Theorem 2

Tag 04EH, permanent. Cite it as [Bourbaki, Tag 04EH].

Commutative Algebra, chapter III, § 2. General results on filtered rings and modules

Let A be a graded ring of type $\mathbf{N}$ , M a graded A-module of type $\mathbf{N}$ and $(A_n)$ and $(M_n)$ their respective graduations. Suppose that there exists an element $a \in A_0$ such that $A_n = A_0 a^n$ and $M_n = a^n M_0$ for all $n > 0$ . Then, if M, is a Noetherian $A_0$ -module, M is a Noetherian A-module.

By virtue of Lemma 1 it is sufficient to prove that every graded submodule N of M is finitely generated. For all $r \geq 0$ , let $N_r = N \cap M_r$ and let L_r be the set of $m \in M_0$ such that $a^r m \in N_r$ . As

a^r A_0 \subset A_r = A_0 a^r, \quad a^r A_0 L_r \subset A_0 a^r L_r \subset A_0 N_r \subset N_r

and hence the L, are sub-A,-modules of M,; moreover,

aN_r \subset N \cap aM_r = N \cap M_{r+1} = N_{r+1}

and hence the sequence $(L_r)_{r \geq 0}$ is increasing. The hypothesis implies that there exists an integer $n \geq 0$ such that $L_r = L_n$ for $r \geq n$ . For each $r \leq n$ , let $(m_{r,s})_{1 \leq s \leq k_r}$ be a system of generators of the A,-module $L_r$ . We shall show that the elements $a^r m_{r,s}$ for $1 \leq s \leq k_r$ , $0 \leq r \leq n$ form a system of generators of the A-module N. As $M_r = a^r M_0$ for all $r$ , $N_r = a^r L_r$ for all $r$ by definition of $L_r$ . Then, for $r \leq n$ ,

N_r = a^r L_r = \sum_{s=1}^{k_r} a^r A_0 m_{r,s} \subset \sum_{s=1}^{k_r} A_0 a^r m_{rs},

and, for $r > n$ ,

N_r = a^r L_n = \sum_{s=1}^{k_n} a^r A_0 m_{n,s} \subset \sum_{s=1}^{k_n} A_0 a^r m_{n,s} \subset \sum_{s=1}^{k_n} A_0 a^{r-n} \cdot (a^n m_{n,s})

which completes the proof (cf. Exercise 10).

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

Theorem 2

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