Proposition 14

Let E and F be two graded A-algebras of type $\mathbf{Z}$, both anticommutative (resp. alternating). Then the skew tensor product $E^g \otimes_A F$ (no. 7) is an anticommutative (resp. alternating) algebra.

A generating system of $E^g \otimes_A F$ consists of the $x \otimes y$, where $x$ (resp. $y$) is a homogeneous element $\neq 0$ of E (resp. F). Consider two such elements $x \otimes y, x' \otimes y'$, with $\deg(x) = p, \deg(y) = q, \deg(x') = p', \deg(y') = q'$, so that $x \otimes y$ is of degree $p + q$ and $x' \otimes y'$ of degree $p' + q'$. Then by definition (no. 7, formula (27)) and by virtue of (36),

and the criterion of Proposition 13 shows that $E^g \otimes_A F$ is anticommutative since $pq' + pp' + qq' - qp' \equiv (p + q)(p' + q')$ (mod. 2). If further E and F are alternating and $p + q$ is odd, one of the numbers $p, q$ is necessarily odd, hence $(x \otimes y)^2 = \pm (x^2) \otimes (y^2) = 0$ and Proposition 13 shows that $E^g \otimes_A F$ is alternating.