Hereditary and extremal orders

Abstract

In this chapter, we consider hereditary orders, those with the simplest kind of module theory; we characterize these orders in several ways, including showing they have an extremal property with respect to their Jacobson radical.

In this chapter, we consider hereditary orders, those with the simplest kind of module theory; we characterize these orders in several ways, including showing they have an extremal property with respect to their Jacobson radical.

1 $\triangleright$ Hereditary and extremal orders

Let R be a Dedekind domain. Then R is hereditary: every submodule of a projective module is again projective. (Hence the name: projectivity is inherited by a submodule.) A noetherian domain is hereditary if and only if every ideal of R is projective, or equivalently, that every submodule of a free R-module is a direct sum of ideals of R. This property is used in the proof of unique factorization of ideals and makes the structure theory of modules over a Dedekind quite nice. (Note, however, that every order in a number field which is not maximal is not hereditary.)

It is important to identify those orders for which projective modules abound. Let B be a simple finite-dimensional F-algebra and let $\mathcal {O}\subseteq B$ be an R-order.

Definition 21.1.1

We say $\mathcal {O}$ is left hereditary if every left $\mathcal {O}$-ideal $I \subseteq \mathcal {O}$ is projective as a left $\mathcal {O}$-module.

We could define also right hereditary, but left hereditary and right hereditary are equivalent for an R-order $\mathcal {O}$, and so we simply say hereditary. We have $\mathcal {O}$ hereditary if and only if every $\mathcal {O}$-submodule of a projective finitely generated $\mathcal {O}$-module is projective—that is to say, projectivity is inherited by submodules. Moreover, being hereditary is a local property.

Maximal orders are hereditary (Theorem 18.1.2), and one motivation for hereditary orders is that many of the results from chapter 18 on the structure of two-sided ideals extend from maximal orders to hereditary orders (Theorem 21.4.9).

Proposition 21.1.2

Suppose $\mathcal {O}$ is hereditary. Then the set of two-sided invertible fractional $\mathcal {O}$-ideals of B forms an abelian group under multiplication, generated by the prime $\mathcal {O}$-ideals.

Hereditary orders are an incredibly rich class of objects, and they may be characterized in a number of equivalent ways (Theorem 21.5.1). We restrict to the complete local case, and suppose now that R is a complete DVR with unique maximal ideal $\mathfrak p$ and residue field $k=R/\mathfrak p$.

Just as maximal orders are defined in terms of containment, we say $\mathcal {O}$ is extremal if whenever $\mathcal {O}' \supseteq \mathcal {O}$ and ${{\,\mathrm{rad}\,}}\mathcal {O}' \supseteq {{\,\mathrm{rad}\,}}\mathcal {O}$, then $\mathcal {O}'=\mathcal {O}$. If $\mathcal {O}$ is not extremal, then

$$\begin{aligned} \mathcal {O}' :=\mathcal {O}{}_{\textsf {\tiny {L}} }({{\,\mathrm{rad}\,}}\mathcal {O}) \supsetneq \mathcal {O}\end{aligned}$$

(21.1.3)

is a superorder. We then have the following main theorem (Theorem 21.5.1).

MainTheorem 21.1.4

Let R be a complete DVR and let $\mathcal {O}\subseteq B$ be an R-order in a simple F-algebra B. Let $J :={{\,\mathrm{rad}\,}}\mathcal {O}$. Then the following are equivalent:

(i):

$\mathcal {O}$ is hereditary;

(ii):

J is projective as a left $\mathcal {O}$-module;

(ii${}^\prime$):

J is projective as a right $\mathcal {O}$-module;

(iii):

$\mathcal {O}{}_{\textsf {\tiny {L}} }(J)=\mathcal {O}$;

(iii${}^\prime$):

$\mathcal {O}{}_{\textsf {\tiny {R}} }(J)=\mathcal {O}$;

(iv):

J is invertible as a (sated) two-sided $\mathcal {O}$-ideal; and

(v):

$\mathcal {O}$ is extremal.

The fact that hereditary orders are the same as extremal orders is quite remarkable, and gives tight control over the structure of hereditary orders: extremal orders are equivalently characterized as endomorphism algebras of flags in a suitable sense, and so we have the following important corollary for quaternion algebras.

Corollary 21.1.5

Suppose further that B is a quaternion algebra. Then an R-order $\mathcal {O}\subseteq B$ is hereditary if and only if either $\mathcal {O}$ is maximal or

$$\begin{aligned} \mathcal {O}\simeq \begin{pmatrix} R &{} R \\ \mathfrak p &{} R \end{pmatrix} \subseteq {{\,\mathrm{M}\,}}_2(F) \simeq B. \end{aligned}$$

It is no surprise that we meet again the order from Example 20.1.2! The reader who is willing to accept Corollary 21.1.5 can profitably move on from this chapter, as the ring of upper triangular matrices is explicit enough to work with in many cases. That being said, the methods we encounter here will be useful in framing investigations of orders beyond the hereditary ones.

2 Extremal orders

In this section, we will see how to extend an order to a superorder using the Jacobson radical, and we will characterize those orders that are extremal with respect to this process.

We work locally throughout this section; let R be a complete DVR with maximal ideal $\mathfrak p ={{\,\mathrm{rad}\,}}(R)$ and residue field $k=R/\mathfrak p$, and let $F={{\,\mathrm{Frac}\,}}R$. Let B be a finite-dimensional separable F-algebra and let $\mathcal {O}\subseteq B$ be an R-order.

21.2.1

Our motivation comes from the following: we canonically associate a superorder as follows. Let $J :={{\,\mathrm{rad}\,}}\mathcal {O}$ and $\mathcal {O}' :=\mathcal {O}{}_{\textsf {\tiny {L}} }(J)$. Then $\mathcal {O}' \supseteq \mathcal {O}$. By Corollary 20.5.5, $J^r \subseteq \mathfrak p \mathcal {O}\subseteq \mathfrak p \mathcal {O}'$ for some $r>0$, and then $J \subseteq {{\,\mathrm{rad}\,}}\mathcal {O}'$.

Definition 21.2.2

An R-order $\mathcal {O}' \subseteq B$radically covers $\mathcal {O}$ if $\mathcal {O}' \supseteq \mathcal {O}$ and ${{\,\mathrm{rad}\,}}\mathcal {O}' \supseteq {{\,\mathrm{rad}\,}}\mathcal {O}$. We say $\mathcal {O}$ is extremal if whenever $\mathcal {O}'$ radically covers $\mathcal {O}$ then $\mathcal {O}'=\mathcal {O}$.

We can think of extremal orders as like maximal orders, but under certain inclusions.

Proposition 21.2.3

An R-order $\mathcal {O}$ is extremal if and only if $\mathcal {O}{}_{\textsf {\tiny {L}} }({{\,\mathrm{rad}\,}}\mathcal {O})=\mathcal {O}$ if and only if $\mathcal {O}{}_{\textsf {\tiny {R}} }({{\,\mathrm{rad}\,}}\mathcal {O})=\mathcal {O}$.

Proof. The argument is due to Jacobinski [Jaci71, Proposition 1].

We first prove ($\Rightarrow$). Suppose $\mathcal {O}$ is extremal, and let $J={{\,\mathrm{rad}\,}}\mathcal {O}$ and $\mathcal {O}'=\mathcal {O}{}_{\textsf {\tiny {L}} }(J)$. By Corollary 20.5.5, J is topologically nilpotent as a $\mathcal {O}$-ideal, so the same is true as a $\mathcal {O}'$ ideal, and $J \subseteq {{\,\mathrm{rad}\,}}\mathcal {O}'$ and $\mathcal {O}'$ radically covers $\mathcal {O}$. Since $\mathcal {O}$ is extremal, we conclude $\mathcal {O}'=\mathcal {O}$. The same argument works on the right.

Next we prove ($\Leftarrow$). Let $J={{\,\mathrm{rad}\,}}\mathcal {O}$, suppose $\mathcal {O}=\mathcal {O}{}_{\textsf {\tiny {L}} }(J)$; let $\mathcal {O}'$ radically cover $\mathcal {O}$, and let $J'={{\,\mathrm{rad}\,}}\mathcal {O}'$. As lattices, we have $\mathfrak p ^s \mathcal {O}' \subseteq J$ for some $s>0$; by Theorem 20.5.1, $(J')^r \subseteq \mathfrak p \mathcal {O}'$ for some $r>0$, so putting these together we have $(J')^t \subseteq J$ for some $t>0$. Suppose $t>1$. Since $\mathcal {O}'$ radically covers, we have $J \subseteq J'$; thus $J(J')^{t-1} \subseteq (J')^t \subseteq J$ and $(J')^{t-1} \subseteq \mathcal {O}{}_{\textsf {\tiny {R}} }(J)=\mathcal {O}$. But then since $((J')^{t-1})^t \subseteq (J')^t \subseteq J$, by Corollary 20.5.5, $(J')^{t-1} \subseteq J$. Continuing in this way, we obtain $t=1$ and $J' \subseteq J$. Therefore $J=J'$ and $\mathcal {O}=\mathcal {O}{}_{\textsf {\tiny {L}} }(J)=\mathcal {O}{}_{\textsf {\tiny {L}} }(J')=\mathcal {O}'$, thus $\mathcal {O}$ is extremal. $\square$

Lemma 21.2.4

Let $\mathcal {O}$ be an R-order and let $\mathcal {O}'\subseteq B$ be an R-order containing $\mathcal {O}$. Let $J' :={{\,\mathrm{rad}\,}}\mathcal {O}'$. Then $\mathcal {O}+J'$ is an R-order that radically covers $\mathcal {O}$. If further $J' \subseteq \mathcal {O}$, then $J' \subseteq J$.

Proof. See Exercise 21.6. $\square$

21.2.5

In view of Lemma 21.2.4, an extremal order is determined by its homomorphic image in a nice k-algebra as follows.

Let $\mathcal {O}$ be an extremal R-order and let $\mathcal {O}'\subseteq B$ be a maximal R-order containing $\mathcal {O}$. Let $J' :={{\,\mathrm{rad}\,}}\mathcal {O}'$. By Lemma 21.2.4, $\mathcal {O}+J'$ is an R-order that radically covers $\mathcal {O}$, so $\mathcal {O}+J'=\mathcal {O}$. Therefore $J' \subseteq \mathcal {O}$. By the second part of Lemma 21.2.4, we immediately conclude $J' \subseteq J :={{\,\mathrm{rad}\,}}\mathcal {O}$. In sum,

$$\begin{aligned} J' = {{\,\mathrm{rad}\,}}\mathcal {O}' \subseteq J = {{\,\mathrm{rad}\,}}\mathcal {O}\subseteq \mathcal {O}. \end{aligned}$$

(21.2.6)

Consider now the reduction map $\rho :\mathcal {O}' \rightarrow \mathcal {O}'/J'$. Since $J' \subseteq \mathcal {O}$, if $A=\rho (\mathcal {O})$ then $\mathcal {O}=\rho ^{-1}(A)$. But since $\mathfrak p \mathcal {O}' \subseteq J'$ and $\mathcal {O}'$ is a maximal R-order, the codomain is a nice, finite dimensional k-algebra, something we will get our hands on in the next section.

Paragraph 21.2.5 has the following consequence.

Lemma 21.2.7

Suppose that B is a division algebra over F and let $\mathcal {O}\subseteq B$ be extremal. Then $\mathcal {O}$ is maximal.

Proof. Recall 13.3.7. The valuation ring $\mathcal {O}' \supseteq \mathcal {O}$ has the unique maximal two-sided ideal $J'={{\,\mathrm{rad}\,}}\mathcal {O}' = \mathcal {O}' \setminus (\mathcal {O}')^\times$, so $\mathcal {O}'/J'$ is a field. We have (21.2.6) $J' \subseteq J$, but then $J/J'=\{0\} \subseteq \mathcal {O}'/J'$ thus $J=J'$. Thus $\mathcal {O}'$ radically covers $\mathcal {O}$, and since $\mathcal {O}$ is extremal, $\mathcal {O}=\mathcal {O}'$. $\square$

Remark 21.2.8. We stop short in our explicit description of local extremal orders in section 21.2: we gave a construction in 21.3.1 only for $B \simeq {{\,\mathrm{M}\,}}_n(F)$. The results extend to $B \simeq {{\,\mathrm{M}\,}}_n(D)$ where D is a division algebra over F by considering lattices in a free left D-module: see Reiner [Rei2003, Theorem 39.14].

3 $*$ Explicit description of extremal orders

We now turn to an explicit description of extremal orders. In Lemma 10.5.4, we saw that maximal orders in a matrix algebra $B={{\,\mathrm{End}\,}}_F(V)$ are endomorphism algebras of lattices. In this section, we extend this to encompass orders that arise from endomorphism algebras of a chain of lattices: these orders are “block upper triangular”, and can be characterized in a number of ways.

21.3.1

Let V be a finite-dimensional F-vector space and let $B={{\,\mathrm{End}\,}}_F(V)$; then V is a simple B-module. Let $M \subseteq V$ be an R-lattice. By Lemma 10.5.4, $\Lambda :={{\,\mathrm{End}\,}}_R(M)$ is a maximal R-order; we have ${{\,\mathrm{rad}\,}}\Lambda = \mathfrak p \Lambda$.

Choosing a basis for M, we get $\Lambda \simeq {{\,\mathrm{M}\,}}_n(R) \subseteq {{\,\mathrm{M}\,}}_n(F) \simeq B$, and ${{\,\mathrm{rad}\,}}\Lambda = {{\,\mathrm{M}\,}}_n(\mathfrak p )$.

Now let $Z :=M \otimes _R k = M/\mathfrak p M$. Then Z is a finite-dimensional vector space over k. Let

$$\begin{aligned} \mathcal E :\{0\}=Z_0 \subsetneq Z_1 \subsetneq \dots \subsetneq Z_{s-1} \subsetneq Z_s = Z \end{aligned}$$

be a (partial) flag, a strictly increasing sequence of k-vector spaces. We define

$$\begin{aligned} \mathcal {O}{}_{\textsf {\tiny {L}} }(\mathcal E ) :=\{\alpha \in \Lambda : \alpha Z_i \subseteq Z_i : i=0,\dots ,s\}. \end{aligned}$$

Equivalently, let $M_i$ be the inverse image of $Z_i$ under the projection $M \rightarrow Z$; then we have a chain

$$\begin{aligned} \mathfrak p M = M_0 \subsetneq M_1 \subsetneq \dots \subsetneq M_{s-1} \subsetneq M_s = M \end{aligned}$$

(21.3.2)

and

$$\begin{aligned} \mathcal {O}{}_{\textsf {\tiny {L}} }(\mathcal E )=\{\alpha \in \Lambda : \alpha M_i \subseteq M_i : i=0,\dots ,s\}. \end{aligned}$$

Lemma 21.3.3

$\mathcal {O}{}_{\textsf {\tiny {L}} }(\mathcal E ) \subseteq \Lambda$ is an R-order with

$$\begin{aligned} {{\,\mathrm{rad}\,}}\mathcal {O}{}_{\textsf {\tiny {L}} }(\mathcal E ) = \{\alpha \in \Lambda : \alpha Z_i \subseteq Z_{i-1}\} = \{\alpha \in \Lambda : \alpha M_i \subseteq M_{i-1}\}. \end{aligned}$$

Proof. That $\mathcal {O}{}_{\textsf {\tiny {L}} }(\mathcal E )$ is an order follows in the same way as the proof of Lemma 10.2.7. For the statement on the radical: let $J=\{\alpha \in \Lambda : \alpha Z_i \subseteq Z_{i-1}\}$. Then $J \subseteq \mathcal {O}{}_{\textsf {\tiny {L}} }(\mathcal E )$ is a two-sided ideal. We have $J^s \subseteq \mathfrak p \Lambda$ pushing along the flag, so $J \subseteq {{\,\mathrm{rad}\,}}\mathcal {O}{}_{\textsf {\tiny {L}} }(\mathcal E )$ by Corollary 20.5.5. Conversely,

$$\begin{aligned} \mathcal {O}{}_{\textsf {\tiny {L}} }(\mathcal E )/J \simeq \bigoplus _{i=1}^{s} {{\,\mathrm{End}\,}}_k(Z_{i}/Z_{i-1}); \end{aligned}$$

each factor is simple, so the sum is (Jacobson) semisimple; therefore $J \subseteq {{\,\mathrm{rad}\,}}\mathcal {O}{}_{\textsf {\tiny {L}} }(\mathcal E )$ and equality holds. $\square$

Example 21.3.4

If we take the trivial flag $\mathcal {O}{}_{\textsf {\tiny {L}} }(\mathcal E ):\{0\}=Z_0 \subsetneq Z_1=Z$, then $\mathcal {O}{}_{\textsf {\tiny {L}} }(\mathcal E )=\Lambda$, so this recovers the construction of maximal orders.

Example 21.3.5

Let $\mathcal E$ be the complete  flag of length $s=n+1=\dim _F V$, where each quotient has $\dim _k (Z_{i+1}/Z_i)=1$. Then there exists a basis $z_1,\dots ,z_n$ of Z so that $Z_i$ has basis $z_1,\dots ,z_{n-i}$; We lift this to basis to $x_1,\dots ,x_n$ of M (by Nakayama’s lemma), and in this basis, we have

$$ \mathcal {O}{}_{\textsf {\tiny {L}} }(\mathcal E )=\begin{pmatrix} R &{} R &{} R &{} \dots &{} R \\ \mathfrak p &{} R &{} R &{} \dots &{} R \\ \mathfrak p &{} \mathfrak p &{} R &{} \dots &{} R \\ \vdots &{} \vdots &{} \vdots &{} \ddots &{} \vdots \\ \mathfrak p &{} \mathfrak p &{} \mathfrak p &{} \dots &{} R \end{pmatrix} $$

consisting of matrices which are upper triangular modulo $\mathfrak p$, and

$$\begin{aligned} {{\,\mathrm{rad}\,}}\mathcal {O}{}_{\textsf {\tiny {L}} }(\mathcal E )=\begin{pmatrix} \mathfrak p &{} R &{} R &{} \dots &{} R \\ \mathfrak p &{} \mathfrak p &{} R &{} \dots &{} R \\ \mathfrak p &{} \mathfrak p &{} \mathfrak p &{} \dots &{} R \\ \vdots &{} \vdots &{} \vdots &{} \ddots &{} \vdots \\ \mathfrak p &{} \mathfrak p &{} \mathfrak p &{} \dots &{} \mathfrak p \end{pmatrix} =\mathcal {O}{}_{\textsf {\tiny {L}} }(\mathcal E )\begin{pmatrix} 0 &{} 1 &{} \dots &{} 0 \\ \vdots &{} \vdots &{} \ddots &{} \vdots \\ 0 &{} 0 &{} \dots &{} 1 \\ \pi &{} 0 &{} \dots &{} 0 \end{pmatrix} \end{aligned}$$

(21.3.6)

where the latter is taken to be a block matrix with lower left entry $\pi$ and top right entry the $(n-1) \times (n-1)$ identity matrix.

Other choices of flag give an order which lie between $\mathcal {O}{}_{\textsf {\tiny {L}} }(\mathcal E )$ and $\Lambda$: we might think of them as being block upper triangular orders.

Now for the punch line of this section.

Proposition 21.3.7

Let $\mathcal {O}\subseteq B$ be an R-order. Then $\mathcal {O}$ is extremal if and only if $\mathcal {O}=\mathcal {O}{}_{\textsf {\tiny {L}} }(\mathcal E )$ for a flag $\mathcal E$.

Proof. Let $\mathcal {O}=\mathcal {O}{}_{\textsf {\tiny {L}} }(\mathcal E )$. Let $J={{\,\mathrm{rad}\,}}\mathcal {O}$; we seek to apply Proposition 21.2.3, so we show that $\mathcal {O}=\mathcal {O}{}_{\textsf {\tiny {L}} }(J)$. By Lemma 21.3.3, we have $J M_i=M_{i-1}$ so $\mathcal {O}{}_{\textsf {\tiny {L}} }(J) M_{i-1}=\mathcal {O}{}_{\textsf {\tiny {L}} }(J) J M_i=J M_i = M_{i-1}$ for $i=1,\dots ,s$. Since $M_0= \mathfrak p M \simeq M$, we conclude $\mathcal {O}{}_{\textsf {\tiny {L}} }(J)=\mathcal {O}{}_{\textsf {\tiny {L}} }(\mathcal E )=\mathcal {O}$ by definition.

Conversely, suppose $\mathcal {O}$ is extremal with $J={{\,\mathrm{rad}\,}}\mathcal {O}$. Let s be minimal so that $J^s=\mathfrak p \mathcal {O}$. We may embed $\mathcal {O}\subseteq \Lambda$ for some $\Lambda$, and we take the flag

$$\begin{aligned} \mathcal E :\{0\}=J^s Z \subsetneq J^{s-1} Z \subsetneq \dots \subsetneq JZ \subsetneq Z. \end{aligned}$$

Then $\mathcal {O}\subseteq \mathcal {O}(\mathcal E )$ and ${{\,\mathrm{rad}\,}}\mathcal {O}(\mathcal E ) \supseteq J$ by construction, so since $\mathcal {O}$ is extremal, we have $\mathcal {O}=\mathcal {O}(\mathcal E )$. $\square$

4 Hereditary orders

We now link the orders in the previous two sections to another important type of order. The theory of extremal and hereditary orders was developed by Brumer [Brum63a, Brum63b], Drozd–Kirichenko [DK68], Harada [Har63a,Har63b,Har63c], Jacobinski [Jaci71], and Hijikata–Nishida [HN94]. An overview of the local and global theory of hereditary orders is given by Reiner [Rei2003, §§2f, 39–40], and Drozd–Kirichenko–Roiter [DKR67] and Hijikata–Nishida [HN94] extend some results from hereditary orders to Bass orders.

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

Definition 21.4.1

We say $\mathcal {O}$ is left hereditary if every left $\mathcal {O}$-ideal $I \subseteq \mathcal {O}$ is projective as a left $\mathcal {O}$-module.

21.4.2

We could similarly define right hereditary, but since an order $\mathcal {O}$ is left and right noetherian, it follows that $\mathcal {O}$ is left hereditary if and only if $\mathcal {O}$ is right hereditary: see Exercise 21.8. When B is a quaternion algebra, the standard involution interchanges and left and right, so the two notions are immediately seen to be equivalent. Accordingly, we say hereditary for either sided notion.

Example 21.4.3

In the generic case $F=R$ and $\mathcal {O}=B$, we note that every semisimple algebra B over a field F is hereditary: by Lemma 7.3.5, every B-module is semisimple hence the direct sum of simple B-modules equivalently maximal left ideals, by Lemma 7.2.7.

21.4.4

By 20.2.6, being hereditary is a local property.

The following lemma motivates the name hereditary: projectivity is inherited by submodules. (Note that since R is noetherian, a finitely generated $\mathcal {O}$-module is noetherian, so every submodule is finitely generated.)

Lemma 21.4.5

Let $\mathcal {O}$ be hereditary, and let P be a finitely generated projective left $\mathcal {O}$-module. Then every submodule $M \subseteq P$ is isomorphic as a left $\mathcal {O}$-module to a finite direct sum of finitely generated left $\mathcal {O}$-ideals; in particular, M is projective.

Proof. We may suppose without loss of generality that $P \simeq \mathcal {O}^r$. We proceed by induction on r; the case $r=1$ holds by definition. Decompose $\mathcal {O}^r = E \oplus \mathcal {O}$ where $E \simeq \mathcal {O}^{r-1}$. From the exact sequence

$$\begin{aligned} 0 \rightarrow \ker \phi \rightarrow M \rightarrow M \cap E \rightarrow 0 \end{aligned}$$

and projectivity, we find that $M \simeq (M \cap E) \oplus \ker \phi$ where $\ker \phi \subseteq \mathcal {O}$ is a left ideal of $\mathcal {O}$. By induction, $M \cap E$ is projective, so the same is true of M. $\square$

Corollary 21.4.6

$\mathcal {O}$ is hereditary if and only if every submodule of a projective $\mathcal {O}$-module is projective.

Proof. The implication $(\Rightarrow )$ is Lemma 21.4.5; for the implication $(\Leftarrow )$, $\mathcal {O}$ is projective (free!) as a left $\mathcal {O}$-module and every left ideal is a $\mathcal {O}$-submodule $I \subseteq \mathcal {O}$, so by hypothesis I is projective. $\square$

Remark 21.4.7. $\mathcal {O}$ is hereditary if and only if every R-lattice $I \subseteq B$ with $\mathcal {O}\subseteq \mathcal {O}{}_{\textsf {\tiny {L}} }(I)$ is projective as a left $\mathcal {O}$-module, after rescaling.

However, a bit of a warning is due. If $I \subseteq B$ is an R-lattice that is projective as a left $\mathcal {O}$-module, then we have shown that I is projective as a left $\mathcal {O}{}_{\textsf {\tiny {L}} }(I)$-module (Lemma 20.3.1), whence left invertible (Theorem 20.3.3) as a lattice. But the converse need not be true; so it is important that in the definition of hereditary we do not require that every left $\mathcal {O}$-fractional ideal I is invertible as a left fractional $\mathcal {O}$-ideal (Definition 16.5.17): the latter carries the extra assumption that I is sated. See also Remark 20.3.8.

Lemma 21.4.8

Let $\mathcal {O}\subseteq B$ be a hereditary R-order and let $\mathcal {O}' \supseteq \mathcal {O}$ be an R-superorder. Then $\mathcal {O}'$ is hereditary.

Proof. Let $I' \subseteq \mathcal {O}'$ be a left $\mathcal {O}'$-ideal. Scaling we may take $I' \subseteq \mathcal {O}$, and it is a left $\mathcal {O}$-ideal. Since $\mathcal {O}$ is hereditary, $I'$ is projective as a left $\mathcal {O}$-module; by Lemma 20.3.1, $I'$ is projective as a left $\mathcal {O}'$-module. $\square$

One of the desirable aspects of hereditary orders is that many of the results from chapter 18 on the structure of two-sided ideals extend from maximal orders to hereditary orders. Indeed, section 18.2 made no maximality hypothesis (we held out as long as we could!).

Theorem 21.4.9

Let R be a Dedekind domain and let $\mathcal {O}$ be a hereditary R-order in a simple F-algebra B. Then the set of two-sided invertible fractional $\mathcal {O}$-ideals of B forms an abelian group under multiplication, generated by the invertible prime $\mathcal {O}$-ideals.

Proof. Proven in the same manner as in Theorem 18.3.4; a self-contained proof is requested in Exercise 21.4. $\square$

Remark 21.4.10. Theorem 21.4.9 is proven by Vignéras [Vig80a, Théorème I.4.5], but there is a glitch in the proof. Let R be a Dedekind domain, let B be a quaternion algebra over $F={{\,\mathrm{Frac}\,}}R$, and let $\mathcal {O}\subseteq B$ be an R-order. Vignéras claims that the two-sided ideals of $\mathcal {O}$ form a group that is freely generated by the prime ideals, and the proof uses that if I is a two-sided ideal then I is invertible. This is false for a general order $\mathcal {O}$ (see Example 16.5.12).

If one restricts to the group of invertible two-sided ideals, the logic of the proof is still flawed. The proof does not use anything about quaternion algebras, and works verbatim for the case where $R=\mathbb Z \subseteq F=\mathbb Q$ and B is replaced by $K=\mathbb Q (\sqrt{d_K})$ and $\mathcal {O}$ is replaced by an order of discriminant $d=d_Kf^2$ that is not maximal, of conductor $f \in \mathbb Z _{>1}$, as in section 16.1. Then the ideal $\mathfrak f =f\mathbb Z +\sqrt{d}\mathbb Z$ is not invertible, but $\mathfrak f \supsetneq (f)$ and (f) is invertible but not maximal, so the group of invertible ideals is not generated by primes.

However, if one supposes that every two-sided ideal is invertible (as a lattice), then the argument can proceed: this is the class of hereditary orders, and is treated in Theorem 21.4.9.

Remark 21.4.11. The module theory for hereditary noetherian prime rings, generalizing hereditary orders, has been worked out by Levy–Robson [LR2011].

5 $*$ Classification of local hereditary orders

We now come to the main theorem of this chapter, relating extremal orders, hereditary orders, their modules and composition series in the local setting.

Theorem 21.5.1

Let R be a complete DVR and let $F :={{\,\mathrm{Frac}\,}}R$. Let B be a finite-dimensional F-algebra, and let $\mathcal {O}\subseteq B$ be an R-order. Let $J :={{\,\mathrm{rad}\,}}\mathcal {O}$. Then the following are equivalent, along with the conditions ${}^\prime$ where ‘left’ is replaced by ‘right’:

• (i) $\mathcal {O}$ is extremal;

• (ii) Every projective indecomposable left $\mathcal {O}$-submodule $P \subseteq B$ is the minimum $\mathcal {O}$-supermodule of JP;

• (iii) Every projective indecomposable left $\mathcal {O}$-module P has a unique composition series;

• (iv) Every projective indecomposable left $\mathcal {O}$-module P has a unique composition series consisting of projectives;

• (v) $\mathcal {O}$ is hereditary;

• (vi) J is projective as a left $\mathcal {O}$-module;

• (vii) If P is a projective indecomposable left $\mathcal {O}$-module, then JP is also projective indecomposable; and

• (viii) J is invertible as a (sated) two-sided $\mathcal {O}$-ideal.

Proof. See Hijikata–Nishida [HN94, §1]. $\square$

Corollary 21.5.2

A maximal order is hereditary.

Proof. We proved this in Theorem 18.1.2, but here is another proof using Theorem 21.5.1: the property of being maximal is local, and a maximal order is extremal. $\square$

To conclude, we classify the lattices of a local hereditary order.

21.5.3

Suppose R is a complete DVR and that $B \simeq {{\,\mathrm{M}\,}}_n(F)$. Suppose $\mathcal {O}\subseteq B$ is a hereditary R-order; then by Theorem 21.5.1, $\mathcal {O}=\mathcal {O}{}_{\textsf {\tiny {L}} }(\mathcal E )$ is extremal, arising from a chain 21.3.2 which by Lemma 21.3.3 is of the form

$$\begin{aligned} \mathfrak p M = M_0 = J^s M \subsetneq J^{s-1} M \subsetneq \dots \subsetneq M_{s-1} = JM \subsetneq M_s = M, \end{aligned}$$

with each quotient $M_i/M_{i+1} \simeq M/JM$ simple.

We claim that the set $M,JM,\dots ,J^{s-1} M$ form a complete set of isomorphism classes of indecomposable left $\mathcal {O}$-modules. Indeed, these modules are all mutually nonisomorphic, because an isomorphism of left $\mathcal {O}$-modules extends to an isomorphism $\phi \in {{\,\mathrm{End}\,}}_B(B) \simeq F$ so is given by (right) multiplication by a power of $\pi$, impossible unless $i \equiv j ~(\text{ mod } ~{s})$. And if N is an indecomposable left $\mathcal {O}$-module, then $FN \simeq F^n$ is ‘the’ simple B-module, so N is isomorphic to a lattice in V. Since J is invertible, we may replace N by $J^r N$ with $r \in \mathbb Z$, to suppose that $M \supseteq N \supsetneq JM$. But $M/JM \simeq \mathcal {O}/J$ is simple as a left $\mathcal {O}$-module, so $N=M$. (See also Reiner [Rei2003, Theorem 39.23].)

Exercises

1.:

Show that a Dedekind domain is hereditary (cf. Exercise 9.5).

2.:

Let $R=\mathbb Z$, let $B=K=\mathbb Q (\sqrt{d})$ with d the discriminant of K, and let $S \subseteq \mathbb Z _K$ be an order. Show that S is hereditary if and only if S is maximal.

3.:

Let R be a DVR with maximal ideal $\mathfrak p =\pi R$ and $F={{\,\mathrm{Frac}\,}}R$ with ${{\,\mathrm{char}\,}}F \ne 2$. Let $B=\displaystyle {\biggl (\frac{1,\pi }{F}\biggr )}$ and $\mathcal {O}=R\langle i,j \rangle$ the standard order. Show directly that ${{\,\mathrm{rad}\,}}\mathcal {O}=\mathcal {O}j = j \mathcal {O}$, and conclude that $\mathcal {O}$ is hereditary (but not a maximal order).

$\triangleright$ 4.:

Give a self-contained proof of Theorem 21.4.9 following Theorem 18.3.4. (Where does the issue with invertibility arise?)

5.:

Let R be a complete DVR and let $\mathcal {O}$ be a hereditary R-order. Show that $\mathcal {O}$ is hereditary if and only if ${{\,\mathrm{rad}\,}}\mathcal {O}$ is an invertible (sated) two-sided $\mathcal {O}$-ideal.

$\triangleright$ 6.:

In this exercise, we prove Lemma 21.2.4 following Reiner [Rei2003, Exercise 39.2]. We adopt the notation from that section, so in particular R be a complete DVR with maximal ideal $\mathfrak p ={{\,\mathrm{rad}\,}}(R)$. Let $\mathcal {O}$ be an R-order and let $\mathcal {O}'\subseteq B$ be an R-order containing $\mathcal {O}$. Let $J'={{\,\mathrm{rad}\,}}\mathcal {O}'$.

(a):

Show that $\mathcal {O}+J'$ is an R-order.

(b):

Show that $\mathcal {O}+J'$ radically covers $\mathcal {O}$. [Hint: let $J={{\,\mathrm{rad}\,}}\mathcal {O}$, and claim that $J+J' \subseteq {{\,\mathrm{rad}\,}}(\mathcal {O}+J')$. For r large, show $J^r \subseteq \mathfrak p \mathcal {O}$ so $(J+J')^r \subseteq \mathfrak p \mathcal {O}' + J'$ and $(J')^r \subseteq \mathfrak p \mathcal {O}'$, and then making r even larger show $(J+J')^{r^3} \subseteq \mathfrak p (\mathcal {O}+ J')$. Conclude using Corollary 20.5.5.]

(c):

If further $J' \subseteq \mathcal {O}$, show that $J' \subseteq J$.

7.:

Let R be a Dedekind domain with $F={{\,\mathrm{Frac}\,}}(R)$, let B be finite-dimensional F-algebra, and let $\mathcal {O}\subseteq B$ be a hereditary order. Let P be a finitely generated projective $\mathcal {O}$-module. Show that P is indecomposable if and only if $V :=P \otimes _R F$ is simple as a B-module.

$\triangleright$ 8.:

Let R be a Dedekind domain, and let $\mathcal {O}\subseteq B$ be an R-order in a finite-dimensional F-algebra. Show that $\mathcal {O}$ is left hereditary (every left $\mathcal {O}$-ideal is projective) if and only if it is right hereditary (every right $\mathcal {O}$-ideal is projective). [See Reiner [Rei2003, Theorem 40.1].]

9.:

Consider the ring

$$\begin{aligned} A :=\left\{ \begin{pmatrix} a &{} 0 \\ b &{} c \end{pmatrix} : a \in \mathbb Z , b,c \in \mathbb Q \right\} . \end{aligned}$$

Show that every submodule of a projective left A-module is projective, but the same is not true on the right.

10.:

Let R be a Dedekind domain. Let B be a separable F-algebra, and let $B \simeq B_1 \times \dots \times B_r$ be its decomposition into simple components, with $B_i=Be_i$ for central idempotents $e_i$. Let $K_i$ be the center of $B_i$, and let $S_i$ be the integral closure of R in $K_i$.

(a):

Let $\mathcal {O}\subseteq B$ be a hereditary R-order. Show that $\mathcal {O}\simeq \mathcal {O}_1 \times \dots \times \mathcal {O}_r$ where $\mathcal {O}_i = \mathcal {O}e_i$, and each $\mathcal {O}_i$ is a hereditary R-order in $B_i$.

(b):

Conversely, if $\mathcal {O}_i \subseteq B_i$ is a hereditary R-order, then $\mathcal {O}_1 \times \dots \times \mathcal {O}_r$ is a hereditary R-order in B.

[Hint: use the fact that hereditary orders are extremal.]

11.:

For the following exercise, we consider integral group rings. Let G be a finite group of order $n=\#G$ and let R be a Dedekind domain with $F={{\,\mathrm{Frac}\,}}R$. Suppose that ${{\,\mathrm{char}\,}}F \not \mid n$. Then $B :=F[G]$ is a separable F-algebra by Exercise 7.16. Let $\mathcal {O}=R[G]$.

(a):

Let $\mathcal {O}' \supseteq \mathcal {O}$ be an R-superorder of $\mathcal {O}$ in B. Show that

$$\begin{aligned} \mathcal {O}\subseteq \mathcal {O}' \subseteq n^{-1} \mathcal {O}. \end{aligned}$$

[Hint: for ell $\alpha =\sum _g a_g g \in \mathcal {O}'$ with $a_g \in F$, show that

$$\begin{aligned} {{\,\mathrm{Tr}\,}}_{B|F}(\alpha g) = na_g \in R. \end{aligned}$$

Conclude that $\mathcal {O}' \subseteq n^{-1} \mathcal {O}$.]

(b):

Show that $\mathcal {O}$ is maximal if and only if $\mathcal {O}$ is hereditary if and only if $n \in R^\times$. [Hint: if $\mathcal {O}$ is hereditary, then $\mathcal {O}$ contains the central idempotent $n^{-1} \sum _{g \in G} g$ by Exercise 21.10.]

(c):

We define theleft conductor of $\mathcal {O}'$ into $\mathcal {O}$ to be the colon ideal

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

(and similarly on right). Prove that

$$\begin{aligned} (\mathcal {O}':\mathcal {O}){}_{\textsf {\tiny {L}} }= \sum _{i=1}^t \frac{n}{n_i}{{\,\mathrm{codiff}\,}}(\mathcal {O}_i'). \end{aligned}$$

12.:

Give an explicit description like Example 21.3.5 for $\mathcal {O}{}_{\textsf {\tiny {L}} }(\mathcal E )$ when $\dim _F V=3,4$.

13.:

Let R be a Dedekind domain, and let $\mathcal {O}\subseteq B$ be an R-order in a finite-dimensional simple F-algebra. Show that $\mathcal {O}$ is maximal if and only if $\mathcal {O}$ is hereditary and ${{\,\mathrm{rad}\,}}\mathcal {O}\subseteq \mathcal {O}$ is a maximal two-sided ideal.