Lemma 3

Let $R$ be a noetherian graded ring with degrees $\geq 0$, $E$ a graded $R$-module of finite type such that $E_n$ is an $R_0$-module of finite length for every $n \in \mathbf{Z}$. The following conditions are equivalent: (i) $E$ is an $R$-module of finite length; (ii) there exists an integer $n_0$ such that $E_n = 0$ for $n \geq n_0$; (iii) every prime ideal of $R$ associated with $E$ contains $R_+ = \bigoplus_{n \geq 1} R_n$.

(i) ⇔ (ii): this is clear. (iii) ⇒ (i): let $p$ be a prime ideal associated with E. If (iii) is satisfied, one has $p = p_0 + R_+$ where $p_0$ is a prime ideal of $R_0$, and the R-module $R/p$ is isomorphic to $R_0/p_0$. By IV, § 3, No. 1, corollary of Prop. 1, the $R_0$-module $R_0/p_0$ is isomorphic to a submodule of one of the $E_k$, hence is of finite length. Consequently, $R/p$ is of finite length. By IV, § 2, No. 5, Prop. 7, $p$ is therefore maximal. In view of the arbitrariness of $p$, the R-module E is of finite length (loc. cit.). (i) ⇒ (iii): let $p$ be a prime ideal associated with E. Then $p$ is graded (IV, § 3, No. 1, Prop. 1) and maximal (IV, § 2, No. 5, Prop. 7), hence contains $R_+$ (§ 6, No. 2, Lemma 1).