Proposition 13

Let E be a graded A-algebra of type $\mathbf{Z}$ and S a generating system of the algebra E consisting of homogeneous elements $\neq 0$; for E to be anticommutative (resp. alternating), it is necessary and sufficient that (36) hold for all $x \in S$ and $y \in S$ (resp. that this condition hold and further that $x^2 = 0$ for all $x$ homogeneous of odd degree belonging to S).

We consider first the case of anticommutative algebras. By Lemma 4, the subalgebra F consisting of the elements all of whose homogeneous components $x \neq 0$ satisfy (36) for all $y \in S$, contains all the elements of S and hence $F = E$. If now $F'$ is similarly the subalgebra of E consisting of the elements all of whose homogeneous components $x \neq 0$ satisfy (36) for every homogeneous element $y \neq 0$, it follows from the above that $F'$ contains all the elements of S and hence $F' = E$, which completes the proof of the proposition in this case.

To prove the proposition in the case of alternating algebras, it can be assumed that E is already anticommutative; it is then immediate that every homogeneous element of odd degree in E is of the form $\sum_i z_ix_i$, where $z_i \in E^+$ and $x_i \in S$ is of odd degree (using the fact that $E^+$ is contained in the centre of E); it follows that $\left( \sum_i z_ix_i \right)^2 = \sum_i z_i^2x_i^2 + \sum_{i<j} z_i z_j (x_ix_j + x_jx_i) = 0$ since $x_i^2 = 0$ by hypothesis and $x_ix_j + x_jx_i = 0$ by (36).