Proposition 6

Tag 0AHE, permanent. Cite it as [Bourbaki, Tag 0AHE].

Let $a, b, c, c'$ be non-zero ideals of $A$ contained in $m$ such that

c' \subset c, \quad ab \subset c, \quad ab^2 \subset c', \quad a^2b \subset c'.

If $x \in G(a)$ and $y \in G(b)$ , then $x^{[-1]}y^{[-1]}x.y, x.y.x^{[-1]}y^{[-1]}$ and $[x,y]$ belong to $G(c')$ and are congruent modulo $G(c')$ .

By § 5, no. 2, Proposition 1, there exist $c_{\alpha\beta} \in A'$ such that

x^{[-1]}y^{[-1]}x.y - [x,y] = \sum_{|\alpha| + |\beta| \geq 3} c_{\alpha\beta} x^\alpha y^\beta.

x = 0

y = 0

, then

x^{[-1]}y^{[-1]}x.y - [x,y] = 0

; hence

c_{0\beta} = c_{\alpha 0} = 0

. On the other hand, the conditions

x \in G(a), \quad y \in G(b), \quad |\alpha| \geq 1, \quad |\beta| \geq 1, \quad |\alpha| + |\beta| \geq 3

imply

c_{\alpha\beta} x^\alpha y^\beta \in G(a^2 b + ab^2) \subset G(c')

and hence

x^{[-1]}y^{[-1]}x.y - [x,y] \in G(c')

. We see similarly that

x.y.x^{[-1]}y^{[-1]} - [x,y] \in G(c').

Finally, by § 5, formula (13),

[x,y] \in G(ab) \subset G(c)

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

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