Theorem 1

The element $Q_M$ of $\mathbf{Z}((T))$ belongs to $\mathbf{Z}[T, T^{-1}]$.

Dividing $H_0$ by the annihilator of the $H_0$-module $M$, one is reduced to the case where $M$ is a faithful $H_0$-module. If $a, b \in \mathbf{Z}$ are such that $M$ is generated as an $H$-module by $M' = \sum_{a \leq i \leq b} M_i$, then $M'$ is a faithful $H_0$-module of finite length; hence the ring $H_0$ is artinian (A, VIII, § 1, No. 3), therefore noetherian (loc. cit., § 9, No. 1). The polynomial ring $H$ is therefore noetherian (loc. cit., § 1, No. 4). If $I$ is empty, then $H = H_0$, and the family of integers $(\operatorname{long}_{H_0}(M_n))_{n \in \mathbf{Z}}$ has finite support, since $M$ is a finitely generated $H_0$-module; whence the theorem in this case. Let us then argue by induction on the cardinal of the set $I$, assumed nonempty; let $j \in I$ and $J = I - \{j\}$. Let $H'$ denote the graded subring of $H$ generated by $H_0$ and the $X_i$ for $i$ in $J$; consider the homothety $(X_j)_M$ of ratio $X_j$ in $M$, its kernel $R$, and its cokernel $S$. For each $n \in \mathbf{Z}$ there is an exact sequence of $H_0$-modules

therefore $R_{n-d_j}$ and $S_n$ are of finite length, and one has

Since $M$ is a finitely generated module over the noetherian ring $H$, the $H$-modules $R$ and $S$ are finitely generated; since they are annihilated by $X_j$, they are finitely generated $H'$-modules. By the induction hypothesis, the elements $P_R \cdot \prod_{i \in J} (1 - T^{d_i})$ and $P_S \cdot \prod_{i \in J} (1 - T^{d_i})$ of $\mathbf{Z}((T))$ therefore belong to $\mathbf{Z}[T, T^{-1}]$; from (4), one derives

that is, $(1 - T^{d_j}) \cdot P_M = P_S - T^{d_j} \cdot P_R$; hence

whence the conclusion.