Indeed, since $\Phi$ is a sesquilinear form, we have $\Phi(y_i, y_j) = \sum_{k,m} a_{ik} \Phi(x_k, x_m) a_{jm}^j$. Hence, if we denote by $A$ the matrix $(a_{ij})$, the matrix $(\Phi(y_i, y_j))$ is equal to $A . (\Phi(x_i, x_j)). {}^t A^j$. Since $\det({}^t A) = \det(A)$, and $\det(A^j) = \det(A)^j$, our assertion is proved.

In particular, if $(e_i)$ and $(e'_i)$ are two bases of $E, D$ and $D'$ the discriminants of $\Phi$ with respect to these bases, and $a$ the determinant of the passing matrix from the basis $(e_i)$ to the basis $(e'_i)$, we have

It follows from prop. 1 that, if $(e_i)$ is a basis of $E$ and $(x_i)$ an arbitrary system of $n$ elements of $E$, $D_\Phi(e_1, \ldots, e_n)$ divides $D_\Phi(x_1, \ldots, x_n)$. In particular the discriminants of $\Phi$ with respect to any two bases of $E$ generate the same principal ideal of $A$.

Let $(E_i)_{i \in I}$ be a finite family of finite-dimensional free $A$-modules, $\Phi_i$ a sesquilinear form for $J$ on $E_i$, and $B_i$ a basis of $E_i$. If $\Phi$ denotes the direct sum of the $\Phi_i$ (§ 1, no 3) and $B$ the basis of $\prod_{i \in I} E_i$ obtained by joining the $B_i$, then obviously

Let $\Phi$ be a sesquilinear form for $J$ on $E$, $h$ a homomorphism of $A$ into a commutative ring $A'$, $\Phi'$ the sesquilinear form on $A' \otimes_A E$ obtained by extension of $\Phi$ (§ 1, no 4) and $(x_1, \ldots, x_n)$ an arbitrary system of elements of $E$. Since $A' \otimes_A E$ is a free $A'$-module, $D_{\Phi'}(1 \otimes x_1, \ldots, 1 \otimes x_n)$ is defined, and we have obviously

a) For a system $(x_1, \ldots, x_n)$ of $n$ elements of $E$ to be free, it is necessary and sufficient that $D_\Phi(x_1, \ldots, x_n)$ be $\neq 0$.

b) For a system $(x_1, \ldots, x_n)$ of $n$ elements of $E$ to be a basis of $E$, it is necessary and sufficient that $D_\Phi(x_1, \ldots, x_n)$ and $D_\Phi(e_1, \ldots, e_n)$ be associated elements in $A$ (cf. Chapter VI, § 1, No. 5).

Let $x_j = \sum_{i=1}^n a_{ji} e_i$ ($a_{ji} \in A$). Let us first prove a). If $D_\Phi(x_1, \ldots, x_n) = 0$, we have $\det(a_{ji}) \cdot \det(a_{ji})' = 0$ (Prop. 1) since $D_\Phi(e_1, \ldots, e_n) \neq 0$ and $A$ is integral; we therefore have $\det(a_{ji}) = 0$, and the vectors $x_j$ are linearly dependent (Chapter III, § 7, No. 1, Th. 1, applied to the vector space $K \otimes_A E$, where $K$ denotes the field of fractions of $A$). Conversely, if these vectors are linearly dependent we have $\det(a_{ji}) = 0$ (ibid.), whence $D_\Phi(x_1, \ldots, x_n) = 0$ (Prop. 1).

Let us now prove b). If $D_\Phi(x_1, \ldots, x_n)$ and $D_\Phi(e_1, \ldots, e_n)$ are associated in $A$, Prop. 1 shows that $\det(a_{ij}) \cdot \det(a_{ij})'$ is invertible in $A$. Thus $\det(a_{ij})$ is itself invertible in $A$; hence the matrix $(a_{ij})$ over $A$ is invertible (Chapter III, § 6, No. 5, Th. 2), and the endomorphism $g$ of $E$ defined by $g(e_i) = x_i$ ($i = 1, \ldots, n$) is an automorphism; consequently $(x_1, \ldots, x_n)$ is a basis of $E$. The converse follows immediately from Prop. 1.

a) The mapping $s_\Phi$ from $E$ into $E^*$ associated with $\Phi$ is bijective.

b) The mapping $d_\Phi$ from $E$ into $E^*$ associated with $\Phi$ is bijective.

c) The element $D_\Phi(S)$ is invertible in $A$.

Indeed condition c) expresses that the matrix of $\Phi$ with respect to $S$ is invertible (Chapter III, § 6, No. 5, Th. 2). Thus c) is equivalent to a) ($§ 1,$ No. 10); similarly c) is equivalent to b).

Indeed, let K be the field of fractions of A, and let $\Phi'$ be the extension of $\Phi$ to the K-vector space $K \otimes_A E$; identify E with a subset of this vector space. The relation $D_{\Phi}(S) \neq 0$ is then equivalent to $D_{\Phi'}(S) \neq 0$ (formula (6)), which itself expresses that $s_{\Phi'}$ is bijective (prop. 3), that is to say that $\Phi'$ is nondegenerate (§ 1, No. 6, prop. 6). Now, for every $x \in K \otimes_A E$, there exists $a \in A$ such that $ax \in E$; consequently, for $\Phi$ to be degenerate, it is necessary and sufficient that $\Phi'$ be so. This proves our assertion.

Indeed, let $A'$ be the algebraic closure of A, and let $B'$ be the algebra $A' \otimes_A B$ over $A'$. If B is separable, $B'$ is semisimple (chap. VIII, § 7, No. 5, cor. of prop. 7) and is therefore the direct sum of n fields isomorphic to $A'$ (chap. VIII, § 6, No. 4, cor. of prop. 9). If $S'$ denotes the canonical basis of $B'$ (identified with ${A'}^n$), one has $D_{B'/A'}(S') = 1$, whence $D_{B'/A'}(S) \neq 0$ (prop. 1) and $D_{B/A}(S) \neq 0$ (formula (6)).

Conversely suppose that one has $D_{B/A}(S) \neq 0$. To prove that B is separable, it is enough to prove that $B'$ is semisimple, that is to say that it admits no nonzero nilpotent element. Now, if $x'$ were a nonzero nilpotent element of $B'$, it could be taken as the first element of a basis $S'$ of $B'$, and one would then have $\mathrm{Tr}_{B'/A'}(x'y') = 0$ for every $y' \in S'$ since a nilpotent endomorphism has its eigenvalues zero (chap. VII, § 5, No. 3, cor. 3 of prop. 8), and hence a zero trace. It would follow that $D_{B'/A'}(S') = 0$, whence $D_{B'/A'}(S) = 0$ (prop. 1) and $D_{B/A}(S) = 0$ (formula (6)), contrary to the assumption.

The first two assertions being evident, it is enough to prove (7). By definition the first member is the determinant of the K-linear endomorphism $u$ of E defined by

Consider the A-linear endomorphism $v$ of E defined by $v(e_i) = \sum_s \Phi(e_i, e_s) e_s$, and the K-linear endomorphism $w$ of E defined by $w(a_j e_i) = (\sum_r \mathrm{Tr}_{A/K}(a_j a_r) a_r) e_i$. We have

since $w(a e_s) = \sum_r \mathrm{Tr}_{A/K}(a a_r) a_r e_s$ for all $a \in A$; thus $u$ is the composite mapping $w \circ v$. Therefore, denoting by $v_K$ the mapping $v$ considered as a K-linear mapping, we have $\det(u) = \det(v_K) \det(w)$. Now $\det(v_K) = N_{A/K}(\det(v))$ (chap. VIII, § 12, no 2, prop. 7), and it is clear that $\det(v) = D_\Phi(e_1, \ldots, e_n)$. On the other hand, since each of the A-modules $Ae_i (i = 1, \ldots, n)$ is stable under $w$, and the determinant of the restriction of $w$ to $Ae_i$ is $\det(\mathrm{Tr}_{A/K}(a_j a_r)) = D_{A/K}(a_1, \ldots, a_q)$, we have $\det(w) = (D_{A/K}(a_1, \ldots, a_q))^n$. Formula (8) therefore reduces to $\det(u) = \det(v_K) \det(w)$, a formula proved above.

Indeed, if we put $\Phi(x, y) = \operatorname{Tr}_{E/A}(xy)$, the K-bilinear form $\Phi'$ of Prop. 6 is $\Phi'(x, y) = \operatorname{Tr}_{E/K}(xy)$ by the transitivity formula for traces (Chap. VIII, § 12, no 2, cor. of Prop. 7).

Exercises. — 1) Let A be an algebra of finite rank over a commutative field K, having a unit element. a) Show that if the radical of A is not zero, the bilinear form $(x, y) \to \operatorname{Tr}_{A/K}(xy)$ on A is degenerate. b) Suppose that K is of characteristic 0. Show that if A is a matrix algebra $M_n(K)$, S the canonical basis of A over K, then $D_{A/K}(S) \neq 0$. c) Deduce from a) and b) that, for an algebra A of finite rank over a field K of characteristic 0 to be absolutely semisimple, it is necessary and sufficient that the bilinear form $(x, y) \to \operatorname{Tr}_{A/K}(xy)$ be non-degenerate (or, what amounts to the same thing, that $D_{A/K}(S) \neq 0$ for every basis S of A over K).

¶ 2) Let B be a ring, A a subring of B containing the unit element of B; B is therefore a (A, A)-bimodule; denote by $^sB$ (resp. $^dB$) the set B considered as a left (resp. right) A-module, by $^sB^*$ (resp. $^dB^*$) the right (resp. left) A-module dual of $^sB$ (resp. $^dB$). For every $x' \in ^sB^*$ and every $b \in B$, $x \to \langle xb, x' \rangle$ is an A-linear form on $^sB$, hence an element of $^sB^*$ which is denoted by $bx'$; the mapping $(b, x') \to bx'$ defines on $^sB^*$ a left B-module structure (cf. chap. III, 2nd ed., App. II, No. 7). a) Let $\varphi$ be a homomorphism of the (A, A)-bimodule B into the (A, A)-bimodule A; in order that the A-bilinear mapping $\Phi : (x, y) \to \varphi(xy)$ from $^sB \times ^dB$ into A be non-degenerate, it is necessary and sufficient that $\varphi(0)$ contain no ideal (left or right) of B distinct from $\{0\}$. One then says that $\varphi$ is a Frobenius homomorphism of B into A. b) Let $\varphi$ be a Frobenius homomorphism of B into A; show that the mapping $d_\varphi$ associated on the right with $\Phi$ is an isomorphism of the left B-module $B_s$ onto a submodule of the left B-module $^sB^*$. Show that $d_\varphi$ is bijective in each of the following two cases: 1° A is a left and right artinian ring satisfying the conditions (N_s) and (N_d) (§ 1, exerc. 11), and $^sB$ and $^dB$ are A-modules of finite length (use exerc. 11 b) of the § 1); 2° A is an artinian commutative and involutive ring (chap. VIII, § 3, exerc. 11), contained in the center of B, and $^sB$ is an A-module of finite length (use exerc. 11 of chap. VIII, § 3).

c) Conversely, show that if $B_s$ and $^sB^*$ are isomorphic, there exists a Frobenius homomorphism of B into A when one is in one of the two cases considered in b) and $^sB$ and $^dB$ have the same length (use exerc. 11 b) of the § 1).

d) With the hypotheses and notations of a), show that the right annihilator (resp. left annihilator) of a left ideal I (resp. of a right ideal r) of B, is the orthogonal $l^0$ (resp. $r^0$) for the form $\Phi$ of the submodule I (resp. r) of $^sB$ (resp. $^dB$).

e) Let $\varphi$ be a Frobenius homomorphism of B into A. Show that if A is an involutive artinian ring (chap. VIII, § 3, exerc. 11) the same is true of B, when one supposes in addition that one or the other of the conditions of b) is satisfied (use b) and d)).

¶ 3) Let A be a commutative field, B a finite rank algebra over A, having a unit element.

a) In order that B be a Frobenius algebra, it is necessary and sufficient that there exist a Frobenius homomorphism of B into A (exerc. 2). (Use exerc. 2 c) and e) above, and exerc. 6 b) of chap. VIII, § 13).

b) Let $\varphi$ be a Frobenius homomorphism of B into A; every A-linear form on B can then be written in one and only one way $x \to \varphi(b'x)$ (resp. $x \to \varphi(xb'')$) where $b'$ and $b''$ belong to B; in order that this form be a Frobenius homomorphism, it is necessary and sufficient that $b'$ (resp. $b''$) be invertible in B.

c) For every $x \in B$, let $x^\sigma$ be the unique element (cf. b)) such that $\varphi(xy) = \varphi(yx^\sigma)$ for every $y \in B$. Show that $x \to x^\sigma$ is an A-automorphism of B. One says that the Frobenius algebra B is symmetric if $\sigma$ is an inner automorphism of B; there is then a Frobenius homomorphism of B into A for which $\sigma$ is the identity (cf. b)). It amounts to the same thing to say that the (B, B)-bimodules B and $^sB^* = ^dB^*$ (which is written B*) are isomorphic (exerc. 2 c)).

d) Let E be a left b-module of finite length, E’ its dual, E’* the dual of E’ considered as vector space over A; E’* is endowed with a structure of left B-module by putting, for $x' \in E'$, $x'' \in {E'}^*$, $b \in B$, $\langle x', bx'' \rangle = \langle x'b, x'' \rangle$ (Chapter III, 2nd ed., App. II, No. 7). For every $x \in E$, let $f_E(x)$ (or simply $f(x)$) be the element of E’* such that $\langle x', f(x) \rangle = \varphi(\langle x, x' \rangle)$ for every $x' \in E'$; now prove that $f$ is a semilinear bijection for the automorphism $\sigma$, of the left B-module E onto the left B-module E’* (use Exercise 10 of Chapter VIII, § 4). For $E = B_s$, one has (with the notations of Exercise 2 b)) $d_\Phi(x^\sigma) = f_{B_s}(x)$ for every $x \in B$.

4) a) Let G be a finite group. Show that the algebra B of the group G over any commutative field A is a symmetric Frobenius algebra (Exercise 3). (Consider the mapping $\varphi$ of B into A which associates, with every element $x = \sum_{s \in G} \xi_s . s$, the element $\varphi(x) = \xi_e$, e denoting the identity element of G).

b) Let E be a finite-dimensional vector space of dimension $n$ over a commutative field A, and B the exterior algebra $\wedge E$. Show that B is a Frobenius algebra. (If $(e_i)_{1 \leq i \leq n}$ is a basis of E, consider the mapping which associates, with every element $x$ of B, the coefficient of $e_1 \wedge e_2 \wedge \ldots \wedge e_n$ in the expression of $x$ using the basis of B corresponding to $(e_i)$). For the Frobenius algebra B to be symmetric, it is necessary and sufficient that $n$ be even or A be of characteristic 2.

¶ 5) a) Show that the tensor product of two Frobenius algebras (resp. Frobenius and symmetric algebras) of finite rank over a commutative field K is a Frobenius algebra (resp. Frobenius and symmetric) (cf. § 1, No. 9, Prop. 9).

b) Let B be an algebra of finite rank over K, L an extension of K of finite degree over K. Show that if the algebra $B_{(L)} = B \otimes_K L$ over L is Frobenius (resp. Frobenius and symmetric), the same is true of B. (Use Exercises 2 c) and 3 c) above and Exercise 2 of Chapter VIII, § 2).

6) Show that every absolutely semisimple algebra B of finite rank over a commutative field A is a symmetric Frobenius algebra. (Reduce to the case where B is simple; use Exercise 5 b) above, as well as Proposition 9 of Chapter VIII, § 12, No. 3).