Theorem 1

Let $A$ be a commutative ring and $M$ a free $A$-*module with basis* $(e_\lambda)_{\lambda \in L}$. For every mapping $\alpha : L \to \mathbf{N}$ of finite support, we write

(product in the commutative algebra $S(M)$). Then, when $\alpha$ runs through the set $\mathbf{N}^{(L)}$ of mappings of $L$ into $\mathbf{N}$, of finite support, the $e^\alpha$ form a basis of the $A$-*module* $S(M)$.

As $M$ is the direct sum of the $M_\lambda = Ae_\lambda$, it suffices to prove the theorem when $L$ is reduced to a single element and then apply Proposition 9. But when $M = Ae$ (e a free element), then $x \otimes y - y \otimes x = 0$ for all $x, y$ in $M$; the ideal $\mathfrak{g}'$ is therefore zero, whence $T(Ae) = S(Ae)$ and the theorem then follows from § 5, no. 5, Theorem 1.

The multiplication table of the basis (14) is given by

(15)

where $\alpha + \beta$ is the mapping $\lambda \mapsto \alpha(\lambda) + \beta(\lambda)$ of $L$ into $\mathbf{N}$. In other words, the basis $(e^\alpha)$ of $S(M)$, with the multiplicative law (15), is canonically isomorphic to the free commutative monoid $N^{(L)}$ derived from $L$; it follows (§ 2, no. 9) that the symmetric algebra $S(M)$ of a free module $M$ with basis whose indexing set is $L$, is canonically isomorphic to the polynomial algebra $A[(X_\lambda)_{\lambda \in L}]$, the canonical isomorphism being obtained by mapping $e_\lambda$ to $X_\lambda$. In particular (§ 2, no. 7, Proposition 7), for every mapping $f : L \to E$ of $L$ into a commutative $A$-algebra $E$, there exists one and only one $A$-algebra homomorphism $\tilde{f} : S(M) \to E$ such that $\tilde{f}(e_\lambda) = f(\lambda)$.

Remark (2). The above results can equally well be obtained as a consequence of the universal properties of polynomial algebras and symmetric algebras, taking account of II, § 1, no. 11, Corollary 3 to Proposition 17.