Integral representation theory

Abstract

In this chapter, we consider a slightly more general framework on the preceding chapters: we consider lattices as projective modules, and relate this to invertibility and representation theory in an integral sense.

In this chapter, we consider a slightly more general framework on the preceding chapters: we consider lattices as projective modules, and relate this to invertibility and representation theory in an integral sense.

1 Projectivity, invertibility, and representation theory

Let R be a Dedekind domain with field of fractions $F={{\,\mathrm{Frac}\,}}R$. Finitely generated, projective R-modules have played an important role throughout this text, and we now seek to understand them in the context of orders.

To this end, let B be a finite-dimensional F-algebra and let $\mathcal {O}\subseteq B$ be an R-order. A left $\mathcal {O}$-lattice M is an R-lattice that is a left $\mathcal {O}$-module, i.e., M is a finitely generated, projective (locally free) R-module that has the structure of a left $\mathcal {O}$-module. We make a similar definition on the right.

We say that a left (or right) $\mathcal {O}$-lattice M is projective if it is a direct summand of a free left (or right) $\mathcal {O}$-module. Projectivity for lattices in B is related to invertibility as follows (Theorem 20.3.3).

Theorem 20.1.1

Let $I \subseteq B$ be an R-lattice. Then I is invertible if and only if I is projective as a left $\mathcal {O}{}_{\textsf {\tiny {L}} }(I)$-module and as a right $\mathcal {O}{}_{\textsf {\tiny {R}} }(I)$-module.

One can also tease apart left and right invertibility if desired; in the quaternion context, these are equivalent anyway because of the standard involution (Main Theorem 20.3.9).

Given our efforts to understand invertible lattices, one may think that Theorem 20.1.1 is all there is to say. However, two issues remain. First, there may be finitely generated (projective) $\mathcal {O}$-modules that are not lattices, and they play a structurally important role for the order $\mathcal {O}$. Second, and this point is subtle: there may be lattices $I \subseteq B$ that are projective as a left $\mathcal {O}$-module, but with $\mathcal {O}{}_{\textsf {\tiny {L}} }(I) \supsetneq \mathcal {O}$; in other words, such lattices are invertible over a larger order, even though they still have good properties as modules over the smaller order.

Example 20.1.2

Let

$$\begin{aligned} \mathcal {O}:=\begin{pmatrix} \mathbb Z &{} \mathbb Z \\ p\mathbb Z &{} \mathbb Z \end{pmatrix} \subseteq B :={{\,\mathrm{M}\,}}_2(\mathbb Q ) \end{aligned}$$

be the order consisting of integral matrices that are upper triangular modulo a prime p. We will exhibit both of the issues above. First, we consider $\mathcal {O}$ as a left $\mathcal {O}$-module: it decomposes as

$$\begin{aligned} \begin{aligned} \mathcal {O}&= \mathcal {O}\begin{pmatrix} 1 &{} 0 \\ 0 &{} 0 \end{pmatrix} \oplus \mathcal {O}\begin{pmatrix} 0 &{} 0 \\ 0 &{} 1 \end{pmatrix} = \begin{pmatrix} \mathbb Z &{} 0 \\ p\mathbb Z &{} 0 \end{pmatrix} \oplus \begin{pmatrix} 0 &{} \mathbb Z \\ 0 &{} \mathbb Z \end{pmatrix} \\&\simeq \begin{pmatrix} \mathbb Z \\ (p) \end{pmatrix} \oplus \begin{pmatrix} \mathbb Z \\ \mathbb Z \end{pmatrix} =:I_1 \oplus I_2. \end{aligned} \end{aligned}$$

(20.1.3)

The two left $\mathcal {O}$-modules $I_1,I_2$ are visibly projective, and they are not isomorphic: intuitively, an isomorphism would have to be multiplication on the left by a $2 \times 2$-matrix that commutes with multiplication $\mathcal {O}$, and so it must be scalar. More precisely, suppose $\phi \in {{\,\mathrm{Hom}}}_{\mathcal {O}}(I_1,I_2)$ is an isomorphism of left $\mathcal {O}$-modules. Extending scalars, we have

$$\begin{aligned} \mathbb Q I_1 = \mathbb Q I_2 = \begin{pmatrix} \mathbb Q \\ \mathbb Q \end{pmatrix} =:V, \end{aligned}$$

and the extension of $\phi$ gives an element in ${{\,\mathrm{Aut}\,}}_B(V)$ where $B={{\,\mathrm{M}\,}}_2(\mathbb Q )={{\,\mathrm{End}\,}}_\mathbb Q (V)$, so commutes with the action of B and is therefore central: which is to say $\phi$ is a scalar matrix, and that is absurd.

The lattice $I={{\,\mathrm{M}\,}}_2(\mathbb Z )$ is invertible as lattice, since it is an order (!); and it is a two-sided fractional $\mathcal {O}$-ideal, but it is not sated. We claim that I is also a projective $\mathcal {O}$-module: this follows from the fact that ${{\,\mathrm{M}\,}}_2(\mathbb Z ) \simeq I_2^{\oplus 2}$ as a left $\mathcal {O}$-module, so ${{\,\mathrm{M}\,}}_2(\mathbb Z )$ is isomorphic to a direct summand of $\mathcal {O}^{\oplus 2}$.

In this chapter, we establish some basic vocabulary of modules in the language of the representation theory of an order. In the case of algebras over a field, we defined a Jacobson radical as a way to measure the failure of the algebra to be semisimple. Similarly, for every ring A, we define the Jacobson radical ${{\,\mathrm{rad}\,}}A$ to be the intersection of all maximal left ideals of A: it again measures the failure of left indecomposable modules to be simple. There is a left-right symmetry to ${{\,\mathrm{rad}\,}}A$, and in fact ${{\,\mathrm{rad}\,}}A \subseteq A$ is a two-sided A-ideal.

Locally, the Jacobson radical plays a key role. Suppose R is a complete DVR with unique maximal ideal $\mathfrak p$. Then $\mathfrak p ={{\,\mathrm{rad}\,}}\mathcal {O}$ since it is the maximal ideal. Moreover, we will see that $\mathfrak p \mathcal {O}\subseteq {{\,\mathrm{rad}\,}}\mathcal {O}$, so $\mathcal {O}/{{\,\mathrm{rad}\,}}\mathcal {O}$ is a finite-dimensional semisimple k-algebra. Much of the structure of $\mathcal {O}$-modules is reflected in the structure of modules over the quotient $\mathcal {O}/{{\,\mathrm{rad}\,}}\mathcal {O}$ (see Lemma 20.6.8).

Remark 20.1.4. In representation theory, generally speaking, to study the action of a group on some kind of object (vector space, simplicial complex, etc.) one introduces some kind of group ring and studies modules over this ring. The major task becomes to classify such modules. For example, let R be a Dedekind domain with $F={{\,\mathrm{Frac}\,}}R$, and let $\mathcal {O}$ be an R-order in a finite-dimensional F-algebra B. A finitely generated integral representation of $\mathcal {O}$ is a finitely generated $\mathcal {O}$-module that is projective as an R-module (in particular, is R-torsion free). The integral representations of $\mathcal {O}$ are quite complicated! Nevertheless, integral representation theory is a beautiful blend of number theory, commutative algebra, and linear algebra. In section 20.6, we will see some of the basic ingredients when R is a DVR, and in section 21.4 in the next chapter we will show that hereditary orders have a tidy integral representation theory. For more on the subject, see the surveys by Reiner [Rei70, Rei76] as well as the massive treatises by Curtis–Reiner [CR81, CR87].

2 Projective modules

As we will need the notion over several different rings, we start more generally: let A be a ring (not necessarily commutative, but with 1). For an introduction to the theory of projective modules and related subjects, see Lam [Lam99, §2] and Curtis–Reiner [CR81, §2], and Berrick–Keating [BK2000, §2].

Definition 20.2.1

Let P be a finitely generated left A-module. Then P is projective as a left A-module if it is a direct summand of a free left A-module.

A finitely generated free module is projective. The notion of projectivity is quite fundamental, as the following proposition indicates.

Proposition 20.2.2

Let P be a finitely generated left A-module. Then the following are equivalent:

• (i) P is projective;

• (ii) There exists a finitely generated left A-module Q such that $P \oplus Q$ is free as a left A-module.

• (iii) Every surjective homomorphism $f:M \rightarrow P$ (of left A-modules) has a splitting $g:P \rightarrow M$ (i.e., $f \circ g={{\,\mathrm{id}\,}}_{P}$);

• (iv) Every diagram

  

  of left A-modules with exact bottom row can be extended as indicated, with $p=f \circ q$; and

• (v) ${{\,\mathrm{Hom}}}_ A (P,-)$ is a (right) exact functor.

Proof. See Lam [Lam99, Chapter 2]. In statement (v), given a short exact sequence

$$\begin{aligned} 0 \rightarrow Q \rightarrow M \rightarrow N \rightarrow 0 \end{aligned}$$

then ${{\,\mathrm{Hom}}}_A(P,-)$ is always left exact, so

$$\begin{aligned} 0 \rightarrow {{\,\mathrm{Hom}}}_ A (P,Q) \rightarrow {{\,\mathrm{Hom}}}_ A (P,M) \rightarrow {{\,\mathrm{Hom}}}_ A (P,N) \end{aligned}$$

(20.2.3)

is exact; the condition for P to be projective is that ${{\,\mathrm{Hom}}}_A(P,-)$ is right exact, so the full sequence

$$\begin{aligned} 0 \rightarrow {{\,\mathrm{Hom}}}_ A (P,Q) \rightarrow {{\,\mathrm{Hom}}}_ A (P,M) \rightarrow {{\,\mathrm{Hom}}}_ A (P,N) \rightarrow 0 \end{aligned}$$

(20.2.4)

is short exact. $\square$

20.2.5

A finite direct sum $P = \bigoplus _i P_i$ of finitely generated A-modules is projective if and only if each summand $P_i$ is projective: indeed, the functor ${{\,\mathrm{Hom}}}_R(P,-)$ is naturally isomorphic to $\prod _i {{\,\mathrm{Hom}}}_R(P_i,-)$, so we apply condition (v) of Proposition 20.2.2.

20.2.6

Localizing Proposition 20.2.2(v), and using the fact that a sequence is exact if and only if it is exact locally (Exercise 20.1(a)), we see that P is projective as a left $\mathcal {O}$-module if and only if $P_{(\mathfrak p )}$ is projective as a left $\mathcal {O}_{(\mathfrak p )}$-module for all primes $\mathfrak p \subseteq R$

Definition 20.2.7

A left $\mathcal {O}$-lattice is an R-lattice M that is a left $\mathcal {O}$-module.

We make a similar definition on the right.

20.2.8

A left $\mathcal {O}$-lattice M is locally free  of rank $r \ge 1$ if $M_\mathfrak p \simeq \mathcal {O}_\mathfrak p ^{\oplus r}$ as left $\mathcal {O}$-modules for all primes $\mathfrak p \subseteq R$. If follows from 20.2.5 and 20.2.6 that a locally free $\mathcal {O}$-lattice is projective.

3 Projective modules and invertible lattices

Now let R be a noetherian domain with $F :={{\,\mathrm{Frac}\,}}R$, let B be a finite-dimensional F-algebra, and let $\mathcal {O}\subseteq B$ be an R-order.

One can extend the base ring of the module while preserving projectivity, as follows.

Lemma 20.3.1

Let $\mathcal {O}\subseteq \mathcal {O}'$ be R-orders in B and let M be a left $\mathcal {O}'$-lattice. If M is projective as a left $\mathcal {O}$-module, then M is projective as a left $\mathcal {O}'$-module.

Proof. Suppose M is projective as a left $\mathcal {O}$-module; then $M \oplus N \simeq \mathcal {O}^r$ for some $r \ge 0$. Tensor with $\mathcal {O}'$ to get

$$\begin{aligned} (\mathcal {O}' \otimes _\mathcal {O}M) \oplus (\mathcal {O}' \otimes _\mathcal {O}N) \simeq (\mathcal {O}')^r. \end{aligned}$$

(20.3.2)

Since multiplication gives an isomorphism of left $\mathcal {O}'$-modules , the result follows. (More generally, see Harada [Har63a, Lemma 1.3].) $\square$

In the commutative case, an R-lattice $\mathfrak a \subseteq F$ is invertible as an R-module if and only if $\mathfrak a$ is projective as a (left and right) R-module. Something is true in this more general context.

Theorem 20.3.3

Let $I \subseteq B$ be an R-lattice.

• (a) $I^{-1}I=\mathcal {O}{}_{\textsf {\tiny {R}} }(I)$ if and only if I is projective as a left $\mathcal {O}{}_{\textsf {\tiny {L}} }(I)$-module, and $II^{-1}=\mathcal {O}{}_{\textsf {\tiny {L}} }(I)$ if and only if I is projective as a right $\mathcal {O}{}_{\textsf {\tiny {R}} }(I)$-module.

• (b) I is projective as a left $\mathcal {O}{}_{\textsf {\tiny {L}} }(I)$-module and a right $\mathcal {O}{}_{\textsf {\tiny {R}} }(I)$-module if and only if I is invertible (as an R-lattice).

The difference between (a) and (b) in Theorem 20.3.3 is the compatibility of the two products.

Proof. We begin with (a). To prove the implication $(\Rightarrow )$, suppose $I^{-1}I=\mathcal {O}{}_{\textsf {\tiny {R}} }(I)$; then there exist $\alpha _i \in I$ and $\alpha _i^* \in I^{-1}$ such that $\sum _i \alpha _i^* \alpha _i = 1$. We may extend the set $\alpha _i$ to generate I as a left $\mathcal {O}{}_{\textsf {\tiny {L}} }(I)$-module by taking $\alpha _i^*=0$ if necessary. We define the surjective map

$$\begin{aligned} \begin{aligned} f:M=\bigoplus _i \mathcal {O}{}_{\textsf {\tiny {L}} }(I) e_i&\rightarrow I \\ e_i&\mapsto \alpha _i. \end{aligned} \end{aligned}$$

(20.3.4)

Consider the map

$$\begin{aligned} g:I&\rightarrow M \\ \beta&\mapsto \sum _i \beta \alpha _i^* e_i; \end{aligned}$$

the map g is defined because for all $\beta \in I$, we have $\beta \alpha _i^* \in II^{-1}$, and as always $II^{-1}I \subseteq I$ so $II^{-1} \subseteq \mathcal {O}{}_{\textsf {\tiny {L}} }(I)$. The map g is a splitting of f since

$$\begin{aligned} (f \circ g)(\beta )=\sum _i \beta \alpha _i^*\alpha _i = \beta \sum _i \alpha _i^*\alpha _i = \beta . \end{aligned}$$

Therefore I is a direct summand of M, so I is projective as a left $\mathcal {O}{}_{\textsf {\tiny {L}} }(I)$-module.

Next we prove $(\Leftarrow )$. There exists a nonzero $r \in I \cap R$ (Exercise 9.2), so to show that I is left invertible, we may replace I with $r^{-1} I$ and therefore suppose that $1 \in I$. Following similar lines as above, let $\alpha _i$ generate I as a left $\mathcal {O}{}_{\textsf {\tiny {L}} }(I)$-module, and consider the surjective map $f:M=\bigoplus _i \mathcal {O}{}_{\textsf {\tiny {L}} }(I) e_i \rightarrow I$ by $e_i \mapsto \alpha _i$. Then since I is projective as a left $\mathcal {O}{}_{\textsf {\tiny {L}} }(I)$-module, this map splits by a map $g:I \rightarrow M$; suppose that $g(1)=(\alpha _i^*)_i$ with $\alpha _i^* \in \mathcal {O}{}_{\textsf {\tiny {L}} }(I)$; then

$$\begin{aligned} (f \circ g)(1)=1=\sum _i \alpha _i^*\alpha _i. \end{aligned}$$

(20.3.5)

For all $\beta \in I$, we have $g(\beta )=(\beta \alpha _i^*)_i \in M$, so $\beta \alpha _i^* \in \mathcal {O}{}_{\textsf {\tiny {L}} }(I)$ for all i; therefore for all $\alpha ,\beta \in I$ we have $\beta \alpha _i^* \alpha \in \mathcal {O}{}_{\textsf {\tiny {L}} }(I)I \subseteq I$, whence $\alpha _i^* \in I^{-1}$ by definition. Thus from (20.3.5) we have $1 \in I^{-1}I$, whence

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

and thus equality holds.

For part (b), the implication ($\Leftarrow$) follows from (a), and the implication ($\Rightarrow$) for compatibility follows from Proposition 16.5.8. $\square$

Remark 20.3.6. The proof of Theorem 20.3.3 follows what is sometimes called the dual basis lemma for a projective module: see Lam [Lam99, (2.9)], Curtis–Reiner [CR81, (3.46)], or Faddeev [Fad65, Proposition 18.2].

20.3.7

Let $\mathcal {O},\mathcal {O}' \subseteq B$ be R-orders. A $\mathcal {O},\mathcal {O}'$-bimodule over R is an abelian group M with a left $\mathcal {O}$-module and a right $\mathcal {O}'$-module structure with the same action by R on the left and right (i.e., acting centrally, so $rm=mr$ for all $r \in R$ and $m \in M$). The R-lattice $I \subseteq B$ is an $\mathcal {O}{}_{\textsf {\tiny {L}} }(I),\mathcal {O}{}_{\textsf {\tiny {R}} }(I)$-bimodule over R.

When the equivalent conditions of Theorem 20.3.3(b) hold, we say that I is projective as a $\mathcal {O}{}_{\textsf {\tiny {L}} }(I),\mathcal {O}{}_{\textsf {\tiny {R}} }(I)$-bimodule over R.

Remark 20.3.8. In Theorem 20.3.3, we only considered an R-lattice I as a module over its left and right orders (i.e., we considered only sated fractional $\mathcal {O},\mathcal {O}'$-ideals), for the reasons explained in 16.5.18.

Although invertible requires working in this way, it is possible for an R-lattice I to be projective as a left $\mathcal {O}$-module but still $\mathcal {O}\subsetneq \mathcal {O}{}_{\textsf {\tiny {L}} }(I)$: for example, if $\mathcal {O}$ is a hereditary order (see Chapter 21) contained properly in a maximal order $\mathcal {O}\subsetneq \mathcal {O}'$, then $\mathcal {O}'$ is projective as a left $\mathcal {O}$-module.

Although this may seem a bit complicated, it is refreshing that for quaternion algebras, all of the sided notions coincide. We recall the equivalences in Main Theorem 16.7.7), building upon them.

MainTheorem 20.3.9

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

(ii):

I is invertible;

(iii):

I is left invertible;

(iii${}^\prime$):

I is right invertible;

(v):

I is projective as a left $\mathcal {O}{}_{\textsf {\tiny {L}} }(I)$-module; and

(v${}^\prime$):

I is projective as a right $\mathcal {O}{}_{\textsf {\tiny {R}} }(I)$-module.

Proof. The equivalences (ii) $\Leftrightarrow$ (ii${}^\prime$) $\Leftrightarrow$ (iii) are from Main Theorem 16.7.7 (proven in Lemma 16.7.5). Theorem 20.3.3(a) gives (v) $\Rightarrow$ (iii) and (v${}^\prime$) $\Rightarrow$ (iii${}^\prime$), and Theorem 20.3.3(b) gives (ii) $\Rightarrow$ (v), (v${}^\prime$). $\square$

Example 20.3.10

Consider again Example 16.5.12. The lattice I has $\mathcal {O}{}_{\textsf {\tiny {L}} }(I)=\mathcal {O}{}_{\textsf {\tiny {R}} }(I)=\mathcal {O}$ (so has the structure of a sated $\mathcal {O},\mathcal {O}$-bimodule) but I is not invertible; from Main Theorem 20.3.9, it follows that I is not projective as a left or right $\mathcal {O}$-module.

4 Jacobson radical

Before proceeding further in our analysis of orders, we pause to extend some notions in sections 7.2 and 7.4 from algebras to rings. We follow Reiner [Rei2003, §6a]; see also Curtis–Reiner [CR81, §5].

Throughout, let A be a ring (not necessarily commutative, but with 1).

Definition 20.4.1

Let M be a left A-module. We say M is irreducible or simple  if $M \ne \{0\}$ and M contains no A-submodules except $\{0\}$ and M. We say M is indecomposable if whenever $M=M_1 \oplus M_2$ with $M_1,M_2$ left A-modules, then either $M_1=\{0\}$ or $M_2=\{0\}$.

20.4.2

We generalize Lemma 7.2.7. If I is a maximal left ideal of A, then A/I is a simple A-module. Conversely, if M is a simple A-module, then for any $x \in M$ nonzero we have $Ax=M$; therefore $M \simeq A/I$ where

$$\begin{aligned} I={{\,\mathrm{ann}\,}}(x) :=\{\alpha \in A : \alpha x =0\}. \end{aligned}$$

Definition 20.4.3

The Jacobson radical ${{\,\mathrm{rad}\,}}A$ is the intersection of all maximal left ideals of A. The ring A is Jacobson semisimple if ${{\,\mathrm{rad}\,}}A =\{0\}$.

Lemma 20.4.4

The Jacobson radical ${{\,\mathrm{rad}\,}}A$ is the intersection of all annihilators of simple left A-modules; ${{\,\mathrm{rad}\,}}A \subseteq A$ is a two-sided A-ideal.

Proof. The same proof as in Lemma 7.4.5 and Corollary 7.4.6 applies, mutatis mutandis. $\square$

Example 20.4.5

If A is a commutative local ring, then ${{\,\mathrm{rad}\,}}A$ is the unique maximal ideal of A.

Example 20.4.6

Let R be a complete DVR with maximal ideal $\mathfrak p ={{\,\mathrm{rad}\,}}R$. Let $F={{\,\mathrm{Frac}\,}}R$ and let D be a division algebra over F. Let $\mathcal {O}\subseteq D$ be the valuation ring, the unique maximal R-order (Proposition 13.3.4). Then $\mathcal {O}$ has a unique two-sided ideal P by 13.3.10, and so ${{\,\mathrm{rad}\,}}\mathcal {O}=P$.

Lemma 20.4.7

$A /{{\,\mathrm{rad}\,}}A$ is Jacobson semisimple.

Proof. Let $J={{\,\mathrm{rad}\,}}A$. Since $JM=\{0\}$ for each simple left A-module M, we may view each such M as a simple left A/J-module. Now let $\alpha \in A$ be such that $\alpha +J \in {{\,\mathrm{rad}\,}}( A /J)$; then $(\alpha +J)M=\{0\}$, so $\alpha M =\{0\}$ and $\alpha \in J$; thus ${{\,\mathrm{rad}\,}}( A /J)=\{0\}$, and A/J is Jacobson semisimple. $\square$

Lemma 20.4.8

We have

$$\begin{aligned} {{\,\mathrm{rad}\,}}A = \{\beta \in A : 1- \alpha _1\beta \alpha _2 \in A ^\times \text { for all }\alpha _1,\alpha _2 \in A \}. \end{aligned}$$

Proof. See Exercise 20.6. $\square$

Corollary 20.4.9

${{\,\mathrm{rad}\,}}A$ is the intersection of all maximal right ideals of A.

Proof. Lemma 20.4.8 gives a left-right symmetric characterization of ${{\,\mathrm{rad}\,}}A$. $\square$

Corollary 20.4.10

If $\phi :A \rightarrow A'$ is a surjective ring homomorphism, then $\phi ({{\,\mathrm{rad}\,}}A) \subseteq {{\,\mathrm{rad}\,}}A'$ and we have an induced surjective homomorphism $A/{{\,\mathrm{rad}\,}}A \rightarrow A'/{{\,\mathrm{rad}\,}}A'$.

Proof. Let $\beta \in {{\,\mathrm{rad}\,}}A$, let $\alpha _1',\alpha _2' \in A'$; since $\phi$ is surjective, there exist preimages $\alpha _1,\alpha _2 \in A$. By Lemma 20.4.8, $1-\alpha _1\beta \alpha _2 \in A^\times$ and

$$\begin{aligned} \phi (1-\alpha _1\beta _1\alpha _2)=1-\alpha _1'\phi (\beta )\alpha _2' \in {A'}^\times , \end{aligned}$$

so by the same lemma, $\phi (\beta ) \in {{\,\mathrm{rad}\,}}A'$. $\square$

Corollary 20.4.11

Let $I \subseteq A$ be a two-sided A-ideal.

• (a) If A/I is Jacobson semisimple, then ${{\,\mathrm{rad}\,}}A \subseteq I$.

• (b) If $I \subseteq {{\,\mathrm{rad}\,}}A$, then $({{\,\mathrm{rad}\,}}A)/I = {{\,\mathrm{rad}\,}}(A/I)$.

Proof. We have a surjection $\phi :A \rightarrow A/I$. For (a), we get $\phi ({{\,\mathrm{rad}\,}}A) \subseteq {{\,\mathrm{rad}\,}}(A/I)=\{0\}$ from Corollary 20.4.10, so ${{\,\mathrm{rad}\,}}A \subseteq I$. For (b), we get ${{\,\mathrm{rad}\,}}(A)/I \subseteq {{\,\mathrm{rad}\,}}(A/I)$ from the surjection, and applying (a) to $(A/I)/({{\,\mathrm{rad}\,}}(A)/I)$ we get ${{\,\mathrm{rad}\,}}(A/I) \subseteq {{\,\mathrm{rad}\,}}(A)/I$. $\square$

Lemma 20.4.12

(Nakayama’s lemma). Let M be a finitely generated left A-module such that $({{\,\mathrm{rad}\,}}A )M=M$. Then $M=\{0\}$.

Proof. If $M \ne \{0\}$, let $x_1,\dots ,x_n$ be a minimal set of generators for M as a left A-module. Since $x_1 \in M = ({{\,\mathrm{rad}\,}}A )M$, we may write

$$\begin{aligned} x_1=\beta _1 x_1 + \dots + \beta _n x_n \end{aligned}$$

with $\beta _i \in {{\,\mathrm{rad}\,}}A$. But then $1-\beta _1 \in A ^\times$, so the generator $x_1$ is redundant, a contradiction. $\square$

Corollary 20.4.13

Let M be a finitely generated left A-module, and let $N \subseteq M$ be a submodule such that $N+({{\,\mathrm{rad}\,}}A)M=M$. Then $N=M$.

Proof. By hypothesis, M/N is finitely generated, and $({{\,\mathrm{rad}\,}}A)(M/N)=M/N$, so by Nakayama’s lemma, $M/N=\{0\}$ and $M=N$. $\square$

Lemma 20.4.14

Let I be a maximal two-sided ideal of A. Then I contains ${{\,\mathrm{rad}\,}}A$.

Proof. If I does not contain ${{\,\mathrm{rad}\,}}A$, then $I+{{\,\mathrm{rad}\,}}A$ is a two-sided ideal of A containing ${{\,\mathrm{rad}\,}}A$ and properly containing I. Since I is maximal, we have $I+{{\,\mathrm{rad}\,}}A = A$. By (the corollary to) Nakayama’s lemma, we get $I= A$, a contradiction. $\square$

5 Local Jacobson radical

Suppose now that R is a complete DVR with fraction field $F={{\,\mathrm{Frac}\,}}R$, maximal ideal $\mathfrak p ={{\,\mathrm{rad}\,}}R$, and residue field $k=R/\mathfrak p$. Let B be a finite-dimensional F-algebra, and let $\mathcal {O}\subseteq B$ be an R-order.

In this setting, we may identify the Jacobson radical via pullback as follows.

Theorem 20.5.1

Let $\phi :\mathcal {O}\rightarrow \mathcal {O}/\mathfrak p \mathcal {O}$ be reduction modulo $\mathfrak p$. Then

$$\begin{aligned} {{\,\mathrm{rad}\,}}\mathcal {O}= \phi ^{-1}({{\,\mathrm{rad}\,}}\mathcal {O}/\mathfrak p \mathcal {O}) \supseteq \mathfrak p \mathcal {O}, \end{aligned}$$

and $({{\,\mathrm{rad}\,}}\mathcal {O})^r \subseteq \mathfrak p \mathcal {O}$ for some $r>0$.

Proof. See Reiner [Rei2003, Theorem 6.15]. $\square$

Corollary 20.5.2

$\mathcal {O}/{{\,\mathrm{rad}\,}}\mathcal {O}$ is a (finite-dimensional) semisimple k-algebra.

Proof. Since ${{\,\mathrm{rad}\,}}\mathcal {O}\supseteq \mathfrak p \mathcal {O}$, we conclude that $\mathcal {O}/\mathfrak p \mathcal {O}$ is a k-algebra; it is Jacobson semisimple by 20.4.7 and hence semisimple by Lemma 7.4.2. $\square$

Definition 20.5.3

A two-sided ideal $J \subseteq \mathcal {O}$ is topologically nilpotent if $J^r \subseteq \mathfrak p \mathcal {O}$ for some $r>0$.

Remark 20.5.4. The order $\mathcal {O}$ as a free R-module has a natural topology induced from the $\mathfrak p$-adic topology on R; J is topologically nilpotent if and only if $J^r \rightarrow \{0\}$ in this topology.

Corollary 20.5.5

Let $I \subseteq \mathcal {O}$ be a two-sided ideal. Then the following are equivalent:

• (a) $I \subseteq {{\,\mathrm{rad}\,}}\mathcal {O}$;

• (b) $I^r \subseteq {{\,\mathrm{rad}\,}}\mathcal {O}$ for some $r>0$; and

• (c) I is topologically nilpotent.

Proof. See Reiner [Rei2003, Exercise 39.1, Exercise 6.3]. $\square$

6 Integral representation theory

We continue our notation that R is a complete DVR. We now turn to some notions in integral representation theory. In this local case, there is a tight connection between the representation theory of $\mathcal {O}$ (viewed in terms of $\mathcal {O}$-modules) and the representation theory of the quotient $\mathcal {O}/\mathfrak p \mathcal {O}$ which is a k-algebra of finite dimension over k, since $\mathcal {O}$ is finitely generated as an R-module.

20.6.1

Recall that a representation of B over F is the same as a left B-module. If M is a finitely-generated left $\mathcal {O}$-module, then $V :=M \otimes _R F$ is a left B-module, and $M \subseteq V$ is an R-lattice. A $\mathcal {O}$-supermodule of M is a left $\mathcal {O}$-module $V \supseteq M' \supseteq M$.

The following result is foundational.

Theorem 20.6.2

(Krull–Schmidt). Every finitely generated left $\mathcal {O}$-module M is expressible as a finite direct sum of indecomposable modules, uniquely determined by M up to $\mathcal {O}$-module isomorphism and reordering.

Proof. Since M is finitely generated over R it is itself noetherian, so the process of decomposing M into direct summands terminates. See Curtis–Reiner [CR81, (6.12)] or Reiner [Rei2003, §6, Exercise 6] for hints that lead to a proof of the second (uniqueness) part. $\square$

Corollary 20.6.3

Let $M=M_1 \oplus \dots \oplus M_r$ be a decomposition into finitely generated indecomposable left $\mathcal {O}$-modules, and let $N \subseteq M$ be a direct summand. Then $N \simeq M_{i_1} \oplus \cdots \oplus M_{i_s}$ for some subset $\{i_1,\dots ,i_s\} \subseteq \{1,\dots ,n\}$.

Proof. By hypothesis, we can write $\bigoplus _{i=1}^n M_i = N \oplus N'$, with $N'$ a finitely generated left $\mathcal {O}$-module. By the Krull–Schmidt theorem (Theorem 20.6.2), if we write $N,N'$ as the direct sums of indecomposable modules, the conclusion follows. $\square$

20.6.4

We saw in 7.2.19 that idempotents govern the decomposition of the F-algebra B into indecomposable left B-modules. The same argument shows that a decomposition

$$\begin{aligned} \mathcal {O}= P_1 \oplus \dots \oplus P_r \end{aligned}$$

(20.6.5)

into a direct sum of indecomposable left $\mathcal {O}$-modules corresponds to an idempotent decomposition $1=e_1+\dots +e_r$, with the $e_i$ a complete set of primitive orthogonal idempotents. Moreover, each $P_i=\mathcal {O}e_i$ is a projective indecomposable left $\mathcal {O}$-module.

Conversely, if P is a projective indecomposable finitely generated left $\mathcal {O}$-module, then $P \simeq P_i$ for some i: taking a set of generators we have a surjective $\mathcal {O}$-module homomorphism $\mathcal {O}^r \rightarrow P$, and since P is projective we have $P \subseteq \mathcal {O}^r$ a direct summand, so Corollary 20.6.3 applies.

Consequently, if P is a projective left $\mathcal {O}$-lattice, then $P \simeq P_1^{\oplus n_1} \oplus \dots \oplus P_r^{\oplus n_r}$ with $n_i \ge 0$ for $i=1,\dots ,r$.

20.6.6

The decomposition of an order into projective indecomposables is a nice way to keep track of other orders, as follows. We extend our notation slightly, and define

$$\begin{aligned} \mathcal {O}{}_{\textsf {\tiny {L}} }(M) :=\{\alpha \in B : \alpha M \subseteq M\} \end{aligned}$$

for every left $\mathcal {O}$-submodule $M \subseteq B$.

Take a decomposition of $\mathcal {O}$ in (20.6.5); since each $P_i$ is a left $\mathcal {O}$-module, extending scalars it is a left B-module, so

$$\begin{aligned} \bigcap _{i=1}^r \mathcal {O}{}_{\textsf {\tiny {L}} }(P_i) = \mathcal {O}. \end{aligned}$$

(20.6.7)

Now let $I \subseteq B$ be an R-lattice with $\mathcal {O}\subset \mathcal {O}{}_{\textsf {\tiny {L}} }(I)$ that is projective as an $\mathcal {O}$-module. By 20.6.4, considering I as a left $\mathcal {O}$-module, we have an isomorphism of left $\mathcal {O}$-modules

with $n_i \ge 0$. We claim that

$$\begin{aligned} \mathcal {O}{}_{\textsf {\tiny {L}} }(I)=\bigcap _{\begin{array}{c} i \\ n_i>0 \end{array}} \mathcal {O}{}_{\textsf {\tiny {L}} }(P_i). \end{aligned}$$

Indeed, we have $\alpha I \subseteq I$ if and only if $\phi (\alpha I) = \alpha \phi (I) \subseteq \phi (I)$, since $\phi$ is a $\mathcal {O}$-module homomorphism so extends to a B-algebra homomorphism, and finally $\alpha \phi (I) \subseteq \phi (I)$ if and only if $\alpha P_i \subseteq P_i$ for all i with $n_i>0$, as in (20.6.7).

We now relate a decomposition of $\mathcal {O}$ into a decomposition of $\mathcal {O}/\mathfrak p \mathcal {O}$.

Lemma 20.6.8

Let $J={{\,\mathrm{rad}\,}}\mathcal {O}$. The association $I \mapsto I/JI$ gives a bijection between isomorphism classes of indecomposable finitely generated projective left $\mathcal {O}$-modules and isomorphism classes of simple finite-dimensional left $\mathcal {O}/J$-modules.

Proof. The proof requires a bit of fiddling with idempotents, but is otherwise straightforward—so it makes a good exercise (Exercise 20.7). $\square$

Corollary 20.6.9

If I is projective indecomposable, then $JI \subseteq I$ is the unique maximal $\mathcal {O}$-submodule of I.

Proof. By Lemma 20.6.8, since I is indecomposable, I/JI is simple so JI is a unique maximal submodule. If $I' \subseteq I$ is another maximal $\mathcal {O}$-submodule, then $JI+I'=I$, and by Nakayama’s lemma $I'=I$, a contradiction. $\square$

We finish our local study over R a complete DVR with composition series for modules over an order.

Definition 20.6.10

Let M be an $\mathcal {O}$-lattice. A composition series for M is a strictly decreasing sequence

$$\begin{aligned} M=M_0 \supseteq M_1 \supseteq M_2 \supseteq \dots \end{aligned}$$

such that $\bigcap _{i=1}^{\infty } M_i=\{0\}$ and each composition factor $M_i/M_{i+1}$ is simple as a $\mathcal {O}$-module.

The length of a composition series is the largest integer r such that $M_r=\{0\}$ if r exists (in which case we call the series finite), and otherwise the length is $\infty$.

20.6.11

If M has a finite composition series, then its length $\ell (M)$ is well-defined, independent of the series. For example, taking $R=F$ and $\mathcal {O}=B$, a finitely generated B-module is a finite-dimensional F-vector space, so every composition series is finite and every B-module V has a well-defined length $\ell (V)$.

20.6.12

Let $N \subseteq M$ be a maximal $\mathcal {O}$-submodule. We claim that $JM \subseteq N$. Otherwise, $N+JM=M$ by maximality, so by Nakayama’s lemma (Corollary 20.4.13), $N=M$, a contradiction.

7 $*$ Stable class group and cancellation

To conclude this chapter, we apply the above results and consider a different way to form of a group of ideal classes; for further reference on the topics of this section, see Curtis–Reiner [CR87, §§49–51] or Reiner [Rei2003, §38].

Let R be a Dedekind domain with field of fractions F.

20.7.1

Recall that the group ${{\,\mathrm{Cl}\,}}R$ records classes of fractional ideals, or what is more relevant here, isomorphism classes of projective modules of rank 1. Here is another way to see the group law on ${{\,\mathrm{Cl}\,}}R$: given two such fractional ideals $\mathfrak a ,\mathfrak b$ up to isomorphism, there is an isomorphism of R-modules

$$\begin{aligned} \mathfrak a \oplus \mathfrak b \simeq R \oplus \mathfrak a \mathfrak b , \end{aligned}$$

and the class of $\mathfrak a \mathfrak b$ is uniquely determined by this isomorphism by 9.3.10.

We now consider an analogous construction to 20.7.1 in the noncommutative setting. Let B be a simple F-algebra and $\mathcal {O}\subseteq B$ an R-order. We begin with a technical lemma.

Lemma 20.7.2

(Weak approximation). Let I be a locally principal left fractional $\mathcal {O}$-ideal and let $\mathfrak a \subseteq R$ be an ideal. Then there exists $\beta \in B^\times$ such that $I\beta \subseteq \mathcal {O}$ and

$$\begin{aligned} (I\beta )_\mathfrak p = \mathcal {O}_\mathfrak p \quad \text { for all }\mathfrak p \mid \mathfrak a . \end{aligned}$$

(20.7.3)

Proof. For each prime $\mathfrak p$, we have $I_\mathfrak p =\mathcal {O}_\mathfrak p \alpha _\mathfrak p$ with $\alpha _\mathfrak p \in B_\mathfrak p ^\times$. Because F is dense in $F_\mathfrak p$, there exists $\beta \in (\mathcal {O}:I){}_{\textsf {\tiny {R}} }$ such that $\alpha _\mathfrak p \beta _\mathfrak p \equiv 1 \pmod \mathfrak{p \mathcal {O}_\mathfrak p }$ for all $\mathfrak p \mid \mathfrak a$. By norms, we have $\beta \in B^\times$. Letting $\mu _\mathfrak p :=\alpha _\mathfrak p \beta _\mathfrak p$, we have $\mu _\mathfrak p - 1 \in \mathfrak p \mathcal {O}_\mathfrak p \supseteq {{\,\mathrm{rad}\,}}\mathcal {O}_\mathfrak p$ by Theorem 20.5.1 so $\mu _\mathfrak p \in \mathcal {O}_\mathfrak p ^\times$ by Lemma 20.4.8. Therefore $(I\beta )_\mathfrak p =\mathcal {O}_\mathfrak p$ for all $\mathfrak p \mid \mathfrak a$. $\square$

Proposition 20.7.4

If $I,I' \subseteq B$ are locally principal left fractional $\mathcal {O}$-ideals, then there exists a locally principal left fractional $\mathcal {O}$-ideal J and an isomorphism

$$\begin{aligned} I \oplus I' \simeq J \oplus \mathcal {O}\end{aligned}$$

(20.7.5)

of left $\mathcal {O}$-modules.

Proof. We may suppose without loss of generality that $I,I' \subseteq \mathcal {O}$. Then we have exact sequences of left $\mathcal {O}$-modules

$$\begin{aligned} 0 \rightarrow I&\xrightarrow {\phi } \mathcal {O}\rightarrow \mathcal {O}/I \rightarrow 0 \\ 0 \rightarrow I'&\xrightarrow {\phi '} \mathcal {O}\rightarrow \mathcal {O}/I' \rightarrow 0 \end{aligned}$$

The module $\mathcal {O}/I$ is R-torsion, annihilated by the (nonzero) R-ideal $\mathfrak a =[\mathcal {O}:I]_R$, and similarly with $I'$, annihilated by $\mathfrak a '=[\mathcal {O}:I']_R$. By weak approximation (Lemma 20.7.2), replacing $I'$ with $I'\beta$ we may suppose that $I'_\mathfrak p = \mathcal {O}_\mathfrak p$ for all $\mathfrak p \mid \mathfrak a$, and hence $\mathfrak a ,\mathfrak a '$ are coprime. Then for all primes $\mathfrak p$ of R, we have either $(\mathcal {O}/I)_\mathfrak p =\{0\}$ so $\phi _\mathfrak p$ is surjective, or correspondingly $\phi '_\mathfrak p$ is surjective.

Now consider the left $\mathcal {O}$-module homomorphism

$$\begin{aligned} \phi +\phi ' :I \oplus I' \rightarrow \mathcal {O}\end{aligned}$$

(20.7.6)

obtained by summing the two natural inclusions. We just showed that $(\phi +\phi ')_\mathfrak p$ is surjective for all primes $\mathfrak p$, so it follows that $\phi +\phi '$ is surjective: the cokernel $M :={{\,\mathrm{coker}\,}}(\phi +\phi ')$ has $M_\mathfrak p =\{0\}$ for all $\mathfrak p$, and $M=\{0\}$. Moreover, since $\mathcal {O}$ is projective as a left $\mathcal {O}$-module, the map $\phi +\phi '$ splits (or note that the map splits locally for every prime $\mathfrak p$, so it splits globally, Exercise 20.1(b)). If we let $J :=\ker (\phi +\phi ')$, we then obtain an isomorphism

$$\begin{aligned} I \oplus I' \simeq J \oplus \mathcal {O}. \end{aligned}$$

(20.7.7)

To conclude, we show that J is locally principal. To this end, we localize at a prime $\mathfrak p$ and note that $I,I'$ are locally principal, so

$$\begin{aligned} I_\mathfrak p \oplus I'_\mathfrak p \simeq \mathcal {O}_\mathfrak{p }^{\oplus 2} \simeq J_\mathfrak p \oplus \mathcal {O}_\mathfrak p . \end{aligned}$$

(20.7.8)

But by the Krull–Schmidt theorem (Theorem 20.6.2) and Exercise 20.8, we can cancel one copy of $\mathcal {O}_\mathfrak p$ from both sides! We conclude that $J_\mathfrak p \simeq \mathcal {O}_\mathfrak p$ as left $\mathcal {O}$-modules and therefore by Lemma 17.3.3 that $J_\mathfrak p$ is (right) principal. $\square$

The candidate binary operation in Proposition 20.7.4 has a simple description in the “coprime” case.

Lemma 20.7.9

Let $I,I' \subseteq \mathcal {O}$ be locally principal integral left $\mathcal {O}$-ideals, and suppose for every prime $\mathfrak p \subseteq R$ either $I_\mathfrak p = \mathcal {O}_\mathfrak p$ or $I'_\mathfrak p =\mathcal {O}_\mathfrak p$. Then

$$\begin{aligned} I\oplus I' \simeq \mathcal {O}\oplus J, \quad \text {where }J=I \cap I'. \end{aligned}$$

Moreover, writing $I_\mathfrak p =\mathcal {O}\alpha _\mathfrak p$ and $I_\mathfrak p '=\mathcal {O}\alpha _\mathfrak p '$, we have

$$\begin{aligned} J_\mathfrak p =\mathcal {O}_\mathfrak p \alpha _\mathfrak p \alpha _\mathfrak p ' = \mathcal {O}_\mathfrak p \alpha _\mathfrak p ' \alpha _\mathfrak p . \end{aligned}$$

By weak approximation (Lemma 20.7.2), the hypothesis of Lemma 20.7.9 can always be arranged to hold for $I,I'$, up to isomorphism (as left $\mathcal {O}$-ideals).

Proof. By hypothesis, if $\phi ,\phi :I,I' \hookrightarrow \mathcal {O}$ are the inclusions, then the map $\phi +\phi ':I \oplus I' \rightarrow \mathcal {O}$ as in (20.7.6) is surjective. We have

$$\begin{aligned} \ker (\phi +\phi ')=\{(\alpha ,\alpha ') \in I \oplus I' : \alpha +\alpha '=0\} \simeq I \cap I' \end{aligned}$$

by projection onto either coordinate, since $\alpha =-\alpha ' \in I \cap I'$. This gives an exact sequence

$$\begin{aligned} 0 \rightarrow I \cap I' \rightarrow I \oplus I' \rightarrow \mathcal {O}\rightarrow 0 \end{aligned}$$

and as above $I \oplus I' \simeq J \oplus \mathcal {O}$ with $J=I \cap I'$. The final statement follows from the hypothesis that either $I_\mathfrak p = \mathcal {O}_\mathfrak p$ or $I'_\mathfrak p =\mathcal {O}_\mathfrak p$, since then $\alpha _\mathfrak p \in \mathcal {O}_\mathfrak p ^\times$ or $\alpha _\mathfrak p ' \in \mathcal {O}_\mathfrak p ^\times$. $\square$

In order to get a well-defined binary operation, we need an equivalence relation: we will need to identify $J,J'$ if $J \oplus \mathcal {O}\simeq J' \oplus \mathcal {O}$. But the copies of $\mathcal {O}$ needed for the axioms start to pile up, so we make the following more general definition.

Definition 20.7.10

Let $J,J' \subseteq B$ be locally principal left $\mathcal {O}$-ideals. We say that J is stably isomorphic to $J'$ if there exists an isomorphism of left $\mathcal {O}$-modules

$$\begin{aligned} J \oplus \mathcal {O}^{\oplus r} \simeq J' \oplus \mathcal {O}^{\oplus r} \end{aligned}$$

for some $r \ge 0$.

Let $[J]{}_{\textsf {\tiny {St}} }$ denote the stable isomorphism class of a left $\mathcal {O}$-ideal J and let ${{\,\mathrm{StCl}\,}}\mathcal {O}$ be the set of stable isomorphism classes of left $\mathcal {O}$-ideals in B.

tClO]${{\,\mathrm{StCl}\,}}\mathcal {O}$stable (or locally free) class group

Proposition 20.7.11

${{\,\mathrm{StCl}\,}}\mathcal {O}$ is an abelian group under the binary operation (20.7.5), written $[I]{}_{\textsf {\tiny {St}} }+[I']{}_{\textsf {\tiny {St}} }= [J]{}_{\textsf {\tiny {St}} }$, with identity $[\mathcal {O}]{}_{\textsf {\tiny {St}} }$.

Accordingly, we call ${{\,\mathrm{StCl}\,}}\mathcal {O}$ the stable class group of $\mathcal {O}$; it is also referred to as the locally free class group of $\mathcal {O}$.

Proof. The operation is well-defined: if $[I_1]{}_{\textsf {\tiny {St}} }=[I_2]{}_{\textsf {\tiny {St}} }$ via $I_1 \oplus \mathcal {O}^{\oplus r} \simeq I_2 \oplus \mathcal {O}^{\oplus r}$ and the same with $[I_1']{}_{\textsf {\tiny {St}} }=[I_2']{}_{\textsf {\tiny {St}} }$, and we perform the binary operation $I_1 \oplus I_1' \simeq J_1 \oplus \mathcal {O}$ and the same with the subscripts ${\phantom {i}}_2$, then

$$\begin{aligned} \begin{aligned} J_1 \oplus \mathcal {O}^{\oplus (r+r'+1)}&\simeq (I_1 \oplus \mathcal {O}^{\oplus r}) \oplus (I_1' \oplus \mathcal {O}^{\oplus r'}) \\&\simeq (I_2 \oplus \mathcal {O}^{\oplus r}) \oplus (I_2' \oplus \mathcal {O}^{\oplus r'}) \\&\simeq J_2 \oplus \mathcal {O}^{\oplus (r+r'+1)} \end{aligned} \end{aligned}$$

(20.7.12)

so $[J_1]{}_{\textsf {\tiny {St}} }=[J_2]{}_{\textsf {\tiny {St}} }$. It is similarly straightforward to verify that the operation is associative and commutative and that $[\mathcal {O}]{}_{\textsf {\tiny {St}} }$ is the identity.

To conclude, we show that ${{\,\mathrm{StCl}\,}}\mathcal {O}$ has inverses. Let $I \subseteq \mathcal {O}$ be a locally principal $\mathcal {O}$-ideal. For each prime $\mathfrak p \subseteq R$, we have $I_\mathfrak p =\mathcal {O}_\mathfrak p \alpha _\mathfrak p$ with $\alpha _\mathfrak p \in B_\mathfrak p ^\times$, and $\alpha _\mathfrak p = 1$ for all but finitely many $\mathfrak p$. Let $I'$ be the R-lattice with $I'_\mathfrak p =\mathcal {O}_\mathfrak p \alpha _\mathfrak p ^{-1}$ for all $\mathfrak p$. Then $I'$ is a left fractional $\mathcal {O}$-ideal, because this is true locally. By weak approximation (Lemma 20.7.2), there exists $\beta \in B^\times$ such that $(I'\beta )_\mathfrak p = \mathcal {O}_\mathfrak p$ for all $\mathfrak p$ such that $I_\mathfrak p \ne \mathcal {O}_\mathfrak p$, i.e., for all $\mathfrak p$ such that $\alpha _\mathfrak p \ne 1$. But now we can perform the group operation as in Lemma 20.7.9: we have $[I]{}_{\textsf {\tiny {St}} }+ [I']{}_{\textsf {\tiny {St}} }= [J]{}_{\textsf {\tiny {St}} }$ where $J=I \cap I'\beta$, and for all $\mathfrak p$ we have

$$\begin{aligned} J_\mathfrak p = \mathcal {O}_\mathfrak p \alpha _\mathfrak p \alpha _\mathfrak p ^{-1} \beta _\mathfrak p = \mathcal {O}_\mathfrak p \beta _\mathfrak p \end{aligned}$$

so $J = \mathcal {O}\beta$ and $J \simeq \mathcal {O}$, so $[I]{}_{\textsf {\tiny {St}} }+ [I']{}_{\textsf {\tiny {St}} }= [\mathcal {O}]{}_{\textsf {\tiny {St}} }$ and $I'$ is an inverse. $\square$

Remark 20.7.13. There is a related group to ${{\,\mathrm{StCl}\,}}\mathcal {O}$, defined as follows. Let A be a ring, and let $\mathcal P (A)$ be the category of finitely generated projective left A-modules under isomorphisms. We define the group $K_0(A)$ to be the free abelian group on the isomorphism classes [P] of objects $P \in \mathcal P (A)$ modulo the subgroup of relations

$$\begin{aligned}{}[P \oplus P'] = [P] + [P'], \quad \,\text {for}\, P,P' \in \mathcal P (A); \end{aligned}$$

equivalently relations $[P]+[P']=[Q]$ for each exact sequence

$$\begin{aligned} 0 \rightarrow P \rightarrow Q \rightarrow P' \rightarrow 0 \end{aligned}$$

since such a sequence splits. The group $K_0(A)$ is sometimes called the projective class group of A. (The group $K_0(A)$ is the Grothendieck group of the category $\mathcal P (A)$.)

Then for $P,Q \in \mathcal P (\mathcal {O})$, we have $[P]=[Q] \in K_0(\mathcal {O})$ if and only if P, Q are stably isomorphic [CR87, Proposition 38.22]. Moreover, there is a natural map

$$\begin{aligned} K_0(\mathcal {O})&\rightarrow K_0(B)\\ [P]&\mapsto [F \otimes _R \mathcal {O}], \end{aligned}$$

and we let $SK_0(\mathcal {O})$ be its kernel, called the reduced projective class group of $\mathcal {O}$. The abelian group $SK_0(\mathcal {O})$ is generated by elements $[P]-[Q]$ where $P,Q \in \mathcal P (\mathcal {O})$ and $F \otimes _R P \simeq F \otimes _R Q$. Finally, we have an isomorphism [CR87, Theorem 49.32]

(20.7.14)

In other words, after all of this work—at least for maximal orders—the reduced projective class group and the stable class group coincide. (For a more general order, one instead compares to a maximal superorder via the natural extension maps ${{\,\mathrm{StCl}\,}}\mathcal {O}\rightarrow {{\,\mathrm{StCl}\,}}\mathcal {O}'$.)

The stable class group was first introduced and studied by Swan [Swa60, Swa62] in this context in the special case where $\mathcal {O}=Z[G]$ is the group ring of a finite group G.

Definition 20.7.15

We say that $\mathcal {O}$ has stable cancellation (or the simplification property) if stable isomorphism implies isomorphism, i.e., if whenever $I,I'$ are locally principal left $\mathcal {O}$-ideals with $I \oplus \mathcal {O}^r \simeq I' \oplus \mathcal {O}^r$ for $r \ge 0$, then in fact $I \simeq I'$.

If we had defined stable isomorphism and cancellation for locally free $\mathcal {O}$-modules, we would arrive at the same groups and condition, so stable cancellation is also called the locally free cancellation.

From now on, suppose that R is a global ring with $F = {{\,\mathrm{Frac}\,}}R$, and $\mathcal {O}\subset B$ is a maximal R-order in a quaternion algebra B over F. We recall section , and the class group $Cl_\Omega R$, where $\Omega \subseteq {{\,\mathrm{Ram}\,}}B$ is the set of real ramified places.

Theorem 20.7.16

(Fröhlich–Swan). Let $R=R_{(S)}$ be a global ring, let B be a quaternion algebra over F, and let $\mathcal {O}\subset B$ be a maximal R-order. Then the reduced norm induces an isomorphism

(20.7.17)

of finite abelian groups.

Proof. See Fröhlich [Frö75, Theorem 2, S X], Swan [Swa80, Theorem 9.4], or Curtis–Reiner [CR87, Theorem 49.32]; we will sketch a proof of a more general version of this theorem in section , when we have idelic methods at our disposal.

20.7.18

Since B is a quaternion algebra, the notions of invertible and locally principal coincide. Then there is a surjective map of sets

$$\begin{aligned} \text {Cls}_{{}_{\textsf {\tiny {L}} }} \mathcal {O}&\rightarrow {{\,\mathrm{StCl}\,}}\mathcal {O}\nonumber \\ [I]{}_{\textsf {\tiny {L}} }&\mapsto [I]{}_{\textsf {\tiny {St}} }. \end{aligned}$$

(20.7.19)

Suppose further that $F={{\,\mathrm{Frac}\,}}R$ is a number field and R is a global ring. Then $\text {Cls}{{}_{\textsf {\tiny {L}} }} \mathcal {O}$ is a finite set, by Main Theorem ; consequently, the stable class group ${{\,\mathrm{StCl}\,}}\mathcal {O}$ is a finite abelian group. However, the map (20.7.19) of sets need not be injective.

The order $\mathcal {O}$ has stable cancellation if and only if the map (20.7.19) is injective (equivalently, bijective).

20.7.20

Suppose that B satisfies the Eichler condition. Then by Eichler’s theorem (Theorem ), the reduced norm also gives a bijection compatible with the surjective map (20.7.19) which must therefore also be a bijection.

What remains, then, is the case where B is definite. We restrict attention to the case where the base field F is a number field, hence a totally real field, and we work with R-orders $\mathcal {O}\subseteq B$, where $R=\mathbb Z _F$ is the ring of integers of F. Vignéras [Vig76b] initiated the classification of definite quaternion orders with stable cancellation, and showed that there are only finitely many such orders. Hallouin–Maire [HM2006] and Smertnig [Sme2015] extended this classification to certain classes of orders, and the complete classification was obtained by Smertnig–Voight [SV2019].

Theorem 20.7.21

(Vignéras, Hallouin–Maire, Smertnig, Smertnig–Voight). Up to isomorphism, there are exactly 316 definite quaternion R-orders with stable cancellation.

The isomorphisms in Theorem 20.7.21 are as R-orders; up to ring isomorphism (identifying Galois conjugates), there are exactly 247.

Example 20.7.22

If $\mathcal {O}$ is a definite maximal quaternion $\mathbb Z$-order, by Theorem 20.7.16 we have ${{\,\mathrm{StCl}\,}}\mathcal {O}= \#{{\,\mathrm{Cl}\,}}^+ \mathbb Z = 1$, so $\mathcal {O}$ has stable cancellation if and only if $\#{{\,\mathrm{Cls}\,}}\mathcal {O}=1$. These orders will be classified in section 25.4: they are the orders of discriminant $D=2,3,5,7,13$. (More generally, if $R=\mathbb Z _F$ has $\#{{\,\mathrm{Cl}\,}}^+ \mathbb Z _F=1$, then a definite, maximal quaternion R-order has stable cancellation if and only if $\#{{\,\mathrm{Cls}\,}}\mathcal {O}=1$.)

Remark 20.7.23. Jacobinski [Jaci68] was the first to consider the stable class group for general orders in the context of his work on genera of lattices; his cancellation theorem states more generally that if B is a central simple algebra over F and B is not a totally definite quaternion algebra, then every R-order $\mathcal {O}\subseteq B$ has stable cancellation. This result was reformulated by Fröhlich [Frö75] in terms of ideles and further developed by Fröhlich–Reiner–Ullom [FRU74]. Swan [Swa80] related cancellation to strong approximation in the context of K-groups.

Brzezinski [Brz83b] also defines the spinor class group of an order, a quotient of its locally free class group; this group measures certain invariants phrased in terms of quadratic forms.

Remark 20.7.24. More generally, a ring A in which every stably free right A-module is free is called a (right) Hermite ring by some authors: for further reference and comparison of terminology, see Lam [Lam2006, Section I.4]. If $\mathcal {O}$ has locally free cancellation, then $\mathcal {O}$ is Hermite; however, the converse does not hold in general—a counterexample is described in detail by Smertnig [Sme2015]. Smertnig–Voight [SV2019] show that there are exactly 375 definite quaternion R-orders with the Hermite property up to isomorphism.

Exercises

Throughout these exercises, let R be a noetherian domain with $F={{\,\mathrm{Frac}\,}}R$, let B be a finite-dimensional F-algebra, let $\mathcal {O}\subseteq B$ be an R-order, and let $J={{\,\mathrm{rad}\,}}\mathcal {O}$.

$\triangleright$ 1.:

Let M, N be left $\mathcal {O}$-lattices.

(a):

Show that a sequence $0 \rightarrow M \rightarrow N \rightarrow M' \rightarrow 0$ of left $\mathcal {O}$-lattices is exact if and only if the sequences $0 \rightarrow M_\mathfrak p \rightarrow N_\mathfrak p \rightarrow M'_\mathfrak p \rightarrow 0$ are exact for all primes $\mathfrak p \subseteq R$. [Hint: Consider the modules measuring the failure of exactness and show they are locally zero, hence zero.]

(b):

Let $\phi :M \rightarrow N$ be a surjective $\mathcal {O}$-module homomorphism. Show that $\phi$ splits (there exists $\psi :N \rightarrow M$ such that $\phi \psi ={{\,\mathrm{id}\,}}_N$) if and only if $\phi _\mathfrak p :M_\mathfrak p \rightarrow N_\mathfrak p$ splits for all primes $\mathfrak p \subseteq R$.

$\triangleright$ 2.:

Suppose R is a DVR and B is a quaternion algebra. Let $J={{\,\mathrm{rad}\,}}\mathcal {O}$. Show that $\overline{J}=J$ and $\mathcal {O}{}_{\textsf {\tiny {L}} }({{\,\mathrm{rad}\,}}\mathcal {O})=\mathcal {O}{}_{\textsf {\tiny {R}} }({{\,\mathrm{rad}\,}}\mathcal {O})$.

3.:

Let R be a complete DVR with $\mathfrak p ={{\,\mathrm{rad}\,}}R$. Show that the $\mathfrak p$-adic topology and the J-adic topology on $\mathcal {O}$ are the same.

4.:

Let R be a DVR with maximal ideal $\mathfrak p$, and let $\mathcal {O}=\begin{pmatrix} R &{} R \\ \mathfrak p &{} R \end{pmatrix} \subseteq B={{\,\mathrm{M}\,}}_2(F)$. Let $I \subseteq B$ be a left fractional $\mathcal {O}$-ideal. Show that either I is invertible as a $\mathcal {O}$-ideal or I is conjugate to ${{\,\mathrm{M}\,}}_2(R)$ by an element of $B^\times$.

5.:

Let $\mathcal {O}$ be a maximal R-order, and let M be a projective left $\mathcal {O}$-lattice. Show that M is indecomposable if and only if FM is a simple left B-module. [Hint: Suppose $W \subseteq FM$ is a left B-submodule of FM, and let $N :=M \cap W$. Show that M/N is a projective $\mathcal {O}$-lattice, so the sequence $0 \rightarrow N \rightarrow M \rightarrow M/N \rightarrow 0$ splits.]

$\triangleright$ 6.:

Let A be a ring (not necessarily commutative, but with 1. In this exercise, we prove Lemma 20.4.8, that

$$\begin{aligned} {{\,\mathrm{rad}\,}}A = \{\beta \in A : 1- \alpha _1\beta \alpha _2 \in A ^\times \text { for all }\alpha _1,\alpha _2 \in A \}. \end{aligned}$$

We first show the inclusion $(\subseteq )$.

(a):

Since ${{\,\mathrm{rad}\,}}A$ is a two-sided ideal, it suffices to show that $1-\beta \in A ^\times$. Show that $A (1-\beta )= A$.

(b):

Let $\alpha \in A$ be such that $\alpha (1-\beta )=1$. Repeating the argument, show that $A (1-(1-\alpha ))= A \alpha = A$.

(c):

Show that $\alpha$ is also a right inverse of $1-\beta$, so $1-\beta \in A^\times$.

Next we show the inclusion $(\supseteq )$.

(d):

Let $\beta \in A$ be such that $1-\alpha \beta \gamma \in A ^\times$ for all $\alpha ,\gamma \in A$. Let M be a simple left A-module. Show that $\alpha M = \{0\}$. Conclude that $\alpha \in {{\,\mathrm{rad}\,}}A$.

$\triangleright$ 7.:

Suppose R is a complete DVR. Prove Lemma 20.6.8: the association $I \mapsto I/JI$ gives a bijection between isomorphism classes of indecomposable finitely generated projective left $\mathcal {O}$-modules and isomorphism classes of simple finite-dimensional left $\mathcal {O}/J$-modules.

$\triangleright$ 8.:

Let R be a complete DVR, and let $I,I',J$ be finitely generated left $\mathcal {O}$-modules such that

$$\begin{aligned} I \oplus J \simeq I' \oplus J \end{aligned}$$

as left $\mathcal {O}$-modules. Prove that $I \simeq I'$ as left $\mathcal {O}$-modules.

9.:

Let $\Lambda ={{\,\mathrm{M}\,}}_n(\mathcal {O})$ with $n \ge 1$. Show that ${{\,\mathrm{StCl}\,}}\Lambda \simeq {{\,\mathrm{StCl}\,}}\mathcal {O}$.

10.:

Show that the Lipschitz order has stable cancellation.

11.:

Let $B=({-1,-3} \mid \mathbb{Q })$ and let $\mathcal {O}=\mathbb Z + \mathbb Z (3i) + \mathbb Z (-1+j)/2 + \mathbb Z (3i+ij)/2$.

(a):

Show that $\mathcal {O}$ is an order with ${{\,\mathrm{discrd}\,}}\mathcal {O}= 9$.

(b):

Show that $\mathcal {O}$ has stable cancellation.