Quaternion ideals and invertibility

Abstract

Much like a space can be understood by studying functions on that space, often the first task to understand a ring A is to understand the ideals of A and modules over A (in other words, to pursue “linear algebra” over A). The ideals of a ring that are easiest to work with are the principal ideals—but not all ideals are principal, and various algebraic structures are built to understand the difference between these two. In this chapter, we consider these questions for the case where A is a quaternion order.

Much like a space can be understood by studying functions on that space, often the first task to understand a ring A is to understand the ideals of A and modules over A (in other words, to pursue “linear algebra” over A). The ideals of a ring that are easiest to work with are the principal ideals—but not all ideals are principal, and various algebraic structures are built to understand the difference between these two. In this chapter, we consider these questions for the case where A is a quaternion order.

1 $\triangleright$ Quaternion ideals

To get warmed up for the noncommutative situation, we consider ideals of quadratic rings. An integer $d \in \mathbb Z$ is a discriminant if $d \equiv 0,1 ~(\text{ mod } ~{4})$. Let S be the quadratic order of nonsquare discriminant $d \in \mathbb Z$, namely,

$$\begin{aligned} S = S(d) :=\mathbb Z \oplus \mathbb Z [(d+\sqrt{d})/2] \subset K=\mathbb Q (\sqrt{d}). \end{aligned}$$

The set of ideals of S has a natural multiplicative structure with identity element S (giving it the structure of a commutative monoid), but we lack inverses and we would surely feel more comfortable with a group structure. So we consider nonzero S-lattices $\mathfrak a \subset K$, and call them fractional ideals of S; equivalently, they are the S-submodules $d^{-1} \mathfrak a \subset K$ with $\mathfrak a \subseteq S$ a nonzero ideal and $d \in \mathbb Z _{>0}$, hence the name fractional ideal (viz. 9.2.4). To get a group structure, we must restrict our attention to the invertible  fractional ideals $\mathfrak a \subset K$, i.e., those such that there exists a fractional ideal $\mathfrak b$ with $\mathfrak a \mathfrak b =S$. The simplest kind of invertible fractional ideals are the principal ones $\mathfrak a =aS$ for $a \in K^\times$, with inverse $\mathfrak a ^{-1}=a^{-1} S$. If a fractional ideal $\mathfrak a$ has an inverse then this inverse is unique, given by

$$\begin{aligned} \mathfrak a ^{-1}=\{x \in K : x\mathfrak a \subseteq S\}; \end{aligned}$$

and for a fractional ideal $\mathfrak a$, we always have $\mathfrak a \mathfrak a ^{-1} \subseteq S$ (but equality may not hold). If $S=\mathbb Z _K$ is the ring of integers (the maximal order) of K, then all nonzero fractional ideals of S are invertible—in fact, this property characterizes Dedekind domains, in that a noetherian commutative ring is a Dedekind domain if and only if every nonzero (prime) ideal is invertible. (See also the summary in section 9.2.)

A fractional ideal $\mathfrak a$ of S is invertible if and only if $\mathfrak a$ is locally principal, i.e., $\mathfrak a \otimes _\mathbb{Z } \mathbb Z _{(p)} = \mathfrak a _{(p)}=a_{(p)} \mathbb Z _{(p)}$ is a principal fractional ideal of the localization $S_{(p)}$ for all primes p. Every locally principal ideal is invertible, and the extent to which the converse holds is something that arises in an important way more generally in algebraic geometry. In the language of commutative algebra, a locally principal S-module is equivalently a projective S-module of rank 1.

Suppose that S is not maximal; then $S(d)=\mathbb Z + f\mathbb Z _K$ for a unique integer $f \in \mathbb Z _{>1}$, the conductor  of S; accordingly, $d=d_Kf^2$, where $d_K \in \mathbb Z$ is the discriminant of $\mathbb Z _K$ (a fundamental discriminant). In this case, there is always an ideal of S that is not invertible. Specifically, consider the ideal

$$\begin{aligned} \mathfrak f =f\mathbb Z + \sqrt{d}\mathbb Z \subseteq S. \end{aligned}$$

(16.1.1)

Then $\mathfrak f$ is a free $\mathbb Z$-module of rank 2 and

$$\begin{aligned} \mathfrak f ^2 = (f\mathbb Z +\mathbb Z \sqrt{d})^2 = f^2\mathbb Z + f\sqrt{d} \mathbb Z = f \mathfrak f \end{aligned}$$

so if $\mathfrak f$ were invertible, then cancelling we would obtain $\mathfrak f = fS$, a contradiction. The source of this example is that $\mathfrak f = f S(d_K)$ since $S(d_K) = \mathbb Z + \sqrt{d_K} \mathbb Z$, so really this fractional ideal belongs to the maximal order $S(d_K)$, not to S. For more on the notion of invertibility for quadratic orders, see Cox [Cox89, §7], with further connections to quadratic forms and class numbers.

We now turn to the quaternionic generalization, where noncommutativity presents some complications. Let B be a quaternion algebra over $\mathbb Q$ and let $\mathcal {O}\subset B$ be an order. To study ideals of $\mathcal {O}$ we must distinguish between left or right ideals, and the product of two (say) right $\mathcal {O}$-ideals need not be again a right $\mathcal {O}$-ideal! To address this, for lattices $I,J \subset B$, we say that I is compatible with J if the right order of I is equal to the left order of J, so that what comes between I and J in the product $I \cdot J$ “matches up”.

A lattice $I \subset B$ is right invertible if there exists a lattice $I' \subset B$ such that

$$II'=\mathcal {O}{}_{\textsf {\tiny {L}} }(I)$$

with a compatible product, and we call $I'$ a right inverse. We similarly define notions on the left, and we say $I \subset B$ is invertible if there is a two-sided inverse $I' \subset B$, so

$$\begin{aligned} II'=\mathcal {O}{}_{\textsf {\tiny {L}} }(I)=\mathcal {O}{}_{\textsf {\tiny {R}} }(I') \text { and } I'I=\mathcal {O}{}_{\textsf {\tiny {L}} }(I')=\mathcal {O}{}_{\textsf {\tiny {R}} }(I) \end{aligned}$$

with both of these products are compatible. If a lattice I has a two-sided inverse, then this inverse is uniquely given by

$$\begin{aligned} I^{-1} :=\{\alpha \in B : I\alpha I \subseteq I\} \end{aligned}$$

(defined so as to simultaneously take care of both left and right): we always have that $II^{-1} \subseteq \mathcal {O}{}_{\textsf {\tiny {L}} }(I)$, but equality is needed for right invertibility, and the same on the left.

Let $\mathcal {O}\subseteq B$ be an order. A left $\mathcal {O}$fractional $\mathcal {O}$ -ideal is a lattice $I \subseteq B$ such that $\mathcal {O}\subseteq \mathcal {O}{}_{\textsf {\tiny {L}} }(I)$; we similarly define on the right. For a maximal order, all lattices are invertible (Proposition 16.6.15(b)).

Proposition 16.1.2

Let $\mathcal {O}\subseteq B$ be a maximal order. Then a left or right fractional $\mathcal {O}$-ideal is invertible.

The simplest kind of invertible lattices arethe principal lattices

$$\begin{aligned} I=\mathcal {O}{}_{\textsf {\tiny {L}} }(I)\alpha =\alpha \mathcal {O}{}_{\textsf {\tiny {R}} }(I) \end{aligned}$$

with $\alpha \in B^\times$: its inverse is $I^{-1}=\alpha ^{-1}\mathcal {O}{}_{\textsf {\tiny {L}} }(I)=\mathcal {O}{}_{\textsf {\tiny {R}} }(I)\alpha ^{-1}$.

The major task of this chapter will be to interrelate these notions in the quaternionic context. Let

$$\begin{aligned} {{\,\mathrm{nrd}\,}}(I) :=\gcd (\{{{\,\mathrm{nrd}\,}}(\alpha ) : \alpha \in I\}), \end{aligned}$$

i.e., ${{\,\mathrm{nrd}\,}}(I)$ is a positive generator of the (finitely generated) subgroup of $\mathbb Q$ generated by ${{\,\mathrm{nrd}\,}}(\alpha )$ for $\alpha \in I$. The main result over $\mathbb Q$ is the following theorem (Main Theorem 16.7.7).

Main Theorem 16.1.3

Let B be a quaternion algebra over $\mathbb Q$ and let $I \subset B$ be an integral lattice. Then the following are equivalent:

(i):

I is locally principal, i.e., $I_{(p)} = I \otimes _\mathbb{Z } \mathbb Z _{(p)}$ is principal for all primes p;

(ii):

I is invertible;

(iii):

I is right invertible;

(iii${}^\prime$):

I is left invertible;

(iv):

${{\,\mathrm{nrd}\,}}(I)^2=[\mathcal {O}{}_{\textsf {\tiny {R}} }(I):I]$; and

(iv${}^\prime$):

${{\,\mathrm{nrd}\,}}(I)^2=[\mathcal {O}{}_{\textsf {\tiny {L}} }(I):I]$.

Accordingly, for I integral, we may define the right absolute norm of I by

$$\begin{aligned} \mathsf{N }(I) :=\#(\mathcal {O}{}_{\textsf {\tiny {R}} }(I)/I) = [\mathcal {O}{}_{\textsf {\tiny {R}} }(I):I] \in \mathbb Z _{\ge 1} \end{aligned}$$

and similarly on the left; by Main Theorem 16.1.3 (iv) $\Leftrightarrow$ (iv${}^\prime$), when I is locally principal, the left and right absolute norms coincide (called then just absolute norm) and are related to the reduced norm by $\mathsf{N }(I) = {{\,\mathrm{nrd}\,}}(I)^2$.

2 Locally principal, compatible lattices

The simplest lattices to understand are those that are principal; but as we saw in section 9.4, lattices over Dedekind domains are inherently local in nature. We are led to consider the more general class of locally principal lattices. We work first with lattices, and later we will keep track of their left and right orders.

Throughout this chapter, let R be a Dedekind domain with field of fractions F, let B be a finite-dimensional algebra over F, and let $I \subseteq B$ be an R-lattice.

Definition 16.2.1

I is principal if there exists $\alpha \in B$ such that

$$\begin{aligned} I=\mathcal {O}{}_{\textsf {\tiny {L}} }(I)\alpha = \alpha \mathcal {O}{}_{\textsf {\tiny {R}} }(I); \end{aligned}$$

we say that I is generated by $\alpha$.

16.2.2

If I is generated by $\alpha \in B$, then since I is a lattice (Definition 9.3.1) we have $IF = B\alpha =B$, so $\alpha \in B^\times$.

16.2.3

If $I=\mathcal {O}{}_{\textsf {\tiny {L}} }(I)\alpha$, then $\mathcal {O}{}_{\textsf {\tiny {R}} }(I)=\alpha ^{-1} \mathcal {O}{}_{\textsf {\tiny {L}} }(I) \alpha$ by Exercise 16.2, so

$$\begin{aligned} I = \alpha (\alpha ^{-1} \mathcal {O}{}_{\textsf {\tiny {L}} }(I) \alpha )= \alpha \mathcal {O}{}_{\textsf {\tiny {R}} }(I). \end{aligned}$$

Therefore it is sufficient to check for a one-sided generator (and if we defined the obvious notions of left principal or right principal, these would be equivalent to the notion of principal).

The notion of principality naturally extends locally.

Definition 16.2.4

An R-lattice I is locally principal if $I_{(\mathfrak p )}=I \otimes _R R_{(\mathfrak p )}$ is a principal $R_{(\mathfrak p )}$-lattice for all primes $\mathfrak p$ of R.

Now let I, J be R-lattices in B. We define the product IJ to be the R-submodule of B generated by the set

$$\begin{aligned} \{\alpha \beta : \alpha \in I,\ \beta \in J\}. \end{aligned}$$

The product IJ is an R-lattice: it is finitely generated as this is true of I, J individually, and there exists a nonzero $r \in I$ (Exercise 9.2) so $rJ \subset IJ$ and thus

$$\begin{aligned} B=FJ=F(rJ) \subseteq F IJ \end{aligned}$$

so equality holds.

When multiplication of two lattices matches up their respective left and right orders, we give it a name.

Definition 16.2.5

We say that I is compatible with J if $\mathcal {O}{}_{\textsf {\tiny {R}} }(I)=\mathcal {O}{}_{\textsf {\tiny {L}} }(J)$.

We will also sometimes just say that the product IJ is compatible to mean that I is compatible with J. The relation “is compatible with” is in general neither symmetric nor transitive.

16.2.6

I has the structure of a right $\mathcal {O}{}_{\textsf {\tiny {R}} }(I)$-module and J the structure of a left $\mathcal {O}{}_{\textsf {\tiny {L}} }(J)$-module. When $\mathcal {O}{}_{\textsf {\tiny {R}} }(I)=\mathcal {O}{}_{\textsf {\tiny {L}} }(J)=\mathcal {O}$, that is, when I is compatible with J, it makes sense to consider the tensor product $I \otimes _\mathcal {O}J$ as an R-module. The multiplication map defined by $\alpha \otimes \beta \mapsto \alpha \beta$ restricts to give an isomorphism as R-lattices. In this way, multiplication of compatible lattices can be thought of as a special case of the tensor product of modules.

We conclude this section with several other basic properties of lattices.

Definition 16.2.7

An R-lattice I is integral if $I^2 \subseteq I$.

In Definition 16.2.7, the product need not be compatible.

Lemma 16.2.8

Let I be an R-lattice. Then the following are equivalent:

(i):

I is integral;

(ii):

For all $\alpha ,\beta \in I$, we have $\alpha \beta \in I$;

(iii):

$I \subseteq \mathcal {O}{}_{\textsf {\tiny {L}} }(I)$, so I is a left ideal of $\mathcal {O}{}_{\textsf {\tiny {L}} }(I)$ in the usual sense;

(iii${}^\prime$):

$I \subseteq \mathcal {O}{}_{\textsf {\tiny {R}} }(I)$; and

(iv):

$I \subseteq \mathcal {O}{}_{\textsf {\tiny {L}} }(I) \cap \mathcal {O}{}_{\textsf {\tiny {R}} }(I)$.

If I is integral, then every element of I is integral over R.

Proof. The equivalence (i) $\Leftrightarrow$ (ii) follows immediately. For (i) $\Leftrightarrow$ (iii), we have $II \subseteq I$ if and only if $I \subseteq \mathcal {O}{}_{\textsf {\tiny {L}} }(I)$ by definition of $\mathcal {O}{}_{\textsf {\tiny {L}} }(I)$, and the same argument gives (i) $\Leftrightarrow$ (iii${}^\prime$), and this then gives (i) $\Leftrightarrow$ (iv). The final statement follows from Lemma 10.3.2. $\square$

In light of Lemma 16.2.8, we need not define notions of left integral or right integral.

For an R-lattice I, there exists nonzero $d \in R$ such that dI is integral, so every R-lattice $I=(dI)/d$ is fractional in the sense that it is obtained from an integral lattice with denominator.

Definition 16.2.9

Let $\mathcal {O}\subseteq B$ be an R-order. A left $\mathcal {O}$ fractional $\mathcal {O}$ -ideal is a lattice $I \subseteq B$ such that $\mathcal {O}\subseteq \mathcal {O}{}_{\textsf {\tiny {L}} }(I)$; similarly on the right.

If $\mathcal {O},\mathcal {O}' \subseteq B$ are R-orders, then a $\mathcal {O},\mathcal {O}'$ fractional $\mathcal {O},\mathcal {O}'$ -idealis a lattice I that is a left fractional $\mathcal {O}$-ideal and a right fractional $\mathcal {O}'$-ideal.

Remark 16.2.10. A left ideal $I \subseteq \mathcal {O}$ in the usual sense is an integral left $\mathcal {O}$-ideal in the sense of Definition 16.2.9 if and only if $IF=B$, i.e., I is a (full) R-lattice. (Same for right and two-sided ideals.) If I is nonzero and B is a division algebra, then automatically I is full and the two notions coincide.

Indeed, suppose $I \subseteq \mathcal {O}$ is a left ideal of $\mathcal {O}$ (in the usual sense). Then $\mathcal {O}\subseteq \mathcal {O}{}_{\textsf {\tiny {L}} }(I)$ so in particular I has the structure of an R-module, and since $\mathcal {O}$ is finitely generated as an R-module and R is noetherian, it follows that I is finitely generated. Consequently, a left ideal $I \subseteq \mathcal {O}$ is a left fractional $\mathcal {O}$-ideal if and only if $IF=B$.

Definition 16.2.11

Let I be a left fractional $\mathcal {O}$-ideal. We say that I is sated (as a left fractional $\mathcal {O}$-ideal) if $\mathcal {O}=\mathcal {O}{}_{\textsf {\tiny {L}} }(I)$. We make a similar definition on the right and for two-sided ideals.

Example 16.2.12

By Lemma 15.6.16, ${{\,\mathrm{codiff}\,}}(\mathcal {O})$ is a two-sided sated $\mathcal {O}$-ideal.

Remark 16.2.13. Our notion of sated is sometimes called proper: we do not use this already overloaded term, as it conflicts with the notion of a proper subset.

3 Reduced norms

Next, we extend the reduced norm to lattices; see also Reiner [Rei2003, §24]. To this end, in this section we suppose that B is semisimple.

Definition 16.3.1

The reduced norm ${{\,\mathrm{nrd}\,}}(I)$ of I is the R-submodule of F generated by the set $\{{{\,\mathrm{nrd}\,}}(\alpha ) : \alpha \in I\}$.

Lemma 16.3.2

The reduced norm ${{\,\mathrm{nrd}\,}}(I)$ is a fractional ideal of F: i.e., it is finitely generated as an R-module.

Proof. We first give a proof when B has a standard involution, and ${{\,\mathrm{nrd}\,}}$ is a quadratic form. Since I is an R-lattice we have $IF=B$; since ${{\,\mathrm{nrd}\,}}(B) \ne \{0\}$, we have ${{\,\mathrm{nrd}\,}}(I) \ne \{0\}$. And I is generated by finitely many $\alpha _i$ as an R-module; the R-module ${{\,\mathrm{nrd}\,}}(I)$ is then generated by the values $a_{ii}={{\,\mathrm{nrd}\,}}(\alpha _i)$ and $a_{ij}={{\,\mathrm{nrd}\,}}(\alpha _i+\alpha _j)-{{\,\mathrm{nrd}\,}}(\alpha _i)-{{\,\mathrm{nrd}\,}}(\alpha _j)$, since then

$$\begin{aligned} {{\,\mathrm{nrd}\,}}\bigl (\textstyle {\sum }_i c_i \alpha _i\bigr ) = \textstyle {\sum }_{i,j} a_{ij} c_ic_j \in \sum _{i,j} R a_{ij} \end{aligned}$$

for all $c_i \in R$.

Now for the general case. Replacing I by rI with $r \in R$ nonzero, we may suppose that I is integral, and hence ${{\,\mathrm{nrd}\,}}(I) \subseteq R$. Since I is a lattice, there exists $r \in I \cap R$ with $r \ne 0$. For all $\mathfrak p$ such that ${{\,\mathrm{ord}\,}}_\mathfrak p (r)=0$, we have $1 \in I_{(\mathfrak p )}$ so ${{\,\mathrm{nrd}\,}}(I_{(\mathfrak p )})=R_{(\mathfrak p )}$. For each of the finitely many primes $\mathfrak p$ that remain, we choose an element $\alpha \in I$ such that ${{\,\mathrm{ord}\,}}_\mathfrak p ({{\,\mathrm{nrd}\,}}(\alpha ))$ is minimal; then ${{\,\mathrm{nrd}\,}}(\alpha )$ generates ${{\,\mathrm{nrd}\,}}(I_{(\mathfrak p )})$, and by the local-global dictionary, these finitely many elements generate ${{\,\mathrm{nrd}\,}}(I)$. $\square$

16.3.3

For a prime $\mathfrak p$ of R we have ${{\,\mathrm{nrd}\,}}(I)_{(\mathfrak p )}={{\,\mathrm{nrd}\,}}(I_{(\mathfrak p )})$, so by the local-global property of lattices (Lemma 9.4.6),

$$\begin{aligned} {{\,\mathrm{nrd}\,}}(I)=\bigcap _\mathfrak{p } {{\,\mathrm{nrd}\,}}(I)_{(\mathfrak p )} = \bigcap _\mathfrak{p } {{\,\mathrm{nrd}\,}}(I_{(\mathfrak p )}). \end{aligned}$$

(16.3.4)

16.3.5

If I is a principal R-lattice generated by $\alpha \in I$ then ${{\,\mathrm{nrd}\,}}(I)={{\,\mathrm{nrd}\,}}(\alpha )R$; more generally, if I is an R-lattice and $\alpha \in B^\times$ then ${{\,\mathrm{nrd}\,}}(\alpha I)={{\,\mathrm{nrd}\,}}(\alpha ){{\,\mathrm{nrd}\,}}(I)$ (Exercise 16.4).

Now suppose that I, J are lattices. Then ${{\,\mathrm{nrd}\,}}(IJ) \supseteq {{\,\mathrm{nrd}\,}}(I){{\,\mathrm{nrd}\,}}(J)$. However, we need not have equality, as the following example indicates.

Example 16.3.6

It is not always true that ${{\,\mathrm{nrd}\,}}(IJ)={{\,\mathrm{nrd}\,}}(I){{\,\mathrm{nrd}\,}}(J)$. For example, if $a \in R$ is neither zero nor a unit, then $I=\begin{pmatrix} aR &{} R \\ aR &{} R \end{pmatrix}$ and $J=\begin{pmatrix} aR &{} aR \\ R &{} R \end{pmatrix}$ are R-lattices in ${{\,\mathrm{M}\,}}_2(F)$ with ${{\,\mathrm{nrd}\,}}(I)={{\,\mathrm{nrd}\,}}(J)=aR$ but $IJ={{\,\mathrm{M}\,}}_2(R)$ and so ${{\,\mathrm{nrd}\,}}(IJ)=R$.

We have $\mathcal {O}{}_{\textsf {\tiny {R}} }(J)={{\,\mathrm{M}\,}}_2(R)=\mathcal {O}{}_{\textsf {\tiny {L}} }(I)$, so J is compatible with I, and ${{\,\mathrm{nrd}\,}}(JI)=a^2R={{\,\mathrm{nrd}\,}}(J){{\,\mathrm{nrd}\,}}(I)$; but

$$\begin{aligned} \mathcal {O}{}_{\textsf {\tiny {R}} }(I)=\begin{pmatrix} R &{} a^{-1}R \\ aR &{} R \end{pmatrix} \quad \text {and} \quad \mathcal {O}{}_{\textsf {\tiny {L}} }(J)=\begin{pmatrix} R &{} aR \\ a^{-1}R &{} R \end{pmatrix}, \end{aligned}$$

so I is not compatible with J.

The issue present in Example 16.3.6 is that the product is not as well-behaved for noncommutative rings as for commutative rings; we need the elements coming between I and J to match up.

Lemma 16.3.7

Suppose that I is compatible with J and that either I or J is locally principal. Then ${{\,\mathrm{nrd}\,}}(IJ)={{\,\mathrm{nrd}\,}}(I){{\,\mathrm{nrd}\,}}(J)$.

Proof. By the local-global property for norms (16.3.4) and since localization commutes with multiplication, i.e.,

$(\mathfrak a \mathfrak b )_{(\mathfrak p )} = \mathfrak a _{(\mathfrak p )} \mathfrak b _{(\mathfrak p )}$ for all (finitely generated) R-modules $\mathfrak a ,\mathfrak b \subseteq F$,

we may localize and suppose that either I or J is principal. Suppose I is (right) principal. Then $I=\alpha \mathcal {O}$ for some $\alpha \in B$ where $\mathcal {O}=\mathcal {O}{}_{\textsf {\tiny {R}} }(I)=\mathcal {O}{}_{\textsf {\tiny {L}} }(J)$. Then

$$\begin{aligned} IJ=(\alpha \mathcal {O})J = \alpha (\mathcal {O}J)=\alpha J \end{aligned}$$

and so ${{\,\mathrm{nrd}\,}}(IJ)={{\,\mathrm{nrd}\,}}(\alpha ){{\,\mathrm{nrd}\,}}(J)={{\,\mathrm{nrd}\,}}(I){{\,\mathrm{nrd}\,}}(J)$ by 16.3.5. The case where J is principal follows in the same way. $\square$

Principal lattices are characterized by reduced norms, as follows.

Lemma 16.3.8

Let I be locally principal and let $\alpha \in I$. Then $\alpha$ generates I if and only if ${{\,\mathrm{nrd}\,}}(\alpha )R={{\,\mathrm{nrd}\,}}(I)$.

Proof. If $I=\alpha \mathcal {O}$ then ${{\,\mathrm{nrd}\,}}(I)={{\,\mathrm{nrd}\,}}(\alpha )R$ by Lemma 16.3.7.

For the converse, let $\mathcal {O}=\mathcal {O}{}_{\textsf {\tiny {R}} }(I)$. We want to show that $I=\alpha \mathcal {O}$, and we know that $I \supseteq \alpha \mathcal {O}$. To prove that equality holds, it suffices to show this locally, so we may suppose that $I=\beta \mathcal {O}$. Then $\alpha =\beta \mu$ with $\mu \in \mathcal {O}$, and ${{\,\mathrm{nrd}\,}}(\alpha )={{\,\mathrm{nrd}\,}}(\beta \mu )={{\,\mathrm{nrd}\,}}(\beta ){{\,\mathrm{nrd}\,}}(\mu )$. By hypothesis, ${{\,\mathrm{nrd}\,}}(\mu ) \in R^\times$, and thus $\mu \in \mathcal {O}^\times$, so $\beta \mathcal {O}=\alpha \mathcal {O}$. $\square$

4 Algebra and absolute norm

The reduced norm of an ideal is related to its algebra norm, as follows. We continue to suppose that B is semisimple, so the definitions of left and right norm coincide.

Definition 16.4.1

The (algebra) norm ${{\,\mathrm{Nm}\,}}_{B|F}(I)$ of I is the R-submodule of F generated by the set $\{{{\,\mathrm{Nm}\,}}_{B|F}(\alpha ) : \alpha \in I\}$.

Remark 16.4.2. The definition of algebra norm by necessity depends on the choice of domain R; indeed, I is an R-lattice.

Proposition 16.4.3

The following are equivalent:

• (i) I is locally principal;

• (ii) ${{\,\mathrm{Nm}\,}}_{B|F}(I)=[\mathcal {O}{}_{\textsf {\tiny {L}} }(I):I]_R$; and

• (iii) ${{\,\mathrm{Nm}\,}}_{B|F}(I)=[\mathcal {O}{}_{\textsf {\tiny {R}} }(I):I]_R$.

If B is simple with $\dim _F B=n^2$, then these are further equivalent to

• (iv) ${{\,\mathrm{nrd}\,}}(I)^n=[\mathcal {O}{}_{\textsf {\tiny {L}} }(I):I]_R$.

• (v) ${{\,\mathrm{nrd}\,}}(I)^n=[\mathcal {O}{}_{\textsf {\tiny {R}} }(I):I]_R$.

Proof. Let $\mathcal {O}=\mathcal {O}{}_{\textsf {\tiny {L}} }(I)$. Let $\alpha \in I$. Right multiplication by $\alpha$ gives an R-module isomorphism (change of basis between two free R-modules), so by Lemma 9.6.4 we have $[\mathcal {O}:\mathcal {O}\alpha ]_R=\det (\alpha )R={{\,\mathrm{Nm}\,}}_{B|F}(\alpha )R$, thinking of $\alpha \in {{\,\mathrm{End}\,}}_F(B)$.

We now prove (i) $\Leftrightarrow$ (ii). We may suppose R is local, so R is a DVR. For all $\alpha \in I$, the following equality holds:

$$\begin{aligned}{}[\mathcal {O}:I]_R [I:\mathcal {O}\alpha ]_R = [\mathcal {O}:\mathcal {O}\alpha ]_R = {{\,\mathrm{Nm}\,}}_{B|F}(\alpha ) R. \end{aligned}$$

(16.4.4)

To show (i) $\Rightarrow$ (ii), if $I=\mathcal {O}\alpha$ then ${{\,\mathrm{Nm}\,}}_{B|F}(I)={{\,\mathrm{Nm}\,}}_{B|F}(\alpha )R$ and so by cancelling $[I:\mathcal {O}\alpha ]_R=R$ in (16.4.4) we obtain (ii). To show (ii) $\Rightarrow$ (i), suppose that ${{\,\mathrm{Nm}\,}}_{B|F}(I)=[\mathcal {O}:I]_R$. Let $\alpha \in I$ be such that ${{\,\mathrm{Nm}\,}}_{B|F}(\alpha )$ has minimal valuation; then ${{\,\mathrm{Nm}\,}}_{B|F}(\alpha )$ generates ${{\,\mathrm{Nm}\,}}_{B|F}(I)$. By (16.4.4), cancelling on both sides $[I:\mathcal {O}\alpha ]_R=R$, and since $\mathcal {O}\alpha \subseteq I$ we conclude $I=\mathcal {O}\alpha$. A similar argument holds on the right, proving (i) $\Leftrightarrow$ (iii). Finally, (iii) $\Leftrightarrow$ (iv) since ${{\,\mathrm{Nm}\,}}_{B|F}(\alpha )={{\,\mathrm{nrd}\,}}(\alpha )^n$, and the same on the right. $\square$

16.4.5

Recalling the proof of Proposition 16.4.3 and the definition of R-index, we always have the containment

$$\begin{aligned} {{\,\mathrm{Nm}\,}}_{B|F}(I) \supseteq [\mathcal {O}{}_{\textsf {\tiny {L}} }(I):I]_R \end{aligned}$$

and the same on the right; by Proposition 16.4.3, equality is equivalent to I being locally principal.

To conclude this section, we suppose for its remainder that F is a local with valuation ring R or a global number field. Then the reduced norm is also related to the absolute norm, an absolute measure of size, as follows.

16.4.6

For a fractional ideal $\mathfrak a$ of R, we define the absolute norm $\mathsf{N }(\mathfrak a )$ to be

$$\begin{aligned} \mathsf{N }(\mathfrak a ) :=[R:\mathfrak a ]_\mathbb{Z }<\infty \end{aligned}$$

(16.4.7)

the index taken as abelian groups, recalling Example 9.6.6. Then

$$\begin{aligned} \mathsf{N }(\mathfrak a )=|{{\,\mathrm{Nm}\,}}_{F/\mathbb Q }(\mathfrak a ) \,| \end{aligned}$$

and if $\mathfrak a \subseteq R$ then

$$\begin{aligned} \mathsf{N }(\mathfrak a )=\#(R/\mathfrak a ), \end{aligned}$$

so this norm is also called the counting norm.

We extend this definition to elements $a \in F^\times$ by defining $\mathsf{N }(a) :=\mathsf{N }(aR)$.

16.4.8

Similarly, if $I \subseteq B$ is a locally principal R-lattice, we define the absolute norm of I to be

$$\begin{aligned} \mathsf{N }(I) :=[\mathcal {O}{}_{\textsf {\tiny {L}} }(I):I]_\mathbb{Z }=[\mathcal {O}{}_{\textsf {\tiny {R}} }(I):I]_\mathbb{Z }, \end{aligned}$$

(16.4.9)

the latter equality by taking $R=\mathbb Z$ in Proposition 16.4.3. If I is integral then

$$\begin{aligned} \mathsf{N }(I)=\#(\mathcal {O}{}_{\textsf {\tiny {L}} }(I)/I)=\#(\mathcal {O}{}_{\textsf {\tiny {R}} }(I)/I). \end{aligned}$$

The absolute norm of I is compatible with the absolute norm on R via

$$\begin{aligned} \mathsf{N }(I)=[\mathcal {O}{}_{\textsf {\tiny {R}} }(I):I]_\mathbb{Z }=[R:{{\,\mathrm{Nm}\,}}_{B|F}(I)]_\mathbb{Z }=\mathsf{N }({{\,\mathrm{Nm}\,}}_{B|F}(I)); \end{aligned}$$

and if B is simple with $\dim _F B=n^2$ then

$$\begin{aligned} \mathsf{N }(I)=\mathsf{N }({{\,\mathrm{Nm}\,}}_{B|F}(I))=\mathsf{N }({{\,\mathrm{nrd}\,}}(I))^n. \end{aligned}$$

(16.4.10)

Remark 16.4.11. The absolute norm may also be defined for a global function field, but there is no canonical ‘ring of integers’ as above.

5 Invertible lattices

We are now in a position to investigate the class of invertible lattices. Let $I \subseteq B$ be an R-lattice.

Definition 16.5.1

I is invertible  if there exists an R-lattice $I' \subseteq B$ that is a (two-sided) inverse  to I, i.e.

$$\begin{aligned} II'=\mathcal {O}{}_{\textsf {\tiny {L}} }(I)=\mathcal {O}{}_{\textsf {\tiny {R}} }(I') \text { and } I'I=\mathcal {O}{}_{\textsf {\tiny {L}} }(I')=\mathcal {O}{}_{\textsf {\tiny {R}} }(I). \end{aligned}$$

(16.5.2)

In particular, both of the products in (16.5.2) are compatible.

16.5.3

If I, J are invertible lattices and I is compatible with J, then IJ is invertible (Exercise 16.10).

16.5.4

If I is a principal lattice, then I is invertible: if $I=\mathcal {O}\alpha$ with $\alpha \in B^\times$ and $\mathcal {O}=\mathcal {O}{}_{\textsf {\tiny {L}} }(I)$, then $I'=\alpha ^{-1} \mathcal {O}$ has

$$\begin{aligned} II'=(\mathcal {O}\alpha )(\alpha ^{-1}\mathcal {O})=\mathcal {O}(\alpha \alpha ^{-1})\mathcal {O}=\mathcal {O}\mathcal {O}=\mathcal {O}\end{aligned}$$

so $I'$ is a right inverse, and

$$\begin{aligned} I'I = (\alpha ^{-1} \mathcal {O})(\mathcal {O}\alpha ) = \alpha ^{-1} \mathcal {O}\alpha = \mathcal {O}{}_{\textsf {\tiny {R}} }(I) \end{aligned}$$

so $I'$ is also a left inverse.

A candidate for the inverse presents itself quite naturally. If $II'=\mathcal {O}{}_{\textsf {\tiny {L}} }(I)$ and $I'I=\mathcal {O}{}_{\textsf {\tiny {R}} }(I)$, then $II'I=I$.

Definition 16.5.5

We define the quasi-inverse of I as

$$\begin{aligned} I^{-1} :=\{\alpha \in B : I\alpha I \subseteq I\}. \end{aligned}$$

(16.5.6)

Lemma 16.5.7

The following statements hold.

• (a) The quasi-inverse $I^{-1}$ is an R-lattice and

  $$\begin{aligned} II^{-1}I \subseteq I. \end{aligned}$$

• (b) If $\mathcal {O}$ is an R-order, then $\mathcal {O}^{-1}=\mathcal {O}$.

Proof. Statement (a) follows as in the proof of Lemma 10.2.7, and the inclusion is by the definition of $I^{-1}$. For statement (b), if $\alpha \in \mathcal {O}$, then $\mathcal {O}\alpha \mathcal {O}\subseteq \mathcal {O}$ since $\mathcal {O}$ is an order; conversely, if $\mathcal {O}\alpha \mathcal {O}\subseteq \mathcal {O}$, then taking $1 \in \mathcal {O}$ on left and right we conclude $\alpha \in \mathcal {O}$. $\square$

We now consider the quasi-inverse as an inverse.

Proposition 16.5.8

The following are equivalent:

• (i) $I^{-1}$ is a (two-sided) inverse for I;

• (ii) $I^{-1}I=\mathcal {O}{}_{\textsf {\tiny {R}} }(I)$ and $II^{-1}=\mathcal {O}{}_{\textsf {\tiny {L}} }(I)$;

• (iii) I is invertible;

• (iv) There is a compatible product $II^{-1}I=I$ and both $1 \in II^{-1}$ and $1 \in I^{-1}I$.

Proof. The implication (i) $\Rightarrow$ (ii) is clear. For (ii) $\Rightarrow$ (i), we need to check the compatibility of the product: but since $I^{-1}I=\mathcal {O}{}_{\textsf {\tiny {R}} }(I)$ we have $\mathcal {O}{}_{\textsf {\tiny {L}} }(I^{-1}) \subseteq \mathcal {O}{}_{\textsf {\tiny {R}} }(I)$, and from the other direction we have the other containment, so these are equal.

The implication (i) $\Rightarrow$ (iii) is clear. For (iii) $\Rightarrow$ (i), suppose that $I'$ is an inverse to I. Then $I = II'I$ so $I' \subseteq I^{-1}$ by definition. Therefore $I \subseteq II^{-1}I \subseteq I$ and equality holds throughout. Multiplying by $I'$ on the left and right then gives

$$\begin{aligned} I^{-1}=(I'I)I^{-1}(II') = I'II' = I'. \end{aligned}$$

Again the implication (i) $\Rightarrow$ (iv) is immediate. To prove (iv) $\Rightarrow$ (ii), we need to show that $II^{-1}=\mathcal {O}{}_{\textsf {\tiny {L}} }(I)$ and $I^{-1}I=\mathcal {O}{}_{\textsf {\tiny {R}} }(I)$; we show the former. By compatibility, $\mathcal {O}{}_{\textsf {\tiny {R}} }(I^{-1})=\mathcal {O}{}_{\textsf {\tiny {L}} }(I)=\mathcal {O}$. If $II^{-1}=J$ then $J = II^{-1}=\mathcal {O}(II^{-1})\mathcal {O}=\mathcal {O}J \mathcal {O}$, so $J \subseteq \mathcal {O}$ is a two-sided ideal of $\mathcal {O}$ containing 1 hence $J=\mathcal {O}$. $\square$

Invertibility is a local property, as one might expect.

Lemma 16.5.9

I is invertible if and only $I_{(\mathfrak p )}$ is invertible for all primes $\mathfrak p$.

Proof. We employ Proposition 16.5.8(iv): We have $II^{-1}I=I$ if and only if

$$\begin{aligned} (II^{-1}I)_{(\mathfrak p )}=I_{(\mathfrak p )}(I^{-1})_{(\mathfrak p )}I_{(\mathfrak p )}=I_{(\mathfrak p )} \end{aligned}$$

for all primes $\mathfrak p$ and e.g. $1 \in II^{-1}$ if and only if $1 \in I_{(\mathfrak p )}I^{-1}_{(\mathfrak p )}$. $\square$

Corollary 16.5.10

If I is locally principal, then I is invertible.

Proof. Combine 16.5.4 with Lemma 16.5.9. $\square$

A compatible product with an invertible lattice respects taking left (and right) orders, as follows.

Lemma 16.5.11

If I is compatible with J and J is invertible, then $\mathcal {O}{}_{\textsf {\tiny {L}} }(IJ)=\mathcal {O}{}_{\textsf {\tiny {L}} }(I)$.

Proof. We always have $\mathcal {O}{}_{\textsf {\tiny {L}} }(I) \subseteq \mathcal {O}{}_{\textsf {\tiny {L}} }(IJ)$ (even without J invertible). To show the other containment, suppose that $\alpha \in \mathcal {O}{}_{\textsf {\tiny {L}} }(IJ)$, so that $\alpha IJ \subseteq IJ$. Multiplying by $J^{-1}$, we conclude $\alpha I \subseteq I$ and $\alpha \in \mathcal {O}{}_{\textsf {\tiny {L}} }(I)$. $\square$

Finally, not every lattice is invertible, and it is helpful to have counterexamples at hand (see also Exercise 16.12).

Example 16.5.12

Let $p \in \mathbb Z$ be prime. Let $B :=\displaystyle {\biggl (\frac{p,p}{\mathbb {Q}}\biggr )}$ and

$$\begin{aligned} \mathcal {O}&:=\mathbb Z \oplus p\mathbb Z i \oplus p\mathbb Z j \oplus \mathbb Z ij \\ I&:=p^2\mathbb Z \oplus \mathbb Z i \oplus \mathbb Z j \oplus \mathbb Z ij. \end{aligned}$$

Then $\mathcal {O}\subset B$ is an order and $\mathcal {O}{}_{\textsf {\tiny {L}} }(I)=\mathcal {O}{}_{\textsf {\tiny {R}} }(I)=\mathcal {O}$. We compute that

$$\begin{aligned} I^{-1}=p\mathbb Z \oplus \mathbb Z i \oplus \mathbb Z j \oplus \mathbb Z ij \end{aligned}$$

(16.5.13)

and

$$\begin{aligned} \mathcal {O}{}_{\textsf {\tiny {L}} }(I^{-1})=\mathcal {O}{}_{\textsf {\tiny {R}} }(I^{-1})=\mathbb Z + \mathbb Z i + \mathbb Z j + \frac{1}{p}\mathbb Z ij = \mathbb Z + \frac{1}{p}\mathcal {O}; \end{aligned}$$

(16.5.14)

so in the product

$$\begin{aligned} II^{-1}=I^{-1}I = p\mathbb Z \oplus p\mathbb Z i \oplus p\mathbb Z j \oplus \mathbb Z ij \subsetneq \mathcal {O}\end{aligned}$$

(16.5.15)

we see I is not invertible and the product is not compatible.

Seen a different way, we have $\overline{I}=I$ and in the compatible product

$$\begin{aligned} I^2 = I\overline{I} = \overline{I}I = p\mathbb Z \oplus p\mathbb Z i \oplus p\mathbb Z j \oplus \mathbb Z ij \end{aligned}$$

(16.5.16)

we have $i,j \in \mathcal {O}{}_{\textsf {\tiny {L}} }(I^2)=\mathcal {O}{}_{\textsf {\tiny {R}} }(I^2)$ but $i,j \not \in \mathcal {O}$; therefore, I is not invertible by Lemma 16.5.11. Indeed,

$$\begin{aligned} \mathcal {O}{}_{\textsf {\tiny {L}} }(I^2)=\mathcal {O}{}_{\textsf {\tiny {R}} }(I^2)=\frac{1}{p}I^2 = \mathbb Z + \frac{1}{p}\mathcal {O}. \end{aligned}$$

Finally, it will convenient to consider invertibility in the context of ideals, labelling left and right orders as follows.

Definition 16.5.17

Let $\mathcal {O},\mathcal {O}' \subseteq B$ be R-orders and let I be a fractional $\mathcal {O},\mathcal {O}'$-ideal. We say I is invertible $\mathcal {O},\mathcal {O}'$ if I is invertible as a lattice and I is sated (i.e., $\mathcal {O}=\mathcal {O}{}_{\textsf {\tiny {L}} }(I)$ and $\mathcal {O}'=\mathcal {O}{}_{\textsf {\tiny {R}} }(I)$).

16.5.18

The condition that I is sated in Definition 16.5.17 is important: we must be careful to work over left and right orders and not some smaller order. Indeed, if I is invertible as an R-lattice then it is invertible as a fractional $\mathcal {O}{}_{\textsf {\tiny {L}} }(I),\mathcal {O}{}_{\textsf {\tiny {R}} }(I)$-ideal, but not for any strictly smaller orders. If $I'$ is an R-lattice and $II'=\mathcal {O}$ for some $\mathcal {O}\subseteq \mathcal {O}{}_{\textsf {\tiny {L}} }(I)$, then multiplying on both sides on the left by $\mathcal {O}{}_{\textsf {\tiny {L}} }(I)$ gives

$$\begin{aligned} \mathcal {O}= II' = \mathcal {O}{}_{\textsf {\tiny {L}} }(I) II' = \mathcal {O}{}_{\textsf {\tiny {L}} }(I) \mathcal {O}= \mathcal {O}{}_{\textsf {\tiny {L}} }(I) \end{aligned}$$

and the same on the right. In other words, if we are going to call out an invertible fractional ideal by labelling actions on left and right, then we require these labels to be the actual orders that make the inverse work.

Remark 16.5.19. Equation 16.1.1 suggested the ‘real issue’ with noninvertible modules for quadratic orders: as an abelian group,

$$\begin{aligned} \mathfrak f = f\mathbb Z + f\sqrt{d} \mathbb Z = f \cdot S(d), \end{aligned}$$

so $\mathfrak f$ is principal and hence certainly invertible as an ideal of S(d)—but not as an ideal of the smaller order S(d). More generally, if $\mathfrak a \subset K=\mathbb Q (\sqrt{d})$ is a lattice in K (free $\mathbb Z$-module of rank 2), we define its multiplicator ring as

$$\begin{aligned} S(\mathfrak a ) :=\{x \in K : x\mathfrak a \subseteq \mathfrak a \}; \end{aligned}$$

the ring $S(\mathfrak a )$ is an order of K and so is also called the order of $\mathfrak a$. In the example above, $S(\mathfrak f )=S(f(\mathbb Z +\sqrt{d_K}\mathbb Z ))=S(d_K) \supsetneq S(d)$. It turns out that every lattice in K is invertible as an ideal of its multiplicator ring [Cox89, Proposition 7.4], and this statement plays an important role in the theory of complex multiplication. (Sometimes, an ideal $\mathfrak a \subseteq S$ is called proper or regular if $S=S(\mathfrak a )$; both terms are overloaded in mathematics, so we will mostly resist this notion.)

Unfortunately, unlike the quadratic case, not every lattice $I \subset B$ is projective as a left module over its left order (or the same on the right): this is necessary, but not sufficient. In Chapter 24, we classify orders $\mathcal {O}$ with the property that every lattice I having $\mathcal {O}{}_{\textsf {\tiny {L}} }(I)=\mathcal {O}$ is projective as an $\mathcal {O}$-module: they are the Gorenstein orders.

Remark 16.5.20. Invertible lattices give rise to a Morita equivalence between their corresponding left and right orders: see Remark 7.2.20.

6 Invertibility with a standard involution

In section 16.6, we follow Kaplansky [Kap69], considering invertibility in the presence of a standard involution. The main result of this chapter is as follows.

Main Theorem 16.6.1

Let R be a Dedekind domain with field of fractions F, and let B be a finite-dimensional F-algebra with a standard involution. Then an R-lattice I is invertible if and only if I is locally principal.

Remark 16.6.2. We can relax the hypothesis that R is a Dedekind domain and instead work with a Prüfer domain , a generalization of Dedekind domains to the non-noetherian context.

We have already seen (Corollary 16.5.10) that the implication $(\Rightarrow )$ in Main Theorem 16.6.1 holds without the hypothesis of a standard involution; the reverse implication is the topic of this section. This implication is not in general true if this hypothesis is removed (but is true again when B is commutative); see Exercise 16.18(a).

Remark 16.6.3. The provenance of the hypothesis that R is a Dedekind domain is the following: if $\mathfrak a \subset R$ is not invertible as an R-module, and $\mathcal {O}\subset B$ is an R-order, then $\mathfrak a \mathcal {O}$ is not invertible as an R-lattice. To make the simplest kind of arguments here, we would like for all (nonzero) ideals $\mathfrak a \subseteq R$ to be invertible, and this is equivalent to the requirement that R is a Dedekind domain (see section 9.2).

Throughout this section, let R be a Dedekind domain with field of fractions F, let B be a finite-dimensional F-algebra, and let $I \subset B$ be an R-lattice. The following concept will be useful in this section.

Definition 16.6.4

We say I is a semi-order if $1 \in I$ and ${{\,\mathrm{nrd}\,}}(I) \subseteq R$.

(For a semi-order I, we necessarily have ${{\,\mathrm{nrd}\,}}(I)=R$ since $1 \in I$.)

Lemma 16.6.5

An R-lattice I is a semi-order if and only if $1 \in I$ and every $\alpha \in I$ is integral over R.

Proof. We have that $\alpha \in I$ is integral over R if and only if ${{\,\mathrm{trd}\,}}(\alpha ) \in R$ and ${{\,\mathrm{nrd}\,}}(\alpha ) \in R$ (by Corollary 10.3.6, since R is integrally closed) if and only if ${{\,\mathrm{nrd}\,}}(\alpha )\in R$ and ${{\,\mathrm{nrd}\,}}(\alpha +1)={{\,\mathrm{nrd}\,}}(\alpha )+{{\,\mathrm{trd}\,}}(\alpha )+1 \in R$. $\square$

In particular, Lemma 16.6.5 implies that an order is a semi-order (by Corollary 10.3.3); we will see that semi-orders behave enough like orders that we can deduce local principality from their structure.

16.6.6

Let $\overline{I} :=\{\overline{\alpha } : \alpha \in I\}$. Then $\overline{I}$ is an R-lattice in B. If I, J are R-lattices then $\overline{IJ} = \overline{J}\,\overline{I}$ (even if this product is not compatible).

If I is a semi-order, then $\overline{I}=I$ (Exercise 16.15). In particular, if $\mathcal {O}$ is an R-order then $\overline{\mathcal {O}}=\mathcal {O}$.

Lemma 16.6.7

We have $\mathcal {O}{}_{\textsf {\tiny {L}} }(I)=\mathcal {O}{}_{\textsf {\tiny {R}} }(\overline{I})$ and $\mathcal {O}{}_{\textsf {\tiny {R}} }(I)=\mathcal {O}{}_{\textsf {\tiny {L}} }(\overline{I})$.

Proof. We have $\alpha \in \mathcal {O}{}_{\textsf {\tiny {L}} }(I)$ if and only if $\alpha I \subseteq I$ if and only if $\overline{\alpha I}=\overline{I}\,\overline{\alpha } \subseteq \overline{I}$ if and only if $\overline{\alpha } \in \mathcal {O}{}_{\textsf {\tiny {R}} }(\overline{I})$ if and only if $\alpha \in \overline{\mathcal {O}{}_{\textsf {\tiny {R}} }(\overline{I})}=\mathcal {O}{}_{\textsf {\tiny {R}} }(\overline{I})$. $\square$

Corollary 16.6.8

If I is a semi-order, then $\mathcal {O}{}_{\textsf {\tiny {L}} }(I)=\mathcal {O}{}_{\textsf {\tiny {R}} }(I)$.

Proof. Apply Lemma 16.6.7 with $\overline{I}=I$.

By Lemma 16.6.7, the standard involution gives a bijection between the set of lattices I with $\mathcal {O}{}_{\textsf {\tiny {L}} }(I)=\mathcal {O}$ and the set of lattices with $\mathcal {O}{}_{\textsf {\tiny {R}} }(I)=\mathcal {O}$.

16.6.9

Suppose that R is a DVR (e.g., a localization of R at a prime ideal $\mathfrak p$). We will show how to reduce the proof of Main Theorem 16.6.1 to that of a semi-order.

Since R is a DVR, the fractional R-ideal ${{\,\mathrm{nrd}\,}}(I) \subseteq R$ is principal, generated by an element with minimal valuation: let $\alpha \in I$ achieve this minimum reduced norm. Then the R-lattice $J=\alpha ^{-1}I$ now satisfies $1 \in J$ and ${{\,\mathrm{nrd}\,}}(J)=R$. Thus J is a semi-order, and J is (locally) principal if and only if I is (locally) principal.

Proof of Main Theorem

The proof is due to Kaplansky [Kap69, Theorem 2]. The statement is local; localizing, we may suppose R is a DVR. By 16.6.9, we reduce to the case where I is a semi-order. In particular, we have $1 \in I$. Let $\alpha _1,\dots ,\alpha _n$ be an R-basis for I.

We claim that

$$\begin{aligned} I^{n+1}=I^n \end{aligned}$$

(16.6.10)

Since $1 \in I$, we have $I^{n} \subseteq I^{n+1}$. It suffices then to prove that a product of $n+1$ basis elements of I lies in $I^n$. By the pigeonhole principle, there must be a repeated term $\alpha _i$ among them. We recall the formula (4.2.16)

$$\begin{aligned} \alpha \beta +\beta \alpha ={{\,\mathrm{trd}\,}}(\beta )\alpha +{{\,\mathrm{trd}\,}}(\alpha )\beta -{{\,\mathrm{trd}\,}}(\alpha \overline{\beta }) \end{aligned}$$

(16.6.11)

for all $\alpha ,\beta \in B$. We can use this relation to “push” the second instance of the repeated element until it meets with its mate, at the expense of terms lying in $I^n$. More precisely, in the R-module $I^2/I$, by (16.6.11),

$$\begin{aligned} \alpha _i\alpha _j\equiv -\alpha _j\alpha _i \pmod {I} \end{aligned}$$

for all i, j; it follows that in $I^{n+1}/I^n$,

$$\begin{aligned} \mu (\alpha _i\alpha _j)\nu \equiv -\mu (\alpha _j\alpha _i)\nu \pmod {I^n} \end{aligned}$$

for all $\mu ,\nu$ appropriate products of basis elements. Therefore we may suppose that the repetition $\alpha _i^2$ is adjacent; but then $\alpha _i$ satisfies a quadratic equation and $\alpha _i^2={{\,\mathrm{trd}\,}}(\alpha _i)\alpha _i-{{\,\mathrm{nrd}\,}}(\alpha _i) \in I$, so in fact the product belongs to $I^n$, and the claim follows.

Now suppose I is invertible; we wish to show that I is principal. From the equality $I^{n+1}=I^n$, we multiply both sides of this equation by $(I^{-1})^n$ and obtain $I=\mathcal {O}=\mathcal {O}{}_{\textsf {\tiny {L}} }(I)=\mathcal {O}{}_{\textsf {\tiny {R}} }(I)$. In particular, I is principal, generated by 1. $\square$

The above proof has the following immediate corollary.

Corollary 16.6.12

An R-lattice I is an R-order if and only if $1 \in I$, every element of I is integral, and I is invertible. In particular, an invertible semi-order is an order.

We conclude with two consequences.

16.6.13

Let I, J be invertible R-lattices such that I is compatible with J. Then ${{\,\mathrm{nrd}\,}}(IJ)={{\,\mathrm{nrd}\,}}(I){{\,\mathrm{nrd}\,}}(J)$, since it is enough to check this locally, and locally both I and J are principal and we have proved the statement in this case (Lemma 16.3.7).

16.6.14

In the presence of a standard involution, we can write the inverse in another way: if I is invertible, then

$$\begin{aligned} \overline{I}I={{\,\mathrm{nrd}\,}}(I) \mathcal {O}{}_{\textsf {\tiny {R}} }(I)\quad \text {and}\quad I\overline{I} = {{\,\mathrm{nrd}\,}}(I) \mathcal {O}{}_{\textsf {\tiny {L}} }(I) \end{aligned}$$

by checking these statements locally (where they follow immediately by computing the norm on a local generator). Since ${{\,\mathrm{nrd}\,}}(I)$ is a fractional R-ideal and thus invertible (R is a Dedekind domain), it follows that if I is invertible, then

$$\begin{aligned} I^{-1} = \overline{I} {{\,\mathrm{nrd}\,}}(I)^{-1}. \end{aligned}$$

In view of 16.6.14, the following important proposition is natural.

Proposition 16.6.15

Let B be a quaternion algebra over F and let $I \subset B$ be an R-lattice. Then the following statements hold.

• (a) We have $I\overline{I}={{\,\mathrm{nrd}\,}}(I)\mathcal {O}$, where $\mathcal {O}\subseteq B$ an R-order satisfying $\mathcal {O}{}_{\textsf {\tiny {L}} }(I) \subseteq \mathcal {O}$, and similarly $\overline{I}I={{\,\mathrm{nrd}\,}}(I)\mathcal {O}'$ with $\mathcal {O}{}_{\textsf {\tiny {R}} }(I) \subseteq \mathcal {O}'$.

• (b) If either $\mathcal {O}{}_{\textsf {\tiny {L}} }(I)$ or $\mathcal {O}{}_{\textsf {\tiny {R}} }(I)$ is maximal, then I is invertible, and both $\mathcal {O}{}_{\textsf {\tiny {L}} }(I)$ and $\mathcal {O}{}_{\textsf {\tiny {R}} }(I)$ are maximal.

Proof. We follow Kaplansky [Kap69, Theorems 6–7]. We again may suppose R is a DVR and I is a semi-order, so $1 \in I$ and ${{\,\mathrm{nrd}\,}}(I)=R$; and $\overline{I}=I$.

First we prove (a). We need to show that $I^2$ is an order. We showed in (16.6.10) (without extra hypothesis) that $I^3=I^4$; with B a quaternion algebra, we will improve this to $I^2=I^3$, whence $(I^2)^2=I^4=I^2$ and consequently $I^2$ is closed under multiplication and hence an R-order.

Let $J=I^3$; then $J^2=(I^3)^2=I^6=I^3=J$, so J is an R-order.

Let $\mathfrak p$ be the maximal ideal of R and consider the 4-dimensional algebra $J/\mathfrak p J$ over $k=R/\mathfrak p$. Then $I/\mathfrak p I \subseteq J/\mathfrak p J$ is a k-subspace containing 1. If $(I/\mathfrak p I)^2=I/\mathfrak p I$, then by dimensions we contradict $(I/\mathfrak p I)^3 = J/\mathfrak p J$; therefore $(I/\mathfrak p I)^2 \supsetneq I/\mathfrak p I$. If $\dim _k(I/\mathfrak p I) \le 2$, then $I/\mathfrak p I$ is a proper k-subalgebra, impossible. Thus $\dim _k(I/\mathfrak p I) \ge 3$ and $\dim _k (I/\mathfrak p I)^2 \ge 4$, and so $(I/\mathfrak p I)^2= J/\mathfrak p J$. By Nakayama’s lemma, it follows that $I^2=J=I^3$. The containments follow directly, e.g. $\mathcal {O}{}_{\textsf {\tiny {L}} }(I) \subseteq \mathcal {O}{}_{\textsf {\tiny {L}} }(I\overline{I})=\mathcal {O}{}_{\textsf {\tiny {L}} }(\mathcal {O})=\mathcal {O}$.

For part (b), applying part (a) we have $\mathcal {O}=\mathcal {O}{}_{\textsf {\tiny {L}} }(I)$ by maximality and the same on the right. But since I is a semi-order, from Corollary 16.6.8 we have $I\overline{I}=\mathcal {O}{}_{\textsf {\tiny {L}} }(I)=\mathcal {O}{}_{\textsf {\tiny {R}} }(I)=\overline{I}I$ so by definition, I has inverse $\overline{I}$. $\square$

7 One-sided invertibility

In this section, we pause to consider one-sided notions of invertibility. We refresh our notation, recalling that R is a Dedekind domain with $F={{\,\mathrm{Frac}\,}}R$ and B is a finite-dimensional algebra over F with $I \subseteq B$ an R-lattice.

Definition 16.7.1

I is right invertible if there exists an R-lattice $I' \subseteq B$, a right inverse , such that the product $II'$ is compatible and $II' = \mathcal {O}{}_{\textsf {\tiny {L}} }(I)$.

A right fractional $\mathcal {O}$-ideal I is right invertible if I is right invertible and sated (viz. 16.5.18).

We similarly define left invertible  and left inverse. Applying the same reasoning as in Lemma 16.5.9, we see that one-sided invertibility is a local property.

Remark 16.7.2. For rings, the (left or) right inverse of an element need not be unique even though a two-sided inverse is necessarily unique. Similarly, left invertibility does not imply right invertibility for lattices in general, and so the one-sided notions can be a bit slippery: see Exercise 16.18(b).

Remark 16.7.3. The compatibility condition in invertibility is important to avoid trivialities. Consider Example 16.3.6: we have $IJ={{\,\mathrm{M}\,}}_2(R)=\mathcal {O}{}_{\textsf {\tiny {L}} }(I)$, and if we let $J=\begin{pmatrix} bR &{} bR \\ R &{} R \end{pmatrix}$ for any nonzero $b \in R$, the equality $IJ={{\,\mathrm{M}\,}}_2(R)$ remains true. Not every author requires compatibility in the definition of (sided) invertibility.

A natural candidate for the right inverse presents itself: if $II'=\mathcal {O}{}_{\textsf {\tiny {L}} }(I)$, then $I'$ maps I into $\mathcal {O}{}_{\textsf {\tiny {L}} }(I)$ on the right. We recall the definition of the colon lattices (Definition 15.6.11). Let $I' :=(\mathcal {O}{}_{\textsf {\tiny {L}} }(I):I){}_{\textsf {\tiny {R}} }$. Then $II' \subseteq \mathcal {O}{}_{\textsf {\tiny {L}} }(I)$ by definition; however, in general equality need not hold and the product need not be compatible. Similarly, since $II^{-1}I \subseteq I$ we have $II^{-1} \subseteq \mathcal {O}{}_{\textsf {\tiny {L}} }(I)$, but again equality need not hold.

The sided version of Proposition 16.5.8 also holds.

Proposition 16.7.4

The following are equivalent:

• (i) $I^{-1}$ is a right inverse for I;

• (ii) I is right invertible;

• (iii) There is a compatible product $II^{-1}I=I$ and $1 \in II^{-1}$.

Similar equivalences hold on the left.

Proof. This is just a sided restriction of the proof of Proposition 16.5.8. For example, to show (ii) $\Rightarrow$ (i), we always have $II^{-1}I \subseteq I$ so $II^{-1} \subseteq \mathcal {O}{}_{\textsf {\tiny {L}} }(I)$; if $I'$ is a right inverse to I, then $II'I=\mathcal {O}{}_{\textsf {\tiny {L}} }(I) I=I$ and $I' \subseteq I^{-1}$, and therefore $II^{-1} \supseteq II'=\mathcal {O}{}_{\textsf {\tiny {L}} }(I)$. Therefore $II^{-1}=\mathcal {O}{}_{\textsf {\tiny {L}} }(I)$ and $I^{-1}$ is a right inverse for I. $\square$

Returning to the setting of the previous section, however, we can show that the one-sided notions of invertibility are equivalent to the two-sided notion.

Lemma 16.7.5

Suppose B has a standard involution. Then an R-lattice I is left invertible if and only if I is right invertible if and only if I is invertible.

Proof. We will show that if I is right invertible then I is left invertible; the other implications follow similarly. By localizing, we reduce to the case where R is a DVR. By the results of 16.6.9, we may suppose that I is a semi-order, so that $\mathcal {O}{}_{\textsf {\tiny {L}} }(I)=\mathcal {O}{}_{\textsf {\tiny {R}} }(I)=\mathcal {O}$ and $I=\overline{I}$. Suppose $II' = \mathcal {O}$. Then $\overline{I'} I = \overline{\mathcal {O}}=\mathcal {O}$, and $\overline{I'}$ is compatible with I since

$$\begin{aligned} \mathcal {O}=\mathcal {O}{}_{\textsf {\tiny {R}} }(I)=\mathcal {O}{}_{\textsf {\tiny {L}} }(I')=\mathcal {O}{}_{\textsf {\tiny {R}} }(\overline{I'}) \end{aligned}$$

as desired. $\square$

Corollary 16.7.6

Suppose R is a Dedekind domain and that B has a standard involution. Then an R-lattice I is right invertible with $II' = \mathcal {O}{}_{\textsf {\tiny {L}} }(I)$ if and only if $I'=(\mathcal {O}{}_{\textsf {\tiny {L}} }(I):I){}_{\textsf {\tiny {R}} }=I^{-1}$.

A similar statement holds for the left inverse; in particular, this shows that a right inverse is necessarily unique.

Proof. The implication $(\Rightarrow )$ is immediate, so we prove $(\Leftarrow )$. Let $\mathcal {O}=\mathcal {O}{}_{\textsf {\tiny {L}} }(I)$. Then

$$\begin{aligned} \mathcal {O}= II' \subseteq I(\mathcal {O}:I){}_{\textsf {\tiny {R}} }\subseteq \mathcal {O}\end{aligned}$$

so equality must hold, and $II' = I(\mathcal {O}:I){}_{\textsf {\tiny {R}} }$. By 16.7.5, I is invertible, and multiplying both sides by $I^{-1}$ gives $I'=(\mathcal {O}:I){}_{\textsf {\tiny {R}} }$. $\square$

We collect the results of this section in the following theorem.

Main Theorem 16.7.7

Let R be a Dedekind domain with $F={{\,\mathrm{Frac}\,}}R$, let B be a quaternion algebra over F, and let $I \subseteq B$ be an R-lattice. Then the following are equivalent:

(i):

I is locally principal;

(ii):

I is invertible;

(iii):

I is left invertible;

(iii${}^\prime$):

I is right invertible;

(iv):

${{\,\mathrm{nrd}\,}}(I)^2=[\mathcal {O}{}_{\textsf {\tiny {L}} }(I):I]_R$; and

(iv${}^\prime$):

${{\,\mathrm{nrd}\,}}(I)^2=[\mathcal {O}{}_{\textsf {\tiny {R}} }(I):I]_R$.

Proof. Main Theorem 16.6.1 proves (i) $\Leftrightarrow$ (ii). For the equivalence (ii) $\Leftrightarrow$ (iii) $\Leftrightarrow$ (iii${}^\prime$), apply Lemma 16.7.5. Finally, the equivalence (i) $\Leftrightarrow$ (iv) $\Leftrightarrow$ (iv${}^\prime$) is supplied by Proposition 16.4.3. $\square$

8 Invertibility and the codifferent

To conclude this chapter, we pick up a remaining thread concerning the (co)different.

Definition 16.8.1

We define the different of $\mathcal {O}$ to be the quasi-inverse of the codifferent:

$$\begin{aligned} {{\,\mathrm{diff}\,}}(\mathcal {O}) :={{\,\mathrm{codiff}\,}}(\mathcal {O})^{-1} = \{ \alpha \in B : \mathcal {O}^\sharp \alpha \mathcal {O}^\sharp \subseteq \mathcal {O}^\sharp \}. \end{aligned}$$

Lemma 16.8.2

The different ${{\,\mathrm{diff}\,}}(\mathcal {O})$ is an integral two-sided $\mathcal {O}$-ideal.

Proof. By Lemma 15.6.16, we have $\mathcal {O}\mathcal {O}^\sharp \mathcal {O}= \mathcal {O}^\sharp$ and so if $\alpha \in {{\,\mathrm{diff}\,}}(\mathcal {O})$ then $\mathcal {O}^\sharp (\mathcal {O}\alpha \mathcal {O}) \mathcal {O}^\sharp = \mathcal {O}^\sharp \alpha \mathcal {O}^\sharp = \mathcal {O}^\sharp$ and ${{\,\mathrm{diff}\,}}(\mathcal {O})$ is a two-sided $\mathcal {O}$-ideal. To prove that ${{\,\mathrm{diff}\,}}(\mathcal {O}) \subseteq \mathcal {O}$, referring to Lemma 15.6.2, starting with $\mathcal {O}^\sharp \alpha \mathcal {O}^\sharp \subseteq \mathcal {O}^\sharp$ taking $1 \in \mathcal {O}^\sharp$ we have $\alpha \mathcal {O}^\sharp \subseteq \mathcal {O}^\sharp$ so $(\alpha \mathcal {O}^\sharp )^\sharp = (\mathcal {O}^\sharp )^\sharp \alpha ^{-1} \supseteq (\mathcal {O}^\sharp )^\sharp$. By Lemma 15.6.5, we have $(\mathcal {O}^\sharp )^\sharp = \mathcal {O}$, so $\mathcal {O}\alpha ^{-1} \supseteq \mathcal {O}$, so $\mathcal {O}\alpha \subseteq \mathcal {O}$ and again taking 1 we get $\alpha \in \mathcal {O}$. $\square$

16.8.3

If ${{\,\mathrm{codiff}\,}}(\mathcal {O})$ is locally principal (section 16.2), then so is ${{\,\mathrm{diff}\,}}(\mathcal {O})$, and by Proposition 16.4.3 we have

$$\begin{aligned} {{\,\mathrm{Nm}\,}}_{B|F}({{\,\mathrm{diff}\,}}(\mathcal {O}))=[\mathcal {O}:{{\,\mathrm{diff}\,}}(\mathcal {O})]_R=[{{\,\mathrm{codiff}\,}}(\mathcal {O}):\mathcal {O}]_R = {{\,\mathrm{disc}\,}}(\mathcal {O}); \end{aligned}$$

so when further B is a quaternion algebra, we have

$$\begin{aligned} {{\,\mathrm{nrd}\,}}({{\,\mathrm{diff}\,}}(\mathcal {O}))={{\,\mathrm{discrd}\,}}(\mathcal {O}). \end{aligned}$$

(\ref {eqn:Inp1n)

Invertibility of ideals is detected by the (co)different [Fad65, Proposition 24.1].

Proposition 16.8.5

If ${{\,\mathrm{codiff}\,}}(\mathcal {O})$ is right invertible, then all sated left fractional $\mathcal {O}$-ideals are right invertible. Similarly, if ${{\,\mathrm{codiff}\,}}(\mathcal {O})$ is left invertible, then all sated right fractional $\mathcal {O}$-ideals are left invertible.

Proof. To get started, we refresh a few things: by Corollary 15.6.13, we have $(II^\sharp )^\sharp = \mathcal {O}{}_{\textsf {\tiny {L}} }(I)=\mathcal {O}$. The product $II^\sharp$ is compatible by Proposition 15.6.6. By Lemma 15.6.5 we have $II^\sharp = \mathcal {O}^\sharp = {{\,\mathrm{codiff}\,}}(\mathcal {O})$.

Now by hypothesis of invertibility, $\mathcal {O}^\sharp (\mathcal {O}^\sharp )^{-1} = \mathcal {O}{}_{\textsf {\tiny {L}} }(\mathcal {O}^\sharp )=\mathcal {O}$ is a compatible product. Therefore the product $I^\sharp (\mathcal {O}^\sharp )^{-1}$ is compatible, and

$$\begin{aligned} I (I^\sharp (\mathcal {O}^\sharp )^{-1}) = (II^\sharp ) (\mathcal {O}^\sharp )^{-1} = \mathcal {O}^\sharp (\mathcal {O}^\sharp )^{-1} =\mathcal {O}. \end{aligned}$$

(16.8.4)

A similar argument holds on the right. $\square$

We have the following corollary of Proposition 16.8.5, phrased in terms of the different.

Corollary 16.8.7

Suppose that B has a standard involution. Then the following are equivalent:

(i):

${{\,\mathrm{codiff}\,}}(\mathcal {O})$ is invertible;

(ii):

${{\,\mathrm{diff}\,}}(\mathcal {O})$ is invertible;

(iii):

All sated left fractional $\mathcal {O}$-ideals I are invertible, with inverse $I^{-1}=I^\sharp {{\,\mathrm{diff}\,}}(\mathcal {O})$; and

(iii$^\prime$):

All sated right fractional $\mathcal {O}$-ideals I are invertible, with inverse $I^{-1}={{\,\mathrm{diff}\,}}(\mathcal {O}) I^\sharp$.

Proof. Combine Proposition 16.8.5 and (16.8.6) with Lemma 16.7.5 and Corollary 16.7.6. $\square$

We conclude with a criterion to determine invertibility; it is not used in the sequel.

Proposition 16.8.8

(Brandt’s invertibility criterion). Let $I \subseteq B$ be an R-lattice. Then I is invertible if and only if

$$\begin{aligned} {{\,\mathrm{nrd}\,}}(I^\sharp ){{\,\mathrm{discrd}\,}}(I) \subseteq {{\,\mathrm{nrd}\,}}(I). \end{aligned}$$

Proof. See Kaplansky [Kap69, Theorem 10] or Brzezinski [Brz82, Theorem 3.4]. $\square$

Exercises

Unless otherwise specified, throughout these exercises let R be a Dedekind domain with field of fractions F, let B be a finite-dimensional F-algebra, and let $I \subseteq B$ be an R-lattice.

• 1. Let $d \in \mathbb Z$ be a nonsquare discriminant, and let $S(d)=\mathbb Z [(d+\sqrt{d})/2]$ be the quadratic ring of discriminant d.

  • (a) Suppose that $d=d_Kf^2$ with $f>1$. Show that the ideal $(f,\sqrt{d})$ of S(d) is not invertible.

  • (b) Consider $d=-12$, and $S=S(-12)=\mathbb Z [\sqrt{-3}]$. Show that every invertible ideal of S is principal (so S has class number 1), but that S is not a PID.

$\triangleright$2.:

Show that if $I=\mathcal {O}{}_{\textsf {\tiny {L}} }(I) \alpha$ with $\alpha \in B^\times$, then $\mathcal {O}{}_{\textsf {\tiny {R}} }(I)=\alpha ^{-1} \mathcal {O}{}_{\textsf {\tiny {L}} }(I) \alpha$.

$\triangleright$3.:

Show that if J is an R-lattice in B and $\mu \in B^\times$, then $\mu J=J$ if and only if $\mu \in \mathcal {O}{}_{\textsf {\tiny {L}} }(J)^\times$.

$\triangleright$4.:

Show that if $\alpha \in B$ then ${{\,\mathrm{nrd}\,}}(\alpha I)={{\,\mathrm{nrd}\,}}(\alpha ){{\,\mathrm{nrd}\,}}(I)$. Conclude that if I is a principal R-lattice, generated by $\alpha \in I$, then ${{\,\mathrm{nrd}\,}}(I)={{\,\mathrm{nrd}\,}}(\alpha )R$.

• 5. Let $\alpha _1,\dots ,\alpha _n$ generate I as an R-module. Give an explicit example where ${{\,\mathrm{nrd}\,}}(I)$ is not generated by ${{\,\mathrm{nrd}\,}}(\alpha _i)$ (cf. Lemma 16.3.2). Moreover, show that for an R-lattice I, there exists a set of R-module generators $\alpha _i$ such that ${{\,\mathrm{nrd}\,}}(I)$ is in fact generated by ${{\,\mathrm{nrd}\,}}(\alpha _i)$.

$\triangleright$6.:

Suppose that R is a Dedekind domain, and let $\mathcal {O}\subseteq B$ be an R-order. Let I be a locally principal right fractional $\mathcal {O}$-ideal. Show that I can be generated as a right $\mathcal {O}$-ideal by two elements, and in fact for $a \in {{\,\mathrm{nrd}\,}}(I)$ nonzero we can write $I=a\mathcal {O}+ \beta \mathcal {O}$ with $\beta \in B^\times$.

• 7. Let F be a number field, let $R \subseteq F$ be a ($\mathbb Z$-)order, and let $\mathfrak a \subseteq R$ be a nonzero ideal. Show that $\mathfrak a$ is projective as an R-module if and only if $\mathfrak a$ is invertible if and only if $\mathfrak a$ is locally principal. [These are all automatic when R is a Dedekind domain 9.4.5.]

• 8. If $I, J \subseteq B$ are R-lattices with $I \subseteq J$, is it true that $I^{-1} \supseteq J^{-1}$?

• 9. Let $I,J,K \subseteq B$ be R-lattices. Show that

  $$\begin{aligned} ((I:J){}_{\textsf {\tiny {L}} }: K){}_{\textsf {\tiny {R}} }= ((I:K){}_{\textsf {\tiny {R}} }: J){}_{\textsf {\tiny {L}} }. \end{aligned}$$

$\triangleright$10.:

Let $I,J \subseteq B$ be R-lattices and suppose that I is compatible with J. Show that IJ is invertible (with $(IJ)^{-1}=J^{-1} I^{-1}$) if both I, J are invertible, but the converse need not hold.

• 11. Let $I,J \subseteq B$ be R-lattices, and suppose that J is invertible. Show that $(I:J){}_{\textsf {\tiny {L}} }= IJ^{-1}$ and $(I:J){}_{\textsf {\tiny {R}} }=J^{-1}I$.

• 12. Let p be prime, let $B=({p,p} \mid {\mathbb {Q}})$, and let $\mathcal {O}:=\mathbb Z \langle i,j \rangle = \mathbb Z \oplus \mathbb Z i \oplus \mathbb Z j \oplus \mathbb Z ij$.

  • (a) Let $I=\{\alpha \in \mathcal {O}: p \mid {{\,\mathrm{nrd}\,}}(\alpha )\}$. Show that $I=p\mathbb Z \oplus \mathbb Z i \oplus \mathbb Z j \oplus \mathbb Z ij$.

  • (b) Show $I=\mathcal {O}i + \mathcal {O}j$, that $\mathcal {O}$ is a two-sided $\mathcal {O}$-ideal, and that $[\mathcal {O}:I]=p$.

  • (c) Show that $I_{(p)} \ne \mathcal {O}_{(p)}\alpha$ for all $\alpha \in I_{(p)}$. [Hint: show that if $\alpha \in I$, then $p^2 \mid [\mathcal {O}:\mathcal {O}\alpha ]$.]

  • (d) Compute that $\mathcal {O}{}_{\textsf {\tiny {L}} }(I) = \mathbb Z + \mathbb Z i + \mathbb Z j + \mathbb Z (ij/p) \supsetneq \mathcal {O}$, and that $I=\mathcal {O}{}_{\textsf {\tiny {L}} }(I)i=\mathcal {O}{}_{\textsf {\tiny {L}} }(I)j$.

  • (e) Compute ${{\,\mathrm{codiff}\,}}(\mathcal {O})$ and ${{\,\mathrm{diff}\,}}(\mathcal {O})$ and show they are invertible.

  [Compare Lemurell [Lem2011, Remark 6.4].]

$\triangleright$13.:

Let K be a separable quadratic field extension of F and let $I \subseteq K$ be an R-lattice. Let $\mathcal {O}=\mathcal {O}{}_{\textsf {\tiny {L}} }(I)=\mathcal {O}{}_{\textsf {\tiny {R}} }(I)$.

(a):

Show that $I\overline{I}=\overline{I}I={{\,\mathrm{nrd}\,}}(I)\mathcal {O}$. [Hint: argue as in Proposition 16.6.15.]

(b):

Conclude that I is invertible as a $\mathcal {O}$-module.

• 14. Show that if $I,J \subseteq B$ are locally principal (hence invertible) R-lattices, then

  $$\begin{aligned}{}[I:J]_R=[J^{-1}:I^{-1}]_R. \end{aligned}$$

$\triangleright$15.:

Let B be an F-algebra with a standard involution $\overline{\phantom {x}}$. Show that if I is a semi-order then $\overline{I}=I$.

• 16. Let R be a Dedekind domain with field of fractions F, let $K \supset F$ be a separable quadratic field extension and let S be an R-order in K. Let $S_K$ be the integral closure of R in K.

  • (a) Show that there exists a (unique) ideal $\mathfrak f =\mathfrak f (S) \subset S_K$ (called the conductor ) such that $S=R + \mathfrak f S_K$.

  • (b) Now let $\mathfrak b \subset K$ be a fractional S-ideal. Show that the following are equivalent:

    • (i) $\mathfrak b$ is a locally principal S-ideal;

    • (ii) $\mathfrak b$ is invertible as a fractional S-ideal, i.e., there exists a fractional ideal $\mathfrak b ^{-1}$ such that $\mathfrak b \mathfrak b ^{-1}=S$ (necessarily $\mathfrak b ^{-1}=(S:\mathfrak b ))$;

    • (iii) There exists $d \in K^\times$ such that $d\mathfrak b + \mathfrak f \cap S = S$; and

    • (iv) $\mathfrak b$ is proper, i.e., $S=\mathcal {O}(\mathfrak b )=\{x \in K : x\mathfrak b \subseteq \mathfrak b \}$.

$\triangleright$17.:

Let $\mathcal {O}\subseteq B$ be an R-order.

(a):

Let $\alpha \in B^\times$. Show that $I=\mathcal {O}\alpha$ is a lattice with $\mathcal {O}{}_{\textsf {\tiny {L}} }(I)=\mathcal {O}{}_{\textsf {\tiny {R}} }(I)=\mathcal {O}$ if and only if $\alpha \in B^\times$ and $\mathcal {O}\alpha = \alpha \mathcal {O}$. Conclude that the set of invertible two-sided principal lattices I with $\mathcal {O}{}_{\textsf {\tiny {L}} }(I)=\mathcal {O}{}_{\textsf {\tiny {R}} }(I)=\mathcal {O}$ forms a group.

(b):

Show that the normalizer of $\mathcal {O}$,

$$\begin{aligned} N_{B^\times }(\mathcal {O})=\{\alpha \in B^\times : \alpha \mathcal {O}\alpha ^{-1} = \mathcal {O}\} \end{aligned}$$

is the group generated by $\alpha \in B^\times$ such that $\mathcal {O}\alpha$ is a two-sided $\mathcal {O}$-ideal.

• 18. The following example is due to Kaplansky [Kap69, pp. 220, 221]. Let R be a DVR with field of fractions F and maximal ideal $\mathfrak p =\pi R$.

  • (a) Consider the R-lattice

    $$ I=\begin{pmatrix} \pi R &{} \pi R &{} R \\ \pi R &{} \pi R &{} R \\ R &{} R &{} R \end{pmatrix} \subset B={{\,\mathrm{M}\,}}_3(F) $$

    Show that I is invertible but is not principal.

  • (b) Consider the R-lattice

    $$ I=\begin{pmatrix} \pi R &{} \pi R &{} R \\ \pi ^2R &{} \pi ^2R &{} R \\ R &{} R &{} R \end{pmatrix} \subset B={{\,\mathrm{M}\,}}_3(F) $$

    Show that I is left invertible but is not right invertible.