Orders

Abstract

In this chapter, continuing with a second background installment, we study when lattices over a domain are closed under a multiplication law: these will be orders, an integral analogue of algebras over fields.

In this chapter, continuing with a second background installment, we study when lattices over a domain are closed under a multiplication law: these will be orders, an integral analogue of algebras over fields.

1 $\triangleright$ Lattices with multiplication

We begin with a brief indication of the theory of orders over the integers. Let B be a finite-dimensional $\mathbb Q$-algebra. An order $\mathcal {O}\subset B$ is a lattice that is also a subring of B (in particular, $1 \in \mathcal {O}$). The property of being an order is a local property for a lattice, i.e., one may check that it is closed under multiplication in every localization $\mathcal {O}_{(p)}$, for p prime.

An order is maximal if it is not properly contained in another order. For example, if we start with the quaternion algebra with $a,b \in \mathbb Z$ nonzero, then the lattice

$$\begin{aligned} \mathcal {O}:=\mathbb Z + \mathbb Z i + \mathbb Z j + \mathbb Z ij \subseteq B \end{aligned}$$

is closed under multiplication, and so defines an order—but it is never a maximal order. ]$\mathcal {O}$order (in an algebra)

An important construction of lattices comes about as follows: if $I \subseteq B$ is a lattice, then

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

is an order, called the left order of I; we similarly define the right order. L]$\mathcal {O}{}_{\textsf {\tiny {L}} }(I)$, $\mathcal {O}{}_{\textsf {\tiny {R}} }(I)$left, right order of a lattice I (in an algebra)

If $\mathcal {O}\subset B$ is an order and $\alpha \in \mathcal {O}$, then $\alpha$ is integral (over $\mathbb Z$), satisfying a monic polynomial with integer coefficients. If B is a quaternion algebra, then $\alpha \in B$ satisfies its reduced characteristic polynomial of degree 2, and $\alpha$ is integral if and only if ${{\,\mathrm{trd}\,}}(\alpha ),{{\,\mathrm{nrd}\,}}(\alpha ) \in \mathbb Z$ (Corollary 10.3.6). When $B=F$ is a number field, the most important order in F is the ring of integers, the set of all integral elements: it is the unique maximal order.

Unfortunately, this construction does not work in the noncommutative setting: the set of all integral elements does not form an order. For one thing, if $\mathcal {O}\subseteq B$ is a maximal order and $\alpha \in B^\times$, then $\alpha \mathcal {O}\alpha ^{-1} \subseteq B$ is a maximal order and when B is noncommutative, we may have $\alpha \mathcal {O}\alpha ^{-1} \ne \mathcal {O}$. But there are more serious problems, as the following example indicates.

Example 10.1.1

Let $B={{\,\mathrm{M}\,}}_2(\mathbb Q )$ and let $\alpha =\begin{pmatrix} 0 &{} 1/2 \\ 0 &{} 0 \end{pmatrix}$ and $\beta =\begin{pmatrix} 0 &{} 0 \\ 1/2 &{} 0 \end{pmatrix}$. Then $\alpha ^2=\beta ^2=0$, so $\alpha ,\beta$ are integral over $R=\mathbb Z$, but $\alpha +\beta$ and $\alpha \beta$ are not integral since ${{\,\mathrm{nrd}\,}}(\alpha +\beta )=-1/4$ and ${{\,\mathrm{trd}\,}}(\alpha \beta )=1/4$. (Such a counterexample does not require the existence of zerodivisors: see Exercise 10.10.)

Understanding orders in quaternion algebras is a major task of this second part of the text. In the simplest case $B={{\,\mathrm{M}\,}}_2(\mathbb Q )$, every maximal order is conjugate (and thus isomorphic) in B to ${{\,\mathrm{M}\,}}_2(\mathbb Z )$. The reader may wish to skip ahead to Chapter 11 to get to know the Hurwitz order before returning to study orders more generally.

2 Orders

Throughout, let R be a domain with field of fractions $F :={{\,\mathrm{Frac}\,}}(R)$, and let B be a finite-dimensional F-algebra. For further reference about orders (as lattices), see Reiner [Rei2003, Chapter 2] and Curtis–Reiner [CR81, §§23, 26].

Definition 10.2.1

An R-order $\mathcal {O}\subseteq B$ is an R-lattice that is also a subring of B.

In particular, if $\mathcal {O}$ is an R-order, then since $\mathcal {O}$ is a subring we have $1 \in \mathcal {O}$, and since $\mathcal {O}$ is an R-module we have $R \subseteq \mathcal {O}$. We will primarily be concerned with R-orders that are projective as R-modules, and call them projective R-orders.

10.2.2

An R-algebra is a ring $\mathcal {O}$ equipped with an embedding $R \hookrightarrow \mathcal {O}$ whose image lies in the center of $\mathcal {O}$. An R-order $\mathcal {O}$ has the structure of an R-algebra, and if $\mathcal {O}$ is an R-algebra that is finitely generated as an R-module, then $\mathcal {O}$ is an R-order of the F-algebra $B=\mathcal {O}\otimes _R F$.

Example 10.2.3

The matrix algebra ${{\,\mathrm{M}\,}}_n(F)$ has the R-order ${{\,\mathrm{M}\,}}_n(R)$. The subring $R[G]=\bigoplus _{g \in G} Rg$ is an R-order in the group ring F[G].

Example 10.2.4

Let $a,b \in R \smallsetminus \{0\}$ and consider the quaternion algebra $B=({a,b} \mid {F})$. Then $\mathcal {O}=R \oplus R i\oplus R j \oplus R k$ is an R-order, because it is closed under multiplication (e.g., $ik=i(ij)=aj \in \mathcal {O}$).

Let $I \subseteq B$ be an R-lattice in the F-algebra B.

10.2.5

An important construction of orders comes as follows. Let

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

(10.2.6)

Lemma 10.2.7

$\mathcal {O}{}_{\textsf {\tiny {L}} }(\mathcal {I}) \subseteq B$ is an R-order.

Proof. Then $\mathcal {O}{}_{\textsf {\tiny {L}} }(I)$ is an R-submodule of B which is a ring. We show it is also an R-lattice. For all $\alpha \in B$, by Lemma 9.3.5(b), there exists nonzero $r \in R$ such that $r(\alpha I) \subseteq I$, hence $\mathcal {O}{}_{\textsf {\tiny {L}} }(I)F=B$. Also by this lemma, there exists nonzero $s \in R$ such that $s=s\cdot 1 \in I$; thus $\mathcal {O}{}_{\textsf {\tiny {L}} }(I) s \subseteq I$ so $\mathcal {O}{}_{\textsf {\tiny {L}} }(I) \subseteq s^{-1}I$. Since R is noetherian and $s^{-1} I$ is an R-lattice so finitely generated, we conclude that $\mathcal {O}{}_{\textsf {\tiny {L}} }(I)$ is finitely generated and is thus an R-lattice.

Definition 10.2.8

The order $\mathcal {O}{}_{\textsf {\tiny {L}} }(I)=\{\alpha \in B : \alpha I \subseteq I\}$ in (10.2.6) is called the left order of I. We similarly define the right order of I by

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

Example 10.2.9

It follows from Lemma 10.2.7 that B has an R-order: the R-span of an F-basis for B defines an R-lattice, so $\mathcal {O}{}_{\textsf {\tiny {L}} }(I)$ is an R-order. (This is a nice way of “clearing denominators” from a multiplication table to obtain an order.)

We can read other properties about lattices from their localizations, such as in the following lemma.

Lemma 10.2.10

Let B be a finite-dimensional F-algebra and let $I \subseteq B$ be an R-lattice. Then the following are equivalent:

• (i) I is an R-order;

• (ii) $I_{(\mathfrak p )}$ is an $R_{(\mathfrak p )}$-order for all primes $\mathfrak p$ of R; and

• (iii) $I_{(\mathfrak m )}$ is an $R_{(\mathfrak m )}$-order for all maximal ideals $\mathfrak m$ of R.

Proof. For (i) $\Rightarrow$ (ii) $\Rightarrow$ (iii), if I is an R-order then $I_{(\mathfrak p )}$ is an $R_{(\mathfrak p )}$-order for all primes $\mathfrak p$, hence a fortiori for all maximal ideals $\mathfrak m$.

To conclude, we prove (iii) $\Rightarrow$ (i), and suppose that $I_{(\mathfrak m )}$ is an $R_{(\mathfrak m )}$-order for all maximal ideals $\mathfrak m$. Then $\bigcap _\mathfrak m I_{(\mathfrak m )} = I$ by Lemma 9.4.6. Thus $1 \in \bigcap _\mathfrak m I_{(\mathfrak m )} = I$, and for all $\alpha ,\beta \in I$ we have $\alpha \beta \in \bigcap _\mathfrak m I_{(\mathfrak m )} = I$, so I is a subring of B and hence an order.

Remark 10.2.11. The hypothesis that R is noetherian is used in Lemma 10.2.7, but it is not actually needed; the fact that $\mathcal {O}{}_{\textsf {\tiny {L}} }(I)$ is an order follows by a process often referred to as noetherian reduction. A basis of B yields a multiplication table, consisting of finitely many elements of F; moreover, we know that I is finitely generated as an R-module. Writing these generators in terms of a basis we can express these generators over the basis using finitely many elements of F. Let $R_0$ be the subring of R generated by these finitely elements, with field of fractions $F_0$, let $B_0$ be the $F_0$-algebra with the same multiplication table as B; let $I_0$ be the $R_0$-submodule generated by the generators for I written over $R_0$. Then $B = B_0 \otimes _{F_0} F$ and $I = I_0 \otimes _{R_0} R$. But now $R_0$ is a finitely generated commutative algebra over its prime ring (the subring generated by 1), so by the Hilbert basis theorem, $R_0$ is noetherian. The argument given then shows that $I_0$ is finitely generated as an $R_0$-module, whence I is finitely generated as an R-module.

Noetherian reduction applies to many results in this text, but non-noetherian rings are not our primary concern; we retain the noetherian hypothesis for simplicity of argument and encourage the interested reader to seek generalizations (when they are possible).

3 Integrality

Orders are composed of integral elements, defined as follows. If $\alpha \in B$, we denote by $R[\alpha ] = \sum _d R\alpha ^d$ the (commutative) R-subalgebra of B generated by $\alpha$.

Definition 10.3.1

An element $\alpha \in B$ is integral over R if $\alpha$ satisfies a monic polynomial with coefficients in R.

Lemma 10.3.2

For $\alpha \in B$, the following are equivalent:

• (i) $\alpha$ is integral over R;

• (ii) $R[\alpha ]$ is a finitely generated R-module;

• (iii) $\alpha$ is contained in a subring A that is finitely generated as an R-module.

Proof. This lemma is standard; the only extra detail here is to note that in (iii) we do not need to assume that the subring A is commutative: (ii) $\Rightarrow$ (iii) is immediate taking $A=R[\alpha ]$, and for the converse, if $A \subseteq B$ is a subring that is finitely generated as an R-module, then $R[\alpha ] \subseteq A$ and since R is noetherian and A is finitely generated as an R-module, it follows that $R[\alpha ]$ is also finitely generated as an R-module.

Corollary 10.3.3

If $\mathcal {O}$ is an R-order, then every $\alpha \in \mathcal {O}$ is integral over R.

10.3.4

We say R is integrally closed (in F) if whenever $\alpha \in F$ is integral over R, then in fact $\alpha \in R$. Inside the field F, the set of elements integral over R (the integral closure of R in F) forms a ring: if $\alpha ,\beta$ are integral over R then $\alpha +\beta$ and $\alpha \beta$ are integral since they lie in $R[\alpha ,\beta ]$ which is a finitely generated submodule of F. The integral closure of R is itself integrally closed.

Lemma 10.3.5

Suppose that R is integrally closed. Then $\alpha \in B$ is integral over R if and only if the minimal polynomial of $\alpha$ over F has coefficients in R.

Proof. Let $f(x) \in R[x]$ be a monic polynomial that $\alpha$ satisfies, and let $g(x) \in F[x]$ be the minimal polynomial of $\alpha$. Let K be a splitting field for g(x), and let $\alpha _1,\dots ,\alpha _n$ be the roots of g(x) in K. Since $g(x) \mid f(x)$, each such $\alpha _i$ is integral over R, and the set of elements in K integral over R forms a ring, so each coefficient of g is integral over R and belongs to F; but since R is integrally closed, these coefficients must belong to R and $g(x) \in R[x]$.

Corollary 10.3.6

If B is an F-algebra with a standard involution, and R is integrally closed, then $\alpha \in B$ is integral over R if and only if ${{\,\mathrm{trd}\,}}(\alpha ),{{\,\mathrm{nrd}\,}}(\alpha ) \in R$.

We may characterize orders in separable algebras as follows.

Lemma 10.3.7

Let $\mathcal {O}\subseteq B$ be a subring of a separable F-algebra B such that $\mathcal {O}{}F=B$. Then $\mathcal {O}$ is an R-order if and only if every $\alpha \in \mathcal {O}$ is integral.

Proof. Let $\mathcal {O}\subseteq B$ be a subring of an F-algebra B such that $\mathcal {O}{}F=B$. Recall from Theorem 7.9.4 that a separable F-algebra is a semisimple F-algebra such that the symmetric bilinear pairing $(\alpha ,\beta ) \mapsto {{\,\mathrm{trd}\,}}(\alpha \beta )$ is nondegenerate.

We need to show that $\mathcal {O}$ is finitely generated. Let $\alpha _1,\dots ,\alpha _n$ be an F-basis for B contained in $\mathcal {O}$. If $\beta \in \mathcal {O}$ then $\beta =\sum _i a_i \alpha _i$ with $a_i \in F$. We have $\beta \alpha _i \in \mathcal {O}$ since $\mathcal {O}$ is a ring, so ${{\,\mathrm{trd}\,}}(\beta \alpha _i)=\sum _j a_j {{\,\mathrm{trd}\,}}(\alpha _j\alpha _i)$ with ${{\,\mathrm{trd}\,}}(\alpha _j\alpha _i) \in R$. Now since B is separable, the matrix $({{\,\mathrm{trd}\,}}(\alpha _i\alpha _j))_{i,j=1,\dots ,n}$ is invertible, say $r=\det ({{\,\mathrm{trd}\,}}(\alpha _i\alpha _j))$, so we can solve these equations for $a_j$ using Cramer’s rule and we find that $a_j \in r^{-1} R$. Consequently $\mathcal {O}\subseteq r^{-1} (R \alpha _1 \oplus \dots \oplus R\alpha _n)$ is a submodule of a finitely generated module so (since R is noetherian) $\mathcal {O}$ is finitely generated.

4 Maximal orders

The integral closure of R in F is the largest ring containing integral elements. Accordingly, we make the following more general definition.

Definition 10.4.1

An R-order $\mathcal {O}\subseteq B$ is maximal  if it is not properly contained in another R-order.

If B is a commutative F-algebra and R is integrally closed in F, then the integral closure S of R in K is integrally closed and therefore S is a maximal R-order in K. However, if B is noncommutative, then the set of elements in B integral over R is no longer necessarily itself a ring, and so the theory of maximal orders is more complicated. (This may seem counterintuitive at first, but certain aspects of the noncommutative situation are quite different!) The problem in the noncommutative setting is that although $R[\alpha ]$ and $R[\beta ]$ may be finitely generated as R-modules for $\alpha ,\beta \in B$, this need not be the case for the R-algebra generated by $\alpha$ and $\beta$.

10.4.2

It follows from Lemma 10.3.7 that a separable F-algebra B has a maximal R-order, as follows. By Lemma 10.2.7, B has an R-order $\mathcal {O}$ (since it has a lattice, taking the R-span of an F-basis), so the collection of R-orders containing $\mathcal {O}$ is nonempty. Given a chain of R-orders containing $\mathcal {O}$, by Lemma 10.3.7 the union of these orders is again an R-order. Since R is noetherian, there exists a maximal element in a chain.

For the rest of this section, we restrict attention and suppose that R is a Dedekind domain. We begin by showing that the property of being a maximal order is a local property.

Lemma 10.4.3

An R-order $\mathcal {O}\subseteq B$ is maximal if and only if $\mathcal {O}_{(\mathfrak p )}$ is a maximal $R_{(\mathfrak p )}$-order for all primes $\mathfrak p$ of R.

Proof. If $\mathcal {O}_{(\mathfrak p )}$ is maximal for each prime $\mathfrak p$ then by Corollary 9.4.7 we see that $\mathcal {O}$ is maximal. Conversely, suppose $\mathcal {O}$ is maximal and suppose that $\mathcal {O}_{(\mathfrak p )} \subsetneq \mathcal {O}_{(\mathfrak p )}'$ is a proper containment of orders for some nonzero prime $\mathfrak p$. Then the set $\mathcal {O}'=\bigl (\bigcap _\mathfrak{q \ne \mathfrak p } \mathcal {O}_\mathfrak q \bigr ) \cap \mathcal {O}'_{(\mathfrak p )}$ is an R-order properly containing $\mathcal {O}$ by Lemma 10.2.10 and Theorem 9.4.9.

Lemma 10.4.4

Let $\mathcal {O}\subset B$ be an R-order. Then for all but finitely many primes $\mathfrak p$ of R, we have that $\mathcal {O}_{(\mathfrak p )} = \mathcal {O}\otimes _R R_{(\mathfrak p )}$ is maximal.

Proof. By 10.4.2, there exists a maximal order $\mathcal {O}' \supseteq \mathcal {O}$. By the local-global principle for lattices (Theorem 9.4.9), we have $\mathcal {O}'_\mathfrak p = \mathcal {O}_\mathfrak p$ for all but finitely many primes $\mathfrak p$.

The structure of (maximal) orders in quaternion algebras over domains of arithmetic interest is the subject of the second part of this text.

5 Orders in a matrix ring

In this section, we study orders in a matrix ring; we restore generality, and let R be a noetherian domain with $F={{\,\mathrm{Frac}\,}}R$.

The matrix ring over F is just the endomorphism ring of a finite-dimension vector space over F, and we seek a similar description for orders as endomorphism rings of lattices (cf. 10.2.5).

Let V be an F-vector space with $\dim _F V = n$ and let $B={{\,\mathrm{End}\,}}_F(V)$. Choosing a basis of V gives an identification $B={{\,\mathrm{End}\,}}_F(V) \simeq {{\,\mathrm{M}\,}}_n(F)$. Given an R-lattice $M \subseteq V$, we define

$$\begin{aligned} {{\,\mathrm{End}\,}}_R(M) :=\{f \in {{\,\mathrm{End}\,}}_F(V) : f(M) \subseteq M\} \subseteq B. \end{aligned}$$

(10.5.1)

The left order (10.2.5) is the special case of (10.5.1) where $M=I \subseteq V=B$.

Example 10.5.2

If $V=Fx_1 \oplus \dots \oplus Fx_n$ and $M=Rx_1 \oplus \dots \oplus Rx_n$, then ${{\,\mathrm{End}\,}}_R(M) \simeq {{\,\mathrm{M}\,}}_n(R)$.

More generally, if M is completely decomposable, i.e. $M=\mathfrak a _1 x_1 \oplus \dots \oplus \mathfrak a _n x_n$ with each $\mathfrak a _i \subseteq F$ invertible fractional ideals, then we have ${{\,\mathrm{End}\,}}_R(M) \subseteq {{\,\mathrm{M}\,}}_n(F)$ the subring of matrices whose ij-entry belongs to the R-module

$$\begin{aligned} \mathfrak a _j\mathfrak a _i^{-1} \simeq {{\,\mathrm{Hom}\,}}_R(\mathfrak a _i,\mathfrak a _j) \subseteq {{\,\mathrm{Hom}\,}}_F(F,F) \simeq F \end{aligned}$$

where the isomorphisms come from multiplication. For example, if $n=2$ then

$$\begin{aligned} {{\,\mathrm{End}\,}}_R(M) \simeq \begin{pmatrix} R &{} \mathfrak a _2\mathfrak a _1^{-1} \\ \mathfrak a _1\mathfrak a _2^{-1} &{} R \end{pmatrix} \subseteq {{\,\mathrm{M}\,}}_2(F). \end{aligned}$$

(Note how the cross terms are aligned correctly in the multiplication!) For example, if $M=Rx_1+ \mathfrak a x_2$, then ${{\,\mathrm{End}\,}}_R(M) \simeq \begin{pmatrix} R &{} \mathfrak a ^{-1} \\ \mathfrak a &{} R \end{pmatrix}$.

Lemma 10.5.3

Let M be an R-lattice of V. Then ${{\,\mathrm{End}\,}}_R(M)$ is an R-order in $B={{\,\mathrm{End}\,}}_F(V)$.

Proof. As in the proof of Lemma 10.2.7, we conclude that ${{\,\mathrm{End}\,}}_R(M) F = B$. Let $\alpha _1,\alpha _2,\dots ,\alpha _n$ be an F-basis for V and let $N=R\alpha _1 \oplus \dots \oplus R\alpha _n$. Thus ${{\,\mathrm{End}\,}}_R(N) \simeq {{\,\mathrm{M}\,}}_n(R)$ is finitely generated as an R-module.

By Lemma 9.3.5 there exists nonzero $r \in R$ such that $rN \subseteq M \subseteq r^{-1} N$. Therefore, if $\phi \in {{\,\mathrm{End}\,}}_R(M)$, so that $\phi (M) \subseteq M$, then

$$\begin{aligned} (r^2 \phi )(N)=r\phi (rN) \subseteq r\phi (M) \subseteq rM \subseteq N \end{aligned}$$

and thus ${{\,\mathrm{End}\,}}_R(M) \subseteq r^{-2} {{\,\mathrm{End}\,}}_R(N)$; since R is noetherian, this implies that ${{\,\mathrm{End}\,}}_R(M)$ is finitely generated as an R-module and ${{\,\mathrm{End}\,}}_R(M)$ is an R-order in B.

Lemma 10.5.4

Let $\mathcal {O}\subseteq B = {{\,\mathrm{End}\,}}_F(V)$ be an R-order. Then $\mathcal {O}\subseteq {{\,\mathrm{End}\,}}_R(M)$ for some R-lattice $M \subseteq V$. In particular, if $\mathcal {O}\subseteq B$ is a maximal R-order, then $\mathcal {O}={{\,\mathrm{End}\,}}_R(M)$ for some R-lattice M.

Proof. Quite generally, if N is any R-lattice in V, then $M=\{x \in N : \mathcal {O}x \subseteq N\}$ is an R-submodule of N with $FM=V$ (as in Lemma 10.2.7), thus M is an R-lattice in V and $\mathcal {O}\subseteq {{\,\mathrm{End}\,}}_R(M)$. If further $\mathcal {O}$ is maximal, then the other containment so equality holds.

Corollary 10.5.5

If R is a PID, then every maximal R-order $\mathcal {O}\subseteq B \simeq {{\,\mathrm{M}\,}}_n(F)$ is conjugate in B to ${{\,\mathrm{M}\,}}_n(R)$.

Proof. The isomorphism $B \simeq {{\,\mathrm{M}\,}}_n(F)$ arises from a choice of basis $x_1,\dots ,x_n$ for V; letting $N=\bigoplus _{i=1}^n R x_i$ we have ${{\,\mathrm{End}\,}}_R(N) \simeq {{\,\mathrm{M}\,}}_n(R)$. The R-order ${{\,\mathrm{M}\,}}_n(R)$ is maximal by Exercise 10.6, since a PID is integrally closed.

By Lemma 10.5.4, we have $\mathcal {O}\subseteq {{\,\mathrm{End}\,}}_R(M)$ for some R-lattice $M \subseteq V$, so if $\mathcal {O}$ is maximal then $\mathcal {O}={{\,\mathrm{End}\,}}_R(M)$. If R is a PID then M is free as an R-module, and we can write $M=Ry_1 \oplus \dots \oplus Ry_n$; the change of basis matrix from $x_i$ to $y_i$ then realizes ${{\,\mathrm{End}\,}}_R(M)$ as a conjugate of ${{\,\mathrm{End}\,}}_R(N) \simeq {{\,\mathrm{M}\,}}_n(R)$.

6 Exercises

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

1.:

Let $\mathfrak c \subseteq R$ be an ideal. Show that

$$ \begin{pmatrix} R &{} R \\ \mathfrak c &{} R \end{pmatrix} = \left\{ \begin{pmatrix} a &{} b \\ c &{} d \end{pmatrix} \in {{\,\mathrm{M}\,}}_2(R) : c \in \mathfrak c \right\} \subseteq {{\,\mathrm{M}\,}}_2(R) $$

is an R-order in ${{\,\mathrm{M}\,}}_2(F)$. Note further that if $\mathfrak c$ is projective (equivalently, locally free) as an R-module, then this R-order is projective as an R-module.

2.:

Let B be a finite-dimensional F-algebra with a standard involution and let $\mathcal {O}\subseteq B$ be an R-order. Verify that ${{\,\mathrm{nrd}\,}}:\mathcal {O}\rightarrow R$ is a quadratic form over R.

3.:

Let $\mathcal {O}, \mathcal {O}' \subseteq B$ be R-orders in an F-algebra B.

(a):

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

(b):

If $\mathcal {O}\subseteq \mathcal {O}'$, show that $\mathcal {O}'^\times \cap \mathcal {O}= \mathcal {O}^\times$.

4.:

Let $\mathcal {O}\subset B$ be an R-order in an F-algebra B and suppose that R is integrally closed. Show that $F \cap \mathcal {O}= R$.

5.:

Let $A_1,\dots ,A_r$ be F-algebras and let $B=A_1 \times \dots \times A_r$. Show that $\mathcal {O}\subseteq B$ is an R-order if and only if $\mathcal {O}$ is an R-lattice in B and $\mathcal {O}\cap A_i$ is an R-order for each i.

6.:

Let R be integrally closed. Show that ${{\,\mathrm{M}\,}}_n(R)$ is a maximal R-order in ${{\,\mathrm{M}\,}}_n(F)$.

7.:

Let $B=({K,b} \mid {F})$ be a quaternion algebra with $b \in R$ and let S be an R-order in K. Let $\mathcal {O}= S + Sj$. Show that $\mathcal {O}$ is an R-order in B.

8.:

Suppose that R is a PID, and let $\mathcal {O}\subseteq B$ be an R-order in the quaternion algebra B. Let $\alpha \in \mathcal {O}$ be such that $S = R[\alpha ]$ is (commutative) domain that is a maximal R-order in its field of fractions.

(a):

Show that $1,\alpha$ extends to an R-basis for $\mathcal {O}$.

(b):

If moreover S is a PID, show that there exists $\beta \in \mathcal {O}$ such that $1,\alpha ,\beta ,\alpha \beta$ is an R-basis for $\mathcal {O}$.

9.:

Let B be an F-algebra with a standard involution and let $\alpha \in B$. Show that if $\alpha$ is integral over R then ${{\,\mathrm{trd}\,}}(\alpha ^n) \in R$ for all $n \in \mathbb Z _{\ge 0}$. Is the converse true?

10.:

Generalize Example 10.1.1: Exhibit a division quaternion algebra B over $\mathbb Q$ and elements $\alpha ,\beta \in B$ such that $\alpha ,\beta$ are integral over $\mathbb Z$ but both $\alpha +\beta$ and $\alpha \beta$ are not.

11.:

Let $\alpha \in {{\,\mathrm{M}\,}}_n(F)$ have characteristic polynomial with coefficients in R. Show that $\alpha$ is conjugate by an element $\beta \in {{\,\mathrm{GL}\,}}_n(F)$ to an element of ${{\,\mathrm{M}\,}}_n(R)$. Explicitly, how do you find such a matrix $\beta$?

$\triangleright$ 12.:

Let $B={{\,\mathrm{M}\,}}_n(F)$ and let $I \subseteq B$ be an R-lattice. Let $I^{\textsf {t} }=\{\alpha ^{\textsf {t} }: \alpha \in I\}$ be the transpose lattice. Show that $\mathcal {O}{}_{\textsf {\tiny {L}} }(I^{\textsf {t} })=\mathcal {O}{}_{\textsf {\tiny {R}} }(I)$.

$\triangleright$ 13.:

Let $I,J \subseteq B$ be R-lattices. Let IJ be the R-submodule of B generated by products $\alpha \beta$ where $\alpha \in I$, $\beta \in J$; i.e.,

$$\begin{aligned} IJ :=\bigl \{ \textstyle {\sum }_{i=1}^k \alpha _i \beta _i : \alpha _i \in I, \beta _i \in J \bigr \}. \end{aligned}$$

(a):

Show that IJ is an R-lattice.

(b):

Let $\mathfrak p$ be a prime of R. Show that products commute with localization in the sense that

$$\begin{aligned} (IJ) \otimes _R R_{(\mathfrak p )} = (I \otimes _R R_{(\mathfrak p )})(J \otimes _R R_{(\mathfrak p )}) \subseteq B_{(\mathfrak p )}=B. \end{aligned}$$

$\triangleright$ 14.:

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

(a):

Show that $\mathcal {O}{}_{\textsf {\tiny {L}} }(\mathcal {O})=\mathcal {O}{}_{\textsf {\tiny {R}} }(\mathcal {O})=\mathcal {O}$.

(b):

Let $\alpha \in B^\times$, and let $\alpha \mathcal {O}= \{\alpha \beta : \beta \in \mathcal {O}\}$. Show that $\alpha \mathcal {O}$ is an R-lattice and that $\mathcal {O}{}_{\textsf {\tiny {L}} }(\alpha \mathcal {O})=\alpha \mathcal {O}\alpha ^{-1}$.

15.:

Let $\mathcal {O}\subseteq B$ be an R-order in an F-algebra B. Let $\gamma \in \mathcal {O}$ and let $N:B^\times \rightarrow F^\times$ be a multiplicative map. Show that $\gamma \in \mathcal {O}^\times$ if and only if $N(\gamma ) \in R^\times$, and in particular, if B has a standard involution, then $\gamma \in \mathcal {O}^\times$ if and only if ${{\,\mathrm{nrd}\,}}(\gamma ) \in R^\times$.