Proposition 7

Let $\Delta$ be a commutative monoid, $M$ a graded $A$-module of type $\Delta$ and $(M_\alpha)_{\alpha \in \Delta}$ its graduation. For every ordered pair $(\alpha, n) \in \Delta \times \mathbf{N}$, let $T^{\alpha, n}(M)$ be the (direct) sum of the submodules $M_{\alpha_1} \otimes M_{\alpha_2} \otimes \cdots \otimes M_{\alpha_n}$ of $T^n(M)$ such that $\sum_{i=1}^n \alpha_i = \alpha$; then $(T^{\alpha, n}(M))_{(\alpha, n) \in \Delta \times \mathbf{N}}$ is the only graduation of type $\Delta \times \mathbf{N}$ compatible with the algebra structure on $T(M)$ and inducing on $M = T^1(M)$ the given gradation.

It has been seen at the beginning of this no. that $T(M)$ is the direct sum of the $T^{\alpha, n}(M)$ and the fact that this is a graduation compatible with the algebra structure follows immediately from the definitions. If $({T'}^{\alpha, n})$ is another graduation of type $\Delta \times \mathbf{N}$ on $T(M)$ compatible with the algebra structure and such that ${T'}^{\alpha, 1}(M) = {T'}^{\alpha, 1}$ for $\alpha \in \Delta$, it follows immediately from the definitions that, for all $n \geq 1$ and all $\alpha \in \Delta$, $T^{\alpha, n}(M) \subset {T'}^{\alpha, n}$; but since $T(M)$ is also the direct sum of the $T^{\alpha, n}(M)$, this implies that ${T'}^{\alpha, n} = T^{\alpha, n}(M)$ (II, § 1, no. 8, Remark).