Corollary

The module $L(X)$ is free and so is each of the submodules $L^\alpha(X)$ where $\alpha \in \mathbf{N}^{(X)}$ and $L^n(X)$ where $n \in \mathbf{N}$. The modules $L^\alpha(X)$ are of finite rank and so are the modules $L^n(X)$ if $X$ is finite.

There exists a Hall set $H$ relative to $X$ (Proposition 11). For all $w \in H$, the element $\Psi'(w)$ of $L(X)$ belongs to one of the modules $L^\alpha(X)$ (with $\alpha \in \mathbf{N}^{(X)}$) and the module $L(X)$ is the direct sum of the submodules $L^\alpha(X)$. Further, for all $\alpha \in \mathbf{N}^{(X)}$, the set of elements of $M(X)$ whose canonical image in $\mathbf{N}^{(X)}$ is equal to $\alpha$ is finite; this shows that each of the modules $L^\alpha(X)$ is free and of finite rank and that $L(X)$ is free. Now $L^n(X) = \sum_{|\alpha|=n} L^\alpha(X)$ and hence $L^n(X)$ is free; when $X$ is finite, the set of $\alpha \in \mathbf{N}^{(X)}$ such that $|\alpha|=n$ is finite and hence $L^n(X)$ is then of finite rank.