Lemma 1

Tag 0673, permanent. Cite it as [Bourbaki, Tag 0673].

The mapping $\delta_A : x \mapsto x \otimes 1 - 1 \otimes x$ is a $K$ -derivation of $A$ into $I$ and $I$ is generated, as a left $A$ -module, by the image of $\delta_A$ .

The first assertion follows from the fact that

(xy) \otimes 1 - 1 \otimes (xy) = (x \otimes 1 - 1 \otimes x) . y + x . (y \otimes 1 - 1 \otimes y)

by (40). On the other hand, if the element

\sum_i x_i \otimes y_i

(for

x_i, y_i

A

) belongs to

I

, by definition

\sum_i x_i y_i = 0

and hence

\sum_i (x_i \otimes y_i) = \sum_i x_i (1 \otimes y_i - y_i \otimes 1)

by (40), which completes the proof of the lemma.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

Cited by

§ 10. Derivations
Proposition 17, § 10

Cites

Nothing.