# Cohen-Macaulay type of orders, generators and ideal classes

Stefano Marseglia  Mathematical Institute, Utrecht University, P.O. Box 80010,
3508 TA, Utrecht, The Netherlands<EMAIL_ADDRESS>
###### Abstract.

In this paper we study the (Cohen-Macaulay) type of orders over Dedekind
domains in étale algebras. We provide a bound for the type, and give formulas
to compute it. We relate the type of the overorders of a given order to the
size of minimal generating sets of its fractional ideals, generalizing known
results for Gorenstein and Bass orders. Finally, we give a classification of
the ideal classes with multiplicator ring of type $2$, with applications to
the computations of the conjugacy classes of integral matrices and the
isomorphism classes of abelian varieties over finite fields.

###### Key words and phrases:

orders, Cohen-Macaulay type, ideal classes, generators

###### 1991 Mathematics Subject Classification:

16H10, 11Y40, 13H10, 15B36, 14K15.

##  1\. Introduction

Cohen-Macaulay rings play a very important role in algebraic geometry. The
associated invariants, such as the (Cohen-Macaulay) type, have been widely
studied in the literature. On the other hand, it seems that a large class of
Cohen-Macaulay rings coming from number theory, namely orders in number
fields, has been so far widely neglected. In this paper we work on filling
this gap, that is, we study orders in terms of their type.

The typical example of an order is the ring of integers ${\mathcal{O}}_{K}$ of
a number field $K$. The order ${\mathcal{O}}_{K}$ is a central object of study
for number theorists since the work of Dedekind and Noether. For example, we
know that all the completions of ${\mathcal{O}}_{K}$ at its maximal ideals
have type $1$. In fact, having (local) type $1$ characterizes a larger class
of Cohen-Macaulay rings, the so-called Gorenstein rings. The study of
Gorenstein rings started with the seminal ‘ubiquity’ paper by Bass [Bas63].
Gorenstein orders in $K$ share several properties with ${\mathcal{O}}_{K}$.
For example, every fractional ideal with Gorenstein multiplicator ring $S$ is
invertible, or, equivalently, locally principal. On the other hand, much less
is known about non-Gorenstein orders, that is, when there are maximal ideals
for which the local type is strictly greater than $1$.

We will now overview the structure of the paper and highlight the main
results. We will work in a more general context than orders in number fields.
We refer the reader to Section 2 for details and missing definitions. Let $Z$
be a Dedekind domain with field of fractions $Q$. Let $K$ be an étale algebra
over $Q$. Let $S$ be an order in $K$. In analogy with the ring of rational
integers $\mathbb{Z}$, the maximal ideals of $S$ will be called primes. Let
${\mathfrak{p}}$ be such a prime. In Section 3, we will study the type of $S$
at ${\mathfrak{p}}$, which is defined as the type of the completion
$\hat{S}_{\mathfrak{p}}$ of $S$ at ${\mathfrak{p}}$, that is,

$\mathrm{type}_{{\mathfrak{p}}}({S})=\dim_{k}\operatorname{Ext}_{\hat{S}_{\mathfrak{p}}}^{1}(k,\hat{S}_{\mathfrak{p}}),$

where $k$ is the residue field of $\hat{S}_{\mathfrak{p}}$. We will also use a
global notion of type by setting

$\mathrm{type}({S})=\max_{\mathfrak{p}}\left\\{\mathrm{type}_{{\mathfrak{p}}}({S})\right\\},$

where ${\mathfrak{p}}$ runs over the primes of $S$. This is finite quantity
because the local types are almost always $1$. See Proposition 3.3. We will
produce formulas to compute these types in terms of the trace dual $S^{t}$ of
$S$. In fact, the presence of such a non-degenerate trace will be crucial for
our investigations in the rest of the paper.

The first main results of the paper about the type are summarized here in Main
Theorem 1. See Proposition 3.1, Corollary 3.6 and Proposition 4.9 for the
complete statements.

######  Main Theorem 1.

Let $S$ be an order in $K$. Then

  * •

We have

$\mathrm{type}_{{\mathfrak{p}}}({S})=\dim_{S/{\mathfrak{p}}}\dfrac{S^{t}}{{\mathfrak{p}}S^{t}}.$

  * •

If $S\neq{\mathcal{O}}_{K}$ then

$\mathrm{type}({S})+1=\max\left\\{\dim_{S/{\mathfrak{p}}}\dfrac{({\mathfrak{p}}:{\mathfrak{p}})}{{\mathfrak{p}}}:{\mathfrak{p}}\text{
is a prime of }S\right\\}.$

  * •

Assume that $\dim_{Q}(K)>1$. Then the type of an order in $K$ is bounded by
$\dim_{Q}(K)-1$. Moreover, this bound is attained: if $p$ is a prime of $Z$
then the order $T=Z+p{\mathcal{O}}_{K}$ satisfies
$\mathrm{type}({T})=\dim_{Q}(K)-1$.

In Section 4, we study the type of an order $S$ together with the type of all
of its overorders $T$, that is, orders such that $S\subseteq
T\subseteq{\mathcal{O}}_{K}$. We relate these numbers with the quantity
$g(S)$, which is defined as the maximum, over the fractional $S$-ideals $I$,
of the size $g_{S}(I)$ of a minimal generating set of $I$ as an $S$-module.
The main result of the section is Theorem 4.7 and it is summarized here as
Main Theorem 2. We invite the reader to compare it to the case when $S$ is a
non-maximal Bass order, that is, such that all the overorders are Gorenstein.
See Proposition 4.6.

######  Main Theorem 2.

Let $S$ be a non-maximal order, $\mathcal{S}$ be the set of overorders $T$ of
$S$ and $d$ a positive integer. Then the following are equivalent:

  * •

$1+\max_{T\in\mathcal{S}}\left\\{\mathrm{type}({T})\right\\}=d$.

  * •

The $S$-module ${\mathcal{O}}_{K}/S$ is minimally generated by $d-1$ elements.

  * •

$g(S)=d$.

The motivation that originally started the investigation in this paper was to
improve our understanding of the isomorphism classes of fractional ideals, and
consequently, to improve the algorithms to compute them, which are described
in [Mar20a]. In Section 5 we recall such results, that where originally stated
for $\mathbb{Z}$-orders in étale $\mathbb{Q}$-algebras, but that also apply to
our more general setting. The Magma [BCP97] implementation of such algorithms
is available at https://github.com/stmar89/AlgEt. As anticipated above, we
completely understand the isomorphism classes of fractional ideals with
Gorenstein multiplicator ring. See Corollary 5.3. Also, as mentioned above,
and stated precisely in Proposition 3.4, the type can be seen as a measure of
how far an order is from being Gorenstein. In Section 6, we focus our
attention to the orders that, from this perspective, are close-to-being
Gorenstein, that is, have type $2$.

For an order $S$ and a prime ${\mathfrak{p}}$ with
$\mathrm{type}_{{\mathfrak{p}}}({S})=2$, we give locally at ${\mathfrak{p}}$ a
classification of the isomorphism classes of fractional $S$-ideals with
multiplicator ring $S$. If we globally have $\mathrm{type}({S})=2$, then we
can patch the local information to effectively produce representatives of the
isomorphism classes of fractional $S$-ideals with multiplicator ring $S$. The
set of such classes is denoted $\operatorname{ICM}_{S}(S)$ in the following
Main Theorem 3. We refer the reader to Theorem 6.2 and Corollary 6.5 for the
complete statements.

######  Main Theorem 3.

Let $S$ be an order in $K$. Then

  * •

Let ${\mathfrak{p}}$ be a prime of $S$ and $I$ a fractional $S$-ideal with
$(I:I)=S$. If $\mathrm{type}_{{\mathfrak{p}}}({S})=2$ then either
$I_{\mathfrak{p}}\simeq S_{\mathfrak{p}}$ or
$I_{\mathfrak{p}}\simeq(S^{t})_{\mathfrak{p}}$.

  * •

Assume that $\mathrm{type}({S})=2$. Let $N$ be the number of primes
${\mathfrak{p}}$ of $S$ such that $\mathrm{type}_{{\mathfrak{p}}}({S})=2$.
Then $|\operatorname{ICM}_{S}(S)|=2^{N}\cdot|\operatorname{Pic}(S)|$.

In Example 6.6, we show how to deduce information about $\mathbb{Z}$-conjugacy
classes of integral matrices and abelian varieties over finite fields from a
computation of isomorphism classes of fractional ideals.

Finally, in Section 7 we make comparisons with other notions of close-to-being
Gorenstein occurring in the recent literature.

### Acknowledgements

The author is grateful to Jonas Bergström and Hendrik W. Lenstra Jr. for
useful discussions and comments on a preliminary version of the paper. The
author thanks Carel Faber for useful comments. The author was supported by NWO
grant VI.Veni.202.107.

##  2\. Preliminaries

###  2.1. Conventions and notation

In the rest of the paper the word ring will mean a non-zero commutative ring
with unity. We will not consider fields as Dedekind Domains. In particular,
Dedekind domains have Krull dimension $1$ and are infinite.

We denote by ${\mathcal{Q}}(R)$ total ring of quotients of a ring $R$ and by
$R^{\times}$ the units of $R$. Given sub-$R$-modules $I$ and $J$ of
${\mathcal{Q}}(R)$ we define $I+J$ and $IJ$ in the natural way, and the colon
as

$(I:_{{\mathcal{Q}}(R)}J)=\left\\{x\in{\mathcal{Q}}(R)\>\ xJ\subseteq
I\right\\}.$

If the total ring of quotient is clear from the context, we will omit it from
the notation and write $(I:J)$ for $(I:_{{\mathcal{Q}}(R)}J)$.

######  Lemma 2.1.

Let $I$ and $J$ be sub-$R$-modules of ${\mathcal{Q}}(R)$ such that
$J\cap{\mathcal{Q}}(R)^{\times}$ is non-empty. Then
$J{\mathcal{Q}}(R)={\mathcal{Q}}(R)$ and the natural $R$-linear morphism

$\displaystyle{\varphi}:(I:J)$
$\displaystyle\longrightarrow\operatorname{Hom}_{R}(J,I)$ $\displaystyle x$
$\displaystyle\longmapsto({\varphi}_{x}:j\mapsto xj)$

is an isomorphism.

###### Proof.

Let $z\in J\cap{\mathcal{Q}}(R)^{\times}$. Then
${\mathcal{Q}}(R)=z{\mathcal{Q}}(R)\subseteq J{\mathcal{Q}}(R)$ and hence we
have the equality ${\mathcal{Q}}(R)=J{\mathcal{Q}}(R)$. Assume that for
$x\in(I:J)$ we have ${\varphi}_{x}=0$. Then $xz=0$ implies that $x=0$ and so
${\varphi}$ is injective. To prove the surjectivity we proceed as follows. For
an $R$-linear morphism $\psi:J\to I$ we get an induced
${\mathcal{Q}}(R)$-linear morphism $\tilde{\psi}:{\mathcal{Q}}(R)\to
I{\mathcal{Q}}(R)$ by setting $\tilde{\psi}(jq)=\psi(j)q$ for every $j\in J$
and $q\in{\mathcal{Q}}(R)$. Set $x=\tilde{\psi}(1)$. For every $j\in J$ we
have

$jx=j\tilde{\psi}(1)=\tilde{\psi}(j)=\psi(j)\in I$

which shows that $x\in(I:J)$ and that ${\varphi}_{x}=\psi$, completing the
proof. ∎

We say that $I$ is invertible if $I(R:I)=R$, and that $I$ is principal if
there exists an element $x\in{\mathcal{Q}}(R)^{\times}$ such that $I=xR$. Note
that if $I$ is principal, then it is invertible. The ring $(I:I)$ is called
the multiplicator ring of $I$. Observe that $R\subseteq(I:I)$.

A version of the following lemma can be found in [Mar20a, Lemma 2.5] with more
restrictive hypothesis, but the same proof.

######  Lemma 2.2.

Let $I$ be an invertible sub-$R$-module of ${\mathcal{Q}}(R)$. Then the
multiplicator ring $(I:I)$ of $I$ equals $R$.

###### Proof.

Using the equality $R=I(R:I)$ we obtain

$R(I:I)=(R:I)I(I:I)=(R:I)I=R.$

Therefore $(I:I)\subseteq R$, and hence $R=(I:I)$. ∎

######  Definition 2.3.

For an finitely generated $R$-module $M$ set

$g_{R}(M)=\min\left\\{d\in\mathbb{Z}_{>0}:M\text{ can be generated by $d$
elements over $R$}\right\\}.$

###  2.2. Orders and Trace form

Let $Z$ be a Dedekind domain with field of fractions $Q$. A $Q$-algebra $K$ is
called étale if there exist an integer $n\geq 0$ and an extension $Q\subseteq
Q^{\prime}$ such that $K\otimes_{Q}Q^{\prime}\simeq Q^{\prime n}$. Note that
an étale $Q$-algebra is commutative, unitary and finite dimensional. In fact,
a commutative finite dimensional $Q$-algebra $K$ is étale if and only if $K$
is isomorphic to a finite product of separable algebraic extensions of $Q$.

Let $K$ be a non-zero étale $Q$-algebra. By a $Z$-lattice in $K$ we mean a
finitely generated sub-$Z$-module $L$ of $K$ that contains a $Q$-basis of $K$,
or equivalently, such that $\dim_{Q}(L\otimes_{Z}Q)=\dim_{Q}(K)$. A $Z$-order
in $K$ is a subring of $K$ which is also a $Z$-lattice in $K$. When the
Dedekind domain $Z$ is clear from the context, we will omit it from the
terminology and talk simply about lattices and orders. An order $S$ is
reduced, Noetherian, integral over $Z$, and has Krull dimension one. Hence
orders are Cohen-Macaulay rings. Note that $K$ is the total ring of quotients
of $S$. By an overorder of $S$, we mean an order $S^{\prime}$ such that
$S\subseteq S^{\prime}\subset K$.

Let ${\mathcal{O}}_{K}$ be the integral closure of $Z$ in $K$. Since $Z$ is a
Dedekind domain and $K$ is étale, one shows that ${\mathcal{O}}_{K}$ is an
order in $K$ and contains all orders in $K$. Therefore we will call
${\mathcal{O}}_{K}$ the maximal order of $K$.

Pick an element $x$ in $K$. After fixing a basis of $K$ over $Q$, we can
represent the multiplication by $x$ on $K$ by a matrix $m_{x}$ with
coefficients in $Q$, whose trace $\operatorname{Tr}(x)$ does not depend on the
choice of the basis. Since $K$ is étale, the induced bilinear form

$K\times K\to Q\qquad(x,y)\mapsto\operatorname{Tr}(xy)$

is non-degenerate.

Let $L$ be a lattice in $K$. The trace dual lattice $L^{t}$ of $L$ is defined
as

$L^{t}=\left\\{x\in K:\operatorname{Tr}(xL)\subseteq Z\right\\}.$

We have a canonical isomorphism of $Z$-modules

(1) $L^{t}\simeq\operatorname{Hom}_{Z}(L,Z),$

induced by $x\mapsto(y\mapsto\operatorname{Tr}(xy))$. Observe that
$(L^{t})^{t}=L$.

Given two lattices $L$ and $L^{\prime}$ in $K$ we define the colon
$(L:L^{\prime})$ by

$(L:L^{\prime})=\left\\{x\in K\ :\ xL^{\prime}\subseteq L\right\\}.$

Note that there $(L:L^{\prime})$ is a lattice and contains a non-zero element
of $Z$. Similarly, the multiplicator ring of $L$ is the order $(L:L)$. Observe
that the isomorphism in (1) is $S$-linear for every subring $S$ of $(L:L)$.

###  2.3. Fractional ideals

Let $Z$ be a Dedekind domain with field of fractions $Q$. Let $S$ be an order
in a non-zero étale $Q$-algebra $K$. A fractional $S$-ideal $I$ is a finitely
generated sub-$S$-module of $K$ which is also a lattice. Note that a finitely
generated sub-$S$-module $I$ of $K$, for example an ideal of $S$, is a
fractional $S$-ideal if and only if $I$ contains a non-zero divisor of $K$.
Given fractional $S$-ideals $I$ and $J$, by Lemma 2.1, we have canonical
identifications

(2)
$\operatorname{Hom}_{S}(J,I)\simeq(I:J)\quad\text{and}\quad\operatorname{End}_{S}(I)\simeq(I:I).$

######  Lemma 2.4.

Let $I$ and $J$ be fractional $S$-ideals.

  1. (i)

$IJ$, $I+J$, $I\cap J$, $(I:J)$ and $I^{t}$ are fractional $S$-ideals.

  2. (ii)

$I$ contains a non-zero element of $Z$.

  3. (iii)

$IJ^{t}=(J:I)^{t}$.

  4. (iv)

If $J\subseteq I$ then the $Z$-modules $I/J$ and $J^{t}/I^{t}$ are isomorphic.

###### Proof.

Part i is clear. For Part ii, since $S$ and $I$ are lattices, then the colon
$(I:S)=I$ contains a non-zero element of $Z$. Part iii is a direct
verification.

We now prove Part iv. Put $M=I/J$. Observe that $M$ is a torsion module over
the Dedekind domain $Z$. Hence $M$ has finite length as a $Z$-module. Note
that $E=Q/Z$ is a minimal injective cogenerator of $Z$, by for example [Lam99,
Examples 19.11.(1)]. It is known that every module of finite length over a
Dedekind domain is isomorphic to its Matlis dual. See for example [Ooi76,
Theorem 1.7]. Hence we have an isomorphism of $Z$-modules

$M\simeq\operatorname{Hom}_{Z}(M,E).$

To conclude the proof, it is enough to observe that we have an $S$-linear
isomorphism

(3) $J^{t}/I^{t}\simeq\operatorname{Hom}_{Z}(M,E)$

induced by

$x+I^{t}\mapsto(y+J\mapsto\operatorname{Tr}(xy)+Z).$

See for example [BF65, page 59]. ∎

The next lemma, which will be used later in the paper, shows that we can
always find generators of a fractional ideal which are non-zero divisors.

######  Lemma 2.5.

Let $I$ be fractional $S$-ideal in $K$, say $I=y_{1}S+\ldots+y_{r}S$. Then
there are elements $x_{0},x_{1},\ldots,x_{r}$ in $I\cap K^{\times}$ which
generate $I$ as an $S$-module.

###### Proof.

By Lemma 2.4.ii, the set $I\cap K^{\times}$ is non-empty, say containing
$x_{0}$. Write $K=K_{1}\times\ldots\times K_{n}$, with $K_{i}$ fields
extension of $Q$, and denote by $\pi_{i}$ the projection $K\twoheadrightarrow
K_{i}$. Observe that $\pi_{i}(x_{0})\neq 0$ for every $i$. Now, for each
generator $y_{j}$ of $I$, set $x_{j}=y_{j}-z_{j}x_{0}$ for $z_{j}\in Z$ such
that

$\pi_{i}(y_{j}-z_{j}x_{0})\neq 0$

for every $i$. Observe that such a $z_{j}$ exists since we have finitely many
constrains while $Z$ is infinite. Moreover $x_{j}S+x_{0}S=y_{j}S+x_{0}S$ which
shows that the elements $x_{0},x_{1},\ldots,x_{r}$ generate $I$ over $S$. ∎

###  2.4. Primes

Let $Z$ be a Dedekind domain with field of fractions $Q$. Let $S$ be an order
in a non-zero étale $Q$-algebra $K$. A prime111In [Rei03] prime ideals of
orders are required to be fractional ideals, that is, they coincide with what
we call primes. of $S$ is a fractional $S$-ideal which is also a prime ideal
of $S$. We now state some lemmas that are well known.

######  Lemma 2.6 ([Rei03, Theorem 22.3]).

The primes of $S$ are precisely the maximal ideals of $S$. If ${\mathfrak{p}}$
is a prime of $S$, then the contraction $p$ of ${\mathfrak{p}}$ in $Z$ is a
prime of $Z$, and the field $S/{\mathfrak{p}}$ is a finite extension of $Z/p$.

######  Lemma 2.7 ([Rei03, Theorem 23.1]).

Let $I\subsetneq S$ be a fractional $S$-ideal. Then $I$ contains a product of
primes of $S$.

######  Lemma 2.8.

Let ${\mathfrak{p}}$ be a prime of $S$. Then $S\subsetneq(S:{\mathfrak{p}})$.
If ${\mathfrak{p}}$ is invertible then $({\mathfrak{p}}:{\mathfrak{p}})=S$,
while if ${\mathfrak{p}}$ is not invertible then
$({\mathfrak{p}}:{\mathfrak{p}})=(S:{\mathfrak{p}})$.

###### Proof.

Let $x$ be a non-zero divisor in ${\mathfrak{p}}$, which exists by Lemma
2.4.ii. By Lemma 2.7, we have

${\mathfrak{p}}\supseteq
xS\supseteq{\mathfrak{p}}_{1}\cdots{\mathfrak{p}}_{r}$

for primes ${\mathfrak{p}}_{i}$ of $S$. We assume that $r$ is minimal. If
$r=1$ then ${\mathfrak{p}}=xS$. Hence $(S:{\mathfrak{p}})=(1/x)S$ which
strictly contains $S$ since $x$ is not a unit of $S$. So assume that $r>1$.
Then there exists $i$ such that ${\mathfrak{p}}_{i}={\mathfrak{p}}$. We assume
that ${\mathfrak{p}}_{1}={\mathfrak{p}}$. By minimality of $r$, there exists
$y\in{\mathfrak{p}}_{2}\cdots{\mathfrak{p}}_{r}\setminus xS$. In particular
$(y/x)\not\in S$. Also, observe that $(y/x)\in(S:{\mathfrak{p}})$ by
construction. Therefore we have concluded the proof of the first part.

We now prove the second. The inclusion
$({\mathfrak{p}}:{\mathfrak{p}})\subseteq(S:{\mathfrak{p}})$ holds by
definition. If ${\mathfrak{p}}$ is invertible then we conclude by Lemma 2.2.
Assume that ${\mathfrak{p}}$ is not invertible, that is,
${\mathfrak{p}}(S:{\mathfrak{p}})\subsetneq S$. Since ${\mathfrak{p}}$ is
maximal by Lemma 2.6, one obtains that
${\mathfrak{p}}(S:{\mathfrak{p}})={\mathfrak{p}}$. Therefore we also have
$(S:{\mathfrak{p}})\subseteq({\mathfrak{p}}:{\mathfrak{p}})$. ∎

######  Lemma 2.9.

Every fractional $S$-ideal $I\subset S$ is contained in finitely many primes
of $S$.

###### Proof.

Observe that the prime ideals of $S$ containing $I$ are necessarily fractional
ideals, hence they are primes by Lemma 2.6. In other words, the ring $S/I$ has
Krull dimension $0$, and since it is Noetherian, it is also Artinian.
Therefore it has only finitely many maximal ideals, which correspond to the
primes of $S$ containing $I$. ∎

###  2.5. Localization

Let $Z$ be a Dedekind domain with field of fractions $Q$. Let $S$ be a
$Z$-order in a non-zero étale $Q$-algebra $K$. Let ${\mathfrak{p}}$ be a prime
ideal of $S$. We will denote by $S_{\mathfrak{p}}$ the localization of $S$ at
${\mathfrak{p}}$ and by $Z_{p}$ the localization of $Z$ at the contraction $p$
of ${\mathfrak{p}}$ in $Z$. Observe that $Z_{p}$ is a overring of $Z$ and has
field of fractions $Q$. On the other hand, in general, the total ring of
quotients of $S_{\mathfrak{p}}$ is not $K$.

######  Lemma 2.10.

The total ring of quotients ${\mathcal{Q}}(S_{\mathfrak{p}})$ is a quotient
$Q$-algebra of $K$. The restriction to $S$ of quotient morphism
$\pi_{\mathfrak{p}}:K\twoheadrightarrow{\mathcal{Q}}(S_{\mathfrak{p}})$
induces the localization morphism $S\to S_{\mathfrak{p}}$.

###### Proof.

Since $S$ is a Noetherian reduced ring, the set of zero-divisors of $S$ is the
union of the finitely many minimal prime ideals $P_{1},\ldots,P_{n}$ of $S$.
Hence the total ring of quotients $K$ of $S$ satisfies

$K\simeq{\mathcal{Q}}(S/P_{1})\times\ldots\times{\mathcal{Q}}(S/P_{n}).$

After possibly reindexing, assume that $P_{1},\ldots,P_{s}$ are the minimal
prime ideals of $S$ which are contained in ${\mathfrak{p}}$. We obtain that

${\mathcal{Q}}(S_{\mathfrak{p}})\simeq{\mathcal{Q}}(S_{\mathfrak{p}}/P_{1}S_{\mathfrak{p}})\times\ldots\times{\mathcal{Q}}(S_{\mathfrak{p}}/P_{s}S_{\mathfrak{p}}).$

Now observe that for each $i=1,\ldots,s$ we have a natural isomorphism
${\mathcal{Q}}(S_{\mathfrak{p}}/P_{i}S_{\mathfrak{p}})\simeq{\mathcal{Q}}(S/P_{i})$.
Hence we get a natural quotient structure

$\pi_{\mathfrak{p}}:K\twoheadrightarrow{\mathcal{Q}}(S_{\mathfrak{p}}).$

The fact that $\pi_{\mathfrak{p}}$ restricted to $S$ induces the localization
morphism is a straightforward verification. ∎

In what follows we will denote by $I_{\mathfrak{p}}$ the localization of a
fractional $S$-ideal $I$ at the prime ${\mathfrak{p}}$ of $S$. Observe that
$I_{\mathfrak{p}}$ can be naturally considered a sub-$S_{\mathfrak{p}}$-module
of ${\mathcal{Q}}(S_{\mathfrak{p}})$. In fact, by Lemma 2.10, we get

$I_{\mathfrak{p}}=\pi_{\mathfrak{p}}(I)S_{\mathfrak{p}}.$

In particular, given fractional $S$-ideals $I$ and $J$ we have that
$J_{\mathfrak{p}}$ contains a non-zero divisor of
${\mathcal{Q}}(S_{\mathfrak{p}})$. Therefore by Lemma 2.1 we have a natural
isomorphism of $S_{\mathfrak{p}}$-modules

$(I_{\mathfrak{p}}:_{{\mathcal{Q}}(S_{\mathfrak{p}})}J_{\mathfrak{p}})\simeq\operatorname{Hom}_{S_{\mathfrak{p}}}(J_{\mathfrak{p}},I_{\mathfrak{p}}).$

######  Lemma 2.11.

Let ${\mathfrak{p}}$ be a prime of $S$. For fractional $S$-ideals $I$ and $J$,
we have the following equalities:

$\displaystyle(I+J)_{\mathfrak{p}}=I_{\mathfrak{p}}+J_{\mathfrak{p}},\quad(IJ)_{\mathfrak{p}}=I_{\mathfrak{p}}J_{\mathfrak{p}},\quad(I\cap
J)_{\mathfrak{p}}=I_{\mathfrak{p}}\cap J_{\mathfrak{p}},\quad\text{and}$
$\displaystyle\quad(I:_{K}J)_{\mathfrak{p}}=(I_{\mathfrak{p}}:_{{\mathcal{Q}}(S_{\mathfrak{p}})}J_{\mathfrak{p}}).$

###### Proof.

The statement for addition, product and intersection follows immediately. For
the colon, we proceed as follows. Let $x_{1},\ldots,x_{r}$ be generators of
$J$ over $S$. By Lemma 2.5 we can assume that $x_{i}\in K^{\times}$ for each
$i$. Then

$(I:J)_{\mathfrak{p}}=\left(\bigcap_{i=1}^{r}(I:x_{i}S)\right)_{\mathfrak{p}}=\bigcap_{i=1}^{r}\frac{1}{\pi_{\mathfrak{p}}(x_{i})}I_{\mathfrak{p}}=\bigcap_{i=1}^{r}\left(I_{\mathfrak{p}}:\pi_{\mathfrak{p}}(x_{i})S_{\mathfrak{p}}\right)=\bigcap_{i=1}^{r}(I_{\mathfrak{p}}:J_{\mathfrak{p}}).$

∎

######  Lemma 2.12.

Let $I$ be a fractional $S$-ideal. Then the following statements are
equivalent:

  1. (i)

$I$ is invertible.

  2. (ii)

for every prime ${\mathfrak{p}}$ of $S$ the localization $I_{\mathfrak{p}}$ is
principal.

  3. (iii)

for every prime ${\mathfrak{p}}$ of $S$ the $S/{\mathfrak{p}}$-vector space
$I/{\mathfrak{p}}I$ has dimension $1$.

###### Proof.

Assume i, that is, $I(S:I)=S$. Then by Lemma 2.11, for every prime
${\mathfrak{p}}$ of $S$ we have

$1=x_{1}y_{1}+\dots+x_{n}y_{n}$

for elements $x_{i}\in I_{\mathfrak{p}}$ and
$y_{i}\in(S_{\mathfrak{p}}:I_{\mathfrak{p}})$. Since $S_{\mathfrak{p}}$ is a
local ring, there exists an index $j$ such that $x_{j}y_{j}$ is a unit in
$S_{\mathfrak{p}}$. Hence

$I_{\mathfrak{p}}\subseteq x_{j}y_{j}I_{\mathfrak{p}}\subseteq
x_{j}S_{\mathfrak{p}}\subseteq I_{\mathfrak{p}},$

that is, $I_{\mathfrak{p}}=x_{j}S_{\mathfrak{p}}$. Note that $x_{j}$ is a non-
zero divisor.

Conversely, assume ii. Then $I_{\mathfrak{p}}$ is invertible for every prime
${\mathfrak{p}}$ of $S$. If $I$ were not invertible, then there would be a
prime ${\mathfrak{p}}$ of $S$ such that $I(S:I)\subseteq{\mathfrak{p}}$. Then
localization at ${\mathfrak{p}}$ together with Lemma 2.11 yields a
contradiction.

The equivalence of ii and iii is a consequence of Nakayama’s Lemma. ∎

######  Lemma 2.13.

Let ${\mathfrak{p}}$ be an invertible prime of $S$. Then for every fractional
$S$-ideal $I$ we have that $I_{\mathfrak{p}}$ is principal. If moreover
$I\subseteq S$ then $I_{\mathfrak{p}}$ equals a power of
${\mathfrak{p}}S_{\mathfrak{p}}$.

###### Proof.

First assume that $I\subseteq S$. If $I=S$ then the statement is true. So
assume that $I\subset S$. Put ${\mathfrak{m}}={\mathfrak{p}}S_{\mathfrak{p}}$
and ${\mathfrak{m}}^{-1}=(S_{\mathfrak{p}}:{\mathfrak{m}})$. By Lemma 2.12 we
have that ${\mathfrak{m}}$ is principal. Also, $I$ contains a product of
primes of $S$ by Lemma 2.7. Hence the localization $I_{\mathfrak{p}}$ contains
a power of ${\mathfrak{m}}$, say $I_{\mathfrak{p}}\supseteq{\mathfrak{m}}^{n}$
with $n$ minimal. We will now show that $I_{\mathfrak{p}}={\mathfrak{m}}^{n}$
by induction. If $n=1$ then $I_{\mathfrak{p}}={\mathfrak{m}}$. On the other
hand, if $n>1$, since ${\mathfrak{m}}$ is invertible in $S_{\mathfrak{p}}$,
then $I_{\mathfrak{p}}{\mathfrak{m}}^{-1}\supseteq{\mathfrak{m}}^{n-1}$. By
induction we obtain
$I_{\mathfrak{p}}{\mathfrak{m}}^{-1}={\mathfrak{m}}^{n-1}$, and hence
$I_{\mathfrak{p}}={\mathfrak{m}}^{n}$. Since $I_{\mathfrak{p}}$ is a power of
a principal ideal, then it is principal as well.

Now consider the general case. Let $x$ be a non-zero divisor in $(S:I)$ which
exists by Lemma 2.4.ii. The previous argument shows that $xI$ is principal,
say $xI=yS$. Then $I=(y/x)S$ and we conclude. ∎

######  Lemma 2.14.

The following statements hold:

  1. (i)

Let ${\mathfrak{p}}$ be a prime of $S$ and
${\mathfrak{f}}=(S:{\mathcal{O}}_{K})$ be the conductor of $S$. Then the
following statements are equivalent:

    1. (a)

${\mathfrak{p}}$ is coprime to ${\mathfrak{f}}$, that is,
${\mathfrak{p}}+{\mathfrak{f}}=S$.

    2. (b)

$S_{\mathfrak{p}}={\mathcal{O}}_{K,{\mathfrak{p}}}$.

    3. (c)

the $S/{\mathfrak{p}}$-vector space
${\mathcal{O}}_{K}/{\mathfrak{p}}{\mathcal{O}}_{K}$ has dimension one.

    4. (d)

${\mathfrak{p}}$ is invertible.

  2. (ii)

There are finitely many primes of $S$ which are not invertible.

###### Proof.

We first prove Part i. Observe that ${\mathfrak{p}}$ is coprime to
${\mathfrak{f}}$ if and only if
${\mathfrak{f}}_{\mathfrak{p}}=S_{\mathfrak{p}}$, which is equivalent to
$S_{\mathfrak{p}}={\mathcal{O}}_{K,{\mathfrak{p}}}$ by Lemma 2.11. Hence
$\ref{invDVR:1}\leavevmode\nobreak\ \Leftrightarrow\leavevmode\nobreak\
\ref{invDVR:2}$. Also, it is an immediate consequence of Nakayama’s Lemma that
$\ref{invDVR:2}\leavevmode\nobreak\ \Leftrightarrow\leavevmode\nobreak\
\ref{invDVR:3}$. Now, assume ib. Then ${\mathfrak{p}}S_{\mathfrak{p}}$ is
invertible. Since all the localizations of ${\mathfrak{p}}$ at the other
primes of $S$ are trivial, by Lemma 2.12 we get that ${\mathfrak{p}}$ is
invertible, that is id holds. Conversely, assume id. By Lemma 2.13
${\mathfrak{f}}_{\mathfrak{p}}$ is invertible. Hence its multiplicator ring
equals $S_{\mathfrak{p}}$ by Lemma 2.2. But it is easy to see that the
multiplicator ring of ${\mathfrak{f}}$ contains, and hence equals,
${\mathcal{O}}_{K}$. Therefore we conclude that
$S_{\mathfrak{p}}={\mathcal{O}}_{K,{\mathfrak{p}}}$, that is, ib holds.

Finally, by Part i we have that the non-invertible primes of $S$ are precisely
the ones containing the conductor ${\mathfrak{f}}$ of $S$, and by Lemma 2.9
there is only finitely many of them, which gives us Part ii. ∎

Noticeably, in Lemma 2.11 we have not mentioned the trace duals. The reason is
that the in general the localization $I_{\mathfrak{p}}$ of a fractional
$S$-ideal $I$ is not a $Z_{p}$-lattice in ${\mathcal{Q}}(S_{\mathfrak{p}})$,
as the next lemma shows.

######  Lemma 2.15.

Let ${\mathfrak{p}}$ be a prime of $S$ and $p$ its contraction in $Z$. The
ring $S_{\mathfrak{p}}$ is a $Z_{p}$-lattice if and only if ${\mathfrak{p}}$
is the unique prime of $S$ above $p$.

###### Proof.

Assume that there exists another prime ${\mathfrak{p}}^{\prime}$ containing
$p$. Pick $x\in{\mathfrak{p}}^{\prime}\setminus{\mathfrak{p}}$, and consider
$1/x\in S_{\mathfrak{p}}$. Then the minimal polynomial of $1/x$ over $Q$ has
denominators in $pZ$ and hence $1/x$ is not integral over $Z_{p}$. Hence
$S_{\mathfrak{p}}$ is not finitely generated over $Z_{p}$.

Conversely, if ${\mathfrak{p}}$ is the unique prime above $p$, then the
natural map

$S\otimes_{Z}Z_{p}\to S_{\mathfrak{p}}$

is an isomorphism. In particular, since $S\otimes_{Z}Z_{p}$ is a finitely
generated $Z_{p}$-module, then $S_{\mathfrak{p}}$ is the same. ∎

###  2.6. Completions

Let $Z$ be a Dedekind domain with field of fractions $Q$. Let $S$ be a
$Z$-order in a non-zero étale $Q$-algebra $K$. For a prime ${\mathfrak{p}}$ of
$S$ we denote by $\hat{S}_{\mathfrak{p}}$ the completion of $S$ at
${\mathfrak{p}}$. Given a fractional $S$-ideal $I$ we put
$\hat{I}_{\mathfrak{p}}=I\otimes_{S}\hat{S}_{\mathfrak{p}}$.

######  Lemma 2.16.

Let ${\mathfrak{p}}$ be a prime of $S$ and $p$ its contraction to $Z$. The
total ring of quotients ${\mathcal{Q}}(\hat{S}_{\mathfrak{p}})$ is an étale
${\mathcal{Q}}(\hat{Z}_{p})$-algebra. The completion $\hat{S}_{\mathfrak{p}}$
is a $\hat{Z}_{p}$-order in ${\mathcal{Q}}(\hat{S}_{\mathfrak{p}})$. Given a
fractional $S$-ideal $I$, we have that $\hat{I}_{\mathfrak{p}}$ is a
fractional $\hat{S}_{\mathfrak{p}}$-ideal.

###### Proof.

By [Bou89, Cor. Ch.III §.13 p.195], since $S\otimes_{Z}\hat{Z}_{p}$ is a
complete semilocal ring, we have an isomorphism

(4)
$S\otimes_{Z}\hat{Z_{p}}\simeq\hat{S}_{{\mathfrak{p}}_{1}}\oplus\ldots\oplus\hat{S}_{{\mathfrak{p}}_{s}},$

where ${\mathfrak{p}}_{1},\ldots,{\mathfrak{p}}_{s}$ are the primes of $S$
above $p$. Applying $-\otimes_{Z}Q$ to this isomorphism shows that
${\mathcal{Q}}(\hat{S}_{{\mathfrak{p}}_{i}})$ is a
${\mathcal{Q}}(\hat{Z}_{p})$-algebra for each $i$, such that the induced trace
map gives a non-degenerate bilinear form. On the other hand, applying
$-\otimes_{S}I$ shows that $\hat{I}_{{\mathfrak{p}}_{i}}$ is a fractional
$\hat{S}_{{\mathfrak{p}}_{i}}$-ideal for each $i$. ∎

######  Lemma 2.17.

Let $I$ be fractional $S$-ideal. Then the following are equivalent.

  1. (i)

$I$ is invertible.

  2. (ii)

$\hat{I}_{\mathfrak{p}}$ is a principal fractional
$\hat{S}_{\mathfrak{p}}$-ideal for every prime ${\mathfrak{p}}$ of $S$.

###### Proof.

Observe that we have a canonical isomorphism

$\frac{I}{{\mathfrak{p}}I}\simeq\frac{\hat{I}_{\mathfrak{p}}}{{\mathfrak{p}}\hat{I}_{\mathfrak{p}}}$

of
$S/{\mathfrak{p}}\simeq\hat{S}_{\mathfrak{p}}/{\mathfrak{p}}\hat{S}_{\mathfrak{p}}$-vector
spaces. The result follows from Lemma 2.12. ∎

The next lemma, together with Equation (1), shows that taking trace duals
commutes with completion. In particular, for a fractional $S$-ideal $I$ and a
prime ${\mathfrak{p}}$ of $S$, the notation $\hat{I}_{\mathfrak{p}}^{t}$ is
not ambiguous.

######  Lemma 2.18.

Given a fractional $S$-ideal $I$ and a prime ${\mathfrak{p}}$ of $S$ with
contraction $p$ in $Z$, we have a natural $\hat{S}_{\mathfrak{p}}$-linear
isomorphism

$\operatorname{Hom}_{Z}(I,Z)\otimes_{S}\hat{S}_{\mathfrak{p}}\simeq\operatorname{Hom}_{\hat{Z}_{p}}(\hat{I}_{\mathfrak{p}},\hat{Z}_{p}).$

###### Proof.

Since $\hat{Z}_{p}$ is a flat $Z$-algebra we get a natural isomorphism

(5)
$\operatorname{Hom}_{Z}(I,Z)\otimes_{Z}\hat{Z}_{p}\simeq\operatorname{Hom}_{\hat{Z}_{p}}(I\otimes_{Z}\hat{Z}_{p},\hat{Z}_{p}).$

By Equation (4) we get the following decompositions of $S$-modules

$\operatorname{Hom}_{Z}(I,Z)\otimes_{Z}\hat{Z}_{p}\simeq\bigoplus_{i=1}^{s}\left(\operatorname{Hom}_{Z}(I,Z)\otimes_{S}\hat{S}_{{\mathfrak{p}}_{i}}\right)$

and

$\operatorname{Hom}_{\hat{Z}_{p}}(I\otimes_{Z}\hat{Z}_{p},\hat{Z}_{p})\simeq\bigoplus_{i=1}^{s}\operatorname{Hom}_{\hat{Z}_{p}}(\hat{I}_{{\mathfrak{p}}_{i}},\hat{Z}_{p}),$

where ${\mathfrak{p}}_{1},\ldots,{\mathfrak{p}}_{s}$ are the primes of $S$
above $p$. Since they are both naturally induced by same isomorphism from
Equation (4), it is not hard to see that the isomorphism in Equation (5)
decompososes into a direct sum of isomorphisms

$\operatorname{Hom}_{Z}(I,Z)\otimes_{S}\hat{S}_{{\mathfrak{p}}_{i}}\simeq\operatorname{Hom}_{\hat{Z}_{p}}(\hat{I}_{{\mathfrak{p}}_{i}},\hat{Z}_{p}),$

as required. ∎

##  3\. Cohen-Macaulay type of an order

We start this section by recalling some definitions. We refer the reader to
[BH93] for more details. Let $R$ be a Cohen-Macaulay local ring with residue
field $k$. A finitely generated $R$-module $C$ is called a canonical module if
$\dim_{k}\operatorname{Ext}_{R}^{i}(k,C)$ is $1$ for $i=\dim R$ and $0$
otherwise. All canonical modules of $R$, if any, are isomorphic. The type of
$R$ is defined as $\dim_{k}\operatorname{Ext}_{R}^{\dim R}(k,R)$. In this
section we will see that it is easy to compute the type for the completion of
an order at a prime. Furthermore, we will define a global notion of type for
an order and we will start studying its properties.

Let $Z$ be a Dedekind domain with field of fractions $Q$. Let $K$ be a non-
zero $Q$-étale algebra and let $S$ be an order in $K$.

######  Proposition 3.1.

Let ${\mathfrak{p}}$ be a prime of $S$. Then

  1. (i)

$\hat{S}^{t}_{\mathfrak{p}}$ is a canonical module of
$\hat{S}_{\mathfrak{p}}$.

  2. (ii)

The type of $\hat{S}_{\mathfrak{p}}$ equals

$\dim_{S/{\mathfrak{p}}}\frac{S^{t}}{{\mathfrak{p}}S^{t}}.$

###### Proof.

Let $p$ be the contraction of ${\mathfrak{p}}$ in $Z$. Since $\hat{Z}_{p}$ is
a DVR, it is easy to see from the definition that
$\hat{Z}_{p}\simeq\operatorname{Hom}_{\hat{Z}_{p}}(\hat{Z}_{p},\hat{Z}_{p})$
is a canonical module of $\hat{Z}_{p}$. Since $Z$ and $S$ have both Krull
dimension $1$, from [BH93, Thm. 3.3.7.(b)] we get that
$C=\operatorname{Hom}_{\hat{Z}_{p}}(\hat{S}_{\mathfrak{p}},\hat{Z}_{p})$ is a
canonical module of $\hat{S}_{\mathfrak{p}}$. To conclude, it is enough to
observe that $C$ is canonically isomorphic to $\hat{S}_{\mathfrak{p}}^{t}$ by
Equation (1). See also [JT15, 3.4]. For the second part, observe that by
Nakayama’s lemma a minimal generating set of $\hat{S}^{t}_{\mathfrak{p}}$ over
$\hat{S}_{\mathfrak{p}}$ has size equal to the dimension of
$\hat{S}^{t}_{\mathfrak{p}}/{\mathfrak{p}}\hat{S}^{t}_{\mathfrak{p}}\simeq
S^{t}/{\mathfrak{p}}S^{t}$ as a vector space over
$\hat{S}_{\mathfrak{p}}/{\mathfrak{p}}\hat{S}_{\mathfrak{p}}\simeq
S/{\mathfrak{p}}$. The statement then follows from [BH93, Prop.
3.3.11.(c).(i)]. ∎

Proposition 3.1 tells us that the terminology used in the following definition
is compatible with the usual notion of type of a local ring.

######  Definition 3.2.

Let ${\mathfrak{p}}$ be a prime of $S$. We say that:

  1. (i)

the (Cohen-Macaulay) type of $S$ at ${\mathfrak{p}}$ is

$\mathrm{type}_{{\mathfrak{p}}}({S}):=\dim_{S/{\mathfrak{p}}}\frac{S^{t}}{{\mathfrak{p}}S^{t}}.$

  2. (ii)

the (Cohen-Macaulay) type of $S$ is

$\mathrm{type}({S}):=\max\left\\{\mathrm{type}_{{\mathfrak{p}}}({S}):{\mathfrak{p}}\text{
is a prime of }S\right\\}.$

Note that $\mathrm{type}({S})$ is a finite quantity since
$\mathrm{type}_{{\mathfrak{p}}}({S})=1$ for almost all primes ${\mathfrak{p}}$
of $S$, as we show in the next Proposition.

######  Proposition 3.3.

Let ${\mathfrak{p}}$ be an invertible prime of $S$. Then

$\mathrm{type}_{{\mathfrak{p}}}({S})=1.$

In particular $\mathrm{type}({S})$ is a finite positive integer.

###### Proof.

By Lemma 2.13, $(S^{t})_{\mathfrak{p}}$ is a principal
$S_{\mathfrak{p}}$-module. This means that the $S/{\mathfrak{p}}$-vector space
$S^{t}/{\mathfrak{p}}S^{t}$ has dimension $1$. Hence we conclude by
Proposition 3.1. Since there are only finitely many non-invertible primes by
Lemma 2.14.ii, we get that $\mathrm{type}({S})$ is finite. ∎

In Proposition 4.9, once more machinery will be available, we will exhibit an
upper bound for the type of the orders in $K$.

The type is a way to encode how far an order is from being Gorenstein. Recall
that a Noetherian ring is Gorenstein if all its localizations have finite
injective dimension. There are several well known equivalent characterizations
of Gorenstein orders. We now recall some of them, which play a role later in
the paper.

######  Proposition 3.4.

The following statements are equivalent:

  * •

$S$ is Gorenstein.

  * •

$\mathrm{type}({S})=1$.

  * •

$S^{t}$ is invertible.

  * •

Every fractional $S$-ideal $I$ with $S=(I:I)$ is invertible.

###### Proof.

By [BH93, Proposition 3.1.19(c)] we have that $S_{\mathfrak{p}}$ is Gorenstein
if and only if $\hat{S}_{\mathfrak{p}}$ is so. Hence, by [BH93, Theorem
3.3.7(a)], $S$ is Gorenstein if and only if, for every prime ${\mathfrak{p}}$
of $S$, the localization $\hat{S}^{t}_{\mathfrak{p}}$ is a principal
fractional $\hat{S}_{\mathfrak{p}}$-ideal, which by Proposition 3.1 is
equivalent to have $\mathrm{type}({S})=1$. The equivalence of the second and
third statements follows from Lemmas 2.17 and 2.12. Finally, the last two
statements are equivalent because $S=(I:I)$ if and only if $II^{t}=S^{t}$,
which is an application of Lemma 2.4.iii. See also [BL94, Proposition 2.7]. ∎

We conclude this section with formulas to compute the type of an order which
will be used later in the paper.

######  Proposition 3.5.

Let ${\mathfrak{p}}$ be a non-invertible prime of $S$. Then

$\mathrm{type}_{{\mathfrak{p}}}({S})+1=\dim_{S/{\mathfrak{p}}}\frac{({\mathfrak{p}}:{\mathfrak{p}})}{{\mathfrak{p}}}.$

###### Proof.

Using Lemmas 2.4.iii and 2.8, we see that
$({\mathfrak{p}}S^{t})^{t}=(S:{\mathfrak{p}})=({\mathfrak{p}}:{\mathfrak{p}})$.
Therefore, the quotient

$\frac{({\mathfrak{p}}S^{t})^{t}}{S}=\frac{({\mathfrak{p}}:{\mathfrak{p}})}{S}$

is isomorphic (as a $Z$-module) to $S^{t}/{\mathfrak{p}}S^{t}$ by Lemma
2.4.iv. Recall that $S/{\mathfrak{p}}$ is a finite field extension of $Z/p$,
where $p$ is the contraction of ${\mathfrak{p}}$ in $Z$, by Lemma 2.6. Hence,
comparing the dimensions over $Z/p$, we conclude that

$\dim_{S/{\mathfrak{p}}}\frac{({\mathfrak{p}}:{\mathfrak{p}})}{{\mathfrak{p}}}=\dim_{S/{\mathfrak{p}}}\left(\frac{({\mathfrak{p}}:{\mathfrak{p}})}{S}\oplus\frac{S}{{\mathfrak{p}}}\right)=\dim_{S/{\mathfrak{p}}}\frac{S^{t}}{{\mathfrak{p}}S^{t}}+1.$

∎

######  Corollary 3.6.

Let $S$ be a non-maximal order. Then

$\mathrm{type}({S})+1=\max\left\\{\dim_{S/{\mathfrak{p}}}\frac{({\mathfrak{p}}:{\mathfrak{p}})}{{\mathfrak{p}}}:{\mathfrak{p}}\text{
is a prime of }S\right\\}.$

###### Proof.

If ${\mathfrak{p}}$ is invertible then $\mathrm{type}_{{\mathfrak{p}}}({S})=1$
by Proposition 3.3. Hence in the definition of the global type for a non-
maximal order it is enough to consider the non invertible primes. Also, by
Lemma 2.2, if ${\mathfrak{p}}$ is invertible then
$S/{\mathfrak{p}}=({\mathfrak{p}}:{\mathfrak{p}})/{\mathfrak{p}}$ . Hence also
in the right hand side of the equality in the statement, we can restrict
ourselves to the non-invertible primes of $S$. Therefore the conclusion
follows from Proposition 3.5. ∎

##  4\. Type, overorders and minimal number of generators.

In this section we exhibit connections between the invertibility and the
minimal number of generators of a fractional ideal, and the type of its
multiplicator ring.

Let $Z$ be a Dedekind domain with field of fractions $Q$ and let $K$ be a non-
zero $Q$-étale algebra. All orders in this section are orders in $K$.

######  Definition 4.1.

For an order $S$, put

$g(S)=\max\left\\{g_{S}(I):I\text{ is a fractional $S$-ideals}\right\\}.$

######  Lemma 4.2.

Let $S$ be an order and $I$ be a fractional $S$-ideal. Then

$g_{S}(I)\leq\max\left\\{2,\dim_{S/{\mathfrak{p}}}(I/{\mathfrak{p}}I):{\mathfrak{p}}\text{
prime of }S\right\\},$

with equality if and only if $g_{S}(I)\neq 1$, that is, $I$ is not principal.

###### Proof.

By Nakayama’s Lemma, for a prime ${\mathfrak{p}}$ of $S$, we have the equality
$g_{S_{\mathfrak{p}}}(I_{\mathfrak{p}})=\dim_{S/{\mathfrak{p}}}(I/{\mathfrak{p}}I)$.
Put $m=\max\left\\{g_{S_{\mathfrak{p}}}(I_{\mathfrak{p}}):{\mathfrak{p}}\text{
a prime of }S\right\\}$.

First we consider the case $m>1$. Construct $b_{1}$ in the following way. Put

$\mathcal{A}=\left\\{{\mathfrak{p}}:g_{S_{\mathfrak{p}}}(I_{\mathfrak{p}})=m\right\\}$

and consider the natural surjection

$\pi:I\to\dfrac{I}{I\cdot\prod_{{\mathfrak{p}}\in\mathcal{A}}{\mathfrak{p}}}\simeq\prod_{{\mathfrak{p}}\in\mathcal{A}}\frac{I}{{\mathfrak{p}}I}.$

Observe that $\mathcal{A}$ is a finite set since
$g_{S_{\mathfrak{p}}}(I_{\mathfrak{p}})=1$ for almost all ${\mathfrak{p}}$ by
Lemmas 2.14ii and 2.13. Pick $b_{1}\in I$ such that $\pi(b_{1})$ is non-zero
in each coordinate. Possibly by replacing $b_{1}$ with $b_{1}+y$ with $y$ a
non-zero divisor in $I\cdot\prod_{{\mathfrak{p}}\in\mathcal{A}}{\mathfrak{p}}$
chosen with a procedure similar to the proof of Lemma 2.5, we can ensure that
$b_{1}$ is in $I\cap K^{\times}$. Now consider the set

$\mathcal{B}=\left\\{{\mathfrak{p}}:I\neq b_{1}S+{\mathfrak{p}}I\right\\}.$

Observe that $\mathcal{B}$ is finite since it coincides with the support of
the finite length $S$-module $I/b_{1}S$. For every
${\mathfrak{p}}\in\mathcal{B}$ define vectors
$b_{1,{\mathfrak{p}}},\ldots,b_{m,{\mathfrak{p}}}$ of $I/{\mathfrak{p}}I$ in
the following way: let $b_{1,{\mathfrak{p}}}$ the image of $b_{1}$ via the
surjection $I\to I/{\mathfrak{p}}I$ and then take vectors
$b_{2,{\mathfrak{p}}},\ldots,b_{m,{\mathfrak{p}}}$, possibly with repetitions,
such that $b_{1,{\mathfrak{p}}},\ldots,b_{m,{\mathfrak{p}}}$ generate
$I/{\mathfrak{p}}I$ over $S/{\mathfrak{p}}$. Note that
$\mathcal{A}\subseteq\mathcal{B}$, so $b_{1,{\mathfrak{p}}}\neq 0$ for all
${\mathfrak{p}}$ such that the dimension of $I/{\mathfrak{p}}I$ is maximal,
that is, equal to $m$. Using the isomorphism

$\frac{I}{I\cdot\prod_{{\mathfrak{p}}\in\mathcal{B}}{\mathfrak{p}}}\simeq\prod_{{\mathfrak{p}}\in\mathcal{B}}\frac{I}{{\mathfrak{p}}I}$

for $i=2,\ldots,m$ pick $b_{i}$ in $I$ such that it reduces to
$b_{i,{\mathfrak{p}}}$ modulo ${\mathfrak{p}}I$ for each prime
${\mathfrak{p}}\in\mathcal{B}$. One checks using localization that
$b_{1},\ldots,b_{m}$ is a set of generators of $I$ over $S$: if
${\mathfrak{p}}$ is in $\mathcal{B}$ then this is the case by construction,
while if ${\mathfrak{p}}$ is not in $\mathcal{B}$ then $b_{1}$ is a local
generator at ${\mathfrak{p}}$.

Now we consider the case when $m=1$. Pick an element $b_{1}$ in $I\cap
K^{\times}$, which is not empty by Lemma 2.4. Define the set $\mathcal{B}$ as
above. In this case, note that ${\mathfrak{p}}\in\mathcal{B}$ if and only if
$b_{1}\in{\mathfrak{p}}I$. Pick $b_{2}\in I$ so that its reduction modulo
${\mathfrak{p}}I$ is non-zero for every ${\mathfrak{p}}\in\mathcal{B}$. Then,
again by localization, one checks that $I=b_{1}S+b_{2}S$. ∎

It is known that, among all fractional $S$-ideals, the one that requires the
biggest number of generators is ${\mathcal{O}}_{K}$.

######  Lemma 4.3 ([Gre82, Corollary 2.2]).

Let $S$ be an order. Then

$g(S)\leq\max\left\\{2,g_{S}({\mathcal{O}}_{K})\right\\}.$

The inequality is strict if and only if $S={\mathcal{O}}_{K}$ and
${\mathcal{O}}_{K}$ is a product of PID, that is, $K$ has class number one.

######  Corollary 4.4.

Let $S$ be a non-maximal order. Then

$g(S)=g_{S}({\mathcal{O}}_{K})=\max\left\\{\dim_{S/{\mathfrak{p}}}({\mathcal{O}}_{K}/{\mathfrak{p}}{\mathcal{O}}_{K}):{\mathfrak{p}}\text{
prime of }S\right\\}.$

###### Proof.

Since $S\neq{\mathcal{O}}_{K}$ then $g_{S}({\mathcal{O}}_{K})\geq 2$. The
first equality follows from Lemma 4.3 and the second follows from Lemma 4.2. ∎

The following proposition allows us to compare the size of minimal generating
sets of the fractional ideals of an order with the type of some specific
overorders.

######  Proposition 4.5.

Let ${\mathfrak{p}}$ be a prime of an order $S$. Then

$\dim_{S/{\mathfrak{p}}}\frac{{\mathcal{O}}_{K}}{{\mathfrak{p}}{\mathcal{O}}_{K}}=\begin{cases}1&\text{if
${\mathfrak{p}}$ is invertible,}\\\
1+\mathrm{type}({S+{\mathfrak{p}}{\mathcal{O}}_{K}})&\text{otherwise.}\end{cases}$

In particular, if $S$ is not maximal then

$g(S)=1+\max\left\\{\mathrm{type}({S+{\mathfrak{p}}{\mathcal{O}}_{K}}):{\mathfrak{p}}\text{
a prime of }S\right\\}.$

###### Proof.

Put ${\mathfrak{P}}={\mathfrak{p}}{\mathcal{O}}$ and consider the overorder
$T=S+{\mathfrak{P}}$ of $S$. Observe that we have inclusions
${\mathfrak{p}}\subseteq S\cap{\mathfrak{P}}\subseteq S$. Since, $1$ is not in
${\mathfrak{P}}$, we have the equality ${\mathfrak{p}}=S\cap{\mathfrak{P}}$.
Hence

(6)
$\frac{T}{{\mathfrak{P}}}=\frac{S+{\mathfrak{P}}}{{\mathfrak{P}}}\simeq\frac{S}{{\mathfrak{P}}\cap
S}=\frac{S}{{\mathfrak{p}}},$

which, in particular, shows that ${\mathfrak{P}}$ is a prime of $T$. Note that
$({\mathfrak{P}}:{\mathfrak{P}})={\mathcal{O}}_{K}$ and so

(7)
$\frac{{\mathcal{O}}_{K}}{{\mathfrak{p}}{\mathcal{O}}_{K}}=\frac{{\mathcal{O}}_{K}}{{\mathfrak{P}}}=\frac{({\mathfrak{P}}:{\mathfrak{P}})}{{\mathfrak{P}}}.$

We already know by Lemma 2.14.i that ${\mathfrak{p}}$ is invertible if and
only if ${\mathcal{O}}_{K}/{\mathfrak{p}}{\mathcal{O}}_{K}$ has dimension $1$
over $S/{\mathfrak{p}}$. So we assume that ${\mathfrak{p}}$ is not invertible.
If ${\mathfrak{P}}$ were invertible then by Lemma 2.2 we would have
$({\mathfrak{P}}:{\mathfrak{P}})=T$. Hence by Equations (6) and (7) we would
have that ${\mathcal{O}}_{K}/{\mathfrak{p}}{\mathcal{O}}_{K}$ has dimension
$1$ as a vector space over $S/{\mathfrak{p}}$. Hence ${\mathfrak{P}}$ is not
invertible. So we can apply Proposition 3.5 which gives us the first equality
in

$\mathrm{type}_{{\mathfrak{P}}}({T})+1=\dim_{T/{\mathfrak{P}}}\frac{({\mathfrak{P}}:{\mathfrak{P}})}{{\mathfrak{P}}}=\dim_{S/{\mathfrak{p}}}\frac{{\mathcal{O}}_{K}}{{\mathfrak{p}}{\mathcal{O}}_{K}},$

while the last equality is given by Equations (6) and (7). To conclude the
proof we need to show that
$\mathrm{type}_{{\mathfrak{P}}}({T})=\mathrm{type}({T})$. Note that
${\mathfrak{P}}$ is a fractional ${\mathcal{O}}_{K}$-ideal inside $T$. So
${\mathfrak{P}}$ is contained in the conductor $(T:{\mathcal{O}}_{K})$, which
is a proper ideal since ${\mathfrak{P}}$ is not invertible. Hence
${\mathfrak{P}}=(T:{\mathcal{O}}_{K})$ by maximality of ${\mathfrak{P}}$. By
Lemma 2.14.i we have that ${\mathfrak{P}}$ is the only non invertible ideal of
$T$. Proposition 3.3 shows then that the only contribution to the global type
of $T$ comes from ${\mathfrak{P}}$, that is,
$\mathrm{type}_{{\mathfrak{P}}}({T})=\mathrm{type}({T})$.

The last statement is an immediate consequence of Corollary 4.4. ∎

In Section 3, we discussed Gorenstein orders. A generalization, which takes
into account all the overorders, is the following. An order which is either
maximal or satisfies the equivalent conditions of Proposition 4.6 is called a
Bass order.

######  Proposition 4.6.

Let $S$ be a non-maximal order. Then the following are equivalent:

  1. (i)

For every overorder $T$ of $S$ we have $\mathrm{type}({T})=1$.

  2. (ii)

Every overorder of $S$ is Gorenstein.

  3. (iii)

Every fractional $S$-ideal $I$ is invertible as a fractional $(I:I)$-ideal.

  4. (iv)

The $S$-module ${\mathcal{O}}_{K}/S$ is cyclic.

  5. (v)

$g(S)=2$.

###### Proof.

The equivalence of the first three statement is a direct consequence of
Proposition 3.4. The equivalence of the last four statements can be found in
[LW85, Theorem 2.1]. ∎

Proposition 3.4 tells us that the type measures the deviation of an order from
being Gorenstein. Following this philosophy, a measure of how an order fails
to being Bass can be given in terms of the types of all the overorders. More
precisely, we have the following generalization of Proposition 4.6.

######  Theorem 4.7.

Let $S$ be a non-maximal order, $\mathcal{S}$ be the set of overorders $T$ of
$S$, and $d$ be a positive integer. Then the following are equivalent:

  1. (i)

$1+\max_{T\in\mathcal{S}}\left\\{\mathrm{type}({T})\right\\}=d$.

  2. (ii)

$\max_{T\in\mathcal{S}}\left\\{\dim_{T/{\mathfrak{P}}}\left(({\mathfrak{P}}:{\mathfrak{P}})/{\mathfrak{P}}\right):{\mathfrak{P}}\text{
a prime of }T\right\\}=d.$

  3. (iii)

$g_{S}({\mathcal{O}}_{K}/S)=d-1$.

  4. (iv)

$\max_{T\in\mathcal{S}}\left\\{\dim_{T/{\mathfrak{P}}}\left({\mathcal{O}}_{K}/{\mathfrak{P}}{\mathcal{O}}_{K}\right):{\mathfrak{P}}\text{
a prime of }T\right\\}=d$.

  5. (v)

$\max_{T\in\mathcal{S}}\left\\{g(T)\right\\}=d$.

  6. (vi)

$g(S)=d$.

###### Proof.

Let $T$ be a non-maximal overorder in $\mathcal{S}$. Then

$\mathrm{type}({T})+1=\max\left\\{\dim_{T/{\mathfrak{P}}}\frac{({\mathfrak{P}}:{\mathfrak{P}})}{{\mathfrak{P}}}:{\mathfrak{P}}\text{
a prime of }T\right\\},$

by Corollary 3.6. This implies the equivalence of i and ii. Corollary 4.4
gives the equivalence of iv and v. Note that for every $T\in\mathcal{S}$,
every fractional $T$-ideal $I$ is also a fractional $S$-ideal. So we have
$g_{T}(I)\leq g_{S}(I)$. Hence
$g(S)=\max_{T\in\mathcal{S}}\left\\{g(T)\right\\}$. In particular, v and vi
are equivalent.

We now show that iii is equivalent to vi. Denote by $M$ the $S$-module
${\mathcal{O}}_{K}/S$. By Lemma 2.14 we have that $M_{\mathfrak{p}}\neq 0$
only for finitely many primes ${\mathfrak{p}}$ of $S$. By Nakayama’s Lemma and
an argument similar to the proof of Lemma 4.2, we have

$g_{S}(M)=\max\left\\{\dim_{S/{\mathfrak{p}}}\left(\dfrac{M}{{\mathfrak{p}}M}\right):\
{\mathfrak{p}}\text{ a prime of }S\right\\}.$

Observe that for every prime ${\mathfrak{p}}$ of $S$ we have

(8)
$\dfrac{M}{{\mathfrak{p}}M}\simeq\dfrac{{\mathcal{O}}_{K}}{S+{\mathfrak{p}}{\mathcal{O}}_{K}}.$

Moreover, we have a short exact sequence of $S/{\mathfrak{p}}$-vector spaces

(9)
$0\to\dfrac{S+{\mathfrak{p}}{\mathcal{O}}_{K}}{{\mathfrak{p}}{\mathcal{O}}_{K}}\to\dfrac{{\mathcal{O}}_{K}}{{\mathfrak{p}}{\mathcal{O}}_{K}}\to\dfrac{{\mathcal{O}}_{K}}{S+{\mathfrak{p}}{\mathcal{O}}_{K}}\to
0.$

As shown in Equation (6), we have that
$(S+{\mathfrak{p}}{\mathcal{O}}_{K})/{\mathfrak{p}}{\mathcal{O}}_{K}$ has
dimension $1$ over $S/{\mathfrak{p}}$. Hence by Equations (8) and (9), we get

$\dim_{S/{\mathfrak{p}}}\left(\dfrac{{\mathcal{O}}_{K}}{{\mathfrak{p}}{\mathcal{O}}_{K}}\right)=\dim_{S/{\mathfrak{p}}}\left(\dfrac{M}{{\mathfrak{p}}M}\right)+1.$

By Corollary 4.4, taking the maximum over the primes of $S$ we get

$g(S)=g_{S}(M)+1,$

as required.

To conclude we show that i and vi are equivalent, that is, that we have an
equality

$g(S)=1+\max_{T\in\mathcal{S}}\left\\{\mathrm{type}({T})\right\\}.$

By Proposition 4.5, we obtain

(10)
$\begin{split}g(S)&=1+\max\left\\{\mathrm{type}({S+{\mathfrak{p}}{\mathcal{O}}_{K}}):\
{\mathfrak{p}}\text{ a prime of }S\right\\}\\\ &\leq
1+\max_{T\in\mathcal{S}}\left\\{\mathrm{type}({T})\right\\}.\end{split}$

For the other direction, pick any non-maximal $T\in\mathcal{S}$ and let
${\mathfrak{P}}$ be prime of $T$. We claim that

(11)
$\dim_{T/{\mathfrak{P}}}\frac{({\mathfrak{P}}:{\mathfrak{P}})}{{\mathfrak{P}}}=g_{T}(({\mathfrak{P}}:{\mathfrak{P}})).$

Indeed, if ${\mathfrak{P}}$ is invertible then by Lemma 2.8 we have that
$T=({\mathfrak{P}}:{\mathfrak{P}})$. On the other hand, if ${\mathfrak{P}}$ is
not invertible then $({\mathfrak{P}}:{\mathfrak{P}})$ is a non-principal
fractional $T$-ideal. Hence Equation (11) follows from Lemma 4.2. Now,
combining Equation (11) and Corollary 3.6, we see

$\mathrm{type}({T})+1=\max\left\\{g_{T}(({\mathfrak{P}}:{\mathfrak{P}})):\
{\mathfrak{P}}\text{ a prime of }T\right\\}$

which is bounded from above by $g(T)$. As we already observed before, we have
$g(T)\leq g(S)$, which gives us

$\mathrm{type}({T})+1\leq g(S),$

as required to complete the proof. ∎

######  Remark 4.8.

In [Mar20b], we showed that an order $S$ in a number field satisfies $g(S)\leq
3$ if and only if every overorder $T$ of $S$ is _super-multiplicative_ , that
is, for every pair of fractional $T$-ideals $I$ and $J$ contained in $T$ we
have

$N(IJ)\geq N(I)N(J),$

where the ideal norm $N(I)$ (resp. $N(J)$ and $N(IJ)$) is defined as the size
of the finite abelian group $T/I$ (resp. $T/J$ and $T/IJ$). In view of Theorem
4.7, we can state that every overorder $T$ of $S$ is supermultiplicative if
and only if for every overorder $T$ of $S$ we have $\mathrm{type}({T})\leq 2$.

We conclude Section 4 by showing that the type of the orders in a non-zero
étale $Q$-algebra $K$ is bounded. Note that the only order in $Q$ is $Z$,
which has type $1$. To avoid this trivial case, we will assume in the next
proposition that $\dim_{Q}(K)>1$.

######  Proposition 4.9.

Assume that $\dim_{Q}(K)>1$. Then the type of an order in $K$ is bounded by
$\dim_{Q}(K)-1$. Moreover, this bound is attained: for any prime $p$ of $Z$
the order $T=Z+p{\mathcal{O}}_{K}$ satisfies
$\mathrm{type}({T})=\dim_{Q}(K)-1$.

###### Proof.

Pick $S$ any order in $K$. If $S$ is Gorenstein then $\mathrm{type}({S})=1$ by
Proposition 3.4. So we assume that $S$ is not Gorenstein. By definition, there
exists a prime ${\mathfrak{p}}$ of $S$ such that
$\mathrm{type}({S})=\mathrm{type}_{{\mathfrak{p}}}({S})$. By Proposition 3.3,
the prime ${\mathfrak{p}}$ is not invertible. Combining Proposition 3.5 and
Lemma 4.2 we obtain
$\mathrm{type}_{{\mathfrak{p}}}({S})=g_{S_{\mathfrak{p}}}(({\mathfrak{p}}:{\mathfrak{p}})_{\mathfrak{p}})-1$.
Also we have the equality

$g_{S_{\mathfrak{p}}}(({\mathfrak{p}}:{\mathfrak{p}})_{\mathfrak{p}})=g_{\hat{S}_{\mathfrak{p}}}(({\mathfrak{p}}:{\mathfrak{p}})\otimes_{S}\hat{S}_{\mathfrak{p}}).$

Let $p$ be the contraction in $Z$ of ${\mathfrak{p}}$. Applying
$({\mathfrak{p}}:{\mathfrak{p}})\otimes_{S}-$ to the isomorphism in Equation
(4), we see that
$({\mathfrak{p}}:{\mathfrak{p}})\otimes_{S}\hat{S}_{\mathfrak{p}}$ is a free
$\hat{Z}_{p}$-module of rank $\leq\dim_{Q}(K)$. In particular
$g_{\hat{S}_{\mathfrak{p}}}(({\mathfrak{p}}:{\mathfrak{p}})\otimes_{S}\hat{S}_{\mathfrak{p}})\leq\dim_{Q}(K)$.
We conclude that $\mathrm{type}({S})\leq\dim_{Q}(K)-1$, as required.

Now we prove the second part. Let $p$ be any prime of $Z$ and define
$T=Z+p{\mathcal{O}}_{K}$ as in the statement. Put $F_{p}=Z/p$. We have a ring
isomorphism

$\frac{T}{p{\mathcal{O}}_{K}}\simeq\frac{Z}{Z\cap p{\mathcal{O}}_{K}}=F_{p},$

where the last equality holds because $Z\cap p{\mathcal{O}}_{K}=p$ by
maximality of $p$. In particular, we get that $p{\mathcal{O}}_{K}$ is a prime
of $T$ with residue field isomorphic to $F_{p}$. Since
$\dim_{Q}(K)=\dim_{F_{p}}({\mathcal{O}}_{K}/p{\mathcal{O}}_{K})>1$ we see that
$T\neq{\mathcal{O}}_{K}$. Thus the inclusion
$p{\mathcal{O}}_{K}\subseteq(T:{\mathcal{O}}_{K})$ is an equality. Hence Lemma
2.14.i tells us that $p{\mathcal{O}}_{K}$ is the unique non-invertible prime
of $T$. Then Proposition 3.3 gives us that
$\mathrm{type}({T})=\mathrm{type}_{p{\mathcal{O}}_{K}}({T})$. Note that the
multiplicator ring of $p{\mathcal{O}}_{K}$ is ${\mathcal{O}}_{K}$. Using
Proposition 3.5, we get

$\mathrm{type}({T})=\dim_{F_{p}}\frac{{\mathcal{O}}_{K}}{p{\mathcal{O}}_{K}}-1=\dim_{Q}(K)-1.$

∎

##  5\.  Ideal classes

Let $Z$ be a Dedekind domain with field of fractions $Q$ and let $K$ be a non-
zero $Q$-étale algebra. Pick an order $S$ in $K$. Recall that two fractional
$S$-ideals $I$ and $J$ are isomorphic as $S$-modules if and only if there
exists $\alpha\in K^{\times}$ such that $\alpha I=J$. The set of isomorphism
classes of invertible fractional $S$-ideal, the _Picard group_ of $S$, which
we denote $\operatorname{Pic}(S)$, has a natural group structure induced by
fractional ideal multiplication. Similarly, the set of isomorphism classes of
all fractional $S$-ideals forms a commutative monoid that we denote by
$\operatorname{ICM}(S)$ and call the _ideal class monoid_ of $S$. The ideal
class monoid has a natural partitioning

$\operatorname{ICM}(S)=\bigsqcup_{T}\operatorname{ICM}_{T}(S),$

where the disjoint union is taken over the overorders $T$ of $S$, and
$\operatorname{ICM}_{T}(S)$ consists of the classes with multiplicator ring
$T$.

We say that two fractional $S$-ideals $I$ and $J$ are _weakly equivalent_ if
$I_{\mathfrak{p}}\simeq J_{\mathfrak{p}}$ for every prime ${\mathfrak{p}}$ of
$S$. This notion was introduced in [DTZ62] for orders in number fields. Denote
by the $\mathcal{W}(S)$ the monoid of weak equivalence classes of fractional
$S$-ideals. As we will see in the next proposition, the multiplicator ring of
a fractional $S$-ideal is an invariant of its weak equivalence class. Hence we
have a partitioning

$\mathcal{W}(S)=\bigsqcup_{T}\mathcal{W}_{T}(S),$

where the disjoint union is taken over the over-orders $T$ of $S$, and
$\mathcal{W}_{T}(S)$ is the subset of $\mathcal{W}(S)$ consisting of the
classes with multiplicator ring equal to $T$.

In the next proposition, we summarize some useful facts about isomorphism and
weak equivalence classes, and how invertible ideals connect the two notions.
More details can be found in [Mar20a].

######  Proposition 5.1.

Let $S$ be an order.

  1. (i)

Let $I$ and $J$ be two fractional $S$-ideals. Then the following are
equivalent

     * •

$I$ and $J$ are weakly equivalent.

     * •

$1\in(I:J)(J:I)$.

     * •

$I$ and $J$ have the same multiplicator ring, say $T$, and there exists an
invertible fractional $T$-ideal $L$ such that $I=LJ$.

  2. (ii)

Let $T$ be an overorder of $S$. Let $T_{0}$ be the smallest overorder of $T$
such that the extension to $T_{0}$ of every fractional $T$-ideal with
multiplicator ring $T$ is invertible. Put ${\mathfrak{f}}_{0}=(T:T_{0})$. Then
every class in $\mathcal{W}_{T}(S)$ admits a representative $I$ such that

${\mathfrak{f}}_{0}\subseteq I\subseteq T_{0}.$

  3. (iii)

Fix an overorder $T$ of $S$. Then there is a free action of
$\operatorname{Pic}(T)$ on $\operatorname{ICM}_{T}(S)$ with quotient
$\mathcal{W}_{T}(S)$.

###### Proof.

The statements for orders over $\mathbb{Z}$ in a non-zero étale
$\mathbb{Q}$-algebras can be found in [Mar20a, Prop 4.1, Prop. 5.1, Thm. 4.6].
The same proofs are valid also for the more general case under consideration
in this paper. Possibly, the only fact used in proving i that is not obvious
in our more general setting is that the extension map induces a surjective
group homomorphism from $\operatorname{Pic}(S)$ onto $\operatorname{Pic}(T)$.
The proof of this for the case $T={\mathcal{O}}_{K}$ can be found in [Wie84,
Sec. 2], and it works in the same way in the general case, as stated in [LW85,
Eq. (4.3.1)]. ∎

######  Remark 5.2.

Part i gives a method to verify if two fractional ideals are weakly equivalent
as long as we have a method to compute products and colons of fractional
ideals, and check inclusion of elements.

From Part i and Part ii, we obtain a method to compute $\mathcal{W}_{T}(S)$ if
the quotient $T_{0}/{\mathfrak{f}}_{0}$ has finitely many sub-$S$-modules and
we have a procedure to loop over all of them. This is the case for example if
the quotient of every two $Z$-lattices is a finite set. Indeed, in that case
we can use algorithms to enumerate subgroups in order to obtain all the
submodules. See for example [FHS19, Sec. 5.2].

Then we can use Part iii to compute $\operatorname{ICM}_{T}(S)$ if we have an
algorithm to compute $\operatorname{Pic}(T)$. Such an algorithm for orders
over $\mathbb{Z}$ and $\mathbb{F}_{q}[t]$ is given in [KP05].

Finally, these steps allow us to compute the whole $\operatorname{ICM}(S)$ if
we can compute the overorders $T$ of $S$. We refer to [HS20] for an effective
method.

Thanks to Proposition 5.1 we can reformulate the properties of being
Gorenstein and Bass for an order in terms of ideal classes.

######  Corollary 5.3.

Let $T$ be an overorder of an order $S$. Then $T$ is Gorenstein if and only if
$|\mathcal{W}_{T}(S)|=1$ if and only if
$\operatorname{Pic}(T)=\operatorname{ICM}_{T}(S)$. The order $S$ is Bass if
and only if

$\operatorname{ICM}(S)=\bigsqcup_{T}\operatorname{Pic}(T),$

where the disjoint union is taken over the overorders $T$ of $S$.

In the next Section we prove an analogous result for orders of type $2$. More
precisely, in Corollary 6.5 we will show that also in this case we completely
understand $\operatorname{ICM}_{T}(S)$ for an overorder of $S$ of
$\mathrm{type}({T})=2$.

##  6\. Orders with CM type 2

Let $Z$ be a Dedekind domain with field of fractions $Q$ and let $K$ be a non-
zero $Q$-étale algebra. Let $S$ be an order in $K$. We start by recalling a
lemma from linear algebra.

######  Lemma 6.1 ([Mar20c, Lemma 4.9]).

Let $U,V$ and $W$ be vector spaces over a field $k$, with $W$ of dimension
$\geq 2$. Let ${\varphi}:U\otimes V\twoheadrightarrow W$ be a surjective
linear map. Then there exists an element $u\in U$ such that
$\dim_{k}{\varphi}(u\otimes V)\geq 2$, or there exists an element $v\in V$
such that $\dim_{k}{\varphi}(U\otimes v)\geq 2$.

In the next Theorem, given an order $S$ with local type $2$, we give a
complete classification of the local isomorphism classes of fractional
$S$-ideals with multiplicator ring $S$.

######  Theorem 6.2.

Let $S$ be an order, ${\mathfrak{p}}$ a prime of $S$ and $I$ a fractional
$S$-ideal with $(I:I)=S$. Assume that $\mathrm{type}_{{\mathfrak{p}}}({S})=2$.
Then either $I_{\mathfrak{p}}\simeq S_{\mathfrak{p}}$ or
$I_{\mathfrak{p}}\simeq(S^{t})_{\mathfrak{p}}$.

###### Proof.

Put $k=S/{\mathfrak{p}}$ and consider the following finite dimensional
$k$-vector spaces:

$U=\frac{I}{{\mathfrak{p}}I},\quad V=\frac{I^{t}}{{\mathfrak{p}}I^{t}},\quad
W=\frac{S^{t}}{{\mathfrak{p}}S^{t}}.$

By Lemma 2.4.iii we have that $S^{t}=(I:I)^{t}=II^{t}$. It follows that the
multiplication map $I\otimes_{S}I^{t}\to S^{t}$ is a surjective $S$-linear
morphism. In particular the induced $k$-linear morphism

$\pi:U\otimes_{k}V\to W$

is a surjective map onto a $2$-dimensional vector space. By Lemma 6.1 there
exists

  1. (i)

$u\in I$ such that $\pi(u+{\mathfrak{p}}I\otimes V)=W$, or

  2. (ii)

$v\in I^{t}$ such that $\pi(U\otimes v+{\mathfrak{p}}I^{t})=W$.

Assume that i holds, that is,

$\pi(u+{\mathfrak{p}}I\otimes
V)=\frac{uI^{t}+{\mathfrak{p}}S^{t}}{{\mathfrak{p}}S^{t}}=W.$

If we write $I^{t}=\sum_{k=1}^{r}\beta_{k}S$ then
$\left\\{u\beta_{k}+{\mathfrak{p}}S^{t}\right\\}$ is a finite set of
generators over $k$ for $W$. By Nakayama’s Lemma we obtain that
$\left\\{u\beta_{k}\right\\}$ gives a set of generators over
$\hat{S}_{\mathfrak{p}}$ for $\hat{S}^{t}_{\mathfrak{p}}$. By identifying $u$
with its image in ${\mathcal{Q}}(\hat{S}_{\mathfrak{p}})$ we see that
$u\hat{I}^{t}_{\mathfrak{p}}=\hat{S}^{t}_{\mathfrak{p}}$, which is equivalent
to $\hat{I}_{\mathfrak{p}}=u\hat{S}_{\mathfrak{p}}$. By [Eis95, Ex. 7.5, p.
203] we get that $I_{\mathfrak{p}}\simeq S_{\mathfrak{p}}$ as an
$S_{\mathfrak{p}}$-module. If ii holds, then by a similar argument we get that
$v\hat{I}_{\mathfrak{p}}=\hat{S}^{t}_{\mathfrak{p}}$ and hence
$I_{\mathfrak{p}}\simeq(S^{t})_{\mathfrak{p}}$. ∎

######  Corollary 6.3.

Let $S$ be a non-Gorenstein order. Then the following are equivalent:

  * •

$\mathrm{type}({S})=2$.

  * •

$\max\left\\{g_{S}(I):\text{ fractional $S$-ideal $I$ with
$S=(I:I)$}\right\\}=2$

###### Proof.

Recall that for every prime ${\mathfrak{p}}$ of $S$, we have
$g_{S_{\mathfrak{p}}}(I_{\mathfrak{p}})=\dim_{S/{\mathfrak{p}}}(I/{\mathfrak{p}}I)$
for every fractional $S$-ideal $I$ by Nakayama’s Lemma. Define

$m=\max\left\\{g_{S}(I):\text{ fractional $S$-ideal $I$ with
$(I:I)=S$}\right\\}.$

Assume that $\mathrm{type}({S})=2$. By Lemma 4.2 this is equivalent to
$g_{S}(S^{t})=2$. Since $(S^{t}:S^{t})=S$ we then have $m\geq 2$. Now, let $I$
be any fractional $S$-ideal with $(I:I)=S$. By Theorem 6.2 we have
$g_{S_{\mathfrak{p}}}(I_{\mathfrak{p}})\leq 2$, and hence by Lemma 4.2, we
also have $g_{S}(I)\leq 2$. Since $I$ was arbitrary we get also $m\leq 2$.
Therefore $m=2$, as required.

For the converse, assume that $m=2$. Since $S$ is not Gorenstein, we have that
$g_{S}(S^{t})>1$ and hence $g_{S}(S^{t})=2$. Hence we get that
$\mathrm{type}({S})=2$. ∎

After a technical lemma, we conclude Section 6 with an application of the
machinery built so far to the computations of isomorphism classes, followed by
Example 6.6.

######  Lemma 6.4.

Let $S$ be an order and let ${\mathfrak{p}}_{1},\ldots,{\mathfrak{p}}_{r}$ be
distinct primes of $S$. Let $I_{1},\ldots,I_{r}$ be fractional $S$-ideals
contained in $S$, possibly with repetitions. Let $m_{1},\ldots,m_{r}$ be non-
negative integers such that

${\mathfrak{p}}_{i}^{m_{i}}\subseteq I_{i,{\mathfrak{p}}_{i}}$

for $1\leq i\leq r$. Put

$J:=\sum_{i=1}^{r}\left(\left(I_{i}+{\mathfrak{p}}_{i}^{m_{i}}\right)\prod_{j\neq
i}{\mathfrak{p}}_{j}^{m_{j}}\right).$

Then for each $1\leq k\leq r$ we have

$J_{{\mathfrak{p}}_{k}}=I_{k,{\mathfrak{p}}_{k}}.$

###### Proof.

Fix an index $k$. For $1\leq i\leq r$, by Lemma 2.11, we have

$\left(I_{i}+{\mathfrak{p}}_{i}^{m_{i}}\right)_{{\mathfrak{p}}_{k}}=\begin{cases}I_{k,{\mathfrak{p}}_{k}},&\text{
if }i=k\\\ S_{{\mathfrak{p}}_{k}},&\text{ if }i\neq k\end{cases}$

and

$\left(\prod_{j\neq
i}{\mathfrak{p}}_{j}^{m_{j}}\right)_{{\mathfrak{p}}_{k}}=\begin{cases}S_{{\mathfrak{p}}_{k}},&\text{
if }i=k\\\ {\mathfrak{p}}_{k}^{m_{k}},&\text{ if }i\neq k.\end{cases}$

Hence

$\displaystyle J_{{\mathfrak{p}}_{k}}$
$\displaystyle=\left(\left(I_{k}+{\mathfrak{p}}_{k}^{m_{k}}\right)\prod_{j\neq
k}{\mathfrak{p}}_{j}^{m_{j}}\right)_{{\mathfrak{p}}_{k}}+\sum_{i\neq
k}\left(\left(I_{i}+{\mathfrak{p}}_{i}^{m_{i}}\right)\prod_{j\neq
i}{\mathfrak{p}}_{j}^{m_{j}}\right)_{{\mathfrak{p}}_{k}}$
$\displaystyle=I_{k,{\mathfrak{p}}_{k}}S_{{\mathfrak{p}}_{k}}+S_{{\mathfrak{p}}_{k}}{\mathfrak{p}}_{k}^{m_{k}}$
$\displaystyle=I_{k,{\mathfrak{p}}_{k}}+{\mathfrak{p}}_{k}^{m_{k}}$
$\displaystyle=I_{k,{\mathfrak{p}}_{k}},$

as required. ∎

######  Corollary 6.5.

Let $S\subseteq T$ be orders such that $\mathrm{type}({T})=2$. Let $N$ be the
number of primes ${\mathfrak{p}}$ of $T$ such that
$\mathrm{type}_{{\mathfrak{p}}}({T})=2$. Then $|\mathcal{W}_{T}(S)|=2^{N}$ and
$|\operatorname{ICM}_{T}(S)|=2^{N}\cdot|\operatorname{Pic}(T)|$.

###### Proof.

The statement follows from Theorem 6.2 and Lemma 6.4. ∎

######  Example 6.6.

Consider the polynomial

$f=x^{6}-6x^{5}+6x^{4}+43x^{3}+96x^{2}-1536x+4096$

Consider the étale $\mathbb{Q}$-algebra $K=\mathbb{Q}[x]/(f)$. Denote by $\pi$
the class of the variable $x$ in $K$. Put $S=\mathbb{Z}[\pi]$ in $K$. There
are $3$ non-invertible primes of $S$: the unique primes
${\mathfrak{p}}_{2},{\mathfrak{p}}_{3}$ and ${\mathfrak{p}}_{5}$ above $2,3$
and $5$, respectively. The dimension of
${\mathcal{O}}_{K}/{\mathfrak{p}}_{i}{\mathcal{O}}_{K}$ over
$S/{\mathfrak{p}}_{i}$, for $i=2,3$ and $5$, are $3,3$ and $2$, respectively.
By Theorem 4.7, all the overorders of $S$ are either Gorenstein or have type
$2$, and there is at least one overorder with type $2$.

One computes that $S$ has $1728$ overorders, $416$ of which are Gorenstein.
Using Corollary 6.5, we can compute that there are $3920$ weak equivalences
classes, and, after computing all the Picard group, we see that there are
$17638816$ isomorphism classes.

We now exhibit two applications of this computation. As described in [Mar20a,
Corollary 8.2], using a generalization of the Latimer-MacDuffee Theorem
[LM33], we have a bijection between the conjugacy classes of
$\mathbb{Z}$-matrices with characteristic polynomial $f$ and
$\operatorname{ICM}(S)$. Hence we deduce that there are $17638816$ conjugacy
classes of $\mathbb{Z}$-matrices with characteristic polynomial $f$.

The second application is about abelian varieties over finite fields. The
polynomial $f$ above defines by Honda-Tate theory an isogeny class of ordinary
abelian varieties over the finite field $\mathbb{F}_{16}$. See the LMFDB
[LMF22] label 3.16.ag_g_br. Using [Mar21, Corollary 4.4], we see that this
isogeny class contains as many isomorphism classes of abelian varieties as the
size of $\operatorname{ICM}(\mathbb{Z}[\pi,16/\pi])$, which is a subset of
$\operatorname{ICM}(S)$. In particular, we check that there are $15112$ such
classes.

##  7\. Comparison with other notions of close-to-being Gorenstein

In the literature there are several conditions that a ring can satisfy that
make it close-to-being Gorenstein. In this section, when applicable, we will
study them in terms of their type. As usual, we will denote by $Q$ the field
of fractions of a Dedekind domain, and by $K$ a non-zero étale $Q$-algebra.

###  7.1. Nearly Gorenstein

Let $R$ be a ring and $M$ an $R$-module. The trace ideal of $M$ is defined as

$\operatorname{tr}M=\sum_{f\in\operatorname{Hom}_{R}(M,R)}f(M).$

See for example [DKT21a, Def. 2.1]. Let $(R,{\mathfrak{m}})$ be a
Cohen–Macaulay local ring with canonical module $C$. Then $R$ is called nearly
Gorenstein if $\operatorname{tr}C$ contains ${\mathfrak{m}}$. Such a notion
was introduced and studied in [HHS19].

If $S$ is an order in $K$ then by Lemma 2.1 we have an identification

$\operatorname{tr}S^{t}=\sum_{f\in(S:S^{t})}f\cdot S^{t}=S^{t}(S:S^{t}).$

Since we are going to work locally, in order to lighten the notation, we set
$R=\hat{S}_{\mathfrak{p}}$ and
${\mathfrak{m}}={\mathfrak{p}}\hat{S}_{\mathfrak{p}}$, where ${\mathfrak{p}}$
is a prime of $S$. In particular, $R^{t}$ is the canonical module of $R$ and
$\mathrm{type}({R})=\mathrm{type}_{{\mathfrak{p}}}({S})$ by Proposition 3.1.
Using this notation, $R$ is nearly Gorenstein if and only if

${\mathfrak{m}}\subseteq R^{t}(R:R^{t}).$

Observe that the inclusion is strict if and only if $R^{t}(R:R^{t})=R$, that
is, $\mathrm{type}({R})=1$, or equivalently $R$ is Gorenstein by Proposition
3.4.

If $R$ is nearly Gorenstein, but not Gorenstein, then the type of $R$ is

$\mathrm{type}({R})=\dim_{R/{\mathfrak{m}}}\frac{R^{t}}{(R^{t})^{2}(R:R^{t})}.$

By Lemma 2.4.iii, we get the following equalities of $R/{\mathfrak{m}}$-vector
spaces

$\frac{R^{t}}{(R^{t})^{2}(R:R^{t})}=\frac{R^{t}}{(R:R^{t})^{t}(R:R^{t})}=\frac{R^{t}}{T^{t}},$

where $T$ is the multiplicator ring of $(R:R^{t})$. Similarly, by multiplying
$m=R^{t}(R:R^{t})$ by $R^{t}$, we get that ${\mathfrak{m}}R^{t}=T^{t}$. Hence
$(R:{\mathfrak{m}})=T$, which by Lemma 2.2 leads to
$T=({\mathfrak{m}}:{\mathfrak{m}})$. We summarize these observations in the
following proposition.

######  Proposition 7.1.

If $R$ is nearly Gorenstein, but not Gorenstein, then

$\mathrm{type}({R})=\dim_{R/{\mathfrak{m}}}\frac{R^{t}}{(R^{t})^{2}(R:R^{t})}.$

Moreover, the multiplicator ring $({\mathfrak{m}}:{\mathfrak{m}})$ of
${\mathfrak{m}}$ equals the multiplicator ring of $(R:R^{t})$.

###  7.2. Almost Gorenstein and Weakly almost Gorenstein

Let $R$ be a local Cohen-Macaulay ring $R$ of Krull dimension $1$ with
normalization $\overline{R}$ and conductor ${\mathfrak{f}}=(R:\overline{R})$.
We say that $R$ is almost Gorenstein if

$\ell_{R}\left(\frac{\overline{R}}{R}\right)=\ell_{R}\left(\frac{R}{{\mathfrak{f}}}\right)+\mathrm{type}({R})-1,$

where $\ell_{R}(-)$ is the length as an $R$-module. The study of almost
Gorenstein rings, where this definition was given, started with [BF97]. It was
extended in [GMP13] and [GTT15]. It is known that almost Gorenstein rings are
nearly Gorenstein, see [HHS19, Prop. 6.1].

Let $S$ be an order in $K$ and ${\mathfrak{p}}$ a prime of $S$. We work
locally, and again we simplify the notation by setting
$R=\hat{S}_{\mathfrak{p}}$,
${\mathfrak{m}}={\mathfrak{p}}\hat{S}_{\mathfrak{p}}$,
$\overline{R}=\hat{\mathcal{O}}_{K,{\mathfrak{p}}}$ and
${\mathfrak{f}}=(R:\overline{R})$.

We can rephrase the definition of the type of $R$ using the length as

$\mathrm{type}({R})=\ell_{R}\left(\frac{R^{t}}{{\mathfrak{m}}R^{t}}\right)=\ell_{R}\left(\frac{(R:{\mathfrak{m}})}{R}\right),$

where the second equality follow from Lemma 2.4.iii and iv. Using that the
length is an additive function on short exact sequences, we get the following
proposition.

######  Proposition 7.2.

The order $R$ is almost Gorenstein if and only if

$\ell_{R}\left(\frac{\overline{R}}{R}\right)=\ell_{R}\left(\frac{(R:{\mathfrak{m}})}{{\mathfrak{f}}}\right)-1.$

Related to the definition of almost Gorenstein, there is the notion of weakly
almost Gorenstein rings  introduced in [DKT21b]. In Krull dimension one, the
two notions coincide, see [DKT21b, Thm. 1.3].

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