Remark 4

Tag 0DV7, permanent. Cite it as [Bourbaki, Tag 0DV7].

Algebra, chapter X, § 4. Produits de torsion

Suppose that each $C^{(i)}$ is endowed with a graded algebra structure compatible with its grading and such that the $d^{(i)}$ are antiderivations (III, p. 117). Let us then endow $\bigotimes_{i \in I} C^{(i)}$ with the left graded tensor product algebra structure of the given structures (III, p. 49). Then $D$ is an antiderivation. Indeed, using the associativity of the tensor product, we may assume that $I = \{1, 2\}$ ; let then $p_1,\ q_1,\ p_2,\ q_2 \in \mathbf{Z},\ x_1 \in (C^{(1)})_{p_1},\ y_1 \in (C^{(1)})_{q_1},\ x_2 \in (C^{(2)})_{p_2},\ y_2 \in (C^{(2)})_{q_2}$ ; we have

(D(x_1 \otimes x_2))(y_1 \otimes y_2) + (-1)^{p_1+p_2} (x_1 \otimes x_2)(D(y_1 \otimes y_2)) =

= (dx_1 \otimes x_2 + (-1)^{p_1} x_1 \otimes dx_2)(y_1 \otimes y_2) +

+ (-1)^{p_1+p_2} (x_1 \otimes x_2)(dy_1 \otimes y_2 + (-1)^{q_1} y_1 \otimes dy_2) =

= (-1)^{p_2 q_1} (dx_1) y_1 \otimes x_2 y_2 + (-1)^{p_1+(p_2-1)q_1} x_1 y_1 \otimes (dx_2) y_2 +

+ (-1)^{p_1+p_2+p_2(q_1-1)} x_1 dy_1 \otimes x_2 y_2 + (-1)^{p_1+p_2+q_1+p_2 q_1} x_1 y_1 \otimes x_2 dy_2

= (-1)^{p_2 q_1} [(dx_1) y_1 + (-1)^{p_1} x_1 dy_1] \otimes x_2 y_2 +

+ (-1)^{p_1+q_1+p_2 q_1} x_1 y_1 \otimes ((dx_2) y_2 + (-1)^{p_2} x_2 dy_2)

= (-1)^{p_2 q_1} [d(x_1 y_1) \otimes x_2 y_2 + (-1)^{p_1+q_1} x_1 y_1 \otimes d(x_2 y_2)]

= (-1)^{p_2 q_1} D(x_1 y_1 \otimes x_2 y_2) = D((x_1 \otimes x_2)(y_1 \otimes y_2)).

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

Cited by

Nothing.

Cites

Nothing.