Proposition 6

— Let $M$ be an $A$-module which is the direct sum of a family $(M_\lambda)_\lambda$, of submodules. If each $M_\lambda$ is a free module, then the canonical homomorphism of $\otimes_{\lambda \in L} \mathbf{TS}(M_\lambda)$ into $\mathbf{TS}(M)$ is an isomorphism.

Let $(e_{i,\lambda})_{i, I_\lambda}$ be a basis of $M_\lambda$. For $\nu \in \mathbf{N}^{(I_\lambda)}$ put $e_{\nu,\lambda} = \prod_{i \in I_\lambda} \gamma_{\nu(i)}(e_{i,\lambda})$. The $e_{\nu,\lambda}$ for $\nu \in \mathbf{N}^{(I_\lambda)}$ form a basis of $TS(M_\lambda)$ (IV, p. 47, Prop. 4 (i)) and $e_{0,\lambda}$ is the unit-element of $\mathbf{TS}(M_\lambda)$. Therefore the elements

where $\nu_\lambda \in \mathbf{N}^{(I_\lambda)}$ and $\nu_\lambda = 0$ except for a finite number of indices, form a basis of $\otimes_A TS(M_\lambda)$. The image of the element (9) under the canonical homomorphism of the proposition is $\prod_{\lambda \in L} e_{\nu_{\lambda,\lambda}}$. If we denote by $(e_i)_i$, the disjoint union of the families $(e_{i,\lambda})_{i, I_\lambda}$, then the above elements are precisely $\prod_{i \in I} \gamma_{\nu(i)}(e_i)$, where $\nu \in \mathbf{N}^{(I)}$, and so they constitute a basis of $\mathbf{TS}(M)$. This proves the proposition.

Under the conditions of Prop. 6, the inverse isomorphism $\mathbf{TS}(M) \to \otimes_{\lambda} \mathbf{TS}(M_\lambda)$ is also called canonical. Frequently $\mathbf{TS}(M)$ is identified with $\otimes_{\lambda} TS(M_\lambda)$ by means of this isomorphism. It should be noted that if $z \in TS(M_\lambda)$ and $z' \in \mathbf{TS}(M_{\mu})$ with $\lambda \neq \mu$, then the element of $\mathbf{TS}(M)$ which we are thus led to denote by $z \otimes z'$ is not the tensor product of $z$ and $z'$ in $\mathbf{T}(M)$ but the symmetric product of $z$ and $z'$.