Proposition 1

Let $M$ be a graded $H$-module. Suppose that $M$ is generated by $M_0$ and that $M_0$ is an $H_0$-module of finite length. Then one has

Moreover, the following conditions are equivalent: (i) $c_M = \mathrm{long}_{H_0}(M_0)$; (ii) $P_M = \mathrm{long}_{H_0}(M_0) \cdot (1 - T)^{-r}$, that is to say $M = M_0$ if $r = 0$ and for $n \in \mathbf{N}$ if $r > 0$; (iii) the canonical homomorphism of $H$-modules is bijective.

Let $R$ denote the kernel of $\varphi$. Since $\varphi$ is surjective, one has

The conditions (i), (ii) and (iii) are respectively equivalent to $c_R = 0$, $P_R = 0$ and $R = 0$. Hence one has (iii) $\Rightarrow$ (ii) $\Rightarrow$ (i), and it is enough to prove that $c_R = 0$ implies $R = 0$. Suppose $R \neq 0$ and let $0 = N^h \subset N^{h-1} \subset ... \subset N^0 = M_0$ be a Jordan-Hölder sequence of the $H_0$-module $M_0$. Let $R^m$ be the intersection of $R$ and of the image of $H \otimes_{H_0} N^m$ in $H \otimes_{H_0} M_0$; there exists an integer $m$ between 1 and $h$ such that $R^m \neq R^{m-1}$. Put $L = R^{m-1}/R^m$; one has $0 \leq c_L \leq c_R$ and it is enough to prove that $c_L \neq 0$. Now, if $k$ is the quotient field of $H_0$ by the maximal annihilator ideal of $N^{m-1}/N^m$, $L$ is identified with a nonzero graded submodule of $k[(X_i)_{i \in I}]$. Hence $L$ contains a submodule isomorphic to a shifted module of $k[(X_i)_{i \in I}]$; since $c_{k[(X_i)_{i \in I}]} = 1$, one therefore has $c_L \geq 1$ (Remarks 2 and 3), which was to be proved.