Two-sided ideals and the Picard group

Abstract

In this chapter, we treat maximal orders like noncommutative Dedekind domains, and we consider the structure of two-sided ideals (and their classes), in a manner parallel to the commutative case.

In this chapter, we treat maximal orders like noncommutative Dedekind domains, and we consider the structure of two-sided ideals (and their classes), in a manner parallel to the commutative case.

1 $\triangleright$ Noncommutative Dedekind domains

Let R be a Dedekind domain with field of fractions F: then by definition R is noetherian, integrally closed, and all nonzero prime ideals of R are maximal. Equivalently, every ideal of R is the product of prime ideals (uniquely up to permutation). To establish this latter property of unique factorization of ideals, there are two essential ingredients: first, every proper ideal contains a finite product of prime ideals, and second, every nonzero prime ideal $\mathfrak p \subseteq R$ is invertible. The first of these uses that R is noetherian and that nonzero prime ideals of R are maximal; the second uses that R is integrally closed.

Here, the theorems are no easier to prove in the case of a quaternion algebra, so we might as well consider them in more generality. Let B be a simple F-algebra and let $\mathcal {O}\subseteq B$ be an R-order.

To draw the closest analogy with Dedekind domains, we suppose that $\mathcal {O}\subset B$ is maximal: this is the noncommutative replacement for integrally closed. Since $\mathcal {O}$ is finitely generated, if $I \subseteq \mathcal {O}$ is a two-sided $\mathcal {O}$-ideal, then I is a finitely generated R-submodule, so the noetherian condition on R automatically implies that every chain of ideals of $\mathcal {O}$ stabilizes. We say a two-sided ideal $P \subseteq \mathcal {O}$ is prime if $P \ne \mathcal {O}$ and for all two-sided ideals $I,J \subseteq \mathcal {O}$, we have

$$\begin{aligned} IJ \subseteq P \quad \Rightarrow \quad I \subseteq P \text { or } J \subseteq P. \end{aligned}$$

Running parallel to the above, we have the following initial lemma.

Lemma 18.1.1

A nonzero two-sided $\mathcal {O}$-ideal is prime if and only it is maximal, and every two-sided $\mathcal {O}$-ideal contains a product of prime two-sided $\mathcal {O}$-ideals.

Completing the analogy with the commutative case, we then have the following theorem.

Theorem 18.1.2

Let R be a Dedekind domain with field of fractions $F={{\,\mathrm{Frac}\,}}R$, let B be a simple F-algebra and let $\mathcal {O}\subseteq B$ be a maximal R-order. Then the following statements hold.

• (a) If $I \subseteq B$ is an R-lattice such that $\mathcal {O}{}_{\textsf {\tiny {L}} }(I)=\mathcal {O}$ or $\mathcal {O}{}_{\textsf {\tiny {R}} }(I)=\mathcal {O}$, then I is invertible and both $\mathcal {O}{}_{\textsf {\tiny {L}} }(I)$ and $\mathcal {O}{}_{\textsf {\tiny {R}} }(I)$ are maximal R-orders.

• (b) Multiplication of two-sided $\mathcal {O}$-ideals is commutative, and every nonzero two-sided $\mathcal {O}$-ideal is the product of finitely many prime two-sided $\mathcal {O}$-ideals, uniquely up to permutation.

Let ${{\,\mathrm{Idl}\,}}(\mathcal {O})$ be the group of invertible two-sided fractional $\mathcal {O}$-ideals. Put another way, Theorem 18.1.2 says that if $\mathcal {O}$ is maximal, then ${{\,\mathrm{Idl}\,}}(\mathcal {O})$ is isomorphic to the free abelian group on the set of nonzero prime two-sided $\mathcal {O}$-ideals under multiplication.

We now consider classes of two-sided ideals, in the spirit of section 17.1. Two candidates present themselves. On the one hand, inside the group ${{\,\mathrm{Idl}\,}}(\mathcal {O})$ of invertible fractional two-sided $\mathcal {O}$-ideals, the principal fractional two-sided $\mathcal {O}$-ideals (those of the form $\mathcal {O}\alpha \mathcal {O}=\mathcal {O}\alpha =\alpha \mathcal {O}$ for certain $\alpha \in B^\times$) form a subgroup ${{\,\mathrm{PIdl}\,}}(\mathcal {O})$, and we could consider the quotient. On the other hand, for a commutative ring S, the Picard group ${{\,\mathrm{Pic}\,}}(S)$ is defined to be the group of isomorphism classes of rank one projective (equivalently, invertible) S-modules under the tensor product. When S is a Dedekind domain, there is a canonical isomorphism ${{\,\mathrm{Cl}\,}}S \cong {{\,\mathrm{Pic}\,}}(S)$.

For simplicity, suppose now that $R=\mathbb Z$. In this noncommutative setting, we analogously define ${{\,\mathrm{Pic}\,}}\mathcal {O}$ to be the group of isomorphism classes of invertible $\mathcal {O}$-bimodules (over $\mathbb Z$) under tensor product. If $I,J \in {{\,\mathrm{Idl}\,}}(\mathcal {O})$, then I, J are isomorphic as $\mathcal {O}$-bimodules if and only if $J=aI$ with $a \in \mathbb Q ^\times$, and this yields an isomorphism

$$\begin{aligned} {{\,\mathrm{Pic}\,}}\mathcal {O}\simeq {{\,\mathrm{Idl}\,}}(\mathcal {O})/\mathbb Q ^\times . \end{aligned}$$

Let

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

be the normalizer of $\mathcal {O}$ in B. By the Skolem–Noether theorem,

$$\begin{aligned} N_{B^\times }(\mathcal {O})/\mathbb Q ^\times \simeq {{\,\mathrm{Aut}\,}}(\mathcal {O}) \end{aligned}$$

is the group of $\mathbb Z$-algebra (or ring) automorphisms of $\mathcal {O}$.

Theorem 18.1.3

Let B be a quaternion algebra over $\mathbb Q$ of discriminant $D :={{\,\mathrm{disc}\,}}B$, and let $\mathcal {O}\subset B$ be a maximal order. Then

$$\begin{aligned} {{\,\mathrm{Pic}\,}}\mathcal {O}\simeq \prod _{p \mid D} \mathbb Z /2\mathbb Z \end{aligned}$$

generated by (unique) prime two-sided $\mathcal {O}$-ideals with reduced norm $p \mid D$, and there is an exact sequence

$$\begin{aligned} \begin{aligned} 0 \rightarrow N_{B^\times }(\mathcal {O})/(\mathbb Q ^\times \mathcal {O}^\times )&\rightarrow {{\,\mathrm{Pic}\,}}\mathcal {O}\rightarrow {{\,\mathrm{Idl}\,}}(\mathcal {O})/{{\,\mathrm{PIdl}\,}}(\mathcal {O}) \rightarrow 0 \\ \alpha (\mathbb Q ^\times \mathcal {O}^\times )&\mapsto [\mathcal {O}\alpha \mathcal {O}]. \end{aligned} \end{aligned}$$

(18.1.4)

In particular, ${{\,\mathrm{Pic}\,}}\mathcal {O}$ is a finite abelian 2-group. As an application of Theorem 18.1.3, we revisit the map (17.1.3):

$$\begin{aligned} {{\,\mathrm{Cls}\,}}\mathcal {O}&\rightarrow {{\,\mathrm{Typ}\,}}\mathcal {O}\\ [I]{}_{\textsf {\tiny {R}} }&\mapsto class of \mathcal {O}{}_{\textsf {\tiny {L}} }(I) \end{aligned}$$

We recall that this map is surjective. The fibers are given by Theorem 18.1.3 (see Proposition 18.5.10): the fiber above the isomorphism class of $\mathcal {O}'$ is in bijection with the set ${{\,\mathrm{PIdl}\,}}(\mathcal {O}') \backslash {{\,\mathrm{Idl}\,}}(\mathcal {O}')$.

Remark 18.1.5. The structure of ${{\,\mathrm{Pic}\,}}\mathcal {O}$ is more complicated when $\mathcal {O}$ is not necessarily a maximal order: in general, the group ${{\,\mathrm{Pic}\,}}\mathcal {O}$ is finite but it may be nonabelian (see Exercise 18.6); worse still, in general the subgroup ${{\,\mathrm{PIdl}\,}}(\mathcal {O})$ may not be a normal subgroup in ${{\,\mathrm{Idl}\,}}(\mathcal {O})$.

2 Prime ideals

Throughout this chapter, let R be a Dedekind domain with field of fractions $F={{\,\mathrm{Frac}\,}}R$, let B be a simple finite-dimensional F-algebra, and let $\mathcal {O}\subseteq B$ be an R-order.

18.2.1

Let $I \subseteq \mathcal {O}$ be a nonzero two-sided ideal. In view of Remark 16.2.10, we see that I is automatically an R-lattice: $IF \subseteq B$ is a two-sided ideal of B, so since B is simple and $I \ne \{0\}$ we must have $IF=B$.

Definition 18.2.2

A two-sided ideal $P \subseteq \mathcal {O}$ is prime if $P \ne \mathcal {O}$ and for all two-sided ideals $I,J \subseteq \mathcal {O}$ we have

$$\begin{aligned} IJ \subseteq P \quad \Rightarrow \quad I \subseteq P \text { or } J \subseteq P. \end{aligned}$$

A two-sided $\mathcal {O}$-ideal $M \subseteq \mathcal {O}$ is maximal if $M \ne \mathcal {O}$ and M is not properly contained in another two-sided ideal.

Example 18.2.3

The zero ideal $P=\{0\}$ is prime: see Exercise 18.2.

18.2.4

Let $P \subseteq \mathcal {O}$ be a two-sided ideal. Then the two-sided $\mathcal {O}/P$-ideals are in bijection with the two-sided $\mathcal {O}$-ideals containing P. If $P \ne \mathcal {O}$, then P is prime if and only if for all two sided $\mathcal {O}/P$-ideals I/P, J/P, we have

$$\begin{aligned} (I/P)(J/P)=\{0\} \quad \Rightarrow \quad I/P=\{0\}\text { or }J/P=\{0\}. \end{aligned}$$

(18.2.5)

Lemma 18.2.6

If M is a maximal two-sided $\mathcal {O}$-ideal, then M is prime.

Proof. Suppose $IJ \subseteq M$. Then $(I+M)(J+M) \subseteq M$. But $I+M \supseteq M$ so either $I+M=M$ or $I+M=\mathcal {O}$ by maximality, and the same is true for J. Since $M \ne \mathcal {O}$ we must have either $I+M=M$ or $J+M=M$, which is to say $I \subseteq M$ or $J \subseteq M$.$\square$

Proposition 18.2.7

• (a) A nonzero two-sided $\mathcal {O}$-ideal is prime if and only it is maximal.

• (b) If $P \subseteq \mathcal {O}$ is a nonzero prime two-sided $\mathcal {O}$-ideal, then $\mathfrak p =P \cap R$ is a nonzero prime ideal of R, and $\mathcal {O}/P$ is a finite-dimensional simple algebra over the field $R/\mathfrak p$.

Proof. We follow Reiner [Rei2003, Theorem 22.3]. The implication ($\Rightarrow$) in (a) follows from Lemma 18.2.6. Conversely, let P be a nonzero prime two-sided $\mathcal {O}$-ideal, and let $\mathfrak p =P \cap R$. We show $\mathfrak p$ is a nonzero prime. By 18.2.1, P is an R-lattice, so $\mathfrak p \ne \{0\}$; since $1 \not \in P$, we have $\mathfrak p \ne R$ and $\mathfrak p$ is nontrivial. If $a,b \in R$, then $ab \in \mathfrak p$ implies $(a\mathcal {O})(b\mathcal {O}) \subseteq P$; since P is prime, we have $a\mathcal {O}\subseteq P$ or $b\mathcal {O}\subseteq P$, so $a \in \mathfrak p$ or $b \in \mathfrak p$.

Now let $J/P={{\,\mathrm{rad}\,}}(\mathcal {O}/P)$ be the Jacobson radical of $\mathcal {O}/P$ (see section 7.4). By Lemma 7.4.8, the ideal J/P is nilpotent; by (18.2.5), we conclude $J/P=\{0\}$. Thus $\mathcal {O}/P$ is semisimple by Lemma 7.4.2 and thus is a product of simple $R/\mathfrak p$-algebras by the Wedderburn–Artin theorem (Main Theorem 7.3.10). But the simple components of $\mathcal {O}/P$ are two-sided ideals that annihilate one another; again by (18.2.5), there can be only one component, and $\mathcal {O}/P$ is simple. Thus $\mathcal {O}/P$ has no nontrivial ideals, and P is maximal.$\square$

Lemma 18.2.8

Every nonzero two-sided ideal of $\mathcal {O}$ contains a (finite) product of nonzero prime ideals.

Proof. If not, then the set of ideals which do not contain such products is nonempty; since $\mathcal {O}$ is noetherian, there is a maximal element M. Since M cannot itself be prime, there exist ideals I, J, properly containing M, such that $IJ \subseteq M$. But both I, J contain products of prime ideals, so the same is true of M, a contradiction.$\square$

We now turn to notions of invertibility.

18.2.9

Let I be an invertible two-sided fractional $\mathcal {O}$-ideal (cf. Definition 16.2.9 and 16.5.17). In particular, $\mathcal {O}{}_{\textsf {\tiny {L}} }(I)=\mathcal {O}{}_{\textsf {\tiny {R}} }(I)=\mathcal {O}$. If J is another invertible two-sided fractional $\mathcal {O}$-ideal, then so is IJ, by Lemma 16.5.11: we have $\mathcal {O}{}_{\textsf {\tiny {L}} }(IJ)=\mathcal {O}{}_{\textsf {\tiny {L}} }(I)=\mathcal {O}$ and $\mathcal {O}{}_{\textsf {\tiny {R}} }(IJ)=\mathcal {O}{}_{\textsf {\tiny {R}} }(J)=\mathcal {O}$. Let ${{\,\mathrm{Idl}\,}}(\mathcal {O})$ be the set of invertible two-sided fractional $\mathcal {O}$-ideals. Then ${{\,\mathrm{Idl}\,}}(\mathcal {O})$ is a group under multiplication with identity element $\mathcal {O}$.

The structure of ${{\,\mathrm{Idl}\,}}(\mathcal {O})$, and quotients under natural equivalence relations, is the subject of this chapter.

3 Invertibility

We now consider invertibility first in the general context of orders, then for maximal orders. The general theory of maximal orders over Dedekind domains in simple algebras was laid out by Auslander–Goldman [AG60]. One of the highlights of this theory are the classification of such orders: they are endomorphism rings of a finitely generated projective module over a maximal order in a division algebra. For a quite general treatment of maximal orders, see the book by Reiner [Rei2003]; in particular, the ideal theory presented here is also discussed in Reiner [Rei2003, §§22–23].

Lemma 18.3.1

Let J be a two-sided $\mathcal {O}$-ideal, not necessarily invertible. If $J \subsetneq \mathcal {O}$, then $J^{-1} \supsetneq \mathcal {O}$.

Proof. The R-lattice $J^{-1}$ has $J^{-1} \supseteq \mathcal {O}$ and $\mathcal {O}{}_{\textsf {\tiny {L}} }(J^{-1}) \subseteq \mathcal {O}{}_{\textsf {\tiny {R}} }(J)=\mathcal {O}$ and the same result holds interchanging left and right.

We follow Reiner [Rei2003, Lemma 23.4] (who calls the proof “mystifying”). Assume for the purposes of contradiction that $J^{-1}=\mathcal {O}$. Since $J \subsetneq \mathcal {O}$, there exists a maximal two-sided $\mathcal {O}$-ideal $M \supseteq J$. Thus $M^{-1} \subseteq J^{-1}=\mathcal {O}$. By Lemma 18.2.6, M is prime. Let $a \in R \cap J^{-1}$ be nonzero. By Lemma 18.2.8, $a\mathcal {O}$ contains a product of prime two-sided $\mathcal {O}$-ideals, so

$$\begin{aligned} M \supseteq a\mathcal {O}\supseteq P_1 P_2 \cdots P_r, \end{aligned}$$

with each $P_i$ prime. We may suppose without loss of generality that $r \in \mathbb Z _{>0}$ is minimal with this property. Since $P_1 \cdots P_r \subseteq M$ and M is prime, we must have $P_i \subseteq M$, so $P_i=M$ by Proposition 18.2.7. Thus

$$\begin{aligned} M \supseteq a\mathcal {O}\supseteq J_1 M J_2 \end{aligned}$$

with $J_1,J_2$ two-sided $\mathcal {O}$-ideals. From $a^{-1} J_1 M J_2 \subseteq \mathcal {O}$, we have $J_1(a^{-1}MJ_2)J_1 \subseteq J_1$, so by definition $a^{-1}MJ_2 J_1 \subseteq \mathcal {O}{}_{\textsf {\tiny {L}} }(J_1)=\mathcal {O}$. Thus $M(a^{-1} J_2J_1)M \subseteq M$ and $a^{-1}J_2J_1 \subseteq M^{-1} \subseteq \mathcal {O}$ so $J_2J_1 \subseteq a\mathcal {O}$. This shows that $a\mathcal {O}$ contains the product $J_2J_1$ of $r-1$ prime two-sided $\mathcal {O}$-ideals, contradicting the minimality of r.$\square$

Using this lemma, we arrive at the following proposition for maximal orders.

Proposition 18.3.2

Let $I \subseteq B$ be an R-lattice such that $\mathcal {O}{}_{\textsf {\tiny {L}} }(I)$ is a maximal R-order. Then I is right invertible, i.e., $II^{-1}=\mathcal {O}{}_{\textsf {\tiny {L}} }(I)$.

Of course, one can also swap left for right in the statement of Proposition 18.3.2. Using the standard involution, we proved Proposition 18.3.2 when B is a quaternion algebra (Proposition 16.6.15(b)).

Proof of Proposition 18.3.2

We follow Reiner [Rei2003, Theorem 23.5]. Let $\mathcal {O}=\mathcal {O}{}_{\textsf {\tiny {L}} }(I)$. Let $J=II^{-1} \subseteq \mathcal {O}$. Then $JI=II^{-1}I \subseteq I$, so $J \subseteq \mathcal {O}{}_{\textsf {\tiny {L}} }(I)=\mathcal {O}$ and J is a two-sided $\mathcal {O}$-ideal. We have

$$\begin{aligned} JJ^{-1}=II^{-1}J^{-1} \subseteq \mathcal {O}, \end{aligned}$$

so $I^{-1}J^{-1} \subseteq I^{-1}$ and therefore $J^{-1} \subseteq \mathcal {O}{}_{\textsf {\tiny {R}} }(I^{-1})$. Additionally,

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

(18.3.3)

but $\mathcal {O}$ is maximal, so equality holds in (18.3.3) and therefore $J^{-1} \subseteq \mathcal {O}$. But $\mathcal {O}\subseteq J^{-1}$ as well, so $J^{-1}=\mathcal {O}$. If $J \subsetneq \mathcal {O}$, then we have a contradiction with Lemma 18.3.1; so $J=\mathcal {O}$, and the proof is complete.$\square$

Putting these ingredients together, we have the following theorem.

Theorem 18.3.4

Let R be a Dedekind domain with $F={{\,\mathrm{Frac}\,}}R$, let B be a simple F-algebra, and let $\mathcal {O}\subseteq B$ be a maximal R-order. Then:

• (a) Multiplication of two-sided ideals is commutative: if I, J are two-sided $\mathcal {O}$-ideals, then $IJ=JI$.

• (b) Every nonzero two-sided $\mathcal {O}$-ideal is invertible and uniquely expressible as a product of prime two-sided ideals in $\mathcal {O}$.

Proof. For (b), invertibility follows from Proposition 18.3.2. For (b) without uniqueness, assume for purposes of contradiction that there is a two-sided ideal of $\mathcal {O}$ that is not the product of prime ideals; then there is a maximal counterexample J. Since J is not prime, there exists a prime Q with $J \subsetneq Q \subsetneq \mathcal {O}$, so $J \subset JQ^{-1} \subsetneq \mathcal {O}$. If $J=JQ^{-1}$, so by cancelling $Q=\mathcal {O}$, a contradiction. Therefore $JQ^{-1}=P_1 \cdots P_r$ is the product of primes by maximality, and $J=P_1 \cdots P_r Q$ is the product of primes, a contradiction.

We now prove (a). If $P,Q \subseteq \mathcal {O}$ are distinct nonzero prime two-sided ideals, and we let $Q'=P^{-1}QP$, then $Q' \subseteq P^{-1}\mathcal {O}P=\mathcal {O}$ is prime and $PQ'=QP \subseteq Q$, so $P \subseteq Q$ or $Q' \subseteq Q$; but equality would hold in each case by maximality, and since $P \ne Q$, we must have $Q'=Q$, and multiplication is commutative.

Finally, uniqueness of the factorization in (b) follows as in the commutative case. If $P_1 \cdots P_r = Q_1 \cdots Q_s$, then $P_1=Q_i$ for some i; multiplying by $P_1^{-1}$ and repeating the argument, we find that $\{P_1, \dots , P_r\}=\{Q_1,\dots ,Q_s\}$, and the result follows.$\square$

Corollary 18.3.5

With hypotheses as in Theorem 18.3.4, the group ${{\,\mathrm{Idl}\,}}(\mathcal {O})$ is isomorphic to the free abelian group on the set of nonzero prime ideals.

With these arguments in hand, we have the following foundational result for quaternion orders.

Theorem 18.3.6

Suppose that R is a Dedekind domain. Let B be a quaternion algebra over F and let $\mathcal {O}\subseteq B$ be a maximal R-order. Then the map

$$\begin{aligned} \begin{aligned} \left\{ \text {Prime two-sided }\mathcal {O}\text {-ideals} \right\}&\leftrightarrow \left\{ \text {Prime ideals of }R\right\} \\ P&\mapsto P \cap R \end{aligned} \end{aligned}$$

(18.3.7)

is a bijection.

Moreover, if R is a global ring, then there is an exact sequence

$$\begin{aligned} \begin{aligned} 0 \rightarrow {{\,\mathrm{Idl}\,}}(R)&\rightarrow {{\,\mathrm{Idl}\,}}(\mathcal {O}) \rightarrow \prod _\mathfrak{p \mid \mathfrak D } \mathbb Z /2\mathbb Z \rightarrow 0 \\ \mathfrak a&\mapsto \mathcal {O}\mathfrak a \mathcal {O}\end{aligned} \end{aligned}$$

(18.3.8)

where $\mathfrak D ={{\,\mathrm{disc}\,}}_R(B)$.

Proof. The map (18.3.7) is defined by Proposition 18.2.7, and it is surjective because $\mathfrak p \mathcal {O}\subseteq P$ is contained in a maximal therefore prime ideal.

Next we show that the map is injective. Let P be a prime ideal, and work with completions at a prime $\mathfrak p$. Then $P_\mathfrak p =P \otimes _R R_\mathfrak p \subseteq \mathcal {O}_\mathfrak p$ is a maximal ideal of $\mathcal {O}_\mathfrak p$. If $B_\mathfrak p \simeq {{\,\mathrm{M}\,}}_2(F_\mathfrak p )$, so $\mathcal {O}_\mathfrak p \simeq {{\,\mathrm{M}\,}}_2(R_\mathfrak p )$, then the only maximal two-sided ideal is $\mathfrak p \mathcal {O}_\mathfrak p$; if instead $B_\mathfrak p$ is a division algebra, then there is a unique maximal two-sided ideal $P_\mathfrak p$ with $P_\mathfrak p ^2=\mathfrak p \mathcal {O}_\mathfrak p$ by Theorem 13.3.11. We can also describe this uniformly, by the proof of Proposition 18.2.7: in all cases, we have $P_\mathfrak p ={{\,\mathrm{rad}\,}}(\mathcal {O}_\mathfrak p )$.

There is a natural group homomorphism

$$\begin{aligned} {{\,\mathrm{Idl}\,}}(R)&\rightarrow {{\,\mathrm{Idl}\,}}(\mathcal {O}) \\ \mathfrak a&\mapsto \mathcal {O}\mathfrak a \mathcal {O}=\mathfrak a \mathcal {O}\end{aligned}$$

This map is injective, since if $\mathfrak a \mathcal {O}=\mathcal {O}$ then $\mathfrak a ^2={{\,\mathrm{nrd}\,}}(\mathfrak a \mathcal {O})={{\,\mathrm{nrd}\,}}(\mathcal {O})=R$, so $\mathfrak a =R$. The cokernel of the map is determined by the previous paragraph.$\square$

Remark 18.3.9. Many of the theorems stated in this section (and chapter) hold more generally for hereditary orders: this notion is pursued in Chapter 21. To see what this looks like in a more general context, see Curtis–Reiner [CR81, §26B]. A very general context in which one can make an argument like in section 18.3 was axiomatized by Asano; for an exposition and several references, see McConnell–Robson [McCR87, Chapter 5].

4 Picard group

We now proceed to consider classes of two-sided ideals. We begin with a natural but abstract definition, in terms of bimodules. (Recall 20.3.7, that a bimodule is over R if the R-action on left and right are equal.)

Definition 18.4.1

The Picard group of $\mathcal {O}$ over R is the group ${{\,\mathrm{Pic}\,}}_R(\mathcal {O})$ of isomorphism classes of invertible $\mathcal {O}$-bimodules over R under tensor product.

Remark 18.4.2. Some authors also write ${{\,\mathrm{Picent}\,}}(\mathcal {O})={{\,\mathrm{Pic}\,}}_{Z(\mathcal {O})}(\mathcal {O})$ when considering the Picard group over the center of $\mathcal {O}$, the most important case. To avoid additional complication, in this section we suppose that B is central over F, so ${{\,\mathrm{Pic}\,}}_R(\mathcal {O})={{\,\mathrm{Picent}\,}}(\mathcal {O})$.

18.4.3

If $I \subseteq B$ is an R-lattice that is a fractional two-sided $\mathcal {O}$-ideal, then I is a $\mathcal {O}$-bimodule over R. Conversely, if I is a $\mathcal {O}$-bimodule over R then $I \otimes _R F \simeq B$ as B-bimodules, and choosing such an isomorphism gives an embedding $I \hookrightarrow B$ as an R-lattice.

Lemma 18.4.4

Let $I,J \subseteq B$ be R-lattices that are fractional two-sided $\mathcal {O}$-ideals. Then I is isomorphic to J as $\mathcal {O}$-bimodules over R if and only if there exists $a \in F^\times$ such that $J=aI$.

Proof. See Exercise 18.9.$\square$

18.4.5

By 18.4.3, there is a natural surjective map

$$\begin{aligned} {{\,\mathrm{Idl}\,}}(\mathcal {O}) \rightarrow {{\,\mathrm{Pic}\,}}_R(\mathcal {O}); \end{aligned}$$

we claim that the kernel of this map is ${{\,\mathrm{PIdl}\,}}(R) \trianglelefteq {{\,\mathrm{Idl}\,}}(\mathcal {O})$. By Lemma 19.5.1, every isomorphism class of invertible $\mathcal {O}$-bimodule is represented by an invertible R-lattice $I \subseteq B$, unique up to scaling by $F^\times$, and if $a \in F^\times$ then $a \mathcal {O}=\mathcal {O}$ if and only if $a \in R \cap \mathcal {O}^\times =R^\times$, so the ideal $aR \in {{\,\mathrm{PIdl}\,}}(R)$ is well-defined. Thus, we obtain a natural isomorphism

(18.4.6)

Equivalently, the sequence

$$\begin{aligned} 1 \rightarrow R^\times \rightarrow F^\times \rightarrow {{\,\mathrm{Idl}\,}}(\mathcal {O}) \rightarrow {{\,\mathrm{Pic}\,}}_R(\mathcal {O}) \rightarrow 1 \end{aligned}$$

is exact. One might profitably take (18.4.6) as the definition of ${{\,\mathrm{Pic}\,}}_R(\mathcal {O})$.

18.4.7

If $\mathcal {O}'$ is locally isomorphic $\mathcal {O}$ (so they are in the same genus), then there is a $\mathcal {O},\mathcal {O}'$-connecting ideal J, and the map

$$\begin{aligned} {{\,\mathrm{Idl}\,}}(\mathcal {O})&\rightarrow {{\,\mathrm{Idl}\,}}(\mathcal {O}') \\ I&\mapsto J^{-1}IJ \end{aligned}$$

is an isomorphism of groups restricting to the identity on ${{\,\mathrm{Cl}\,}}R$, so from (18.4.6) we obtain an isomorphism

$$\begin{aligned} {{\,\mathrm{Pic}\,}}_R(\mathcal {O}) \simeq {{\,\mathrm{Pic}\,}}_R(\mathcal {O}'), \end{aligned}$$

analogous to Lemma 17.4.11.

Our remaining task in this section is to examine the structure of ${{\,\mathrm{Pic}\,}}_R(\mathcal {O})$, and to this end we suppose that B is a quaternion algebra over F.

18.4.8

Suppose that $\mathcal {O}$ is a maximal R-order with $F={{\,\mathrm{Frac}\,}}R$ a global field. Then taking the quotient by ${{\,\mathrm{PIdl}\,}}(R)$ in the first two terms in (18.3.8) yields an exact sequence

$$\begin{aligned} 0 \rightarrow {{\,\mathrm{Cl}\,}}R \rightarrow {{\,\mathrm{Pic}\,}}_R(\mathcal {O}) \rightarrow \prod _\mathfrak{p \mid \mathfrak D } \mathbb Z /2\mathbb Z \rightarrow 0. \end{aligned}$$

(18.4.9)

Although this sequence need not split, it does show that the Picard group of the maximal order $\mathcal {O}$ is not far from the class group ${{\,\mathrm{Cl}\,}}R$, the difference precisely measured by the primes that ramify in B.

In general, for a quaternion R-order $\mathcal {O}$ we have the following result.

Proposition 18.4.10

${{\,\mathrm{Pic}\,}}_R(\mathcal {O})$ is a finite group.

Proof. If $\mathcal {O}$ is maximal, we combine (18.4.9) with the finiteness of ${{\,\mathrm{Cl}\,}}R$ and the fact that there are only finitely many primes $\mathfrak p$ dividing the discriminant $\mathfrak D$.

Now let $\mathcal {O}$ be an R-order. Then there exists a maximal R-order $\mathcal {O}' \supseteq \mathcal {O}$. We argue as in Exercise 17.3. We define a map of sets:

$$\begin{aligned} {{\,\mathrm{Pic}\,}}_R(\mathcal {O})&\rightarrow {{\,\mathrm{Pic}\,}}_R(\mathcal {O}') \\ [I]&\mapsto [\mathcal {O}'I\mathcal {O}'] \end{aligned}$$

The class up to scaling by $F^\times$ is well-defined, and $I' :=\mathcal {O}'I\mathcal {O}' \supseteq I$ an R-lattice with left and right orders containing $\mathcal {O}'$, but since $\mathcal {O}'$ is maximal these orders equal $\mathcal {O}'$ and $I'$ is invertible.

By the first paragraph, by finiteness of ${{\,\mathrm{Pic}\,}}_R(\mathcal {O}')$, after rescaling we may suppose $I'$ is one of finitely many possibilities. But there exists nonzero $r \in R$ such that $r\mathcal {O}' \subset \mathcal {O}$, so

$$\begin{aligned} I'=\mathcal {O}'I\mathcal {O}' \subseteq (r^{-1}\mathcal {O})I(r^{-1}\mathcal {O}) = r^{-2} I \subseteq r^{-2} I' \end{aligned}$$

so $r^2 I' \subseteq I \subseteq I'$; since $I'/r^2 I'$ is a finite group, this leaves only finitely many possibilities for I.$\square$

Remark 18.4.11. The study of the Picard group is quite general. It was studied in detail by Fröhlich [Frö73]; see also Curtis–Reiner [CR87, §55].

5 Classes of two-sided ideals

In this section, we compare the Picard group to the group of “ideals modulo principal ideals”.

Let ${{\,\mathrm{PIdl}\,}}(\mathcal {O}) \le {{\,\mathrm{Idl}\,}}(\mathcal {O})$ be the subgroup of principal two-sided fractional $\mathcal {O}$-ideals (invertible by 16.5.4). Let

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

be the normalizer of $\mathcal {O}$ in $B^\times$.

Lemma 18.5.1

There is an exact sequence of groups

$$\begin{aligned} \begin{aligned} 1 \rightarrow \mathcal {O}^\times \rightarrow N_{B^\times }(\mathcal {O})&\rightarrow {{\,\mathrm{PIdl}\,}}(\mathcal {O}) \rightarrow 1 \\ \alpha&\mapsto \mathcal {O}\alpha \mathcal {O}. \end{aligned} \end{aligned}$$

(18.5.2)

Proof. We have $\alpha \in N_{B^\times }(\mathcal {O})$ if and only if $\alpha \mathcal {O}=\mathcal {O}\alpha$ if and only if $\mathcal {O}\alpha \mathcal {O}$ is a principal two-sided fractional $\mathcal {O}$-ideal, as in Exercise 16.17; this gives a surjective group homomorphism $N_{B^\times }(\mathcal {O}) \rightarrow {{\,\mathrm{PIdl}\,}}(\mathcal {O})$. The kernel is the set of $\alpha \in B^\times$ such that $\alpha \mathcal {O}=\mathcal {O}$, and this normal subgroup is precisely $\mathcal {O}^\times$.$\square$

Proposition 18.5.3

There is an isomorphism of groups

(18.5.4)

If ${{\,\mathrm{PIdl}\,}}(\mathcal {O}) \trianglelefteq {{\,\mathrm{Idl}\,}}(\mathcal {O})$ is normal, then the isomorphism (18.5.4) induces a natural exact sequence

$$\begin{aligned} \begin{aligned} 0 \rightarrow N_{B^\times }(\mathcal {O})/(F^\times \mathcal {O}^\times )&\rightarrow {{\,\mathrm{Pic}\,}}_R(\mathcal {O}) \rightarrow {{\,\mathrm{Idl}\,}}(\mathcal {O})/{{\,\mathrm{PIdl}\,}}(\mathcal {O}) \rightarrow 0 \\ \alpha F^\times \mathcal {O}^\times&\mapsto \text {class of }\mathcal {O}\alpha \mathcal {O}. \end{aligned} \end{aligned}$$

(18.5.5)

Proof. There is an isomorphism $N_{B^\times }(\mathcal {O})/\mathcal {O}^\times \simeq {{\,\mathrm{PIdl}\,}}(\mathcal {O})$ by (18.5.2). The image of $F^\times \le N_{B^\times }(\mathcal {O})$ in ${{\,\mathrm{PIdl}\,}}(\mathcal {O})$ under this map consists of two-sided ideals of the form $\mathcal {O}a \mathcal {O}$ with $a \in F^\times$; we have $\mathcal {O}a \mathcal {O}= \mathcal {O}$ if and only if $a \in \mathcal {O}^\times$ if and only if $a \in R^\times$, so this image is isomorphic to the group ${{\,\mathrm{PIdl}\,}}(R)$ of principal fractional R-ideals via the map $aR \mapsto \mathcal {O}a \mathcal {O}$. The first isomorphism follows. The exact sequence (18.5.5) is then just rewriting the natural sequence

$$ 0 \rightarrow {{\,\mathrm{PIdl}\,}}(\mathcal {O})/{{\,\mathrm{PIdl}\,}}(R) \rightarrow {{\,\mathrm{Idl}\,}}(\mathcal {O})/{{\,\mathrm{PIdl}\,}}(R) \rightarrow {{\,\mathrm{Idl}\,}}(\mathcal {O})/{{\,\mathrm{PIdl}\,}}(\mathcal {O}) \rightarrow 0. $$

Remark 18.5.6 The moral of Proposition 18.5.3 is that, unlike the commutative case where the two notions coincide, the two notions of “isomorphism classes of invertible bimodules” and “ideals modulo principal ideals” are in general different for a quaternion order. These notions coincide precisely when $N_{B^\times }(\mathcal {O})/F^\times \simeq \mathcal {O}^\times /R^\times$, or equivalently (by the Skolem–Noether theorem) that every R-algebra automorphism of $\mathcal {O}$ is inner, which is to say ${{\,\mathrm{Aut}\,}}_R(\mathcal {O})={{\,\mathrm{Inn}\,}}_R(\mathcal {O})=\mathcal {O}^\times /R^\times$.

18.5.7

Unfortunately, the subgroup ${{\,\mathrm{PIdl}\,}}(\mathcal {O}) \le {{\,\mathrm{Idl}\,}}(\mathcal {O})$ need not be normal in general (Exercise 17.11), so statements like Proposition 18.5.3 depend on the order $\mathcal {O}$ having good structural properties. If $\mathcal {O}$ is a maximal order, then ${{\,\mathrm{Idl}\,}}(\mathcal {O})$ is abelian, so the result holds in this case.

In general, from the proof but using cosets one still obtains the equality

$$\begin{aligned} \#({{\,\mathrm{Idl}\,}}(\mathcal {O})/{{\,\mathrm{PIdl}\,}}(\mathcal {O})) \cdot \#(N_{B^\times }(\mathcal {O})/(F^\times \mathcal {O}^\times )) = \#{{\,\mathrm{Pic}\,}}_R(\mathcal {O}). \end{aligned}$$

(18.5.8)

Remark 18.5.9 If $\mathcal {O},\mathcal {O}'$ are connected, then ${{\,\mathrm{Pic}\,}}_R(\mathcal {O}) \simeq {{\,\mathrm{Pic}\,}}_R(\mathcal {O}')$ by 18.4.7 but this isomorphism need not respect the exact sequence (18.5.5). Each order $\mathcal {O}$ “balances” the contribution of this group between the normalizer $N_{B^\times }(\mathcal {O})/(F^\times \mathcal {O}^\times )$ and the quotient ${{\,\mathrm{Idl}\,}}(\mathcal {O})/{{\,\mathrm{PIdl}\,}}(\mathcal {O})$—and these might be of different sizes for $\mathcal {O}'$. We will return to examine more closely this structure in section 28.9, when strong approximation allows us to be more precise in measuring the discrepancy.

We conclude with an application to the structure of (right) class sets. We examine from Lemma 17.4.13 the fibers of the surjective map (17.4.14)

$$\begin{aligned} {{\,\mathrm{Cls}\,}}\mathcal {O}&\rightarrow {{\,\mathrm{Typ}\,}}\mathcal {O}\\ [I]&\mapsto class of \mathcal {O}{}_{\textsf {\tiny {L}} }(I). \end{aligned}$$

Refreshing our notation, let B be a central simple F-algebra and let $\mathcal {O}\subset B$ an R-order.

Proposition 18.5.10

The map $I \mapsto [I]$ induces a bijection

$$\begin{aligned} {{\,\mathrm{PIdl}\,}}(\mathcal {O}) \backslash {{\,\mathrm{Idl}\,}}(\mathcal {O}) \leftrightarrow \{[I] \in {{\,\mathrm{Cls}\,}}\mathcal {O}: \mathcal {O}{}_{\textsf {\tiny {L}} }(I) \simeq \mathcal {O}\}. \end{aligned}$$

Proof. Let $\mathcal {O}'$ be an order of the same type as $\mathcal {O}$. Since (17.4.14) is surjective, there exists $[I] \in {{\,\mathrm{Cls}\,}}\mathcal {O}$ such that $\mathcal {O}{}_{\textsf {\tiny {L}} }(I) \simeq \mathcal {O}'$. We are free to replace $\mathcal {O}'$ by an isomorphic order, so we may suppose $\mathcal {O}{}_{\textsf {\tiny {L}} }(I)=\mathcal {O}'$. For all $[I'] \in {{\,\mathrm{Cls}\,}}\mathcal {O}$ with $\mathcal {O}{}_{\textsf {\tiny {L}} }(I') \simeq \mathcal {O}'$ (running over the fiber), since $\mathcal {O}{}_{\textsf {\tiny {L}} }(\alpha I')=\alpha \mathcal {O}' \alpha ^{-1}$ for $\alpha \in B^\times$ we may suppose without loss of generality that the representative $I'$ has $\mathcal {O}{}_{\textsf {\tiny {L}} }(I')=\mathcal {O}'$.

We then define a map

$$\begin{aligned} \begin{aligned} {{\,\mathrm{PIdl}\,}}(\mathcal {O}') \backslash {{\,\mathrm{Idl}\,}}(\mathcal {O}')&\rightarrow \{[I'] \in {{\,\mathrm{Cls}\,}}\mathcal {O}: \mathcal {O}{}_{\textsf {\tiny {L}} }(I')=\mathcal {O}'\} \\ J'&\mapsto [J'I] \end{aligned} \end{aligned}$$

(18.5.11)

The map is surjective, because if $J'=I'I^{-1}$ then $\mathcal {O}{}_{\textsf {\tiny {L}} }(J')=\mathcal {O}{}_{\textsf {\tiny {R}} }(J')=\mathcal {O}$, so $J'$ is a two-sided invertible $\mathcal {O}'$-ideal. It is injective because if $[J'I]=[K'I]$ for $J',K' \in {{\,\mathrm{Idl}\,}}(\mathcal {O}')$ then $K'=\alpha ' J'$ with $\alpha ' \in B^\times$, but further we need $\mathcal {O}{}_{\textsf {\tiny {L}} }(K')=\alpha '\mathcal {O}'\alpha '^{-1}=\mathcal {O}'$, so in fact $[J'I]=[K'I]$ if and only if $\alpha ' \in N_{B^\times }(\mathcal {O}')$, and the result then follows from Lemma 18.5.1.$\square$

We have the following corollaries.

Corollary 18.5.12

We have

$$ \#{{\,\mathrm{Cls}\,}}\mathcal {O}= \sum _{[\mathcal {O}'] \in {{\,\mathrm{Typ}\,}}\mathcal {O}} [{{\,\mathrm{Idl}\,}}(\mathcal {O}') : {{\,\mathrm{PIdl}\,}}(\mathcal {O}')] = \#{{\,\mathrm{Pic}\,}}_R \mathcal {O}\sum _{[\mathcal {O}'] \in {{\,\mathrm{Typ}\,}}\mathcal {O}} \frac{1}{z_{\mathcal {O}'}} $$

where $z_{\mathcal {O}'}=[N_{B^\times }(\mathcal {O}'):F^\times \mathcal {O}'^\times ]$.

Proof. For the first equality, combine Lemma 17.4.13 and Proposition 18.5.10, computing the size of the fibers. For the second, substitute (18.5.8) and use 18.4.7.$\square$

Corollary 18.5.13

Let $\mathcal {O}_i$ be representatives of ${{\,\mathrm{Typ}\,}}\mathcal {O}$. For each i, let $I_i$ be a connecting $\mathcal {O}_i,\mathcal {O}$-ideal, and let $J_{i,j}$ be representatives of ${{\,\mathrm{PIdl}\,}}(\mathcal {O}_i) \backslash {{\,\mathrm{Idl}\,}}(\mathcal {O}_i)$. Then the set $\{J_{i,j} I_i\}_{i,j}$ is a complete set of representatives for ${{\,\mathrm{Cls}\,}}\mathcal {O}$.

Proof. We choose representatives and take the fibers of the map (17.4.14).$\square$

Remark 18.5.14 When ${{\,\mathrm{PIdl}\,}}(\mathcal {O}) \trianglelefteq {{\,\mathrm{Idl}\,}}(\mathcal {O})$, then in Proposition 18.5.10 we have written the class set ${{\,\mathrm{Cls}\,}}\mathcal {O}$ as a disjoint union of abelian groups. The fact that the bijection is noncanonical is due to the fact that we choose a connecting ideal, so without making choices we obtain only a disjoint union of principal homogeneous spaces (i.e., torsors) under the groups ${{\,\mathrm{PIdl}\,}}(\mathcal {O}')\backslash {{\,\mathrm{Idl}\,}}(\mathcal {O}')$.

Exercises

Unless otherwise specified, let R be a Dedekind domain with field of fractions $F={{\,\mathrm{Frac}\,}}R$, let B be a simple finite-dimensional F-algebra, and let $\mathcal {O}\subseteq B$ be an R-order.

• 1. Show that the following are equivalent:

  • (i) $\mathcal {O}$ is a maximal R-order;

  • (ii) $\mathcal {O}{}_{\textsf {\tiny {L}} }(I)=\mathcal {O}{}_{\textsf {\tiny {R}} }(I)=\mathcal {O}$ for all fractional two-sided $\mathcal {O}$-ideals I; and

  • (iii) $\mathcal {O}{}_{\textsf {\tiny {L}} }(I)=\mathcal {O}{}_{\textsf {\tiny {R}} }(I)=\mathcal {O}$ for all two-sided $\mathcal {O}$-ideals $I \subseteq \mathcal {O}$.

• 2. Show that the zero ideal is a prime ideal of $\mathcal {O}$.

• 3. Let $J \subseteq \mathcal {O}$ be a nonzero two-sided ideal of $\mathcal {O}$ in the ring-theoretic sense: J is an additive subgroup closed under left and right multiplication by $\mathcal {O}$. Show that J is an R-lattice.

• 4. Let R be a DVR with maximal ideal $\mathfrak p$, and let $\mathcal {O}=\begin{pmatrix} R &{} R \\ \mathfrak p &{} R \end{pmatrix} \subseteq B={{\,\mathrm{M}\,}}_2(F)$. Show that the two-sided ideal $\mathfrak p {{\,\mathrm{M}\,}}_2(R) \subseteq \mathcal {O}$ is not a prime ideal.

• 5. Let $R :=\mathbb Z [\sqrt{-6}]$ and $F :=\mathbb Q (\sqrt{-6})$. Let $B :=({2,\sqrt{-6}} \mid {F})$.

  • (a) Show that $2R=\mathfrak p _2^2$ and $3R=\mathfrak p _3^2$ for primes $\mathfrak p _2,\mathfrak p _3 \subseteq R$.

  • (b) Show that ${{\,\mathrm{Ram}\,}}(B)=\{\mathfrak{p }_2,\mathfrak{p }_3\}$.

  • (c) Let $\mathcal {O}$ be a maximal order in B. Show that there is a unique two-sided ideal $P_2$ such that $P_2^2=\mathfrak p _2\mathcal {O}$.

  • (d) Prove that $[P_2] \in {{\,\mathrm{Pic}\,}}_R(\mathcal {O})$ has order 4, and conclude that the sequence (18.4.9) does not split.

  • (e) Show that we may take

    $$\begin{aligned} \mathcal {O}=R + \mathfrak p _2^{-1}(\sqrt{-6}+i) + Rj + \mathfrak p _2^{-1}(\sqrt{-6}+i)j \end{aligned}$$

    as the maximal order, and then that I is generated by i and $\sqrt{-6}ij/2$, and finally that $I^2=(\sqrt{-6}+i)/2$.

• 6. Let $B={{\,\mathrm{M}\,}}_n(F)$ with $n \ge 2$, let $\mathcal {O}={{\,\mathrm{M}\,}}_n(R)$, let $\mathfrak p \subseteq R$ be prime with $k=R/\mathfrak p$, and let $\mathcal {O}(\mathfrak p ) = R + \mathfrak p \mathcal {O}$.

  • (a) Show that $\mathcal {O}(\mathfrak p )$ is an order of reduced discriminant $\mathfrak p ^3$.

  • (b) Show that $\mathcal {O}^\times \simeq {{\,\mathrm{GL}\,}}_n(R)$ normalizes $\mathcal {O}(\mathfrak p ) \subseteq \mathcal {O}$, so that

    $$\begin{aligned} \mathcal {O}(\mathfrak p )^\times \trianglelefteq \mathcal {O}^\times \simeq {{\,\mathrm{GL}\,}}_n(R), \end{aligned}$$

    and that the map

    $$\begin{aligned} \mathcal {O}^\times&\hookrightarrow {{\,\mathrm{Idl}\,}}(\mathcal {O}) \\ \gamma&\mapsto \mathcal {O}\gamma = \gamma \mathcal {O}\end{aligned}$$

    induces an injective group homomorphism ${{\,\mathrm{PGL}\,}}_n(k) \hookrightarrow {{\,\mathrm{Idl}\,}}(\mathcal {O})$. Conclude that ${{\,\mathrm{Idl}\,}}(\mathcal {O})$ is not an abelian group.

• 7. Show that Theorem 18.3.4 holds more generally for B a semisimple F-algebra (but still $\mathcal {O}\subseteq B$ maximal). [Hint: Decompose B into a product of simple F-algebras.]

• 8. Let $\mathcal {O}$ be maximal, and let $P_1,\dots ,P_r \subseteq \mathcal {O}$ be distinct prime two-sided ideals. Let

  $$\begin{aligned} I :=\prod _{i=1}^r P_i^{e_i} \quad \text {and}\quad J :=\prod _{i=1}^r P_i^{f_i} \end{aligned}$$

  with $e_i,f_i \in \mathbb Z$.

  • (a) Prove that $I \subseteq \mathcal {O}$ if and only if $e_i \ge 0$ for all $i=1,\dots ,n$, and in this case there is a ring isomorphism

    $$\begin{aligned} \mathcal {O}/I \simeq \bigoplus _{i=1}^r \mathcal {O}/P_i^{e_i}. \end{aligned}$$

  • (b) Prove that $I \supseteq J$ if and only if $e_i \le f_i$ for all i.

  • (c) Show $I+J = \prod _{i=1}^r P_i^{\min (e_i,f_i)}$ and $I \cap J = \prod _{i=1}^r P_i^{\max (e_i,f_i)}$.

$\triangleright$ 9.:

Prove Lemma 18.4.4: Show that fractional two-sided $\mathcal {O}$-ideals $I,J \subseteq B$ are isomorphic as $\mathcal {O}$-bimodules over R if and only if there exists $a \in F^\times$ such that $J=aI$. [Hint: Peek at Lemma 19.5.1.]

• 10. Let $K \supseteq F$ be a finite, separable extension and let S be the integral closure of R in K. Show that the map $I \mapsto I \otimes _R S$ defines a group homomorphism ${{\,\mathrm{Pic}\,}}\mathcal {O}\rightarrow {{\,\mathrm{Pic}\,}}(\mathcal {O}\otimes _R S)$.