Discriminants

Abstract

Discriminants measure volume and arithmetic complexity, and they simultaneously encode ramification. We devote this chapter to their study.

Discriminants measure volume and arithmetic complexity, and they simultaneously encode ramification. We devote this chapter to their study.

1 $\triangleright$ Discriminantal notions

Let $x_1,\dots ,x_n \in \mathbb R ^n$, and let A be the matrix with columns $x_i$. Then the parallelopiped with edges from the origin to $x_i$ has volume $|\det (A) \,|$. We can compute this volume in another way:

$$\begin{aligned} \det (A)^2=\det (A^{\textsf {t} }A) = \det (M) \end{aligned}$$

(15.1.1)

where M has ijth entry equal to the ordinary dot product $x_i\cdot x_j$.

The absolute discriminant of a number field is a volume and a measure of arithmetic complexity, as follows. If $x_1,\dots ,x_n$ is a $\mathbb Z$-basis for $\mathbb Z _F$ and $\iota :F \hookrightarrow F \otimes _\mathbb{Q } \mathbb R \simeq \mathbb R ^n$ (normalized with an extra factor of $\sqrt{2}$ at the complex places), then the volume of $\mathbb Z _F$ in this embedding is the absolute determinant of the matrix with columns $\iota (x_i)$, and its square is defined to be the absolute discriminant of F. Replacing the dot product in the definition of M in (15.1.1) with the trace form $(x,y) \mapsto {{\,\mathrm{Tr}\,}}_{F/\mathbb Q }(xy)$, we see that the absolute discriminant is a positive integer. A prime p is ramified in F if and only if it divides the discriminant, so this volume also records arithmetic properties of F.

More generally, whenever we have a symmetric bilinear form $T :V \times V \rightarrow F$ on a finite-dimensional F-vector space V, there is a volume defined by the determinant $\det (T(x_i,x_j))_{i,j}$: and when T arises from a quadratic form Q, this is volume is the discriminant of Q (up to a normalizing factor of 2 in odd degree, see 6.3.1). In particular, if B is a finite-dimensional algebra over F, there is a bilinear form

$$ \begin{aligned} B \times B&\rightarrow F \\ (\alpha ,\beta )&\mapsto {{\,\mathrm{Tr}\,}}_{B|F}(\alpha \beta ) \end{aligned} $$

(or, when B is semisimple, the bilinear form associated to the reduced trace ${{\,\mathrm{trd}\,}}$) and so we obtain a discriminant—a “squared” volume—measuring in some way the complexity of B. As in the commutative case, discriminants encode ramification.

In this chapter, we establish basic facts about discriminants, including how they behave under inclusion (measuring index) and localization. To illustrate, let B be a quaternion algebra over $\mathbb Q$ and let $\mathcal {O}\subset B$ be an order. We define the discriminant  of $\mathcal {O}$ to be

$$\begin{aligned} {{\,\mathrm{disc}\,}}(\mathcal {O}) :=|\det ({{\,\mathrm{trd}\,}}(\alpha _i\alpha _j))_{i,j} \,| \in \mathbb Z _{>0} \end{aligned}$$

(15.1.2)

where $\alpha _1,\dots ,\alpha _4$ is a $\mathbb Z$-basis for $\mathcal {O}$. For example, if $B=\displaystyle {\biggl (\frac{a,b}{\mathbb {Q}}\biggr )}$ with $a,b \in \mathbb Z \smallsetminus \{0\}$, then the standard order $\mathcal {O}=\mathbb Z +\mathbb Z i+\mathbb Z j + \mathbb Z k$ has

$$\begin{aligned} {{\,\mathrm{disc}\,}}(\mathcal {O})=(4ab)^2; \end{aligned}$$

indeed, this is the discriminant of the quadratic form $\langle 1,-a,-b,ab \rangle$, the reduced norm restricted to $\mathcal {O}$. If $a,b<0$, i.e. B is definite, then the reduced norm is a Euclidean norm on $B_\infty = B \otimes _\mathbb Q \mathbb R \simeq \mathbb H$; normalizing with an extra factor $\sqrt{2}$, the discriminant is square of the covolume of the lattice $\mathcal {O}\subset B_\infty$. For example, the Lipschitz order $\mathbb Z \langle i,j \rangle$ (11.1.1) has ${{\,\mathrm{disc}\,}}(\mathbb Z \langle i,j\rangle )=4^2$, the square of the covolume of the lattice $(\sqrt{2}\mathbb Z )^4 \subseteq \mathbb R ^4$.

If $\mathcal {O}' \supseteq \mathcal {O}$, then ${{\,\mathrm{disc}\,}}(\mathcal {O})=[\mathcal {O}':\mathcal {O}]^2{{\,\mathrm{disc}\,}}(\mathcal {O}')$; in particular $\mathcal {O}'=\mathcal {O}$ if and only if ${{\,\mathrm{disc}\,}}(\mathcal {O}')={{\,\mathrm{disc}\,}}(\mathcal {O})$. It follows that the discriminant of an order is always a square, so we define thereduced discriminant ${{\,\mathrm{discrd}\,}}(\mathcal {O})$ to be the positive integer square root, and ${{\,\mathrm{discrd}\,}}(\mathcal {O})^2={{\,\mathrm{disc}\,}}(\mathcal {O})$. The discriminant of an order measures how far the order is from being a maximal order. We will show (Theorem 15.5.5) that $\mathcal {O}$ is a maximal order if and only if ${{\,\mathrm{discrd}\,}}(\mathcal {O})={{\,\mathrm{disc}\,}}B$, where ${{\,\mathrm{disc}\,}}B$ is the (squarefree) product of primes ramified in B.

In an extension of Dedekind domains, the different of the extension is an ideal whose norm is the discriminant of the extension (see Neukirch [Neu99, §III.2]). The different is perhaps not as popular as its discriminant cousin, but it has many nice properties, including easy-to-understand behavior under base extension. Similar conclusions holds in the noncommutative context (presented in section 15.6).

2 Discriminant

For further reference on discriminants, see Reiner [Rei2003, §10, §14].

Let R be a noetherian domain and let $F={{\,\mathrm{Frac}\,}}R$. Let B be a semisimple algebra over F with $\dim _F B=n$. For elements $\alpha _1,\dots ,\alpha _n \in B$, we define

$$\begin{aligned} d(\alpha _1,\dots ,\alpha _n) :=\det ({{\,\mathrm{trd}\,}}(\alpha _i\alpha _j))_{i,j=1,\dots ,n}. \end{aligned}$$

(15.2.1)

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

Definition 15.2.2

The discriminant of I is the R-submodule ${{\,\mathrm{disc}\,}}(I) \subseteq F$ generated by the set

$$\begin{aligned} \{d(\alpha _1,\dots ,\alpha _n) : \alpha _1,\dots ,\alpha _n \in I\}. \end{aligned}$$

15.2.3

If $I=\mathcal {O}$, then for $\alpha _1,\dots ,\alpha _n \in \mathcal {O}$ we have $\alpha _i\alpha _j \in \mathcal {O}$ and so ${{\,\mathrm{trd}\,}}(\alpha _i\alpha _j) \in R$ for all i, j. Thus $d(\alpha _1,\dots ,\alpha _n) \in R$ and therefore ${{\,\mathrm{disc}\,}}(\mathcal {O}) \subseteq R$.

Remark 15.2.4. When working over $\mathbb Z$, it is common to take the discriminant instead to be the positive generator of the discriminant as an ideal; passing between these should cause no confusion.

Although Definition 15.2.2 may look unwieldly, it works as well in the commutative case as in the noncommutative case. Right away, we see that if $\mathcal {O}\subseteq \mathcal {O}'$ are R-orders, then ${{\,\mathrm{disc}\,}}(\mathcal {O}') \mid {{\,\mathrm{disc}\,}}(\mathcal {O})$.

The function d itself transforms in a nice way under a change of basis, as follows.

Lemma 15.2.5

Let $\alpha _1,\dots ,\alpha _n \in B$ and suppose $\beta _1,\dots ,\beta _n \in B$ are of the form $\beta _i=\sum _{j=1}^n m_{ij} \alpha _j$ with $m_{ij} \in F$. Let $M=(m_{ij})_{i,j=1,\dots ,n}$. Then

$$\begin{aligned} d(\beta _1,\dots ,\beta _n)=\det (M)^2 d(\alpha _1,\dots ,\alpha _n). \end{aligned}$$

(15.2.6)

Proof. By properties of determinants, if $\beta _1,\dots ,\beta _n$ are linearly dependent (over F) then $d(\beta _1,\dots ,\beta _n)=0$ and either $\alpha _1,\dots ,\alpha _n$ are also linearly dependent or $\det (M)=0$, and in either case the equality (15.2.6) holds trivially.

So suppose that $\beta _1,\dots ,\beta _n$ are linearly independent, then $\alpha _1,\dots ,\alpha _n$ are also linearly independent and the matrix M, a change of basis matrix, is invertible. By Gaussian reduction, we can write M as a product of elementary matrices (a matrix that coincides with the identity matrix except for a single off-diagonal entry), permutation matrices (a matrix interchanging rows suffices), and a diagonal matrix; it is enough to check that the equality holds when M is a matrix of one of these forms. And for such a matrix, the equality can be checked in a straightforward manner using the corresponding property of determinants. $\square$

Corollary 15.2.7

If I is free as an R-module, and $\alpha _1,\dots ,\alpha _n$ is an R-basis for I, then

$$\begin{aligned} {{\,\mathrm{disc}\,}}(I) = d(\alpha _1,\dots ,\alpha _n)R. \end{aligned}$$

Proof. The matrix M writing any other $\beta _1,\dots ,\beta _n \in I$ in terms of the basis has $M \in {{\,\mathrm{M}\,}}_n(R)$ so $\det (M) \in R$, and therefore $d(\beta _1,\dots ,\beta _n) \in d(\alpha _1,\dots ,\alpha _n)R$ by Lemma 15.2.5. $\square$

15.2.8

More generally, if I is completely decomposable with

$$\begin{aligned} I=\mathfrak a _1 \alpha _1 \oplus \dots \oplus \mathfrak a _n \alpha _n \end{aligned}$$

such as in (9.3.7), then from (15.2.6)

$$\begin{aligned} {{\,\mathrm{disc}\,}}(I)=(\mathfrak a _1 \cdots \mathfrak a _n)^2 d(\alpha _1,\dots ,\alpha _n). \end{aligned}$$

More generally, the discriminant is well-behaved under automorphisms because the reduced trace is so.

Corollary 15.2.9

If is an F-algebra automorphism, then ${{\,\mathrm{disc}\,}}(\phi (I))={{\,\mathrm{disc}\,}}(I)$.

Proof. By Proposition 7.8.6, we have ${{\,\mathrm{trd}\,}}(\phi (\alpha \beta ))={{\,\mathrm{trd}\,}}(\alpha \beta )$ for all $\alpha ,\beta \in B$. Therefore, for all $\alpha _1,\dots ,\alpha _n \in B$ we have $d(\phi (\alpha _1),\dots ,\phi (\alpha _n))=d(\alpha _1,\dots ,\alpha _n)$; the result ${{\,\mathrm{disc}\,}}(\phi (I))={{\,\mathrm{disc}\,}}(I)$ follows. $\square$

Our primary interest will be in the case $I=\mathcal {O}$.

Example 15.2.10

Suppose ${{\,\mathrm{char}\,}}F \ne 2$. Let $B :=({a,b} \mid {F})$ with $a,b \in R$. Let $\mathcal {O}:=R \oplus Ri \oplus Rj \oplus Rij$ be the standard order. Then ${{\,\mathrm{disc}\,}}(\mathcal {O})$ is the principal R-ideal generated by

$$ d(1,i,j,ij)=\det \begin{pmatrix} 2 &{} 0 &{} 0 &{} 0 \\ 0 &{} 2a &{} 0 &{} 0 \\ 0 &{} 0 &{} 2b &{} 0 \\ 0 &{} 0 &{} 0 &{} -2ab \end{pmatrix} = -(4ab)^2. $$

The calculation when ${{\,\mathrm{char}\,}}F = 2$ is requested in Exercise 15.1.

Example 15.2.11

Let $B :={{\,\mathrm{M}\,}}_n(F)$ and $\mathcal {O}:={{\,\mathrm{M}\,}}_n(R)$. Then ${{\,\mathrm{disc}\,}}(\mathcal {O})=R$ (Exercise 15.2).

15.2.12

Let $B :=({K,b} \mid {F})$ be a quaternion algebra over F with $b \in R$ and let S be an R-order in K. Let $\mathcal {O}:=S \oplus Sj$; then $\mathcal {O}$ is an R-order in B by Exercise 10.7. We have ${{\,\mathrm{disc}\,}}(\mathcal {O})=b^2 {{\,\mathrm{disc}\,}}(S)^2$, by Exercise 15.4.

In particular, let F be a nonarchimedean local field, let R be its valuation ring and $\mathfrak p =R\pi$ its maximal ideal, and let B be a division quaternion algebra over F. Then by Theorem 13.3.11, we have $B \simeq ({K,\pi } \mid {F})$ with $K \supseteq F$ an unramified separable quadratic extension of F. The valuation ring S of K has ${{\,\mathrm{disc}\,}}(S)=R$, so the valuation ring $\mathcal {O}=S \oplus Sj$ of B has discriminant ${{\,\mathrm{disc}\,}}(\mathcal {O})=\mathfrak p ^2$.

15.2.13

Equation (15.2.6) and the fact that $I_{(\mathfrak p )} = I \otimes _{R} R_{(\mathfrak p )}$ implies the equality

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

on localizations and for the same reason an equality for the completions ${{\,\mathrm{disc}\,}}(I_\mathfrak p )={{\,\mathrm{disc}\,}}(I)_\mathfrak p$. In other words, the discriminant respects localization and completion and can be computed locally. Therefore, by the local-global principle (Lemma 9.4.6),

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

Lemma 15.2.14

If B is separable as an F-algebra and I is projective as an R-module, then ${{\,\mathrm{disc}\,}}(I)$ is a nonzero projective fractional ideal of R.

Proof. Since I is an R-lattice, there exist elements $\alpha _1,\dots ,\alpha _n$ which are linearly independent over F. Since B is separable, by Theorem 7.9.4, ${{\,\mathrm{trd}\,}}$ is a nondegenerate bilinear pairing on B so ${{\,\mathrm{disc}\,}}(I)$ is a nonzero ideal of R. It follows from Lemma 15.2.5 that ${{\,\mathrm{disc}\,}}(I)$ is finitely generated as an R-module, since this is true of I: we apply d to all subsets of a set of generators for I as an R-module. To show that ${{\,\mathrm{disc}\,}}(I)$ is projective, by 9.2.1 we show that ${{\,\mathrm{disc}\,}}(I)$ is locally principal. Let $\mathfrak p$ be a prime ideal of R. Since I is a projective R-module, its localization $I_{(\mathfrak p )}$ is free; thus from Corollary 15.2.7, we conclude that ${{\,\mathrm{disc}\,}}(I)_{(\mathfrak p )}={{\,\mathrm{disc}\,}}(I_{(\mathfrak p )})$ is principal over $R_{(\mathfrak p )}$ and generated by ${{\,\mathrm{disc}\,}}(\alpha _1,\dots ,\alpha _n)$ for an $R_{(\mathfrak p )}$-basis $\alpha _1,\dots ,\alpha _n$ of I, as desired. $\square$

We conclude this section comparing lattices by their index and discriminant as follows. We recall the definition of index (section 9.6).

Lemma 15.2.15

Let $I,J \subseteq B$ be projective R-lattices. Then

$$\begin{aligned} {{\,\mathrm{disc}\,}}(I)=[J:I]_R^2 {{\,\mathrm{disc}\,}}(J). \end{aligned}$$

Moreover, if $I \subseteq J$, then ${{\,\mathrm{disc}\,}}(I)={{\,\mathrm{disc}\,}}(J)$ if and only if $I=J$.

Proof. For the first statement, we argue locally, and combine (15.2.6) and Lemma 9.6.4. For the second statement, clearly ${{\,\mathrm{disc}\,}}(J) \subseteq {{\,\mathrm{disc}\,}}(I)$, and if $I=J$ then equality holds; and conversely, from ${{\,\mathrm{disc}\,}}(I)=[J:I]_R^2 {{\,\mathrm{disc}\,}}(J)={{\,\mathrm{disc}\,}}(J)$ we conclude $[J:I]_R=R$, hence $J=I$ by Proposition 9.6.8. $\square$

Remark 15.2.16. We defined the discriminant for semisimple algebras so that it is given in terms of the reduced trace. This definition extends to an arbitrary finite-dimensional F-algebra B, replacing the reduced trace by the algebra trace ${{\,\mathrm{Tr}\,}}_{B|F}$. If B is a central simple F-algebra of dimension $n^2$, then $n {{\,\mathrm{trd}\,}}= {{\,\mathrm{Tr}\,}}_{B|F}$ so when $n \in F^\times$ one can recover the discriminant as we have defined it here from the more general definition; but if $n = 0 \in F$ then the discriminant of B computed with the algebra trace will be zero.

3 Quadratic forms

Essentially the same definition of discriminant (Definition 15.2.2) applies to quadratic modules, as follows. We recall 6.3.1, where the discriminant was defined in all characteristics.

Let $Q:M \rightarrow L$ be a quadratic module over R (Definition 9.7.3) with ${{\,\mathrm{rk}\,}}M=n$ and associated bilinear map $T :M \times M \rightarrow L$.

15.3.1

Let $x_1,\dots ,x_n \in M$ and . If n is even, we define

$$\begin{aligned} d(x_1,\dots ,x_n;f) :=\det (f(T(x_i,x_j)))_{i,j=1,\dots ,n}. \end{aligned}$$

(15.3.2)

If n is odd, then by specializing the universal determinant as in 6.3.4, we define

$$\begin{aligned} d(x_1,\dots ,x_n;f) :=(\det /2)(f(T(x_i,x_j)))_{i,j=1,\dots ,n}. \end{aligned}$$

(15.3.3)

Thediscriminant of Q is then the ideal ${{\,\mathrm{disc}\,}}(Q) \subseteq R$ generated by the set

(15.3.4)

15.3.5

If M, L are free with R-basis $x_1,\dots ,x_n$ and e, respectively, then letting the dual to e with $f(e)=1$ gives

$$\begin{aligned} {{\,\mathrm{disc}\,}}(Q)=d(x_1,\dots ,x_n;f)R. \end{aligned}$$

In particular, since M, L are projective and therefore locally free over R, the discriminant of Q is locally free and hence a projective R-ideal.

Lemma 15.3.6

The discriminant of a quadratic module is well-defined up to similarity.

Proof. Let $Q:M \rightarrow L$ and $Q':M' \rightarrow L'$ be quadratic modules over R similar by and . It suffices to check the invariance locally, so to this end we may suppose that the modules are free; choose a basis $M=\sum _{i=1}^n Rx_i$ and $L=Re$, and let $x_i'=g(x_i)$ and $e'=h(e)$. Then $M'=\sum _{i=1}^n Rx_i'$ and $L'=Re'$. Let $f,f'$ be dual to $e,e'$; then postcomposing Q and $Q'$ by $f,f'$ we may suppose $L=L'=R$ and h is the identity.

We then have $Q'(g(x))=Q(x)$ for all $x \in M$, so the same is true of the associated bilinear forms $T,T'$. But then $d(x_1',\dots ,x_n')=d(x_1,\dots ,x_n)$, and by 15.3.5 this implies ${{\,\mathrm{disc}\,}}(Q)={{\,\mathrm{disc}\,}}(Q')$ as ideals of R. $\square$

15.3.7

Let B be a finite-dimensional F-algebra with a standard involution. Then the reduced norm is a quadratic form on B with associated bilinear form $T(\alpha ,\beta )={{\,\mathrm{trd}\,}}(\alpha \overline{\beta })$. Although the bilinear form differs by the presence of this standard involution from the definition of discriminant in (15.2.1), the resulting discriminants are the same (up to $R^\times$): see Exercise 15.13.

Lemma 15.3.8

The quadratic module Q is nonsingular if and only if ${{\,\mathrm{disc}\,}}(Q)=L$.

In particular, suppose that $M \simeq R^n$ is free with basis $e_i$ and $L=R$, and let $[T] :=(T(e_i,e_j))_{i,j} \in {{\,\mathrm{M}\,}}_n(R)$ be the Gram matrix in this basis. Then Q is nonsingular if and only if $\det ([T]),(\det /2)([T]) \in R^\times$ according as n is even or odd.

Proof. The map $T :M \rightarrow {{\,\mathrm{Hom}\,}}_R(M,L)$ is an isomorphism if and only if it is an isomorphism in every localization, so we may suppose that Q is free, with $M=R^n$ and $L=R$, which is to say we may prove the second statement in the case where R is local, with maximal ideal $\mathfrak p$ and residue field $k :=R/\mathfrak p$. Let be the reduction of Q; its Gram matrix is . Over the field k, we have that is nonsingular if and only if it is nondegenerate if and only if $\det [T],(\det /2)([T]) \ne 0$ according as n is even or odd; since R is local, these are equivalent to asking that these values are in $R^\times$. An application of Nakayama’s lemma then implies the result. $\square$

4 Reduced discriminant

In this section, we extract a square root of the discriminant for quaternion orders. Indeed, in Example 15.2.10, we saw that the discriminant of the standard R-order $\mathcal {O}\subseteq B=({a,b} \mid {F})$ is ${{\,\mathrm{disc}\,}}(\mathcal {O})=(4ab)^2R$, a square. If $\mathcal {O}'$ is another projective R-order, then ${{\,\mathrm{disc}\,}}(\mathcal {O}')=[\mathcal {O}:\mathcal {O}']_R^2{{\,\mathrm{disc}\,}}(\mathcal {O})$ by Lemma 15.2.15, so in fact the discriminant of every R-order is the square of an R-ideal.

In fact, there is a way to define this square root directly, inspired by vector calculus.

15.4.1

If $u,v,w \in \mathbb R ^3$ then $|u\cdot (v \times w) \,|$, the absolute value of the so-calledmixed product (orscalar triple product orbox product), is the volume of the parallelopiped defined by u, v, w; identifying $\mathbb R ^3 \simeq \mathbb H ^0$ as in section 2.4, from (2.4.10) we can write

$$\begin{aligned} 2u\cdot (v \times w)=u\cdot (vw-wv)=-{{\,\mathrm{trd}\,}}(u(vw-wv)). \end{aligned}$$

For example, $2=-2i \cdot (j \times k)=-i\cdot (jk-kj)={{\,\mathrm{trd}\,}}(ijk)$.

More generally (and carefully attending to the factors of 2) we make the following definition. Let B be a quaternion algebra over F.

15.4.2

For $\alpha _1,\alpha _2,\alpha _3 \in B$, we define

$$\begin{aligned} m(\alpha _1,\alpha _2,\alpha _3)&:={{\,\mathrm{trd}\,}}( (\alpha _1\alpha _2-\alpha _2\alpha _1)\overline{\alpha _3}) \\&= \alpha _1\alpha _2\overline{\alpha _3}-\alpha _2\alpha _1\overline{\alpha _3} - \alpha _3\overline{\alpha _2}\,\overline{\alpha _1}+\alpha _3 \overline{\alpha _1}\,\overline{\alpha _2}. \end{aligned}$$

Lemma 15.4.3

The form $m :B \times B \times B \rightarrow F$ is an alternating trilinear form which is well-defined as a form on B/F.

Proof. The form is alternating because for all $\alpha _1,\alpha _2 \in B$ we have $m(\alpha _1,\alpha _1,\alpha _2)=0$ and

$$ m(\alpha _1,\alpha _2,\alpha _1) = {{\,\mathrm{trd}\,}}((\alpha _1\alpha _2-\alpha _2\alpha _1)\overline{\alpha _1}) = {{\,\mathrm{trd}\,}}({{\,\mathrm{nrd}\,}}(\alpha _1)\alpha _2)-{{\,\mathrm{trd}\,}}(\alpha _2 {{\,\mathrm{nrd}\,}}(\alpha _1))=0 $$

and similarly $m(\alpha _1,\alpha _2,\alpha _2)=0$. The trilinearity follows from the linearity of the reduced trace. Finally, from these two properties, the descent to B/F follows from the computation $m(1,\alpha _1,\alpha _2)=0$ for all $\alpha _1,\alpha _2 \in B$.

(Alternatively, one can check that the pairing descends to B/F first, so that the involution becomes $\overline{\alpha +F}=-\alpha +F$, and then the alternating condition is immediate.) $\square$

Definition 15.4.4

Let $I \subseteq B$ be an R-lattice. Thereduced discriminant  of I is the R-submodule ${{\,\mathrm{discrd}\,}}(I)$ of F generated by

$$\begin{aligned} \{m(\alpha _1,\alpha _2,\alpha _3) : \alpha _1,\alpha _2,\alpha _3 \in I\}. \end{aligned}$$

15.4.5

If $\alpha _i,\beta _i \in B$ with $\beta _i=M\alpha _i$ for some $M \in {{\,\mathrm{M}\,}}_3(F)$, then

$$\begin{aligned} m(\beta _1,\beta _2,\beta _3)=\det (M) m(\alpha _1,\alpha _2,\alpha _3) \end{aligned}$$

(15.4.6)

by Exercise 15.10. It follows that if $I \subseteq J$ are projective R-lattices in B, then

$$\begin{aligned} {{\,\mathrm{discrd}\,}}(I)=[J:I] {{\,\mathrm{discrd}\,}}(J). \end{aligned}$$

Lemma 15.4.7

If I is a projective R-lattice in B, then ${{\,\mathrm{disc}\,}}(I)={{\,\mathrm{discrd}\,}}(I)^2$.

Proof. First, we claim that

$$\begin{aligned} m(i,j,ij)^2 = -d(1,i,j,ij). \end{aligned}$$

If ${{\,\mathrm{char}\,}}F \ne 2$, then ${{\,\mathrm{disc}\,}}(1,i,j,ij)=-(4ab)^2$ by Example 15.2.10 and

$$\begin{aligned} m(i,j,ij)={{\,\mathrm{trd}\,}}((ij-ji)\overline{ij})={{\,\mathrm{trd}\,}}(2ij(\overline{ij}))=4ab, \end{aligned}$$

as claimed. See Exercise 15.1 for the case ${{\,\mathrm{char}\,}}F = 2$. This computation verifies the result for the order $\mathcal {O}=R \oplus Ri \oplus Rj \oplus Rij$.

The lemma now follows using (15.2.6) and (15.4.6), for it shows that

$$\begin{aligned} m(\alpha _1,\alpha _2,\alpha _3)^2 = -d(1,\alpha _1,\alpha _2,\alpha _3) \end{aligned}$$

for all $\alpha _1,\alpha _2,\alpha _3 \in B$, and the latter generate ${{\,\mathrm{discrd}\,}}(I)$ by Exercise 15.7. $\square$

The notions in this section extend more generally to an arbitrary algebra B with a standard involution.

5 Maximal orders and discriminants

We now relate discriminants to maximal orders. Throughout this section, we suppose that R is a Dedekind domain. We record the following important principle.

Lemma 15.5.1

Let $\mathcal {O}\subseteq \mathcal {O}'$ be R-orders. Then $\mathcal {O}=\mathcal {O}'$ if and only if ${{\,\mathrm{disc}\,}}\mathcal {O}={{\,\mathrm{disc}\,}}\mathcal {O}'$.

Proof. In the nontrivial direction, by Lemma 15.2.15 we have

$$\begin{aligned} {{\,\mathrm{disc}\,}}\mathcal {O}= [\mathcal {O}':\mathcal {O}]_R^2 {{\,\mathrm{disc}\,}}(\mathcal {O}') \end{aligned}$$

so ${{\,\mathrm{disc}\,}}\mathcal {O}={{\,\mathrm{disc}\,}}\mathcal {O}'$ if and only if $\mathcal {O}=\mathcal {O}'$. $\square$

First, we ensure the existence of maximal orders (cf. 10.4.2) using the discriminant.

Proposition 15.5.2

There exists a maximal R-order $\mathcal {O}\subseteq B$, and every order $\mathcal {O}$ is contained in a maximal R-order $\mathcal {O}' \subseteq B$.

Proof. The algebra B has at least one R-order $\mathcal {O}$ as the left- or right-order of a lattice 10.2.5. If $\mathcal {O}$ is not maximal, then there exists an order $\mathcal {O}' \supsetneq \mathcal {O}$ with ${{\,\mathrm{disc}\,}}(\mathcal {O}') \supsetneq {{\,\mathrm{disc}\,}}(\mathcal {O})$ by Lemma 15.5.1. If $\mathcal {O}'$ is maximal, we are done; otherwise, we can continue in this way to obtain orders $\mathcal {O}=\mathcal {O}_1 \subsetneq \mathcal {O}_2 \subsetneq \dots$ and an ascending chain of ideals ${{\,\mathrm{disc}\,}}(\mathcal {O}_1) \subsetneq {{\,\mathrm{disc}\,}}(\mathcal {O}_2) \subsetneq \dots$ of R; but since R is noetherian, the latter stabilizes after finitely many steps, and the resulting order is then maximal, by Lemma 15.2.15. $\square$

Using the discriminant as a measure of index, we can similarly detect when orders are maximal. We recall (10.4.3) that the property of being maximal is a local property, so we begin with the local matrix case.

Lemma 15.5.3

Suppose that R is a DVR, and let $\mathcal {O}\subseteq B :={{\,\mathrm{M}\,}}_n(F)$ be an R-order. Then $\mathcal {O}$ is maximal if and only if ${{\,\mathrm{disc}\,}}\mathcal {O}= R$.

Proof. First, suppose $\mathcal {O}$ is maximal. Then by Corollary 10.5.5, we conclude $\mathcal {O}\simeq {{\,\mathrm{M}\,}}_n(R)$ (conjugate in B). By Corollary 15.2.9, we have ${{\,\mathrm{disc}\,}}\mathcal {O}={{\,\mathrm{disc}\,}}{{\,\mathrm{M}\,}}_n(R)$; we computed in Example 15.2.11 that ${{\,\mathrm{disc}\,}}{{\,\mathrm{M}\,}}_n(R)=R$, as claimed. The converse follows by taking $\mathcal {O}'$ a maximal order containing $\mathcal {O}$ (furnished by Proposition 15.5.2) and applying Lemma 15.5.1. $\square$

Example 15.5.4

By 15.2.12, if F is a nonarchimedean local field with valuation ring R and B is a division quaternion algebra over F, then the valuation ring $\mathcal {O}\subset B$ is the unique maximal order (Theorem 13.3.11) with ${{\,\mathrm{disc}\,}}\mathcal {O}= \mathfrak p ^2$ and ${{\,\mathrm{discrd}\,}}\mathcal {O}= \mathfrak p$. Arguing as in Lemma 15.5.3, we find that an R-order in B is maximal if and only if it has reduced discriminant $\mathfrak p$.

Maximality can be detected over global rings in terms of discriminants, as follows.

Theorem 15.5.5

Let R be a global ring with field of fractions F, let B be a quaternion algebra over F, and let $\mathcal {O}\subseteq B$ be an R-order. Then $\mathcal {O}$ is maximal if and only if

$$\begin{aligned} {{\,\mathrm{discrd}\,}}(\mathcal {O})={{\,\mathrm{disc}\,}}_R(B). \end{aligned}$$

(15.5.6)

Proof. Suppose that $\mathcal {O}$ is maximal. Then $\mathcal {O}_\mathfrak p$ is maximal for all primes $\mathfrak p$ of R. If $B_\mathfrak p \simeq {{\,\mathrm{M}\,}}_2(F_\mathfrak p )$ is split, then by Lemma 15.5.3, ${{\,\mathrm{discrd}\,}}\mathcal {O}_\mathfrak p =R_\mathfrak p$; if $B_\mathfrak p$ is a division algebra, then ${{\,\mathrm{discrd}\,}}\mathcal {O}_\mathfrak p =\mathfrak p R_\mathfrak p$. Since discriminants are defined locally, we conclude that

$$\begin{aligned} {{\,\mathrm{discrd}\,}}(\mathcal {O})=\prod _\mathfrak{p \in {{\,\mathrm{Ram}\,}}B \smallsetminus {{{\texttt {\textit{S}}}}}} \mathfrak p = {{\,\mathrm{disc}\,}}_R(B) \end{aligned}$$

if R as a global ring is the ring of ${{{\texttt {\textit{S}}}}}$-integers.

In the other direction, if (15.5.6) holds, we choose $\mathcal {O}' \supseteq \mathcal {O}$ be a maximal R-superorder and conclude that ${{\,\mathrm{disc}\,}}(\mathcal {O})={{\,\mathrm{disc}\,}}_R(B)^2={{\,\mathrm{disc}\,}}(\mathcal {O}')$ so $\mathcal {O}=\mathcal {O}'$ is maximal by Lemma 15.5.1. $\square$

Example 15.5.7

We recall Example 14.2.13, giving an explicit description of quaternion algebras $B=({a,b} \mid {\mathbb {Q}})$ of prime discriminant $D=p$. We now exhibit an explicit maximal order in each of these algebras.

For $p=2$, we have $B=({-1,-1} \mid {\mathbb {Q}})$ and take $\mathcal {O}\subseteq B$ the Hurwitz order.

For $p \equiv 3 ~(\text{ mod } ~{4})$, we took $B=({-p,-1} \mid {\mathbb {Q}})$. The order $\mathcal {O}:=\mathbb Z \langle (1+i)/2, j \rangle = S \oplus Sj$ with $S :=\mathbb Z [(1+i)/2]$ has ${{\,\mathrm{discrd}\,}}\mathcal {O}= p$ by 15.2.12, so $\mathcal {O}$ is maximal by Theorem 15.5.5.

For $p \equiv 1 ~(\text{ mod } ~{4})$, we had $B=({-p,-q} \mid {\mathbb {Q}})$ where $q \equiv 3 ~(\text{ mod } ~{4})$ is prime and $\displaystyle {\biggl (\displaystyle {\frac{q}{p}}\biggr )=-1}$, so that by qudaratic reciprocity $\displaystyle {-p}{q}=-\displaystyle {q}{p}=1$. In this case, let $c \in \mathbb Z$ be such that $c^2 \equiv -p ~(\text{ mod } ~{q})$. Then

$$\begin{aligned} \mathcal {O}:=\mathbb Z \oplus \mathbb Z \frac{1+j}{2} \oplus \mathbb Z \frac{i(1+j)}{2} \oplus \mathbb Z \frac{(c+i)j}{q} \end{aligned}$$

is a maximal order: one checks that $\mathcal {O}$ is closed under multiplication (in particular, the basis elements are integral), and then that ${{\,\mathrm{disc}\,}}\mathcal {O}=p$. The order $\mathbb Z \langle i,(1+j)/2 \rangle \subseteq \mathcal {O}$ has the larger reduced discriminant pq, hence the need for a denominator q in the fourth element.

For further discussion of explicit maximal orders over $\mathbb Z$, see Ibukiyama [Ibu82, pp. 181–182] or Pizer [Piz80a, Proposition 5.2]. For a more general construction, see Exercise 15.5.

6 Duality

To round out the chapter, we relate the discriminant and trace pairings to the dual and the different. For a detailed, general investigation of the dual in the context of other results for orders, see Faddeev [Fad65].

We continue with the hypothesis that R is a domain with $F={{\,\mathrm{Frac}\,}}R$. Let B be an F-algebra with $n :=\dim _F B<\infty$. As the trace pairing will play a significant role in what follows, we suppose throughout that B is separable (in particular, semisimple) as an F-algebra with reduced trace ${{\,\mathrm{trd}\,}}$. Let I, J be R-lattices in B.

Definition 15.6.1

The dual  of I (over R, with respect to ${{\,\mathrm{trd}\,}}$) is

$$\begin{aligned} I^{\sharp } :=\{\alpha \in B : {{\,\mathrm{trd}\,}}(\alpha I) \subseteq R \}= \{\alpha \in B : {{\,\mathrm{trd}\,}}(I\alpha ) \subseteq R \}. \end{aligned}$$

Some properties of the dual are evident.

Lemma 15.6.2

• (a) If $I \subseteq J$ then $I^{\sharp } \supseteq J^{\sharp }$.

• (b) For all $\beta \in B^\times$, we have $(\beta I)^\sharp = I^\sharp \beta ^{-1}$.

• (c) If $\mathfrak p \subseteq R$ is prime, then $(I_{(\mathfrak p )})^\sharp =(I^\sharp )_{(\mathfrak p )}$ and the same with the completion.

Proof. For parts (a) and (b), see Exercise 15.15. The proof of part (c) is similarly straightforward. $\square$

15.6.3

Suppose that I is free over R with basis $\alpha _1,\dots ,\alpha _n$. Since the trace pairing on B is nondegenerate (Theorem 7.9.4), there exists a dual basis $\alpha _i^{\sharp } \in B$ to $\alpha _i$ under the reduced trace ${{\,\mathrm{trd}\,}}$, so that ${{\,\mathrm{trd}\,}}(\alpha _i^\sharp \alpha _j)=0,1$ according as $i \ne j$ or $i=j$.

Then $I^\sharp$ is free over R with basis $\alpha _1^\sharp ,\dots ,\alpha _n^\sharp$: if $\beta =b_1\alpha _1^\sharp +\dots +b_n\alpha _n^\sharp$ with $b_1,\dots ,b_n \in F$, then $\beta \in I^\sharp$ if and only if ${{\,\mathrm{trd}\,}}(\alpha _i \beta )=b_i \in R$ for all i.

Lemma 15.6.4

$I^{\sharp }$ is an R-lattice in B.

Proof. Let $\alpha _1,\dots ,\alpha _n \in I$ be an F-basis for B, and let $J=\sum _i R\alpha _i \subseteq I$. Then there exists nonzero $r \in R$ such that $rI \subseteq J$, so $J \subseteq I \subseteq r^{-1}J$. Let $\alpha _1^\sharp ,\dots ,\alpha _n^\sharp \in B$ be the dual basis as in 15.6.3. It follows that $J^\sharp = \sum _i R \alpha _i^\sharp$ is an R-lattice, and consequently by Lemma 15.6.2(a)–(b) we have $r J^\sharp \subseteq I^\sharp \subseteq J^\sharp$; since R is noetherian, $I^\sharp$ is an R-lattice. $\square$

From now on, we suppose that R is a Dedekind domain; in particular, I is then projective as an R-module.

Lemma 15.6.5

The natural inclusion $I \hookrightarrow (I^\sharp )^\sharp \subseteq B$ is an equality.

Proof. If $\alpha \in I$ and $\beta \in I^\sharp$ then ${{\,\mathrm{trd}\,}}(\alpha \beta ) \subseteq R$ and $\alpha \in (I^\sharp )^\sharp$. To show that the map is an equality, we argue locally, so we may suppose that I is free over R with basis $\alpha _i$; then by applying 15.6.3 twice, $(I^\sharp )^\sharp$ has basis $(\alpha _i^\sharp )^\sharp =\alpha _i$, and equality holds. $\square$

Proposition 15.6.6

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

Proof. First the inclusion ($\subseteq$). Let $\alpha \in \mathcal {O}{}_{\textsf {\tiny {R}} }(I)$; then $I\alpha \subseteq I$, so $I^{\sharp } I \alpha \subseteq I^{\sharp } I$ and

$$\begin{aligned} {{\,\mathrm{trd}\,}}(\alpha I^\sharp I) = {{\,\mathrm{trd}\,}}(I^{\sharp } I \alpha ) \subseteq {{\,\mathrm{trd}\,}}(I^\sharp I) \subseteq R \end{aligned}$$

hence $\alpha I^\sharp \subseteq I^\sharp$ and $\alpha \in \mathcal {O}{}_{\textsf {\tiny {L}} }(I^\sharp )$. Thus $\mathcal {O}{}_{\textsf {\tiny {R}} }(I) \subseteq \mathcal {O}{}_{\textsf {\tiny {L}} }(I^{\sharp }) \subseteq \mathcal {O}{}_{\textsf {\tiny {R}} }((I^\sharp )^\sharp )=\mathcal {O}{}_{\textsf {\tiny {R}} }(I)$ by Lemma 15.6.5, so equality holds. A similar argument works on the other side. $\square$

The name dual is explained by the following lemma.

Proposition 15.6.7

Let ${{\,\mathrm{trd}\,}}(I)=\mathfrak a \subseteq F$. Then the map

(15.6.8)

is an isomorphism of $\mathcal {O}{}_{\textsf {\tiny {R}} }(I),\mathcal {O}{}_{\textsf {\tiny {L}} }(I)$-bimodules over R.

Proof. For $\beta \in I$, let $\phi _\beta :I \rightarrow \mathfrak a$ be defined by $\phi _\beta (\alpha ) = {{\,\mathrm{trd}\,}}(\alpha \beta )$ for $\alpha \in I$. The map $\beta \mapsto \phi _\beta \in {{\,\mathrm{Hom}\,}}_R(I,\mathfrak a )$ from (15.6.8) is an R-module homomorphism. Moreover, it a map of $\mathcal {O}{}_{\textsf {\tiny {R}} }(I),\mathcal {O}{}_{\textsf {\tiny {L}} }(I)$-bimodules: if $\gamma \in \mathcal {O}{}_{\textsf {\tiny {L}} }(I)$ then $\gamma \in \mathcal {O}{}_{\textsf {\tiny {R}} }(I^\sharp )$ by Lemma 15.6.6, with induced map

$$\begin{aligned} \phi _{\beta \gamma }(\alpha )={{\,\mathrm{trd}\,}}(\alpha \beta \gamma )={{\,\mathrm{trd}\,}}(\gamma \alpha \beta ) = \phi _\beta (\gamma \alpha ) = (\gamma \phi _\beta )(\alpha ) \end{aligned}$$

(15.6.9)

and similarly on the other side.

Finally, we prove that the map (15.6.8) is also an isomorphism. Extending scalars to F, the trace pairing gives an isomorphism of F-vector spaces

$$\begin{aligned} {{\,\mathrm{Hom}\,}}_R(I,\mathfrak a ) \otimes _R F \simeq {{\,\mathrm{Hom}\,}}_F(B,F)&\simeq B \\ \beta&\mapsto \phi _\beta \end{aligned}$$

because the pairing is nondegenerate (as B is separable). So immediately the map is injective; and it is surjective, because if $\phi \in {{\,\mathrm{Hom}\,}}_R(I,\mathfrak a )$ then $\phi =\phi _\beta$ for some $\beta \in B$, but then $\phi (\alpha )={{\,\mathrm{trd}\,}}(\alpha \beta ) \in R$ for all $\alpha \in I$, so $\beta \in I^\sharp$ by definition. $\square$

Remark 15.6.10. The content of Proposition 15.6.7 is that although one can always construct the module dual, the trace pairing concretely realizes this module dual as a lattice. (And we speak of bimodules in the proposition because ${{\,\mathrm{Hom}\,}}_R(I,\mathfrak a )$ does not come equipped with the structure of R-lattice in B.) This module duality, and the fact that I is projective over R, can be used to give another proof of Lemma 15.6.5.

The dual asks for elements that pair integrally under the trace. We might also ask for elements that multiply one lattice into another, as follows.

Definition 15.6.11

Let I, J be R-lattices. Theleft colon lattice of I with respect to J is the set

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

and similarly theright colon lattice is

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

Note that $(I:I){}_{\textsf {\tiny {L}} }=\mathcal {O}{}_{\textsf {\tiny {L}} }(I)$ is the left order of I (and similarly on the right). The same proof as in Lemma 10.2.7 shows that $(I:J){}_{\textsf {\tiny {L}} }$ and $(I:J){}_{\textsf {\tiny {R}} }$ are R-lattices.

Lemma 15.6.12

We have

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

Proof. We have $\beta \in (IJ)^\sharp$ if and only if ${{\,\mathrm{trd}\,}}(\beta IJ) \subseteq R$ if and only if $\beta \alpha \in J^\sharp$ for all $\alpha \in I$ if and only if $\beta \in (J^\sharp :I){}_{\textsf {\tiny {L}} }$. A similar argument works on the other side, considering ${{\,\mathrm{trd}\,}}(IJ \beta )$ instead. $\square$

Corollary 15.6.13

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

Proof. Combining Lemmas 15.6.5 and 15.6.12,

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

and similarly on the right. $\square$

Definition 15.6.14

The level  of I is the fractional ideal ${{\,\mathrm{lvl}\,}}(I)={{\,\mathrm{nrd}\,}}(I^\sharp ) \subseteq F$.

We now relate the above duality to the discriminant.

Definition 15.6.15

Thecodifferent of $\mathcal {O}$ is

$$\begin{aligned} {{\,\mathrm{codiff}\,}}(\mathcal {O}) :=\mathcal {O}^{\sharp }. \end{aligned}$$

Lemma 15.6.16

$\mathcal {O}{}_{\textsf {\tiny {L}} }({{\,\mathrm{codiff}\,}}(\mathcal {O}))=\mathcal {O}{}_{\textsf {\tiny {R}} }({{\,\mathrm{codiff}\,}}(\mathcal {O}))=\mathcal {O}$ and $\mathcal {O}\subseteq {{\,\mathrm{codiff}\,}}(\mathcal {O})$.

Proof. By Proposition 15.6.6, $\mathcal {O}=\mathcal {O}{}_{\textsf {\tiny {R}} }(\mathcal {O})=\mathcal {O}{}_{\textsf {\tiny {L}} }({{\,\mathrm{codiff}\,}}(\mathcal {O}))$ and similarly on the right. And $\mathcal {O}\subseteq {{\,\mathrm{codiff}\,}}(\mathcal {O})$ since ${{\,\mathrm{trd}\,}}(\mathcal {O}\mathcal {O})={{\,\mathrm{trd}\,}}(\mathcal {O}) \subseteq R$. $\square$

The major role played by the codifferent is its relationship to the discriminant, as follows.

Lemma 15.6.17

${{\,\mathrm{disc}\,}}(\mathcal {O})=[{{\,\mathrm{codiff}\,}}(\mathcal {O}):\mathcal {O}]_R$.

Proof. For a prime $\mathfrak p \subseteq R$ we have ${{\,\mathrm{disc}\,}}(\mathcal {O})_{(\mathfrak p )}={{\,\mathrm{disc}\,}}(\mathcal {O}_{(\mathfrak p )})$ and $[\mathcal {O}_{(\mathfrak p )}^\sharp : \mathcal {O}_{(\mathfrak p )}]_{R_{(\mathfrak p )}}=([\mathcal {O}^\sharp :\mathcal {O}]_R)_{(\mathfrak p )}$, and so to establish the equality we may argue locally. Since $\mathcal {O}_{(\mathfrak p )}$ is free over $R_{(\mathfrak p )}$, we reduce to the case where $\mathcal {O}$ is free over R, say $\mathcal {O}=\sum _i R \alpha _i$. Then $\mathcal {O}^\sharp =\sum _i R \alpha _i^\sharp$ with $\alpha _1^\sharp ,\dots ,\alpha _n^\sharp \in B$ the dual basis, as in 15.6.3.

The ideal ${{\,\mathrm{disc}\,}}(\mathcal {O})$ is principal, generated by $d(\alpha _1,\dots ,\alpha _n)=\det ({{\,\mathrm{trd}\,}}(\alpha _i\alpha _j))_{i,j}$; at the same time, the R-index $[\mathcal {O}^\sharp :\mathcal {O}]_R$ is generated by $\det (\delta )$ where $\delta$ is the change of basis from $\alpha _i^{\sharp }$ to $\alpha _i$. But $\delta$ is precisely the matrix $({{\,\mathrm{trd}\,}}(\alpha _i\alpha _j))_{i,j}$ (Exercise 15.14), and the result follows. $\square$

Remark 15.6.18. In certain circumstances, it is preferable to work with an integral ideal measuring the discriminant, so instead of the codifferent instead a different: we will want to take a kind of inverse. We study this in the next chapter: see section 16.8.

Exercises

Unless otherwise specified, let R be a noetherian domain with field of fractions F.

• 1. Let ${{\,\mathrm{char}\,}}F=2$ and let $\displaystyle {\biggl [\frac{a,b}{F}\biggr )}$ be a quaternion algebra over F with $a,b \in R$ and $b \ne 0$. Show that $\mathcal {O}=R + Ri + Rj + Rij$ is an R-order in B and compute the (reduced) discriminant of $\mathcal {O}$.

• 2. Let $B={{\,\mathrm{M}\,}}_n(F)$ and $\mathcal {O}={{\,\mathrm{M}\,}}_n(R)$ with $n \ge 1$. Show that ${{\,\mathrm{disc}\,}}(\mathcal {O})=R$. [Hint: Compute directly on a basis $\{e_{ij}\}_{i,j}$ of matrix units, which satisfy $e_{ij}e_{i'j'} = e_{ij'}$ if $j=i'$, otherwise zero.]

• 3. Suppose R is a global ring, so F is a global field; let B be a quaternion algebra over F and let $\mathcal {O}\subseteq B$ be an R-order. Prove that for all primes $\mathfrak p \subseteq R$, we have $\mathcal {O}_\mathfrak p \simeq {{\,\mathrm{M}\,}}_n(R_\mathfrak p )$ if and only if $\mathfrak p \not \mid {{\,\mathrm{disc}\,}}\mathcal {O}$.

• 4. Let $B :=({K,b} \mid {F})$ be a quaternion algebra over a field F with $b \in F^\times$. Let $S \subseteq K$ be an R-order with $\mathfrak d :={{\,\mathrm{disc}\,}}(S)$; let $\mathfrak b \subseteq K$ be a fractional S-ideal (which can be but need not be invertible), and finally let $\mathcal {O}:=S \oplus \mathfrak b j$.

  • (a) Show that $\mathcal {O}$ is an R-order if and only if ${{\,\mathrm{Nm}\,}}_{K|F} \mathfrak b \subseteq b^{-1} R$.

  • (b) Compute that ${{\,\mathrm{discrd}\,}}\mathcal {O}= \mathfrak d ({{\,\mathrm{Nm}\,}}_{K|F} \mathfrak b )b$.

• 5. In this exercise, we consider a construction of maximal orders as crossed products in the simplest case over $\mathbb Q$, continuing Exercise 14.9. Let $B :=\displaystyle {\biggl (\frac{q^\diamondsuit ,b}{\mathbb {Q}}\biggr )}$ be a quaternion algebra of discriminant D, where $b \in \mathbb Z$ is squarefree with $D \mid b$ and q is an odd prime with $q^\diamondsuit =\pm q \equiv 1 \pmod {4}$, the minus sign if B is indefinite.

  Let $K :=\mathbb Q (\sqrt{q^\diamondsuit })$ be the quadratic field of discriminant $q^\diamondsuit$. Let $S \subseteq K$ be the ring of integers of K, so ${{\,\mathrm{disc}\,}}S=q^\diamondsuit$.

  • (a) Show that for all odd primes $p \mid (b/D)$, we have $\biggl (\displaystyle {\frac{q^{\diamondsuit }}{p}}\biggr )=1$. Conclude there exists an ideal $\mathfrak b \subseteq S$ such that ${{\,\mathrm{Nm}\,}}\mathfrak b = b/D$.

  • (b) Let $\mathfrak q \subseteq S$ be the unique prime above q, and let

    $$\begin{aligned} \mathcal {O}:=S \oplus (\mathfrak q \mathfrak b )^{-1} j \end{aligned}$$

    Show that $\mathcal {O}$ is a maximal order in B.

  • (c) Let $c \in \mathbb Z$ satisfy $c^2 \equiv q^{\diamondsuit } ~(\text{ mod } ~{4b/D})$. Show that the order $\mathcal {O}$ in (b) can be written

    $$\begin{aligned} \mathcal {O}= \mathbb Z \oplus \mathbb Z \frac{1+i}{2} \oplus \mathbb Z j \oplus \mathbb Z \frac{D(c+i)j}{2bq}. \end{aligned}$$

• 6. Let B be a separable F-algebra with $\dim _F B=n$. Show that $\alpha _1,\dots ,\alpha _n \in B$ are linearly independent over F if and only if $d(\alpha _1,\dots ,\alpha _n) \ne 0$.

• 7. Let $\mathcal {O}$ be an R-order. Show that ${{\,\mathrm{disc}\,}}(\mathcal {O})$ is generated by

  $$\begin{aligned} \{d(1,\alpha _1,\dots ,\alpha _{n-1}) : \alpha _1,\dots ,\alpha _{n-1} \in \mathcal {O}\}. \end{aligned}$$

• 8. Let I be an R-lattice in B over F, let K be a finite extension field of F, and let S be a domain containing R with field of fractions K. Show that

  $$\begin{aligned} {{\,\mathrm{disc}\,}}(I \otimes _R S) = {{\,\mathrm{disc}\,}}(I) \otimes _R S = {{\,\mathrm{disc}\,}}(I)S. \end{aligned}$$

• 9. Let ${{\,\mathrm{char}\,}}F \ne 2$ and let B be a quaternion algebra over F. Let $\alpha ,\beta \in B$ be such that $F(\alpha ) \cap F(\beta ) = F$. Recall the discriminant form $\Delta$ (Exercise 4.3), and let

  $$\begin{aligned} s :={{\,\mathrm{trd}\,}}(\alpha \beta ) - \frac{{{\,\mathrm{trd}\,}}(\alpha ){{\,\mathrm{trd}\,}}(\beta )}{2}. \end{aligned}$$

  Show that

  $$\begin{aligned} d(1,\alpha ,\beta ,\alpha \beta ) = -(s^2-4\Delta (\alpha )\Delta (\beta ))^2 = -\Delta (\alpha \beta )^2. \end{aligned}$$

  [Hint: reduce to the case where ${{\,\mathrm{trd}\,}}(\alpha )={{\,\mathrm{trd}\,}}(\beta )=0$, noting the invariance of s.]

• 10. Let B be a quaternion algebra over F. Define $m:B \times B \times B \rightarrow F$ by $m(\alpha _1,\alpha _2,\alpha _3) :={{\,\mathrm{trd}\,}}([\alpha _1,\alpha _2]\overline{\alpha _3})$ for $\alpha _i \in B$. If $\beta _i=M\alpha _i$ for some $M \in {{\,\mathrm{M}\,}}_3(F)$, show that

  $$\begin{aligned} m(\beta _1,\beta _2,\beta _3)=\det (M) m(\alpha _1,\alpha _2,\alpha _3). \end{aligned}$$

• 11. Let B be a quaternion algebra over F. Give another proof that

  $$\begin{aligned} m(\alpha _1,\alpha _2,\alpha _3)^2=d(1,\alpha _1,\alpha _2,\alpha _3) \end{aligned}$$

  (cf. Brzezinski [Brz82, Lemma 1.1(a)]) for all $\alpha _i \in B$ as follows:

  • (a) Suppose $B={{\,\mathrm{M}\,}}_2(F)$. Show that the matrix units

    $$ e_{12}=\begin{pmatrix} 0 &{} 1 \\ 0 &{} 0 \end{pmatrix}, \quad e_{21}=\begin{pmatrix} 0 &{} 0 \\ 1 &{} 0 \end{pmatrix}, \quad e_{22}=\begin{pmatrix} 1 &{} 0 \\ 0 &{} 0 \end{pmatrix} $$

    span B/F, and $m(e_{12},e_{21},e_{22})^2=d(1,e_{12},e_{21},e_{22})$. Conclude using Exercise 15.10.

  • (b) Reduce to (a) in general by taking a splitting field for B.

• 12. Suppose $R=R_{({{{\texttt {\textit{S}}}}})}$ is a global ring with $2 \in R^\times$. Let $K \supset F$ be a quadratic field extension and $S\subseteq K$ an R-order. Let ${{\,\mathrm{Ram}\,}}(K)$ be the set of places of F that are ramified in K. Show that S is maximal if and only if its discriminant is equal to

  $$\begin{aligned} {{\,\mathrm{disc}\,}}_R(S)=\prod _{\begin{array}{c} \mathfrak p \in {{\,\mathrm{Ram}\,}}(K) \setminus {{{\texttt {\textit{S}}}}} \end{array}} \mathfrak p \subseteq R \end{aligned}$$

  in analogy with Theorem 15.5.5.

• 13. Let B be a finite-dimensional F-algebra with a standard involution. Compare

  $$\begin{aligned} \det ({{\,\mathrm{trd}\,}}(\alpha _i\alpha _j))_{i,j} \quad \text {with} \quad \det ({{\,\mathrm{trd}\,}}(\alpha _i\overline{\alpha _j}))_{i,j} \end{aligned}$$

  for $\alpha _i \in B$, and show that defining the discriminant of an order $\mathcal {O}\subseteq B$ with respect to either pairing gives the same result.

$\triangleright$ 14.:

Let B be a semisimple F-algebra with $\dim _F B=n$, let I be an R-lattice that is free over R with basis $\alpha _1,\dots ,\alpha _n$, and let $\alpha _1^\sharp ,\dots ,\alpha _n^\sharp \in B$ be the dual basis, so ${{\,\mathrm{trd}\,}}(\alpha _i^\sharp \alpha _j)=1,0$ according as $i=j$ or not. Show that the change of basis matrix from $\{\alpha _i^\sharp \}_i$ to $\{\alpha _i\}_i$ is given by $({{\,\mathrm{trd}\,}}(\alpha _i\alpha _j))_{i,j}$.

• 15. Let $I \subseteq B$ be an R-lattice in a separable algebra B.

  • (a) If $J \subseteq B$ is an R-lattice with $I \subseteq J$, show that $I^{\sharp } \supseteq J^{\sharp }$.

  • (b) Show that for all $\beta \in B^\times$, we have $(\beta I)^\sharp = I^\sharp \beta ^{-1}$

• 16. Let R be a DVR with maximal ideal $\mathfrak p =\pi R$, and let $\mathcal {O}:=\begin{pmatrix} R &{} R \\ \mathfrak p ^e &{} R \end{pmatrix}$ for $e \ge 0$. Compute the codifferent ${{\,\mathrm{codiff}\,}}(\mathcal {O})$: in particular, show that ${{\,\mathrm{codiff}\,}}(\mathcal {O})$ is a principal two-sided $\mathcal {O}$-ideal, and find a generator. Verify Lemma 15.6.17.

• 17. Let R be a noetherian domain with $F={{\,\mathrm{Frac}\,}}R$. Let B be a central simple algebra over F. Let $\mathcal {O}\subseteq B$ be an R-order. We say $\mathcal {O}$ is Azumaya  if $\mathcal {O}$ is R-simple, which is to say every two-sided ideal $I \subseteq \mathcal {O}$ is of the form $\mathfrak a \mathcal {O}=\mathcal {O}\mathfrak a$ with $\mathfrak a = I \cap R \subseteq R$.

  • (a) Show that $\mathcal {O}$ is Azumaya if and only if every R-algebra homomorphism $\mathcal {O}\rightarrow A$ is either the zero map or injective.

  • (b) Show that $\mathcal {O}$ is Azumaya if and only if $\mathcal {O}/\mathfrak m \mathcal {O}$ is a central simple algebra over the field $R/\mathfrak m$ for all maximal ideals $\mathfrak m$ of R.

  • (c) Suppose that B is a quaternion algebra. Show that the quaternion order $\mathcal {O}$ is Azumaya if and only if ${{\,\mathrm{disc}\,}}\mathcal {O}=R$. Conclude that the only Azumaya quaternion algebra over the valuation ring R of a local field is ${{\,\mathrm{M}\,}}_2(R)$, and that the only Azumaya quaternion algebra over $\mathbb Z$ is ${{\,\mathrm{M}\,}}_2(\mathbb Z )$.

[See Auslander and Goldman [AG60] or Milne [Milne80, §IV.1].]

• 18. Let G be a finite group of order $n=\#G$ and let R be a 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 17.6.

  • (a) Consider the algebra trace ${{\,\mathrm{Tr}\,}}_{B|F}$ and its associated bilinear form. Show that in the basis of F[G] given by the elements of G that the trace pairing is the scalar matrix n.

  • (b) Now write $B \simeq B_1 \times \dots \times B_r$ as a product of simple F-algebras. Let $K_i$ be the center of $B_i$, and let $\dim _{K_i} B_i=n_i^2$. Show that ${{\,\mathrm{Tr}\,}}|_{B_i}=n_i {{\,\mathrm{trd}\,}}$. Let $\mathcal {O}=R[G]$, and suppose that $\mathcal {O}\simeq \mathcal {O}_1 \times \dots \times \mathcal {O}_r$. Show that

    $$\begin{aligned} {{\,\mathrm{codiff}\,}}(\mathcal {O}) = n_1^{-1} \mathcal {O}_1 \times \dots \times n_r^{-1} \mathcal {O}_r. \end{aligned}$$