{1}\frac{1}{2^{r}}e_{1}+x_{2}\frac{1}{2^{r}}e_{2}+\dots+x_{n}e_{n}+L)=\frac{s_{11}x_{1}^{2}+2s_{12}x_{1}x_{2}+s_{22}x_{2}^{2}}{2^{r+1}}+\mathbb{Z}$.
We must have $s_{11}$ is even, since otherwise the submodule generated by
$\frac{1}{2^{r}}e_{1}+L$ would itself be a finite quadratic module, which
implies that $\underline{D}_{2^{r}}^{a,b,c}$ has an orthogonal direct sum
decomposition by Lemma 2.3. Likewise, we have $s_{22}$ is even. On the other
hand, $s_{12}=s_{21}$ must be odd, since $B_{Q}$ is nondegenerate. Thus,
$D_{\underline{L}}$ is isometrically isomorphic to
$\underline{D}_{2^{r}}^{s_{11}/2,s_{12},s_{22}/2}$. Using Lemma 3.15, we
deduce that $s_{11}s_{22}\equiv 4ac\mod 8$.

Conversely, suppose that $L$ has a basis $(e_{1},e_{2},\dots,e_{n})$ such that
the Gram matrix of $\underline{L}$ with respect to this basis is of the form
$\mathop{\mathrm{diag}}(2^{r},2^{r},1,\dots,1)\cdot S$, where
$S=(s_{ij})_{1\leq i,j\leq n}\in GL_{n}(\mathbb{Z})$ satisfies the conditions
listed in the lemma. Then
$(\frac{1}{2^{r}}e_{1},\frac{1}{2^{r}}e_{2},e_{3},\dots,e_{n})$ is a basis of
$L^{\sharp}$, by Proposition 3.5. Using (2.3) and the conditions on $S$, we
obtain
$Q(x_{1}\frac{1}{2^{r}}e_{1}+x_{2}\frac{1}{2^{r}}e_{2}+\dots+x_{n}e_{n}+L)=\frac{s_{11}/2\cdot
x_{1}^{2}+s_{12}x_{1}x_{2}+s_{22}/2\cdot x_{2}^{2}}{2^{r}}+\mathbb{Z}$. Thus
$D_{\underline{L}}$ is isometrically isomorphic to
$\underline{D}_{2^{r}}^{s_{11}/2,s_{12},s_{22}/2}$. Again by Lemma 3.15, we
have $D_{\underline{L}}$ is isometrically isomorphic to
$\underline{D}_{2^{r}}^{a,b,c}$. ∎

######  Theorem 4.7.

Let $r$ be a positive integer, and set $\varepsilon=(-1)^{r+1}$.

  1. 1.

Let $v$ be a solution of the congruence $2^{r}v^{2}\equiv-\varepsilon\mod 3$,
and set $x=\frac{2^{r}v^{2}+\varepsilon}{3}$. Then the discriminant module of
the even nondegenerate lattice given by the Gram matrix

$\begin{pmatrix}2^{r+1}&2^{r}&0&0\\\ 2^{r}&2^{r+1}&2^{r}v&0\\\
0&2^{r}v&2x&1\\\ 0&0&1&\varepsilon\cdot 2\end{pmatrix}$ (4.22)

is isometrically isomorphic to $\underline{B}_{2^{r}}$. Moreover, there is no
such lattice of smaller rank.

  2. 2.

The discriminant module of the even nondegenerate lattice given by the
following Gram matrix

$\begin{pmatrix}0&2^{r}\\\ 2^{r}&0\end{pmatrix}$ (4.23)

is isometrically isomorphic to $\underline{C}_{2^{r}}$. Moreover, there is no
such lattice of smaller rank.

###### Proof.

We omit the proof of the statement in Part 2, and focus on proving the
statement in Part 1.

First of all, there is no even nondegenerate lattice of rank $1$ whose
discriminant module is isomorphic to $\underline{B}_{2^{r}}$, since such
discriminant module must be a cyclic group, while $\underline{B}_{2^{r}}$ is
not. Secondly, it can be verified by Condition 2 of Lemma 4.6 that there is
also no such lattice of rank $2$. And then we shall prove there is no such
lattice of rank $3$ by contradiction. Assume that there is a lattice with such
property. Then its Gram matrix with respect to some basis must be (by Lemma
4.6)

$\begin{pmatrix}2^{r}&0&0\\\ 0&2^{r}&0\\\
0&0&1\end{pmatrix}\cdot\begin{pmatrix}2x&y&v\\\ y&2z&w\\\
2^{r}v&2^{r}w&2t\end{pmatrix}$

where $x,y,z,v,w,t$ are all integers such that $x,y,z$ are odd, and the
determinant of the second matrix in the above expression is $\pm 1$. But we
see that the determinant must be even by expansion with respect the third row,
which is a contradiction. Finally, we shall verify that the given matrix in
Part 1 of the theorem actually provides an even nondegenerate lattice whose
discriminant module is isomorphic to $\underline{B}_{2^{r}}$. Note that this
matrix is a product of $\mathop{\mathrm{diag}}(2^{r},2^{r},1,1)$ and

$S=\begin{pmatrix}2&1&0&0\\\ 1&2&v&0\\\ 0&2^{r}v&2x&1\\\
0&0&1&\varepsilon\cdot 2\end{pmatrix},$ (4.24)

and that $v$ exists and $x$ is an integer. By direct calculation we see that
$\det(S)=1$, and $S$ satisfies the conditions stated in Lemma 4.6. Hence the
discriminant module of this lattice is isomorphic to $\underline{B}_{2^{r}}$
by Lemma 4.6, which concludes the proof. ∎

######  Corollary 4.8.

Any finite quadratic module is isometrically isomorphic to the discriminant
module of some even nondegenerate lattice.

###### Proof.

This follows immediately from Theorem 3.19, Theorem 4.2, Theorem 4.5, and
Theorem 4.7. ∎

######  Example 4.9.

We look for lattices whose discriminant module is isometrically isomorphic to
$\underline{A}_{m}$ defined in (3.9) for $m$ being an odd integer greater than
$1$. Suppose that $m=p_{1}^{r_{1}}p_{2}^{r_{2}}\dots p_{k}^{r_{k}}$ with
$p_{1},p_{2},\dots,p_{k}$ different odd primes, and $r_{1},r_{2},\dots,r_{n}$
positive integers. Set $m_{i}=m/p_{i}^{r_{i}}$, then $m_{i}$’s are relatively
prime. Find integers $M_{i}$ such that $m_{1}M_{1}+\dots m_{k}M_{k}=1$. It can
be verified that $\underline{A}_{m}$ is isometrically isomorphic to the
orthogonal direct sum of the finite quadratic modules
$\underline{A}_{p_{i}^{r_{i}}}^{M_{i}}$ with $1\leq i\leq k$, and the
isomorphism is given by the map
$x+m\mathbb{Z}\mapsto(x+p_{1}^{r_{1}}\mathbb{Z},x+p_{2}^{r_{2}}\mathbb{Z},\dots,x+p_{k}^{r_{k}}\mathbb{Z})$.
To find an even nondegenerate lattice with discriminant module
$\underline{A}_{m}$, we first construct lattices with Gram matrix (4.16),
(4.17), (4.18), or (4.19) for $\underline{A}_{p_{i}^{r_{i}}}^{M_{i}}$, and
then form there orthogonal direct sum. Specially, if $m$ is a square, then
$\genfrac{(}{)}{}{}{M_{i}}{p_{i}}=\genfrac{(}{)}{}{}{m_{i}}{p_{i}}=1$. Hence
only the Gram matrix (4.16) would occur. As a consequence, the even
nondegenerate lattice with Gram matrix

$\mathop{\mathrm{diag}}\left(\begin{pmatrix}2p_{1}^{r_{1}}&p_{1}^{r_{1}}\\\
p_{1}^{r_{1}}&\frac{1}{2}\left(p_{1}^{r_{1}}-\genfrac{(}{)}{}{}{-1}{p_{1}}^{r_{1}}\right)\end{pmatrix}\dots,\begin{pmatrix}2p_{k}^{r_{k}}&p_{k}^{r_{k}}\\\
p_{k}^{r_{k}}&\frac{1}{2}\left(p_{k}^{r_{k}}-\genfrac{(}{)}{}{}{-1}{p_{k}}^{r_{k}}\right)\end{pmatrix}\right)$
(4.25)

has discriminant module $\underline{A}_{m}$. The rank of this lattice is $2k$,
which is not necessarily the least rank of such lattices.

##  5 Discriminant modules of positive definite even lattices

We have proved the theorem of Wall (Corollary 4.8) that any finite quadatic
module is a discriminant module of an even nondegenerate lattice, by giving
concrete constructions for those indecomposable modules. In this section, we
aim to find positive definite even lattices for arbitrary finite quadratic
modules.

To be aware of the possible least rank before constructing such lattices, we
shall need the Milgram’s formula. Before stating the theorem, we recall some
notions. Let $\underline{M}=(M,Q)$ be a finite quadratic module, we define the
$\sigma$-invariant of $\underline{M}$ as

$\sigma(\underline{M})=\frac{1}{\sqrt{|M|}}\sum_{x\in
M}\exp(-2\uppi\mathrm{i}Q(x)).$ (5.26)

For an even nondegenerate lattice $\underline{L}=(L,B)$, the
$\sigma$-invariant of $\underline{L}$ is defined to be the $\sigma$-invariant
of $D_{\underline{L}}$. And we define the signature of $\underline{L}$
(denoted by $\mathop{\mathrm{sign}}(\underline{L})$) to be the signature of
the real orthogonal geometry $(\mathbb{R}\otimes L,B)$, i.e., the positive
index of inertia minus the negative index of inertia of $(\mathbb{R}\otimes
L,B)$. These two quantities are related by

######  Theorem 5.1.

Let $\underline{L}=(L,B)$ be an even nondegenerate lattice, and
$D_{\underline{L}}$ be its discriminant module. Then

$\sigma(\underline{L})=\sigma(D_{\underline{L}})=\exp\left(-2\uppi\mathrm{i}\left(\frac{\mathop{\mathrm{sign}}(\underline{L})}{8}\right)\right).$
(5.27)

This identity, which is essentially on evaluation of Gauss sums, was first
discovered and proved by Milgram. An easier proof (comparing to the original
proof of Milgram) can be founed in [6, Appendix 4], where the authors show
that the $\sigma$-invariant actually depends only on the ambient rational
inner product space, and is multiplicative with respect to the orthogonal
direct sum of rational inner product spaces, and then they prove the identity
by evaluating a special Gauss sum using one-dimensional Fourier analysis.

Thus, we can evaluate the $\sigma$-invariant of any finite quadratic module
$\underline{M}$ by first finding an even nondegenetate lattice whose
discriminant module is just $\underline{M}$, and then using Milgram’s formula.
One can also evaluate $\sigma(\underline{M})$ directly, as Strömberg has
worked out this in [13, Section 3]. We list these results for the reader’s
convenience.

######  Proposition 5.2.

Let $p$ be a prime, and $a$ an integer with $(p,a)=1$, and let $r$ be a
positive integer. Then

$\displaystyle\sigma(\underline{A}_{p^{r}}^{a})$
$\displaystyle=\genfrac{(}{)}{}{}{2a}{p^{r}}\exp\left(-2\uppi\mathrm{i}\frac{1-p^{r}}{8}\right)\quad\text{for
}p>2,$ $\displaystyle\sigma(\underline{A}_{2^{r}}^{a})$
$\displaystyle=\genfrac{(}{)}{}{}{a}{2^{r}}\exp\left(-2\uppi\mathrm{i}\frac{a}{8}\right)\quad\text{for
}p=2,$ $\displaystyle\sigma(\underline{B}_{2^{r}})$ $\displaystyle=(-1)^{r},$
$\displaystyle\sigma(\underline{C}_{2^{r}})$ $\displaystyle=1.$

As has been pointed out, these identities can be deduced by evaluating the
inertia of the matrices (4.16), (4.17), (4.18), (4.19), (4.20), (4.21),
(4.22), (4.23), and using Theorem 5.1 (All inertia can be worked out by
standard procedure of completing squares over the reals for these matrices, so
we omit the tedious details.).

######  Remark 5.3.

In Part 4 of Theorem 4.2, we have constructed an even nondegenerate lattice of
rank $4$ whose discriminant module is $\underline{A}_{p^{r}}^{a}$, under
certain conditions. Also we have showed that there is no such lattice of rank
$1$ and $2$. We can now assert: the lattice constructed there is of the least
rank. This can be proved by the formula of $\sigma$-invariants just stated in
Proposition 5.2.

We now turn to the task of finding positive definite even lattices of the
least positive rank with given finite quadratic modules.

######  Lemma 5.4.

Let $\underline{L}=(L,B_{L})$ be a positive definite even lattice of rank
$n\in\mathbb{Z}_{\geq 1}$, and $D_{\underline{L}}$ be its discriminant module.
Let $\underline{M}=(M,Q_{M})$ be a finite quadratic module. Let $p$ be a prime
and $a$ be an integer with $(a,p)=1$, and let $r$ be a positive integer.
Suppose that $D_{\underline{L}}$ and $\underline{M}$ are isometrically
isomorphic.

  1. 1.

If $\underline{M}$ is trivial, i.e., $|M|=1$, then $n\equiv 0\mod 8$.

  2. 2.

If $\underline{M}=\underline{A}_{p^{r}}^{a}$, with $r$ even and $p>2$, then
$n\equiv 0\mod 8$.

  3. 3.

If $\underline{M}=\underline{A}_{p^{r}}^{a}$, with $r$ odd and $p>2$, then
$n\equiv 3-\genfrac{(}{)}{}{}{-1}{p}-2\genfrac{(}{)}{}{}{a}{p}\mod 8$.

  4. 4.

If $\underline{M}=\underline{A}_{2^{r}}^{a}$, with $r$ even and $(a,2)=1$,
then $n\equiv a\mod 8$.

  5. 5.

If $\underline{M}=\underline{A}_{2^{r}}^{a}$, with $r$ odd and $(a,2)=1$, then
$n\equiv a+2-2\genfrac{(}{)}{}{}{a}{2}\mod 8$.

  6. 6.

If $\underline{M}=\underline{B}_{2^{r}}$, then $n\equiv 2-2(-1)^{r}\mod 8$.

  7. 7.

If $\underline{M}=\underline{C}_{2^{r}}$, then $n\equiv 0\mod 8$.

###### Proof.

This follows immediately from Theorem 5.1 and Proposition 5.2. ∎

######  Remark 5.5.

Let $\underline{L}=(L,B)$ be an even nondegenerate lattice. If
$D_{\underline{L}}$ is the zero module, i.e., $L=L^{\sharp}$, then
$\underline{L}$ is called a unimodular lattice. The name comes from the fact
that the volumn of a closed fundamental domain of $(\mathbb{R}\otimes L)/L$ in
the real space $\mathbb{R}\otimes L$ is $1$ under a natural selection of
Lebesgue measure. The investigation of this kind of lattices has a long
history and constitutes an important area in number theory. In the indefinite
case, one has a perfect structure theorem, while in the definite case, one has
related the theory on such lattices to theories on many other disciplines,
such as packing of equal balls in Euclidean spaces. The answer to the question
of finding a positive definite even lattice with the zero discriminant module
and least positive rank is well-known: The lattice
$\Gamma_{8}:=\left\\{(x_{1},\dots,x_{8})\in\mathbb{R}^{8}\middle|x_{1},\dots,x_{8}\text{
are all in }\mathbb{Z}\text{ or }1/2+\mathbb{Z},\,x_{1}+\dots+x_{8}\in
2\mathbb{Z}\right\\}$ satisfies the required conditions. See [4] for a
complete discussion. In the rest of this section, we focus on finding positive
definite even lattices for indecomposable (nonzero) finite quadratic modules.

Now we proceed to deal with the modules $\underline{A}_{p^{r}}^{a}$ with $r$
even.

######  Theorem 5.6.

Let $p$ be an odd prime, and $a$ be an integer with $(a,p)=1$, let $r$ be an
even positive integer.

  1. 1.

If $p\neq 7$, then the discriminant module of the positive definite even
lattice given by the following Gram matrix

$\begin{pmatrix}2qp^{r}&p^{r}v&0&0&0&0&0&0\\\ p^{r}v&2x&1&0&0&0&0&0\\\
0&1&2&1&0&0&0&0\\\ 0&0&1&2&1&0&0&0\\\ 0&0&0&1&2&1&0&0\\\ 0&0&0&0&1&2&1&0\\\
0&0&0&0&0&1&2&1\\\ 0&0&0&0&0&0&1&2\end{pmatrix}$ (5.28)

is isometrically isomorphic to $\underline{A}_{p^{r}}^{a}$, where $q\equiv
4\mod 7$ is a prime such that
$\genfrac{(}{)}{}{}{q}{p}=\genfrac{(}{)}{}{}{a}{p}$, $v$ is a solution of the
congruence $7p^{r}v^{2}\equiv-1\mod 4q$, and
$x=\frac{7p^{r}v^{2}+12q+1}{28q}$.

  2. 2.

If $p=7$, and $\genfrac{(}{)}{}{}{a}{7}=1$, then the discriminant module of
the positive definite even lattice given by the following Gram matrix

$\begin{pmatrix}8\cdot 7^{r}&7^{r}v&0&0&0&0&0&0\\\ 7^{r}v&2x&1&0&0&0&0&0\\\
0&1&2&1&0&0&0&0\\\ 0&0&1&2&1&0&0&0\\\ 0&0&0&1&2&1&0&0\\\ 0&0&0&0&1&2&1&0\\\
0&0&0&0&0&1&2&1\\\ 0&0&0&0&0&0&1&2\end{pmatrix}$ (5.29)

is isometrically isomorphic to $\underline{A}_{7^{r}}^{a}$, where $v$ is a
solution of the congruence $7^{r-1}v^{2}\equiv-1\mod 16$, and
$x=\frac{7^{r}v^{2}+7}{16}$.

  3. 3.

If $p=7$, and $\genfrac{(}{)}{}{}{a}{7}=-1$, then the discriminant module of
the positive definite even lattice given by the following Gram matrix

$\begin{pmatrix}10\cdot 7^{r}&7^{r}v&0&0&0&0&0&0\\\ 7^{r}v&2x&5&0&0&0&0&0\\\
0&5&4&1&0&0&0&0\\\ 0&0&1&2&1&0&0&0\\\ 0&0&0&1&2&1&0&0\\\ 0&0&0&0&1&2&1&0\\\
0&0&0&0&0&1&2&1\\\ 0&0&0&0&0&0&1&2\end{pmatrix}$ (5.30)

is isometrically isomorphic to $\underline{A}_{7^{r}}^{a}$, where $v$ is a
solution of the congruences $7^{r}v^{2}\equiv 1\mod 20$, and
$x=\frac{7^{r}v^{2}+79}{20}$.

Moreover, in each of the above cases, $8$ is the least rank of positive
definite even lattices whose discriminant module is isomorphic to
$\underline{A}_{p^{r}}^{a}$.

###### Proof.

Before proving the theorem itself, we state an auxiliary property. We define a
sequence of matrices $F_{l}$ for $l\in\mathbb{Z}_{\geq 1}$ as follows. The
matrix $F_{l}$ is a matrix of size $l\times l$, and the $(i,j)$-entry $f_{ij}$
is equal to $2$, if $i=j$, is equal to $1$, if $i-j=\pm 1$, and is equal to
$0$ otherwise. One shows by induction that $\det(F_{l})=l+1$. In particular,
$\det(F_{5})=6$, and $\det(F_{6})=7$. Also note that $F_{l}$’s are all
positive definite matrices.

Now proceed to prove the theorem. First of all, we can prove that $v$ exists
in three cases by elementary number theory, and hence $x$ is an integer.
Secondly, divide the first row of each of the matrices (5.28), (5.29), and
(5.30) by $p^{r}$, and denote the result by $S$. Then it can be verified that
$S$ is an integral matrix and satisfies the three conditions posed in Lemma
4.1. Thus, to conclude that the discriminant modules of these lattices are
isomorphic to $\underline{A}_{p^{r}}^{a}$, it remains to prove that
$\det(S)=\pm 1$ in three cases. Using the determinant of $F_{5}$ and $F_{6}$
just stated, and using the expansion according to the first row, we find the
determinant of the matrix $S$ obtaining from (5.28) is $28qx-7p^{r}v^{2}-12q$,
which equals $1$ since $x=\frac{7p^{r}v^{2}+12q+1}{28q}$. Similarly the
matrices $S$ obtaining from (5.29) and (5.30) also have determinant $1$.
Thirdly, the lattices given by these Gram matrices are positive definite. To
prove this, note that by Theorem 5.1 and Proposition 5.2 the signatures are
$0$ or $\pm 8$. But in all three cases, each of the ambient real orthogonal
geometries has a positive definite subspace of dimension $5$ (with Gram matrix
$F_{5}$), hence $8$ is the only possible signature, i.e., these lattices are
positive definite. This assertion can also be proved by completing squares
directly. Finally, we see that an even nondegenerate nonzero lattice of rank
less than $8$ can not be positive definite by Part 2 of Lemma 5.4. ∎

Next we consider $\underline{A}_{p^{r}}^{a}$ with $r$ odd.

######  Theorem 5.7.

Let $p$ be an odd prime, and $a$ be an integer with $(a,p)=1$, let $r$ be an
odd positive integer.

  1. 1.

If $\genfrac{(}{)}{}{}{-1}{p}=1$ and $\genfrac{(}{)}{}{}{a}{p}=1$, then let
$q$, $v$, and $x$ be as in Part 1 of Theorem 5.6, and we have the discriminant
module of the positive definite even lattice given by the Gram matrix (5.28)
is isometrically isomorphic to $\underline{A}_{p^{r}}^{a}$.

  2. 2.

If $\genfrac{(}{)}{}{}{-1}{p}=1$ and $\genfrac{(}{)}{}{}{a}{p}=-1$, then let
$q$, $v$, and $x$ be as in Part 4 of Theorem 4.2, and we have the discriminant
module of the positive definite even lattice given by the Gram matrix (4.19)
is isometrically isomorphic to $\underline{A}_{p^{r}}^{a}$.

  3. 3.

If $\genfrac{(}{)}{}{}{-1}{p}=-1$ and $\genfrac{(}{)}{}{}{a}{p}=1$, then the
discriminant module of the positive definite even lattice given by the Gram
matrix (4.16) in Part 1 of Theorem 4.2 is isometrically isomorphic to
$\underline{A}_{p^{r}}^{a}$.

  4. 4.

If $\genfrac{(}{)}{}{}{-1}{p}=-1$ and $\genfrac{(}{)}{}{}{a}{p}=-1$, then the
discriminant module of the positive definite even lattice given by the Gram
matrix

$\begin{pmatrix}2qp^{r}&p^{r}v&0&0&0&0\\\ p^{r}v&2x&1&0&0&0\\\ 0&1&2&1&0&0\\\
0&0&1&2&1&0\\\ 0&0&0&1&2&1\\\ 0&0&0&0&1&2\end{pmatrix}$ (5.31)

is isometrically isomorphic to $\underline{A}_{p^{r}}^{a}$, where $q\equiv
3\mod 5$ is a prime such that $\genfrac{(}{)}{}{}{q}{p}=-1$, and $v$ is a
solution of the congruence $5p^{r}v^{2}\equiv-1\mod 4q$, and
$x=\frac{5p^{r}v^{2}+8q+1}{20q}$.

Moreover, in each of the above cases, the rank of the given lattice is the
least among all positive definite even lattices whose discriminant module is
isomorphic to $\underline{A}_{p^{r}}^{a}$.

###### Proof.

We have proved the assertions of Part 1, Part 2, and Part 3 in Theorem 4.2 and
Theorem 5.6, except for the fact that the lattices given by the Gram matrices
in Part 2 and Part 3 are positive definite, which can be verified directly.

To prove the statement in Part 4, we use Lemma 4.1, as in the proof of Theorem
5.6. We omit the details.

Finally, from Part 3 of Lemma 5.4, we see that the rank of the given lattice
is actually the least among all positive definite even lattices whose
discriminant module is isomorphic to $\underline{A}_{p^{r}}^{a}$ in each of
the four cases. ∎

We now turn to the modules $\underline{A}_{2^{r}}^{a}$ with $r$ even.

######  Theorem 5.8.

Let $r$ be an even positive integer, and $a$ an odd integer.

  1. 1.

If $a\equiv 1\mod 8$, then the discriminant module of the positive definite
even lattice given by the $1\times 1$ Gram matrix $(2^{r})$ is isometrically
isomorphic to $\underline{A}_{2^{r}}^{a}$.

  2. 2.

If $a\equiv 3\mod 8$, then the discriminant module of the positive definite
even lattice given by the Gram matrix (4.21) with $\varepsilon=-1$ in Part 3
of Theorem 4.5 is isometrically isomorphic to $\underline{A}_{2^{r}}^{a}$.

  3. 3.

If $a\equiv 5\mod 8$, then the discriminant module of the positive definite
even lattice given by the Gram matrix

$\begin{pmatrix}5\cdot 2^{r}&2^{r}v&0&0&0\\\ 2^{r}v&2x&1&0&0\\\ 0&1&2&1&0\\\
0&0&1&2&1\\\ 0&0&0&1&2\end{pmatrix}$ (5.32)

is isometrically isomorphic to $\underline{A}_{2^{r}}^{a}$, where $v$ is a
solution of the congruence $2^{r-1}v^{2}\equiv-2\mod 5$, and
$x=\frac{2^{r-1}v^{2}+2}{5}$.

  4. 4.

If $a\equiv 7\mod 8$, then the discriminant module of the positive definite
even lattice given by the Gram matrix

$\begin{pmatrix}7\cdot 2^{r}&2^{r}v&0&0&0&0&0\\\ 2^{r}v&2x&1&0&0&0&0\\\
0&1&2&1&0&0&0\\\ 0&0&1&2&1&0&0\\\ 0&0&0&1&2&1&0\\\ 0&0&0&0&1&2&1\\\
0&0&0&0&0&1&2\end{pmatrix}$ (5.33)

is isometrically isomorphic to $\underline{A}_{2^{r}}^{a}$, where $v$ is a
solution of the congruence $2^{r-1}v^{2}\equiv-3\mod 7$, and
$x=\frac{2^{r-1}v^{2}+3}{7}$.

Moreover, in each of the above cases, the rank of the given lattice is the
least among all positive definite even lattices whose discriminant module is
isomorphic to $\underline{A}_{2^{r}}^{a}$.

###### Proof.

The statements in Part 1 and Part 2 have been proved in Theorem 4.5, except
for the fact that the lattices defined in these two cases are positive
definite, which can be verified directly (e.g., by completing squares). To
prove statements in Part 3 and Part 4, we argue as in the proof of Theorem
5.6, using Lemma 4.4 instead of Lemma 4.1. Finally, to prove the last
assertion in the theorem, use Part 4 of Lemma 5.4. ∎

Now turn to the modules $\underline{A}_{2^{r}}^{a}$ with $r$ odd.

######  Theorem 5.9.

Let $r$ be an odd integer greater than $1$, and $a$ be an odd integer.

  1. 1.

If $a\equiv 1\mod 8$, then the conclusion is the same as in Part 1 of Theorem
5.8.

  2. 2.

If $a\equiv 3\mod 8$, then the discriminant module of the positive definite
even lattice given by the Gram matrix

$\begin{pmatrix}19\cdot 2^{r}&2^{r}v&0&0&0&0&0\\\ 2^{r}v&2x&1&0&0&0&0\\\
0&1&2&1&0&0&0\\\ 0&0&1&2&1&0&0\\\ 0&0&0&1&2&1&0\\\ 0&0&0&0&1&2&1\\\
0&0&0&0&0&1&2\end{pmatrix}$ (5.34)

is isometrically isomorphic to $\underline{A}_{2^{r}}^{a}$, where $v$ is a
solution of the congruence $2^{r-1}v^{2}\equiv-8\mod 19$, and
$x=\frac{2^{r-1}v^{2}+8}{19}$.

  3. 3.

If $a\equiv 5\mod 8$, then the discriminant module of the positive definite
even lattice given by the Gram matrix

$\begin{pmatrix}5\cdot 2^{r}&2^{r}v&0&0&0&0&0&0&0\\\
2^{r}v&2x&7&0&0&0&0&0&0\\\ 0&7&4&1&0&0&0&0&0\\\ 0&0&1&2&1&0&0&0&0\\\
0&0&0&1&2&1&0&0&0\\\ 0&0&0&0&1&2&1&0&0\\\ 0&0&0&0&0&1&2&1&0\\\
0&0&0&0&0&0&1&2&1\\\ 0&0&0&0&0&0&0&1&2\end{pmatrix}$ (5.35)

is isometrically isomorphic to $\underline{A}_{2^{r}}^{a}$, where $v$ is a
solution of the congruence $2^{r-1}v^{2}\equiv 1\mod 5$, and
$x=\frac{2^{r-1}v^{2}+39}{5}$.

  4. 4.

If $a\equiv 7\mod 8$, then the conclusion is the same as in Part 4 of Theorem
5.8.

Moreover, in each of the above cases, the rank of the given lattice is the
least among all positive definite even lattices whose discriminant module is
isomorphic to $\underline{A}_{2^{r}}^{a}$.

###### Proof.

We omit the proof, since it is similar to that of the previous theorem. ∎

Finally, we shall deal with the modules $\underline{B}_{2^{r}}$ and
$\underline{C}_{2^{r}}$.

######  Theorem 5.10.

Let $r$ be a positive integer.

  1. 1.

If $r$ is odd, then the conclusion for $\underline{B}_{2^{r}}$ is the same as
in Part 1 of Theorem 4.7 with $\varepsilon=1$. The lattice defined there is
actually positive definite.

  2. 2.

If $r$ is even, then the discriminant module of the positive definite even
lattice given by the Gram matrix

$\begin{pmatrix}2\cdot 2^{r}&2^{r}&0&0&0&0&0&0\\\ 2^{r}&10\cdot
2^{r}&2^{r}v&0&0&0&0&0\\\ 0&2^{r}v&2x&1&0&0&0&0\\\ 0&0&1&2&1&0&0&0\\\
0&0&0&1&2&1&0&0\\\ 0&0&0&0&1&2&1&0\\\ 0&0&0&0&0&1&2&1\\\
0&0&0&0&0&0&1&2\end{pmatrix}$ (5.36)

is isometrically isomorphic to $\underline{B}_{2^{r}}$, where $v$ is a
solution of the congruence $2^{r}v^{2}\equiv-8\mod 19$, and
$x=\frac{2^{r}v^{2}+8}{19}$.

  3. 3.

The discriminant module of the positive definite even lattice given by the
Gram matrix

$\begin{pmatrix}2\cdot 2^{r}&2^{r}&0&0&0&0&0&0\\\ 2^{r}&4\cdot
2^{r}&2^{r}v&0&0&0&0&0\\\ 0&2^{r}v&2x&1&0&0&0&0\\\ 0&0&1&2&1&0&0&0\\\
0&0&0&1&2&1&0&0\\\ 0&0&0&0&1&2&1&0\\\ 0&0&0&0&0&1&2&1\\\
0&0&0&0&0&0&1&2\end{pmatrix}$ (5.37)

is isometrically isomorphic to $\underline{C}_{2^{r}}$, where $v$ is a
solution of the congruence $2^{r}v^{2}\equiv-3\mod 7$, and
$x=\frac{2^{r}v^{2}+3}{7}$.

Moreover, in each of the above cases, the rank of the given lattice is the
least among all positive definite even lattices whose discriminant module is
isomorphic to $\underline{B}_{2^{r}}$ (in Case 1 and 2) or
$\underline{C}_{2^{r}}$ (in Case 3).

###### Proof.

To prove the statement in Part 1, we need only to vefiry that the lattice
mentioned is positive definite, which is obtained by completing squares. To
prove statements in another two parts, first note that the congruences in both
parts are solvable, so the quantity $x$ is integer in both cases. Then use
Lemma 4.6 to conclude that matrices (5.36) and (5.37) give lattices whose
discriminant modules are $\underline{B}_{2^{r}}$ and $\underline{C}_{2^{r}}$
respectively. ∎

As in Corollary 4.8, we can combine the constructions in this section to
obtain a theorem about arbitrary finite quadratic modules.

######  Corollary 5.11.

Any finite quadratic module is isometrically isomorphic to the discriminant
module of some positive definite even lattice.

###### Proof.

We proceed similarly as in the proof of Corollary 4.8, using Theorem 3.19,
Theorem 5.6, Theorem 5.7, Theorem 5.8, Theorem 5.9, and Theorem 5.10 instead.
∎

##  6 Some Remarks

### Finite symmetric bilinear modules.

There is a parallel theory of finite symmetric bilinear modules, besides that
of finite quadratic modules. Here a finite symmetric bilinear module means a
finite $\mathbb{Z}$-module equipped with a nondegenerate symmetric bilinear
form taking values in $\mathbb{Q}/\mathbb{Z}$. All of the methods described in
Section 2 also apply to such structures. Analogous versions of Corollary 3.9
and Proposition 3.10 remain true for finite symmetric bilinear modules. The
indecomposable ones still have the form like those in Definition 3.11, that is
(by abuse of language),

$\displaystyle\underline{A}_{p^{r}}^{a}$
$\displaystyle:=\left(\mathbb{Z}/p^{r}\mathbb{Z},\,(\overline{x},\overline{y})\mapsto\frac{a}{p^{r}}xy+\mathbb{Z}\right)\,\text{for
}p>2,$ $\displaystyle\underline{A}_{2^{r}}^{a}$
$\displaystyle:=\left(\mathbb{Z}/2^{r}\mathbb{Z},\,(\overline{x},\overline{y})\mapsto\frac{a}{2^{r}}xy+\mathbb{Z}\right)\,\text{for
}p=2,$ $\displaystyle\underline{B}_{2^{r}}$
$\displaystyle:=\left((\mathbb{Z}/2^{r}\mathbb{Z})^{2},\,\left((\overline{x_{1}},\overline{y_{1}}),(\overline{x_{2}},\overline{y_{2}})\right)\mapsto\frac{2x_{1}x_{2}+x_{1}y_{2}+x_{2}y_{1}+2y_{1}y_{2}}{2^{r}}+\mathbb{Z}\right),$
$\displaystyle\underline{C}_{2^{r}}$
$\displaystyle:=\left((\mathbb{Z}/2^{r}\mathbb{Z})^{2},\,\left((\overline{x_{1}},\overline{y_{1}}),(\overline{x_{2}},\overline{y_{2}})\right)\mapsto\frac{x_{1}y_{2}+x_{2}y_{1}}{2^{r}}+\mathbb{Z}\right),$

where $p$ is a prime, $a$ is an integer not divisible by $p$, and
$\overline{x}$ denotes $x+p^{r}\mathbb{Z}$. One also has a structure theorem,
which asserts that any finite symmetric bilinear module can be decomposed as
an orthogonal direct sum of the modules just listed. See [14, §5] for a proof
and further discussion. We claim that the methods using in our proof of
Theorem 3.19 also apply to the finite symmetric bilinear module case, with
some minor adaptation.

We would like to explain why we have focused on finite quadratic modules
instead of bilinear ones. Let $\underline{L}=(L,B)$ be an integral lattice,
and $L^{\sharp}$ be its dual (c.f. Definition 3.4). Suppose $L$ is odd, i.e.,
there is a vector $v\in L$ such that $Q(v)=2^{-1}B(v,v)\in 2^{-1}+\mathbb{Z}$.
Then the construction of discriminant modules in Example 3.7 is problematic,
since the quantity $\underline{Q}(v+L)$, where $\underline{Q}$ denotes the
quadratic map on $L^{\sharp}/L$, depends on the choice of a representative of
the coset $v+L$. Nevertheless, the bilinear map
$\underline{B}(x+L,y+L)=B(x,y)+\mathbb{Z}$ is well-defined on $L^{\sharp}/L$,
no matter whether $L$ is even or odd. Futhermore, the structure
$(L^{\sharp}/L,\underline{B})$ is a finite symmetric bilinear module. Thus, it
seems more nature to study integral lattices using “discriminant bilinear
module”. However, only to finite quadratic modules, one can associate Weil
representations which are important objects as explained in Introduction. This
is the reason we focus on finite quadratic modules.

### The question of uniqueness.

We have not mentioned whether the decompositon in Theorem 3.19 is unique in
some sense. Of course, the primary decompositon given in Lemma 3.16 is unique.
So the real question is: given some finite quadratic module of cardinality
$p^{r}$, where $p$ is a prime, and $r$ is a positive integer, does the
decomposition of this module into indecomposable modules unique up to
isometric isomorphism? Unless the given module is itself indecomposable, the
answer is always no. Wall succeeded in getting all relations among different
decompositions when $p>2$, and he also get some but not all relations when
$p=2$. See [14, §5]. Here we only give a simple example to illustrate this.
Consider the $\mathbb{Z}$-module $(\mathbb{Z}/3\mathbb{Z})^{2}$ equipped with
a quadratic map
$Q(x+3\mathbb{Z},y+3\mathbb{Z})=\frac{x^{2}+y^{2}}{3}+\mathbb{Z}$. It is easy
to see this can be decomposed as an orthogonal direct sum of two modules each
of which is isometrically isomorphic to $\underline{A}_{3}^{1}$. Now let $G$
be the submodule generated by $(1+3\mathbb{Z},1+3\mathbb{Z})$, then G equipped
with the quadratic map inherited from $(\mathbb{Z}/3\mathbb{Z})^{2}$ is
isometrically isomorphic to $\underline{A}_{3}^{2}$. A direct calculation
shows that $G^{\perp}$ is also isometrically isomorphic to
$\underline{A}_{3}^{2}$. Hence $\mathbb{Z}$-module
$(\mathbb{Z}/3\mathbb{Z})^{2}$ has another essentially different decomposition
as an orthogonal direct sum of two copies of $\underline{A}_{3}^{2}$.

## Acknowledgment

Thanks are due to Prof. Hai-Gang Zhou, my Ph. D. supervisor, who gave me
useful suggestion, and led me into the beautiful area of modular forms and
Jacobi forms. Thanks are also due to Prof. Nils-Peter Skoruppa, a top expert
in the field of Jacobi forms, who taught me knowledge and skill revelant to
the content of this paper.

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