b) Every left or right ideal of A of finite codimension contains a two-sided ideal of finite codimension.

Formulas (3) and (4) prove that if E is a left ideal of A, then $E'$ is a right A-submodule of $A^*$; if V is a right A-submodule of $A^*$, then $V'$ is a left ideal of A. The cases of right ideals and of two-sided ideals are treated analogously. This proves a).

Let $\mathfrak{a}$ be a left ideal of A of finite codimension. The left A-module $E = A_s/\mathfrak{a}$ is finite-dimensional over K, and so is End$_K(E)$. The kernel of the homomorphism $a\mapsto a_E$ from A to End$_K(E)$ is therefore a two-sided ideal of A of finite codimension contained in $\mathfrak{a}$. Passing to the opposite ring $A^o$, we see that every right ideal of A of finite codimension contains a two-sided ideal of finite codimension.

When F is a right A-module and G is a K-vector space, we defined in II, §4, No. 1, p. 268, Proposition 1, a) an isomorphism $\gamma$ of K-vector spaces from Hom$_K(F\otimes_AE,G)$ to Hom$_A(E$, Hom$_K(F,G))$. This isomorphism sends $\varphi : F\otimes_AE\rightarrow G$ to the homomorphism $e\mapsto (f\mapsto \varphi (f\otimes e))$. Through the canonical isomorphism from $A_d\otimes_AE$ to E (II, §3, No. 4, p. $249$)$,\gamma$ can be identified with $c''_E$ when $F = A_d$ and G = K.

Analogously, the isomorphism $\beta$ defined in II, §4, No. 1, p. 268, Proposition 1, b) specializes to an isomorphism $\beta$ from Hom$_K(E^*\otimes_AA_s,K)$ to Hom$_A(E^*$, Hom$_K(A_s,K))$. When E is finite-dimensional, E can be canonically identified with Hom$_K(E^*,K)$ and $E^*$ with $E^*\otimes_AA_s$; the homomorphism $\beta$ is then identified with $c'_E$.

a) The set of coefficients of E is the union of the images of the A-linear mappings from E to $A^*$.

b) Suppose, moreover, that E has finite dimension $n$ over K. Then the left A-module $\Theta_E(A)$ is isomorphic to a quotient of $E^n$, the A-module E is isomorphic to a submodule of $\Theta_E(A)^n$, and every element of $\Theta_E(A)$ is a coefficient of $E^n$.

Assertion a) follows from the surjectivity of $c''_E$.

Let us prove b). Let $(e^*_1, . . . , e^*_n)$ be a basis of $E^*$ over K. Since $\Theta_E(A)$ is generated by the coefficients of E, the A-linear mapping

from $E^n$ to $\Theta_E(A)$ is surjective. By a), every element of $\Theta_E(A)$ is a coefficient of $E^n$. Moreover, the A-linear mapping

from E to $\Theta_E(A)^n$ is injective; this gives b).

Since End$_K(V)$ is finite-dimensional over $K,\mathfrak{a}$ is a two-sided ideal of A of finite codimension. The vector space $\Theta_{\pi}(A)$ is a finite-dimensional subspace of $A^*$, and its orthogonal in A is equal to $\mathfrak{a}$; by Theorem 7 of II, §7, No. 5, p. 301, the space $\Theta_{\pi}(A)$ is therefore the orthogonal of $\mathfrak{a}$ in $A^*$. Moreover, since $\mathfrak{a}$ is the kernel of $\pi$, the transpose of the mapping $\pi$ defines an isomorphism from the dual of $\pi (A)$ to the orthogonal of $\mathfrak{a}$ in $A^*$ (II, §7, No. 5, p. 299, Corollary of Theorem 5).

By definition, Θ(A) is the union of the orthogonals of the two-sided ideals of A of finite codimension. We therefore have $\Theta_{\pi}(A)\subset \Theta (A)$ for every finite-dimensional linear representation $\pi$ of A. Conversely, let $f$ be an element of Θ(A), and let $\mathfrak{a}$ be a two-sided ideal of finite codimension, contained in the kernel of $f$ (VIII, p. 376, Definition 2). Denote by $\pi$ the linear representation of A in $A/\mathfrak{a}$ deduced from the left regular representation in A by passing to the quotient. Let $x$ be the class of 1 (mod $\mathfrak{a}$), and let $x^*$ be the linear form on $A/\mathfrak{a}$ deduced from $f$. We have $f=c_{\pi}(x, x^*)$, so that $f$ is a coefficient of $\pi$.

Take note that if $(V, \pi )$ is a linear representation of A that is not finite-dimensional over K, then the space $\Theta_{\pi}(A)$ is not necessarily contained in Θ(A).

b) For every $S\in \mathscr{S}_K$, the isotypical component of Θ(A) of type S is equal to $\Theta_S(A)$. The left A-module $\Theta^{ss}(A)$ is the direct sum of the submodules $\Theta_S(A)$, where S runs through $\mathscr{S}_K$.

c) For every S in $\mathscr{S}_K$, the right A-module $S^*$ is simple, and $\Theta_S(A)$ is the isotypical component of type $S^*$ of the right A-module Θ(A). The mapping that sends S to cl($S^*$) is a bijection from $\mathscr{S}_K$ to the set of classes of simple $A^o$-modules of finite dimension over K.

d) Viewed as a right A-module, Θ(A) has socle $\Theta^{ss}(A)$.

Let E be a semisimple A-module of finite dimension over K. Every coefficient of E belongs to the image of an A-linear mapping from E to $A^*$ (VIII, p. 378, Proposition 2) and therefore belongs to $\Theta^{ss}(A)$. Conversely, let $f\in \Theta^{ss}(A)$. Then the A-module $Af$ is finite-dimensional over K and is semisimple. By the remark in VIII, p. 367$,f$ is a coefficient of $Af$.

For every S in $\mathscr{S}_K$, the isotypical component of Θ(A) of type S is generated by the images of the A-linear mappings from S to $A^*$; it is therefore equal to $\Theta_S(A)$ by Proposition 2, a) of VIII, p. 378. On the other hand, if S is a simple A-module that is not finite-dimensional over K, then the isotypical component of Θ(A) of type S is zero: indeed, every simple submodule of Θ(A) is monogenous and therefore finite-dimensional over K. Since the socle of Θ(A) is the direct sum of its isotypical components (VIII, p. 65, Proposition 4), this proves b).

Let S be a simple A-module of finite dimension over K. Since the K-vector space S is not reduced to 0, the same holds for $S^*$. Let E be a submodule of the right A-module $S^*$; its orthogonal $E'$ in S is an A-submodule of S. Since S is simple, we have either $E'= 0$, in which case $E = S^*$, or $E'= S$, in which case E = 0. Hence $S^*$ is a simple right A-module.

We have $\Theta (A^o) = \Theta (A)$ (VIII, p. 376); we identify right A-modules with left $A^o$-modules. Since every vector space of finite dimension over K is isomorphic to its bidual, the above proves that the mapping $S\mapsto$ cl($S^*$) is a bijection from $\mathscr{S}_K$ to the set of classes of simple $A^o$-modules of finite dimension over K. Now, for S in $\mathscr{S}_K$, the isotypical component of $\Theta (A^o)$ of type $S^*$ is equal to $\Theta_{S^*}(A^o)$ by assertion b) applied to the algebra $A^o$, and we have $\Theta_{S^*}(A^o) = \Theta_S(A)$. Assertions c) and d) follow immediately.

This follows from Theorem 1, b) because we have $\Theta_E(A)\not= 0$ for every nonzero A-module E.

Let $S_1$ and $S_2$ be simple A-modules of finite dimension over K with a common nonzero coefficient. We then have $\Theta_{S_1}(A)\cap \Theta_{S_2}(A)\not= 0$. Now, $\Theta_{S_1}(A)$ is isotypical of type $S_1$, and $\Theta_{S_2}(A)$ is isotypical of type $S_2$. So $S_1$ is isomorphic to $S_2$.

By Theorem 1, $\Theta^{ss}(A)$ is an $(A$, A)-sub-bimodule of $A^*$, the direct sum of the $(A$, A)-bimodules $\Theta_S(A)$ for S running through $\mathscr{S}_K$. These bimodules are pairwise nonisomorphic because they are already nonisomorphic as left A-modules.

Fix an S in $\mathscr{S}_K$. Denote by D the opposite field of End$_A$(S), and view S as a right D-module and $S^*$ as a left D-module. With these conventions, S is an $(A$, D)-bimodule, $S^*$ is a $(D$, A)-bimodule, and $S\otimes_DS^*$ is an $(A$, A)-bimodule.

b) The $(A,A)$-bimodule $\Theta_S(A)$ is simple.

The mapping $\gamma_S: S\otimes_KS^*\rightarrow \Theta_S(A)$ characterized by the formula $\gamma_S(x\otimes x^*) =c_S(x, x^*)$ is $(A$, A)-linear and surjective and satisfies $\gamma_S(xd\otimes x^*)$ $=\gamma_S(x\otimes dx^*)$ for $x\in S,x^*\in S^*$, and $d\in D$. It therefore defines a surjective $(A$, A)-linear mapping $\lambda_S: S\otimes_DS^*\rightarrow \Theta_S(A)$ characterized by formula (6).

Let us prove that $S\otimes_DS^*$ is a simple $(A$, A)-bimodule. By Corollary 2 of VIII, p. 63, every $(A$, A)-sub-bimodule of $S\otimes_DS^*$ is of the form $S\otimes_DH$, where H is a $(D$, A)-sub-bimodule of $S^*$. Since $S^*$ is a simple right A-module (VIII, p. 380, Theorem 1, c)), we have H = 0 or $H = S^*$; the assertion follows.

The homomorphism $\lambda_S$ is $(A$, A)-linear and nonzero; since $S\otimes_DS^*$ is a simple bimodule, $\lambda_S$ is injective (VIII, p. 47, Proposition 2, a)).

(i) The character Tr$_S$ of S is not zero.

(ii) There exists an element $d\in D$ such that Tr$_{D/K}(d)\not= 0$.

(iii) The extension Z of K is separable, and the characteristic $p$ of K does not divide the degree [D : Z].

The right D-vector space S is finite-dimensional. Let $(e_1, . . . , e_n)$ be a basis of S over D, and let $u$ be an element of End$_D(S)$. Let $(d_{ij})$ be the matrix of $u$ with respect to the basis $(e_1, . . . , e_n)$. We have $u(e_j) =\sum^n_{i=1}e_id_{ij}$ for $j\in [1, n]$. Denote by $u_K$ the mapping $u$ viewed as an endomorphism of the K-vector space S and by $(u_{ij})$ the matrix of $u_K$ with respect to the decomposition $(e_iD)$ of the K-vector space S as a direct sum (II, §10, No. 5, p. 346). We have Tr($u_K$) $=\sum_i$ Tr($u_{ii}$). Moreover, $u_{ii}$ is the endomorphism of the K-vector space $e_iD$ defined by $u_{ii}(e_id) =e_id_{ii}d$ for $d\in D$, so its trace is equal to Tr$_{D/K}(d_{ii})$. We have thus proved the equality

(9) Tr($u_K$) $=$ Tr$_{D/K}(\sum_id_{ii})$.

By Burnside’s theorem (VIII, p. 83, Corollary 1 of Proposition 4), the mapping $a\mapsto a_S$ from A to End$_D(S)$ is surjective. The equivalence of properties (i) and (ii) therefore follows from formula (9).

If the extension Z of K is separable, then there exists an element $d\in Z$ such that Tr$_{Z/K}(d)\not= 0$ (V, §8, No. 2, p. 49, Corollary of Proposition 1). Moreover, we have Tr$_{D/K}(d) = [D : Z]$ Tr$_{Z/K}(d)$ (III, §9, No. 4, p. 548, Corollary); consequently, if $p$ does not divide [D : Z], then we have Tr$_{D/K}(d)\not=$ 0. This proves that (iii) implies (ii).

If the extension Z of K is not separable, then we have Tr$_{Z/K}(x) = 0$ for every $x\in Z$, and therefore Tr$_{D/K}(d) =$ Tr$_{Z/K}$(Tr$_{D/Z}(d)$) $= 0$ for every $d\in D$.

Now suppose that $p$ divides the degree [D : Z]. It then divides the reduced degree of D over Z. For every $d\in D$, we have Tr$_{D/Z}(d) = 0$ (VIII, p. 344, Remark), and therefore Tr$_{D/K}(d) =$ Tr$_{Z/K}$(Tr$_{D/Z}(d)$) $= 0$. This completes the proof of the implication (ii) $\Rightarrow$ (iii).

Since the field is perfect, the extension Z of K is separable (V, §15, No. 5, p. 125, Theorem 3). Property (iii) then follows from Corollary 3 of VIII, p. 323.

Let F be a finite subset of $\mathscr{S}_0$, and let S be an element of F. By assumption, there exists an element $a\in A$ such that Tr$_S(a)\not= 0$. By Corollary 1 of Proposition 4 (VIII, p. 83), there exists an element $b\in A$ such that $b_S=a_S$ and $b_T= 0$ for every $T\in F-\{S\}$. We have Tr$_S(b)\not= 0$ and Tr$_T(b) = 0$ for $T\in F-\{S\}$. The family (Tr$_S$)$_{S\in F}$ is therefore free. Proposition 6 follows.

b) Suppose that the field K is perfect of nonzero characteristic $p$. Then the homomorphism $\theta_K$ is injective, and the kernel of $\theta$ is $pR_K(A)$. Let E be a semisimple A-module of finite dimension over K. The linear form Tr$_E$ is zero if and only if there exists an A-module F such that E is isomorphic to $F^p$.

Let $\mathscr{S}_K$ be the set of classes of simple A-modules of finite dimension over K. The elements [S], where S runs through $\mathscr{S}_K$, form a basis of the $\mathbf{Z}$-module $R_K$(A), so the elements $1\otimes [S]$ form a basis of the K-vector space $K\otimes_{\mathbf{Z}}R_K(A)$ (VIII, p. 195).

Suppose that the field K is perfect. By Proposition 6 and the corollary of VIII, p. 383, the elements $\theta ([S]) =\theta_K(1\otimes [S]) =$ Tr$_S$ of $A^*$ are linearly independent over K. It follows that the homomorphism $\theta_K$ is injective and that the kernel of the homomorphism $\theta$ consists of the elements $\sum_{S\in\mathscr{S}_K}n_S[S]$ of $R_K(A)$ such that $n_S\cdot 1_K= 0$ for every $S\in \mathscr{S}_K$. The kernel of $\theta$ is therefore equal to $pR_K$(A), where $p$ is the characteristic of K. In particular, if K has characteristic 0, then the homomorphism $\theta$ is injective.

The last assertion of a) follows from the corollary of VIII, p. 190.

Suppose that the assumptions of b) are satisfied, and let E be a semisimple A-module of finite dimension over K. If there exists an A-module F such that E is the direct sum of $p$ submodules isomorphic to F, then the module F is semisimple and finite-dimensional over K and we have Tr$_E=p$ Tr$_F= 0$. Conversely, suppose Tr$_E= 0$. We have $[E]\in pR_K$(A), so the multiplicity of every simple module $S\in \mathscr{S}_K$ in E is a multiple of $p$ (VIII, p. 190, Proposition 7); we can write it as $p n_S$ with $n_S\in \mathbf{N}$. The family $(n_S)$ has finite support. Set $F =\oplus_{S\in\mathscr{S}_K}S^{n_S}$. We then have $[E] = [F^p]$, so that E and $F^p$ are isomorphic (VIII, p. 190, Corollary).

Let E be an A-module of finite dimension over K. Let $a\in A$. Denote by $\chi_E(a; T)$ the determinant of the endomorphism $1_E+ Ta_E$ of the K[T]-module $E[T] = K[T]\otimes_KE$ (III, §8, No. 11, p. 541). We have the relation

(10) $\chi_E(a; T)\equiv 1 +$ Tr$_E(a)T$ (mod $T^2K[T]$)

(loc. cit., formula (49)). Since this polynomial has constant term equal to 1, it is an invertible formal power series (IV, §4, No. 4, p. 30). Moreover, if

is an exact sequence of A-modules, all finite-dimensional over K, then we have $\chi_E(a; T) =\chi_{E'}(a; T)\chi_{E''}(a,T)$ (III, §8, No. 7, p. 534, formula (31)). By the definition of the Grothendieck group $R_K(A)$ (VIII, p. 194) and its universal property (VIII, p. 186, Proposition 4), there exists a unique homomorphism $\chi_a$ from the group $R_K(A)$ to the multiplicative group 1 + TK[[T]] such that $\chi_a([E]) =\chi_E(a; T)$ for every A-module E of finite dimension over K. We deduce from formula (10) the relation

(11) $\chi_a(x)\equiv 1 +\theta (x)(a)T$ (mod $T^2K[[T]]$)

for $x\in R_K(A)$ and $a\in A$.

If E and $E'$ are two A-modules of finite dimension over K with isomorphic semisimplifications, then we have $\chi_E(a; T) =\chi_{E'}(a; T)$.

Let $x$ be an element of $R_K(A)$ such that $\chi_{\mathscr{A}}(x) = 1$. By (11), we have $\theta (x)(a) = 0$ for every $a\in \mathscr{A}$, and therefore $\theta (x) = 0$ because $\theta (x)$ is a K-linear form on A and $\mathscr{A}$ generates the K-vector space A. If the characteristic of K is zero, then this implies $x= 0$ (VIII, p. 384, Corollary of Proposition 6), and therefore the result in this case. Suppose from now on that the characteristic $p$ of K is nonzero.

Let us first treat the case when K is algebraically closed. By the above and loc. cit., we then have $x\in pR_K(A)$. Let $y\in R_K(A)$ be such that $x=py$. For every element $a$ of $\mathscr{A}$, we have $\chi_a(y)^p=\chi_a(py) =\chi_a(x) = 1$. So we have $(\chi_a(y)-1)^p= 0$, and therefore $\chi_a(y) = 1$ because the ring K[[T]] is an integral domain. So $y$ belongs to the kernel of the endomorphism $\chi_{\mathscr{A}}$. It follows by induction that $x$ belongs to $p^n R_K(A)$ for every integer $n \geq 1$. Since $R_K(A)$ is a free $\mathbf{Z}$-module, this implies $x = 0$, and therefore the injectivity of $\chi_{\mathscr{A}}$ in this case.

If $K$ is no longer supposed to be algebraically closed, then we choose an algebraic closure $\overline{K}$ of $K$ and consider the diagram of groups and group homomorphisms