Proposition 10

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 $S^{\alpha, n}(M)$ be the submodule of $S^n(M)$ the direct sum of the submodules $\bigotimes_{\lambda \in J} S^{n_\lambda}(M_{\alpha_\lambda})$, where $(n_\lambda)_{\lambda \in L}$ runs through the set of families of integers $\geq 0$ such that $\sum_{\lambda \in L} n_\lambda = n$, $J$ is its support and, for each $(n_\lambda), (\alpha_\lambda)_{\lambda \in J}$ runs through the set of families of $\Delta^J$ such that $\sum_{\lambda \in J} \alpha_\lambda = \alpha$. Then $(S^{\alpha, n}(M))_{(\alpha, n) \in \Delta \times \mathbf{N}}$ is the only graduation of type $\Delta \times \mathbf{N}$ compatible with the algebra structure on $S(M)$ and which induces on $M = S^1(M)$ the given graduation.

The fact that $S(M)$ is the direct sum of the $S^{\alpha, n}(M)$ follows from the Corollary to Proposition 9; the rest of the proof is identical with the end of the proof of § 5, no. 5, Proposition 7.

Suppose more particularly that $\Delta = \mathbf{Z}$ and let $S(M)$ be given the total graduation (of type $\mathbf{Z}$) (II, § 11, no. 1) corresponding to the graduation of type $\mathbf{Z} \times \mathbf{N}$ (and hence also of type $\mathbf{Z} \times \mathbf{Z}$) defined above; the homogeneous elements of degree $n \in \mathbf{Z}$ under this graduation are therefore those of the direct sum of the $S^{q, m}(M)$ for $q + m = n$. Let $f$ be a homogeneous linear mapping of degree 0 of the graded $A$-module $M$ into a commutative graded $A$-algebra $F$ of type $\mathbf{Z}$; then the algebra homomorphism $g : S(M) \to F$ such that $f = g \circ \phi_M'$ is a homomorphism of graded algebras of type $\mathbf{Z}$, as follows from the formula $g(x_1 x_2 \ldots x_n) = f(x_1) f(x_2) \ldots f(x_n)$ for homogeneous $x_i$ in $M$, from the hypothesis on $f$ and from the definition of the graduation of type $\mathbf{Z}$ on $S(M)$.