Theorem 2

Tag 005P, permanent. Cite it as [Bourbaki, Tag 005P].

A generating module is balanced.

Let A be a ring, and let M be a generating A-module; by definition, there exist an integer $n\geqslant 0$ , elements $x_1, . . . , x_n$ of M, and elements $x^*_1, . . . , x^*_n$ of the dual $M^*$ of M satisfying $\sum^n_{i=1}\langle x_i, x^*_i\rangle = 1$ . Recall (II, §4, No. 2, p. 271) that we define a group homomorphism $\theta : M^*\otimes_AM\rightarrow$ End $_A(M)$ by the formula $\theta (x^*\otimes y)(x) =\langle x, x^*\rangle y$ . If $u$ is an element of the bicommutant of M, then it commutes with End $_A(M)$ ; therefore, for every $y\in M$ , we have

u(y) =u(\sum_{i=1}^n\langle x_i, x^*_i\rangle y)=\sum_{i=1}^nu(\theta (x^*_i\otimes y)(x_i))

=\sum_{i=1}^n\theta (x^*_i\otimes y)(u(x_i)) =(\sum_{i=1}^n\langle u(x_i), x^*_i\rangle )y

Consequently, $u$ belongs to $A_M$ , and M is balanced.

Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.

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Cites

II, §4, No. 2, p. 271