Theorem 2

Let $X$ be a finite set and $n = \mathrm{Card}(X)$.

(a) For every integer $r \geqslant 1$, the $K$-module $L^r(X)$ is free of rank

where $\mu$ is the Möbius function.

(b) For all $\alpha \in \mathbf{N}^X - \{0\}$, the $K$-module $L^\alpha(X)$ (§ 2, no. 6) is free of rank

We already know that the modules $L^r(X)$, where $r \in \mathbf{N}$, and $L^\alpha(X)$, where $\alpha \in \mathbf{N}^X$, are free (§ 2, no. 11, Corollary to Theorem 1). Consider the multi-gradation $(A^\alpha(X))_{\alpha \in \mathbf{N}^X}$ of $A(X)$ defined by the canonical homomorphism $\phi$ of $Mo(X)$ into $\mathbf{N}^X$ (Algebra, Chapter III, § 3, no. 1, Example 3); then $A^\alpha(X) \cap L(X) = L^\alpha(X)$ by Remark 3 of no. 1. For $\alpha \in \mathbf{N}^X$, the $K$-module $A^\alpha(X)$ admits as basis the set of words in which each letter $x$ of $X$ appears $\alpha(x)$ times. Let $d(\alpha)$ be the number of these words, that is the rank of $A^\alpha(X)$; we shall calculate in two different ways the formal power series

defined by

(1) We have

whence

(2) For all $\alpha \in \mathbf{N}^X - \{0\}$, let $(e_{\alpha,j})_{1 \leq j \leq c(\alpha)}$ be a basis of $L^\alpha(X)$ and give the set $I$ of ordered pairs $(\alpha, j)$ such that $\alpha \in \mathbf{N}^X - \{0\}$ and $1 \leq j \leq c(\alpha)$ a total ordering. By Theorem 1 of no. 1 and the Poincaré–Birkhoff–Witt Theorem (Chapter I, § 2, no. 7, Corollary 3 to Theorem 1), the elements

where the index $m$ runs through $\mathbf{N}^{(I)}$, form a basis of $A(X)$. Each $y_m$ is of multidegree $\sum_{(\alpha, j) \in I} m(\alpha, j) \alpha$. Let $u(m)$ denote this multidegree. It follows that whence finally Comparing (18) and (19), we obtain Lemma 1 then gives (b).

If we now substitute the same indeterminate $U$ for the $T_x$ for $x \in X$ in formula (20), we obtain

Applying Lemma 1 again, we deduce (a).