§ 8. Norms and traces

Throughout this paragraph K denotes a field.

1. Recall

Let A be an algebra of finite degree $n$ over K. For each $x \in A$ we denote by $L_x$ the linear mapping $a \mapsto xa$ of A into itself. We recall (III, p. 543) that the trace of $L_x$ is called the trace of x relative to A and is written $\mathrm{Tr}_{A/K}(x)$ ; likewise the determinant of $L_x$ is called the norm of x relative to A and is denoted by $N_{A/K}(x)$ . The discriminant of a sequence $(x_1, ..., x_n)$ of n elements of A is by definition the determinant $D_{A/K}(x_1, ..., x_n)$ of the matrix $(\mathrm{Tr}_{A/K}(x_i x_j))_{1 \leq i,j \leq n}$ (III, p. 549).

Let K' be an extension of K and let $A' = A_{(K')}$ be the K'-algebra derived from A by extension of scalars. We have the formulae

(1)

\mathrm{Tr}_{A'/K'}(1 \otimes x) = \mathrm{Tr}_{A/K}(x) . 1 , \quad N_{A'/K'}(1 \otimes x) = N_{A/K}(x) . 1

for all

x \in A

(III, p. 544). For every sequence

(x_1, ..., x_n)

of elements of A we have

(2)

D_{A'/K'}(1 \otimes x_1, ..., 1 \otimes x_n) = D_{A/K}(x_1, ..., x_n) . 1 ,

as follows from the first formula (1).

2. Norms and traces in etale algebras

Let A be an etale algebra of (finite) degree $n$ over K. By definition there exist then an extension L of K and distinct homomorphisms $u_1, ..., u_n$ of A into L with the following properties.

a) every homomorphism of $A$ into $L$ is equal to one of the $u_i$ (V, p. 29, Cor.); b) there exists an $L$ -algebra isomorphism $u : A_{(L)} \to L^n$ such that

u(1 \otimes x) = (u_1(x), \ldots, u_n(x)) \quad \text{for all } x \in A .

Moreover, every algebraically closed extension $L$ of $K$ has the above properties (V, p. 30, Prop. 2).

We fix $L, u_1, \ldots, u_n$ in what follows. Let $x \in A$ ; we shall prove the formulae

\text{Tr}_{A/K}(x) \cdot 1 = \sum_{i=1}^n u_i(x) , \quad \text{N}_{A/K}(x) \cdot 1 = \prod_{i=1}^n u_i(x) .

Let $v$ be multiplication by $l \otimes x$ in $A_{(L)}$ ; with respect to the basis of $A_{(L)}$ which is the image under $u^{-1}$ of the canonical basis of $L^n$ , the matrix of the linear mapping $v$ is diagonal, with diagonal elements $u_1(x), \ldots, u_n(x)$ . We conclude that

Tr_{A_{(L)}/L}(1 \circ x) \cdot 1 = \sum_{r=1}^n u_i(x) , \quad \text{whence} \quad \text{Tr}_{A/K}(x) \cdot 1 = \sum_{i=1}^n u_i(x) \text{ by (1)} ;

the case of norms is treated similarly.

Further let $(x_1, \ldots, x_n)$ be a sequence of elements of $A$ , let U be the matrix

(u_i(x_j))_{1 \leq i,j \leq n}

and let $(t_{ij}) = 'U . U$ . By the first formula (3) we have

\text{Tr}_{A/K}(x_i x_j) \cdot 1 = \sum_{k=1}^n u_k(x_i x_j) = \sum_{k=1}^n u_k(x_i) u_k(x_j) = t_{ij} ;

passing to determinants, we obtain

D_{A/K}(x_1, \ldots, x_r) \cdot 1 = [\det u_i(x_j)]^2 .

Proposition 1

— Let $A$ be a commutative algebra of finite degree over $K$ . Then the following conditions are equivalent: a) The algebra $A$ is etale. b) There exists a basis of $A$ whose discriminant is non-zero. c) For each $x \neq 0$ in $A$ there exists $y$ in $A$ such that $\text{Tr}_{A/K}(xy) \neq 0$ . *Further, when these conditions are satisfied, the discriminant of any basis of $A$ is non-zero.

We shall show that when $A$ is assumed etale, the discriminant of $A$ with respect to any basis $(x_1, \ldots, x_n)$ of $A$ over $K$ is non-zero; this will in particular establish the implication $a) \Rightarrow b)$ . By (4), with the above notation, it suffices to show that the matrix U is invertible, or equivalently, that the system of linear equations

\sum_{i=1}^n \lambda_i u_i(x_j) = 0 \quad (\text{for } 1 \leq j \leq n)

has only the solution $\lambda_1 = \cdots = A, = 0$ in L. Now the relation (5) implies $\sum_{i=1}^n \lambda_i u_i(x) = 0$ for all $x \in A$ , whence $\lambda_i = 0$ for $1 \leq i \leq n$ , by the theorem on the linear independence of homomorphisms (V, p. 27, Th. 1).

The equivalence of b) and c) is a consequence of the following general lemma:

Lemma 1

Let V be a vector space of finite dimension over K and B a bilinear form on V x V. Let $(v_1, \ldots, v_n)$ be a basis of V over K and $\Delta = \det B(v_i, v_j)$ . Then $A \neq 0$ if and only if, for each $x \neq 0$ in V there exists y in V such that $B(x, y) \neq 0$ .

We have $A \neq 0$ if and only if the system of linear equations

\sum_{i=1}^n \lambda_i B(v_i, v_j) = 0 \quad (1 \leq j \leq n)

has only the solution $A, = \cdots = A, = 0$ in K. If we put $x = \sum_{i=1}^n \lambda_i v_i$ , the above system is equivalent to $B(x, v_j) = 0$ for $1 \leq j \leq n$ , or also, since $(v_1, \ldots, v_n)$ is a basis of V over K, to $B(x, y) = 0$ for all $y \in V$ , whence the lemma.

Let us show that condition c) implies that A is reduced. Let x be a nilpotent element of A : for any $y \in A$ the element xy is nilpotent, and so the endomorphism $L_{xy}$ of the vector space A is nilpotent. Now the following lemma implies that $\mathrm{Tr}(xy) = 0$ for all $y \in A$ , whence $x = 0$ under hypothesis c).

Lemma 2

Let V be a vector space of finite dimension over K and $u$ a nilpotent endomorphism of V, then $\mathrm{Tr}(u) = 0$ .

For each integer $n \geq 0$ let $V_n$ be the image of $u^n$ . Since $u$ is nilpotent, there exists an integer $r \geq 0$ such that $V_0 = V, V_r = 0$ and $V_i \neq V_{i+1}$ for $0 \leq i < r - 1$ . Let $d_i$ be the dimension of $V_{i-1}$ (for $1 \leq i \leq r$ ). There exists a basis $(x_1, \ldots, x_d)$ of V such that the vectors $x_j$ with $d - d_i < j \leq d$ form a basis of $V_{i-1}$ (for $1 \leq i \leq r$ ). We have $u(V_{i-1}) \subset V_i$ and so the diagonal elements of the matrix of $u$ for the basis $(x_1, \ldots, x_d)$ are zero. Thus we have $\mathrm{Tr}(u) = 0$ and the lemma follows.

Finally let us show that b) implies a). Let $(x_1, \ldots, x_n)$ be a basis of A over K such that $D_{A/K}(x_1, \ldots, x_n) \neq 0$ . Let K' be an extension of K, A' the K'-algebra derived from A by extension of scalars and $x'_i = 1 \otimes x_i$ for $1 \leq i \leq n$ . By Formula (2) (V, p. 47) we have $D_{A'/K'}(x'_1, \ldots, x'_n) \neq 0$ . Applying the preceding result to A' we see that A' is reduced, hence the algebra A is etale (V, p. 34, Th. 4).

Corollary

— Let E be an extension of finite degree of K. For E to be separable it is necessary and sufficient that there exist a in E such that $\mathrm{Tr}_{E/K}(a) \neq 0$ .

The condition is necessary by Prop. 1. Conversely, assume that there exists $a \in E$ such that $\mathrm{Tr}_{E/K}(a) \neq 0$ . Given $x \neq 0$ in E, if we put $y = a x'$ , then $\mathrm{Tr}_{E/K}(xy) \neq 0$ . Now Prop. 1 shows that E is an etale algebra over K, hence a separable extension of K.

3. Norms and traces in extensions of finite degree

The transitivity formulae in algebras (III, p. 548) imply the following proposition in the case of extensions of finite degree.

Proposition 2

— Let F be an extension of finite degree of K and E a subextension of F. Then for every $x \in F$ we have

\text{Tr}_{F/K}(x) = \text{Tr}_{E/K}(\text{Tr}_{F/E}(x)) \\ \text{N}_{F/K}(x) = \text{N}_{E/K}(\text{N}_{F/E}(x)) .

Corollary

— Put $m = [F : E]$ ; then for every $x \in E$ we have

\text{Tr}_{F/K}(x) = m \cdot \text{Tr}_{E/K}(x) \\ \text{N}_{F/K}(x) = \text{N}_{E/K}(x)^m .

Proposition 3

— Let E be an extension of finite degree n of K and x an element of E, of degree d over K. Write $f(X) = X^d + \sum_{i=1}^d a_i X^{d-i}$ for the minimal polynomial of X over K. Then we have

\text{Tr}_{E/K}(x) = - \frac{n}{d} a_1 \\ \text{N}_{E/K}(x) = (((-1)^d a_d)^{n/d} = (-1)^n a_d^{n/d} .

Prop. 3 follows directly from the Cor. of Prop. 2 and the lemma:

Lemma 3

Let R be a commutative ring, $f(X) = X^d + \sum_{i=1}^d a_i X^{d-i}$ a monic polynomial of $\mathbf{R}[X]$ , A the R-algebra $\mathbf{R}[X]/(f)$ and x the residue class of X in A. Then $\text{Tr}_{A/R}(x) = -a_1$ and $\text{N}_{A/R}(x) = (-1)^d a_d$ .

By the Cor. (IV, p. 11) the sequence $(1, x, ..., x^{d-1})$ is a basis of A ; further we have

x \cdot 1 = x , \quad x \cdot x = x^2 , \ldots , x \cdot x^{d-2} = x^{d-1} ,

x \cdot x^{d-1} = -a_d \cdot 1 - a_{d-1} \cdot x - \ldots - a_1 \cdot x^{d-1}

The matrix which expresses the multiplication by x with respect to the basis $(1, x, ..., x^{d-1})$ of $A$ is thus of the following form (we have taken $d = 5$ to fix the ideas):

\begin{pmatrix} 0 & 0 & 0 & 0 & -a \\ 1 & 0 & 0 & 0 & -a \\ 0 & 1 & 0 & 0 & -a_3 \\ 0 & 0 & 1 & 0 & -a_2 \\ 0 & 0 & 0 & 1 & -a_1 \end{pmatrix}

The trace of this matrix is clearly $-a$ ; the determinant may be calculated by expanding by the first row, and we then find

(-1)^{d-1} (-a_d) = (-1)^d a_d .

For the rest of this No. we denote by E an extension of finite degree of K and by $x$ an element of E. We shall indicate how to calculate the norm and trace of $x$ in various cases.

a) Case of a separable extension: suppose that E is separable of degree $n$ over K, denote by $\Omega$ an algebraic closure of K and by $\sigma_1, \ldots, \sigma_n$ the $n$ distinct $K$ -homomorphisms of E into R. By Formula (3) ( $V$ , p. 48) we have in R

\text{Tr}_{E/K}(x) = \sum_{i=1}^n \sigma_i(x) , \quad N_{E/K}(x) = \prod_{i=1}^n \sigma_i(x) .

b) Case of a p-radical extension: suppose that K is of characteristic $p > 0$ and that the extension E is p-radical; there exists an integer $e \geq 0$ such that $[E:K] = p^e$ ( $V$ , p. 26, Prop. 4). If $f$ is the height of $x$ over K, the minimal polynomial of $x$ over K is $X^{p^f} - x^{p^f}$ ( $V$ , p. 24, Prop. 1). By Prop. 3 we have $N_{E/K}(x) = (x^{p^f})^{p^e/p^f}$ , whence

N_{E/K}(x) = x^{p^e} = x^{[E:K]}

For the trace we find $\text{Tr}_{E/K}(x) = -p^{e-f} a$ , where $a$ is the coefficient of $X^{p^f-1}$ in the polynomial $X^{p^f} - x^{p^f}$ ; in other words, we have

\text{Tr}_{E/K}(x) = p^e \cdot x = [E:K] x = \begin{cases} x & \text{if } [E:K] = 1 \\ 0 & \text{if } [E:K] > 1 . \end{cases}

c) General case: we can summarize the calculation of norm and trace in the following proposition:

Proposition 4

Let p be the characteristic exponent of K and E an extension of finite degree of K. Let $\sigma_1, \ldots, \sigma_n$ be the distinct K-homomorphisms of E into an algebraic closure $\Omega$ of K, and let $p^e = [E:K]_i$ . For each $x \in E$ we have in $\Omega$

\text{Tr}_{E/K}(x) = p^e \cdot \sum_{i=1}^n \sigma_i(x) , \quad N_{E/K}(x) = \left( \prod_{i=1}^n \sigma_i(x) \right)^{p^e} .

Let E, be the relative separable closure of K in E; then $E_s$ is a separable extension of degree $n$ of K and $\sigma_1, \ldots, \sigma_n$ induce distinct K-homomorphisms of E, into R; further, E is ap-radical extension of E, of degreep $^e$ ( $V$ , p. 44, Prop. 13 and p. 46). So Prop. 4 follows from the Formulae (6), (7), (13), (14) and (12).

Exercises

See the exercises for § 8.