§ 9. Norms and traces

Throughout this paragraph, $K$ will denote a commutative ring and $A$ a unital associative $K$ -algebra. Every $A$ -module will be assumed to be given the $K$ -module structure obtained by restricting the scalars to $K$ .

1. Norms and traces relative to a module

Definition 1

Let $M$ be an $A$ -module admitting a finite basis as a $K$ -module. For all $a \in A$ , let $a_M$ be the endomorphism $x \mapsto ax$ of the $K$ -module $M$ . The trace, determinant and characteristic polynomial of $a_M$ are called respectively the trace, norm and characteristic polynomial of $z$ relative to $M$ .

The trace and norm of $a$ are therefore elements of $K$ , denoted respectively by $\operatorname{Tr}_{M/K}(a)$ and $N_{M/K}(a)$ ; the characteristic polynomial of $a$ is an element of $K[X]$ , denoted by $\mathrm{Pc}_{M/K}(a; X)$ . We omit $K$ in the above notation when there is no risk of confusion.

From the properties of the trace and determinant of an endomorphism (II, § 4, no. 3 and § 8, no. 1) we obtain the relations

\operatorname{Tr}_M(a + a') = \operatorname{Tr}_M(a) + \operatorname{Tr}_M(a')

(2) $\mathrm{Tr}_M(aa') = \mathrm{Tr}_M(a'a)$ (3) $N_M(aa') = N_M(a)N_M(a')$

for all $a, a'$ in $A$ .

Let $(e_i)_{1 \leq i \leq n}$ be a basis of the $K$ -module $M$ and $(m_{ij}(a))$ the matrix of the endomorphism $a_M$ with respect to this basis. The functions $m_{ij}$ are linear forms on the $K$ -module $A$ and

(4) $\mathrm{Tr}_M(a) = \sum_{i=1}^n m_{ii}(a)$ (5) $N_M(a) = \det(m_{ij}(a))$ (6) $\mathrm{Pc}(a; X) = \det(\delta_{ij}X - m_{ij}(a)).$

It follows from the method of calculating a determinant (§ 8, no. 11, formula (50)) that

(7) $\mathrm{Pc}_M(a; X) = X^n + c_1 X^{n-1} + \cdots + c_n$

where

(8) $c_1 = -\mathrm{Tr}_M(a), \quad c_n = (-1)^n N_M(a).$

For $\lambda \in K$ ,

(9) $\mathrm{Tr}_M(\lambda) = n.\lambda, \quad N_M(\lambda) = \lambda^n, \quad \mathrm{Pc}_M(\lambda; X) = (X - \lambda)^n.$

Let $K'$ be a commutative $K$ -algebra. Let $M' = K' \otimes_K M$ and $A' = K' \otimes_K A$ , so that $M'$ has an $A'$ -module structure (§ 4, Example 2). As a $K'$ -module $M'$ admits the basis consisting of the $1 \otimes e_i$ ( $1 \leq i \leq n$ ) and the matrix of $a_M$ with respect to $(e_i)$ is equal to the matrix of $(1 \otimes a)_M'$ with respect to $(e'_i)$ . Then

(12) $\mathrm{Tr}_{M'}(1 \otimes a) = \mathrm{Tr}_M(a).1, \quad N_{M'}(1 \otimes a) = N_M(a).1$ $\mathrm{Pc}_{M'}(1 \otimes a; X) = \mathrm{Pc}_M(a; X).1$

for all $a \in A$ , where 1 denotes the unit element of $K'$ . If in particular we take $K' = K[X]$ , then

(13) $\mathrm{Pc}_{M/K}(a; X) = N_{M[X]/K[X]}(X - a).$

2. Properties of norms and traces relative to a module

If $M$ and $M'$ are two isomorphic $A$ -modules with finite bases over $K$ , then, for all $a \in A$ ,

(14) $\mathrm{Tr}_{M'}(a) = \mathrm{Tr}_M(a), \quad N_{M'}(a) = N_M(a), \quad \mathrm{Pc}_{M'}(a; X) = \mathrm{Pc}_M(a; X)$

for if $f$ is an isomorphism of $M$ onto $M'$ , the matrix of $a_M$ with respect to a basis $B$ of $M$ over $K$ is the same as the matrix of $a_{M'}$ with respect to the basis $f(B)$ of $M'$ .

Proposition 1

Let $M = M_0 \supset M_1 \supset \ldots \supset M_r = \{0\}$ be a decreasing sequence of submodules of an $A$ -module $M$ such that each of the $K$ -modules $P_i = M_{i-1}/M_i$ ( $1 \leq i \leq r$ ) admits a finite basis. Then the $K$ -module $M$ admits a finite basis and

\operatorname{Tr}_M(a) = \sum_{i=1}^r \operatorname{Tr}_{P_i}(a), \qquad N_M(a) = \prod_{i=1}^r N_{P_i}(a)

(15)

P_{C_M}(a; X) = \prod_{i=1}^r P_{C_{P_i}}(a; X).

Let $B'_i$ be a basis of $P_i$ over $K$ ; then a system of representatives $B_i$ of $B'_i$ (mod. $M_i$ ) is a basis of a supplementary submodule of the $K$ -module $M_i$ in the $K$ -module $M_{i-1}$ (II, § 1, no. 11, Proposition 21). The union $B$ of the $B_i$ ( $1 \leq i \leq r$ ) is a basis of $M$ over $K$ . Let $X_{tt}$ be the matrix of the endomorphism $a_{P_i}$ with respect to the basis $B'_i$ . It is immediate that the matrix of $a_M$ with respect to $B$ is of the form

\begin{pmatrix} X_{rr} & X_{r,r-1} & \cdots & X_{r,1} \\ 0 & X_{r-1,r-1} & \cdots & X_{r-1,1} \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & X_{11} \end{pmatrix}

and the proposition follows from formulae (4), (5) and (6) of no. 1 and formula (31) of § 8, no. 6.

Proposition 2

Let $A, A'$ be two $K$ -algebras, $M$ an $A$ -module and $M'$ an $A'$ -module. Suppose that $M$ and $M'$ are free $K$ -modules of respective dimensions $n$ and $n'$ and consider $M \otimes_K M'$ as an $(A \otimes_K A')$ -module ( $§ 4$ , no. 3, Example 2). Then, for $a \in A$ and $a' \in A'$ ,

\operatorname{Tr}_{M \otimes M'}(a \otimes a') = \operatorname{Tr}_M(a) \operatorname{Tr}_{M'}(a')

(16)

N_{M \otimes M'}(a \otimes a') = (N_M(a))^{n'} (N_{M'}(a'))^n.

(17)

Formula (16) follows from II, § 4, no. 4, formula (26) and formula (17) from § 8, no. 6, formula (33).

3. Norm and trace in an algebra

Definition 2

Let $A$ be a $K$ -algebra which is a finite-dimensional free $K$ -module. For every element $a \in A$ , the trace (resp. norm,\footnote{† This notion should not be confused with the notion for norm in an algebra over a valued field (\emph{General Topology}), IX, § 3, no. 7.} resp. characteristic polynomial) of $a$ relative to $A$ and $K$ is the trace (resp. determinant, resp. characteristic polynomial) of the endomorphism $x \mapsto ax$ of the $K$ -module $A$ .

The trace, norm and characteristic polynomial of $a \in A$ relative to $A$ and $K$ are denoted by $\mathrm{Tr}_{A/K}(a)$ , $N_{A/K}(a)$ and $\mathrm{Pc}_{A/K}(a; X)$ ; we omit $K$ and even $A$ from this notation when there is no risk of confusion. Note that the trace (resp. norm, characteristic polynomial) of $a \in A$ is just the trace (resp. norm, characteristic polynomial) of $a$ relative to the $A$ -module $A_s$ .

Suppose that $A$ is the product $A_1 \times A_2 \times \cdots \times A_m$ of a finite number finite-dimensional algebras over $K$ which are free $K$ -modules. Using the above remark and Proposition 1 of no. 2, we have, for every element

a = (a_1, \ldots, a_m) \in A,

\mathrm{Tr}_{A/K}(a) = \sum_{i=1}^m \mathrm{Tr}_{A_i/K}(a_i), \qquad N_{A/K}(a) = \prod_{i=1}^m N_{A_i/K}(a_i)

(18)

\mathrm{Pc}_{A/K}(a; X) = \prod_{i=1}^m \mathrm{Pc}_{A_i/K}(a_i; X).

Similarly, Proposition 2 of no. 2 shows that if $A$ and $A'$ are two algebras, which are free $K$ -modules, of finite dimensions $n, n'$ respectively over $K$ , then, for $a \in A, a' \in A'$ .

\mathrm{Tr}_{A \otimes A'}(a \otimes a') = \mathrm{Tr}_A(a) \, \mathrm{Tr}_{A'}(a')

(19)

N_{A \otimes A'}(a \otimes a') = (N_A(a))^{n'} (N_{A'}(a'))^n.

(20)

Finally let $A$ be a finite-dimensional algebra over $K$ which is a free $K$ -module, $h$ a homomorphism of $K$ into a commutative ring $K'$ and $A' = A_{(K')}$ the $K'$ -algebra derived from $A$ by extending the scalars by means of $h$ . It follows from formula (12) of no. 1 that, for all $a \in A$ ,

\mathrm{Tr}_{A'/K'}(1 \otimes a) = h(\mathrm{Tr}_{A/K}(a)), \qquad N_{A'/K'}(1 \otimes a) = h(N_{A/K}(a))

(21)

\mathrm{Pc}_{A'/K'}(1 \otimes a; X) = \bar{h}(\mathrm{Pc}_{A/K}(a; X))

where $\bar{h}$ is the homomorphism $K[X] \to K'[X]$ derived from $h$ . In particular, for $K' = K[X]$ , we obtain, writing $A[X] = A \otimes_K K[X]$ ,

\mathrm{Pc}_{A/K}(a; X) = N_{A[X]/K[X]}(X - a).

(22)

More generally, if $K'$ is a commutative $K$ -algebra and $A' = A \otimes_K K'$ , then, for all $x \in A'$ ,

\mathrm{Pc}_{A/K}(a; x) = N_{A'/K'}(x - a).

Example

(1) Let $A$ be a quadratic algebra over $K$ of type $(\alpha, \beta)$ and $(e_1, e_2)$ a basis of type $(\alpha, \beta)$ (§ 2, no. 3). For $x = \xi e_1 + \eta e_2$ , $\mathrm{Tr}_{A/K}(x) = 2\xi + \beta \eta$ and $N_{A/K}(x) = \xi^2 + \beta \xi \eta - \alpha \eta^2$ ; these functions are therefore identical with the Cayley trace and norm of $x$ (§ 2, no. 24).

(2) Let $A$ be a quaternion algebra over $K$ . A direct calculation allows us to verify that $\mathrm{Tr}_{A/\mathbf{K}}(x) = 2T(x)$ and $N_{A/\mathbf{K}}(x) = (N(x))^2$ , where T and N are the Cayley trace and norm (§ 2, no. 4).

(3) Let $A = \mathbf{M}_n(\mathbf{K})$ and let the canonical basis $(E_{ij})$ of A (II, § 10, no. 3) be arranged in lexicographic order. Then it is immediately seen that for every matrix $X = \sum_{i,j} \xi_{ij} E_{ij}$ the matrix (of order $n^2$ ) of the endomorphism $Y \mapsto XY$ is of the form

\begin{pmatrix} X & 0 & \ldots & 0 \\ 0 & X & \ldots & 0 \\ \ldots & \ldots & \ldots & \ldots \\ 0 & 0 & \ldots & X \end{pmatrix}

whence $\mathrm{Tr}_{A/\mathbf{K}}(X) = n \cdot \mathrm{Tr}(X)$ and $N_{A/\mathbf{K}}(X) = (\det(X))^n$ .

4. Properties of norms and traces in an algebra

Proposition 3

Let A be a K-algebra admitting a finite basis. For an element $a \in A$ to be invertible, it is necessary and sufficient that its $N_{A/\mathbf{K}}(a)$ be invertible in $\mathbf{K}$ .

If $a$ admits an inverse $a'$ in $A$ , then

N_{A/\mathbf{K}}(a) N_{A/\mathbf{K}}(a') = N_{A/\mathbf{K}}(aa') = N_{A/\mathbf{K}}(1) = 1

by formula (3) of no. 1. Conversely, if $N_{A/\mathbf{K}}(a)$ is invertible, the endomorphism $h : x \mapsto ax$ is bijective (§ 8, no. 2, Theorem 1). Then there exists $a' \in A$ such that $aa' = 1$ ; then $h(a'a - 1) = aa'a - a = (aa' - 1)a = 0$ , whence $aa' = 1$ since $h$ is injective. Hence $a'$ is the inverse of $a$ .

Proposition 4

Let A be a K-algebra admitting a finite basis. For all $a \in A$ , $\mathrm{Pc}_{A/\mathbf{K}}(a; a) = 0$ .

This follows immediately from the Cayley-Hamilton theorem (§ 8, no. 11, Proposition 20).

Proposition 5

Let A be a K-algebra and m a two-sided ideal of A. Suppose that $A_0 = A/m$ is a free K-module of finite dimension n, that there exists an integer $r > 0$ such that $m^r = \{0\}$ and that $m^{i-1}/m^i$ is a free $A_0$ -module of finite dimension $s_i$ for $1 \leq i \leq r$ . Let $s = s_1 + \cdots + s_r$ and for all $a \in A$ let $a_0$ denote the class of $a$ mod. m. Then A is a free K-module of dimension $ns$ and, for all $a \in A$ ,

\mathrm{Tr}_A(a) = s \cdot \mathrm{Tr}_{A_0}(a_0), \qquad N_A(a) = (N_{A_0}(a_0))^s

\mathrm{Pc}_A(a; X) = (\mathrm{Pc}_{A_0}(a_0; X))^s.

By virtue of II, § 1, no. 13, Proposition 25, $m^{i-1}/m^i$ is a free K-module of dimension $ns_i$ . Hence Proposition 1 of no. 2 can be applied with $P_i = m^{i-1}/m^i$ ;

this shows in the first place that $A$ is a free $K$ -module of dimension $n(s_1 + \cdots + s_r) = ns$ . Moreover, the hypothesis implies that the $A$ -module $P_i$ is isomorphic to a direct sum of $s_i$ submodules isomorphic to the $A$ -module $A_0$ ; by Proposition 1 of no. 2, therefore $N_{P_i}(a) = N_{A_0}(a)^{s_i}$ ; finally therefore

N_A(a) = N_{A_0}(a)^s.

In this formula

N_{A_0}(a)

is defined by considering

A_0

as a left

A

-module and it is equal to the determinant of the

K

-linear mapping

x \mapsto ax

A_0

into itself; but, as

ax = a_0x

for

x \in A_0

N_{A_0}(a) = N_{A_0}(a_0)

, which completes the proof of formula (23) for the norm. The two others are shown analogously.

Proposition 6

Let $A$ be a commutative $K$ -algebra admitting a finite basis over $K$ and $V$ an $A$ -module admitting a finite basis over $A$ . Then $V$ admits a finite basis over $K$ and for every $A$ -endomorphism $u$ of $V$ , if $u_K$ is the mapping $u$ considered as a $K$ -endomorphism of $V$ ,*

\begin{align*} \operatorname{Tr}(u_K) &= \operatorname{Tr}_{A/K}(\operatorname{Tr}(u)), \qquad \det(u_K) = N_{A/K}(\det(u)) \\ \operatorname{Pc}(u_K; X) &= N_{A[X]/K[X]}(\operatorname{Pc}(u; X)). \end{align*}

(24)

Let $(a_i)_{1 \leq i \leq m}$ be a basis of $A$ over $K$ and $(e_j)_{1 \leq j \leq n}$ a basis of $V$ over $A$ ; then $(a_i e_j)$ is a basis of $V$ over $K$ (II, § 1, no. 13, Proposition 25). On the other hand the third of formulae (24) can be deduced from the second applied to the endomorphism $X - \bar{u}$ of the $A[X]$ -module $A[X] \otimes_A V$ (§ 8, no. 10). It will therefore suffice to show the first two formulae in (24). We shall first establish the following lemma:

*Lemma 1. Let $X_{ij}$ ( $1 \leq i \leq n,\ 1 \leq j \leq n$ ) be $n^2$ indeterminates, $X$ the square matrix $(X_{ij})$ of order $n$ and $D(X_{11}, \ldots, X_{nn}) \in \mathbf{Z}[X_{11}, \ldots, X_{nn}]$ the determinant $\det(X)$ . On the other hand let $A$ be a commutative ring, $M_{ij}$ ( $1 \leq i \leq n,\ 1 \leq j \leq n$ ) $n^2$ matrices of order $m$ over $A$ , which are pairwise permutable, and $M$ the square matrix of order $mn$ over $A$ which can be expressed as a square matrix of matrices (II, § 10, no. 7)

M = \begin{pmatrix} M_{11} & M_{12} & \cdots & M_{1n} \\ M_{21} & M_{22} & \cdots & M_{2n} \\ \cdots & \cdots & \cdots & \cdots \\ M_{n1} & M_{n2} & \cdots & M_{nn} \end{pmatrix}

Then the determinant of

M

is equal to the determinant of the square matrix

D(M_{11}, \ldots, M_{nn})

of order

m

We proceed by induction on $n$ , the cases $n = 0$ and $n = 1$ being trivial.

Let $Z$ be a new indeterminate and $N_{ij}$ the matrix $M_{ij} + \delta_{ij} Z I_m$ ( $\delta_{ij}$ the Kronecker index). If $D^{ij}(X_{11}, \ldots, X_{nn})$ is the cofactor of $X_{ij}$ in the matrix $X$ (§ 8, no. 6), then

\text{(25)} \quad X_{ji} D^{ki}(X_{11}, \ldots, X_{nn}) = \delta_{jk} D(X_{11}, \ldots, X_{nn})

(§ 8, no. 6, formula (28)). We write $N'_{ij} = D^{ij}(N_{11}, \ldots, N_{nn})$ , which is a square matrix of order $m$ over $\mathbf{A}[Z]$ and consider the product $N . U$ , where

U = \begin{pmatrix} N'_{11} & 0 & \cdots & 0 \\ N'_{12} & I_m & \cdots & 0 \\ \cdots & \cdots & \cdots & \cdots \\ N'_{1n} & 0 & \cdots & I_m \end{pmatrix},

N = \begin{pmatrix} N_{11} & N_{12} & \cdots & N_{1n} \\ N_{21} & N_{22} & \cdots & N_{2n} \\ \cdots & \cdots & \cdots & \cdots \\ N_{n1} & N_{n2} & \cdots & N_{nn} \end{pmatrix}

Performing this product in blocks (II, § 10, no. 5) and using formulae (25), we obtain

N . U = \begin{pmatrix} P & N_{12} & \cdots & N_{1n} \\ 0 & N_{22} & \cdots & N_{2n} \\ \cdots & \cdots & \cdots & \cdots \\ 0 & N_{n2} & \cdots & N_{nn} \end{pmatrix}

where we have written $P = D(N_{11}, \ldots, N_{nn})$ . Let

Q = \begin{pmatrix} N_{22} & \cdots & N_{2n} \\ \cdots & \cdots & \cdots \\ N_{n2} & \cdots & N_{nn} \end{pmatrix}

which is a matrix of order $m(n-1)$ ; then (§ 8, no. 6, formula (31)) (det $N$ ) (det $U$ ) = (det $P$ )(det $Q$ ) and det $U = \det N'_{11}$ . But by the induction hypothesis, det $Q = \det(D^{11}(N_{11}, \ldots, N_{nn})) = \det N'_{11}$ and by virtue of the definition of the $N_{ij}$ , clearly det $Q$ is a polynomial in $\mathbf{A}[Z]$ of degree $m(n-1)$ , whose term in $Z^{m(n-1)}$ has coefficient 1 ; it follows immediately that det $Q$ is not a divisor of zero in the graded algebra $\mathbf{A}[Z]$ . We therefore conclude that det $N = \det(D(N_{11}, \ldots, N_{nn}))$ in $\mathbf{A}[Z]$ ; if we substitute 0 for $Z$ in these polynomials, then det $M = \det(D(M_{11}, \ldots, M_{nn}))$ .

Having shown this lemma, the K-module V is the direct sum of the K-modules $Ae_j$ ( $1 \leq j \leq n$ ); we write $u(e_j) = \sum_{k=1}^n c_{jk} e_k$ . For every element $xe_j \in Ae_j$ , with $x \in A$ , the component of $u(xe_j)$ in $Ae_k$ is $xc_{jk} e_k$ ; it follows that the matrix of $u_k$ with respect to the basis $(a_i e_j)$ of the K-module V can be expressed in the form of a square matrix of matrices $(M_{jk})$ , where $M_{jk}$ is the matrix of the K-linear mapping $xe_j \mapsto xc_{jk} e_k$ of $Ae_j$ into $Ae_k$ with respect to the bases $(a_i e_j)_{1 \leq i \leq m}$ and $(a_i e_k)_{1 \leq i \leq m}$ of these two K-modules (II, § 10, no. 5). If for all $t \in A$ , $M(t)$ denotes the matrix, with respect to the basis $(a_i)_{1 \leq i \leq m}$ of A over K, of the endomorphism $x \mapsto xt$ of A, then $M_{jk} = M(c_{jk})$ ; as $t \mapsto M(t)$ is a ring homomorphism the matrices $M_{jk}$ are permutable with one another. Then

\det u_K = \det(D(M_{11}, \ldots, M_{nn}))

by Lemma 1. But as

t \mapsto M(t)

is a ring homomorphism,

D(M_{11}, \ldots, M_{nn})

is the matrix of the K-endomorphism

x \mapsto x.\det(c_{jk})

of A with respect to basis

(a_i)

; by definition its determinant is therefore

N_{A/K}(\det(u))

, which proves the second formula of (24). On the other hand,

\operatorname{Tr}(u_K) = \sum_{j=1}^n \operatorname{Tr}(M_{jj}) = \sum_{j=1}^n \operatorname{Tr}_{A/K}(c_{jj}) = \operatorname{Tr}_{A/K}\left( \sum_{j=1}^n c_{jj} \right) = \operatorname{Tr}_{A/K}(\operatorname{Tr}(u))

and the proof of Proposition 6 is complete.

Corollary

Let A be a commutative K-algebra admitting a finite basis over K and B an A-algebra admitting a finite basis over A. Then B admits a finite basis over K, and, for all $b \in B$ ("transitivity formulae")*

\begin{align*} \operatorname{Tr}_{B/K}(b) &= \operatorname{Tr}_{A/K}(\operatorname{Tr}_{B/A}(b)), \quad N_{B/K}(b) = N_{A/K}(N_{B/A}(b)) \\ P_{C_{B/K}}(b; X) &= N_{A[X]/K[X]}(P_{C_{B/A}}(b; X)). \end{align*}

(26) This follows immediately from Proposition 6, setting

V = B

and

u(x) = bx

Remark

Suppose that the homomorphism $\lambda \mapsto \lambda . 1$ of K into A is injective and let K be identified with its image in A; suppose that A admits a finite basis $(e_i)_{1 \leq i \leq n}$ as a K-module. Let s be an automorphism of A such that $s(K) = K$ . Let a be an element of A; then, by transporting the structure

\begin{align*} \operatorname{Tr}_{A/K}(s(a)) &= s(\operatorname{Tr}_{A/K}(a)) \\ N_{A/K}(s(a)) &= s(N_{A/K}(a)). \end{align*}

(27) (28) Consider also a derivation D of A ( $§ 10$ , no. 2) such that $D(K) \subset K$ and write $D(e_i) = \sum_{j=1}^n e_j \mu_{ji}$ where $\mu_{ji} \in K$ ; write

ae_i = \sum_{j=1}^n e_j \lambda_{ji} \quad \text{with} \quad \lambda_{ji} \in K.

Then

D(a)e_i + aD(e_i) = D(ae_i) = \sum_{j=1}^n (D(e_j)\lambda_{ji} + e_j D(\lambda_{ji})).

It follows that

D(a)e_i = \sum_{j=1}^n e_j v_{ji}

with

v_{ji} = D(\lambda_{ji}) + \sum_{k=1}^n (\mu_{jk}\lambda_{ki} - \lambda_{jk}\mu_{ki})

. As

\sum_{i,k} (\mu_{ik}\lambda_{ki} - \lambda_{ik}\mu_{ki}) = 0

, therefore

\operatorname{Tr}_{A/\mathbf{K}}(D(a)) = \sum_{i=1}^n D(\lambda_{ii})

, in other words

\operatorname{Tr}_{A/\mathbf{K}}(D(a)) = D(\operatorname{Tr}_{A/\mathbf{K}}(a)). *

5. Discriminant of an algebra

Definition 3

Let A be a K-algebra admitting a finite basis of n elements. The discriminant of a sequence $(x_1, \ldots, x_n)$ of n elements of A, with respect to K, denoted by $D_{A/\mathbf{K}}(x_1, \ldots, x_n)$ , is the discriminant of the square matrix*

(\operatorname{Tr}_{A/\mathbf{K}}(x_i x_j))_{1 \leq i \leq n, 1 \leq j \leq n}.

Consider first a basis

(e_i)_{1 \leq i \leq n}

of A over K and write

e_i e_j = \sum_{k=1}^n c_{ij k} e_k \quad \text{with} \quad c_{ij k} \in \mathbf{K}.

Then

\operatorname{Tr}_{A/\mathbf{K}}(e_i) = \sum_{s=1}^n c_{iss}

, whence

\operatorname{Tr}_{A/\mathbf{K}}(e_i e_j) = \sum_{k,s} c_{ij k} c_{kss}

and therefore

D_{A/\mathbf{K}}(e_1, \ldots, e_n) = \det \left( \left( \sum_{k,s} c_{ij k} c_{kss} \right)_{1 \leq i \leq n, 1 \leq j \leq n} \right).

Now let

(x_i)_{1 \leq i \leq n}, (x'_i)_{1 \leq i \leq n}

be two sequences of n elements of A and suppose that there exists a square matrix of order n,

M = (m_{ij})

, with coefficients in K, such that

x_i = \sum_{j=1}^n m_{ij} x'_j

for

1 \leq i \leq n

. We write

T = (\operatorname{Tr}_{A/\mathbf{K}}(x_i x_j))_{1 \leq i \leq n, 1 \leq j \leq n}, \qquad T' = (\operatorname{Tr}_{A/\mathbf{K}}(x'_i x'_j))_{1 \leq i \leq n, 1 \leq j \leq n}.

Then

\operatorname{Tr}_{A/\mathbf{K}}(x_i x_j) = \sum_{p,q} m_{ip} m_{jq} \operatorname{Tr}_{A/\mathbf{K}}(x'_p x'_q)

, whence

T = M \cdot T' \cdot {}^t M

; the rule of multiplication of determinants therefore gives

\det T = \det M \cdot \det T' \cdot \det {}^t M = (\det M)^2 \det T'

whence finally

D_{A/\mathbf{K}}(x_1, \ldots, x_n) = (\det M)^2 D_{A/\mathbf{K}}(x'_1, \ldots, x'_n).

The above formula shows in particular that the discriminants of two bases of A over K differ by the square of an invertible element of K and therefore generate the same (principal) ideal of K. This ideal $\Delta_{A/K}$ is called the discriminant ideal of A over K; by formula (32) the discriminant of every sequence of n elements of A which differ only in the order of the terms have the same discriminant, for the determinant of a permutation matrix is equal to $\pm 1$ .

Example

(1) If A is a quadratic algebra of type $(\alpha, \beta)$ over K, then (in the notation of § 2, no. 3) $\operatorname{Tr}(e_1) = 2, \operatorname{Tr}(e_2) = \beta$ ,

\operatorname{Tr}(e_2^2) = \alpha \operatorname{Tr}(e_1) + \beta \operatorname{Tr}(e_2) = 2\alpha + \beta^2,

whence $D_{A/K}(e_1, e_2) = \beta^2 + 4\alpha$ .

(2) Let $A = K[X]/K[X]P$ , where $P(X) = X^3 + pX + q$ , so that if x is the image of X in A, 1, x, $x^2$ form a basis of A over K and $x^3 = -px - q$ . It is seen immediately that $\operatorname{Tr}(1) = 3, \operatorname{Tr}(x) = 0, \operatorname{Tr}(x^2) = -2p$ , taking account of the relation $x^3 = -px - q, \operatorname{Tr}(x^3) = -3q$ and $\operatorname{Tr}(x^4) = 2p^2$ , whence easily $D_{A/K}(1, x, x^2) = -4p^3 - 27q^2$ .

(3) Let A be a quaternion algebra of type $(\alpha, \beta, \gamma)$ over K and $(1, i, j, k)$ a basis of A of type $(\alpha, \beta, \gamma)$ ; taking account of § 3, no. 5, formula (30), it is easily found that $\operatorname{Tr}(1) = 4, \operatorname{Tr}(i) = 2\beta, \operatorname{Tr}(j) = \operatorname{Tr}(k) = 0$ , then

D_{A/K}(1, i, j, k) = -16\gamma^2(\beta^2 + 4\alpha)^2.

(4) Let $A = M_n(K)$ and consider the canonical basis $(E_{ij})_{1 \leq i \leq n, 1 \leq j \leq n}$ of A over K (II, § 10, no. 3). It is immediate that $\operatorname{Tr}_{A/K}(E_{ij}) = 0$ if $j \neq i$ and $\operatorname{Tr}_{A/K}(E_{ii}) = n$ for all i; it follows without difficulty that the matrix $(\operatorname{Tr}(E_{ij}E_{hk}))$ of order $n^2$ is of the form $n.P$ , where P is a permutation matrix, whence $D_{A/K}((E_{ij})) = \pm n^{n^2}$ .

Exercises

See the exercises for § 9.