Proposition 9

Let Θ be a linear subspace of Θ(A). The following properties are equivalent:

(i) Θ is an $(A,A)$-sub-bimodule of Θ(A).

(ii) Θ is a subcogebra of Θ(A).

(iii) There exists a hereditary set $\mathscr{C}$ of classes of A-modules of finite dimension over K such that $\Theta = \Theta_{\mathscr{C}}(A)$.

When these properties hold, the set $\mathscr{C}$ mentioned in (iii) is uniquely determined.

The last assertion follows from the corollary of VIII, p. 388: the set $\mathscr{C}$ consists of the classes of A-modules E of finite dimension over K such that $\Theta_E(A)$ is contained in Θ.

(iii) $\Rightarrow$ (ii): Let $\mathscr{C}$ be a hereditary set of classes of A-modules of finite dimension over K. Then $\Theta_{\mathscr{C}}(A)$ is the union of the directed family $(\Theta_E(A))_{E\in\mathscr{C}}$. Since $\Theta_E(A)$ is a subcogebra of Θ(A) for every $E\in \mathscr{C}$ (VIII, p. 390, formula (19)), the same holds for $\Theta_{\mathscr{C}}(A)$.

(ii) $\Rightarrow$ (i): Let $f\in \Theta (A)$. Let $f'_1, . . . , f'_n, f''_1, . . . , f''_n$ be elements of Θ(A) satisfying $c(f) =\sum f'_i\otimes f''_i$. For $a,b$ in A, we have $f(ab) =\sum f'_i(a)f''_i(b)$, and therefore

(VIII, p. 375, formulas (3) and (4)). Consequently, a subcogebra of Θ(A) is an $(A$, A)-sub-bimodule.

(i) $\Rightarrow$ (iii): Suppose that Θ is an $(A$, A)-sub-bimodule of Θ(A); let $\mathscr{C}$ be the set of classes of A-modules E of finite dimension over K such that $\Theta_E(A)$ is contained in Θ. The set $\mathscr{C}$ is hereditary (VIII, p. 377, Remark 2), and we have $\Theta_{\mathscr{C}}(A)\subset \Theta$ by construction. Let $f\in \Theta$ and $E = Af$. Then E is finite-dimensional over K. Consequently, every linear form on E is of the form $u_a:g\mapsto g(a)$ with $a$ in A (II, §7, No. 5, p. 302, Corollary 2). Now, for $a, b, x\in A$, we have

so that $c_E(af, u_b) =af b$. We therefore have $\Theta_E(A)\subset \Theta$. Consequently, the A-module E is of type $\mathscr{C}$, and $f$ is one of its coefficients. We therefore have $\Theta \subset \Theta_{\mathscr{C}}(A)$ and, finally, $\Theta = \Theta_{\mathscr{C}}(A)$.