Lemma 4

Let $b$ be an element of $H$, homogeneous of degree $> 0$, and belonging to none of the minimal elements $p$ of $\mathrm{Supp}(M)$ such that $\dim(H/p) = \dim_H(M)$. Then $\dim_H(M/bM) = \dim_H(M) - 1$.

Put $d = \dim_H(M)$. By the definition of $b$, we have $\dim_H(M/bM) < d$. By prop. 4, we have

and formula (8) of § 3, no. 2 yields the inequality Finally, we have by prop. 4. Hence $\dim_H(M/bM) \geq d - 1$, whence lemma 4.

Let us take up again the proof of Prop. 6, b). We may suppose $\dim_H(M) > 0$. Let us remark that every minimal element of $\mathrm{Supp}(M)$ is graded (apply Lemma 1 of No. 2 to the quotient of $H$ by the annihilator of $M$). By Prop. 8 of III, § 1, No. 4, there therefore exists a homogeneous element $b$ of $H$, of degree $> 0$, not belonging to any of the minimal elements $p$ of $\mathrm{Supp}(M)$ such that $\dim(H/p) = \dim_H(M)$. By Lemma 4, one has $\dim_H(M/bM) = \dim_H(M) - 1$. Suppose further that $H$ is generated by $H_1$ as an $H_0$-algebra and that the residue field $k$ of $H_0$ is infinite. For every minimal element $p$ of $\mathrm{Supp}(M)$ such that $\dim(H/p) = \dim_H(M)$, consider the vector subspace $V_p = (p \cap H_1) \otimes_{H_0} k$ of the vector space $V = H_1 \otimes_{H_0} k$ over $k$. If one had $V_p = V$, one would have $\mathfrak{p} \cap H_1 = H_1$ (II, § 3, No. 2, Prop. 4), whence $H_1 \subset \mathfrak{p}$ and $\dim_H(M) = \dim(H/\mathfrak{p}) \leq \dim(H/H_{\geq 1}) = 0$, which is not the case. Since $k$ is supposed infinite, the union of the $V_p$ is distinct from $V$; if $b \in H_1$ is such that $b \otimes 1$ does not belong to any of the $V_p$, one has $\dim(M/bM) = \dim_H(M) - 1$.

Proceeding by induction on $d = \dim_H(M)$, one then constructs a sequence $(b_1, ..., b_d)$ of elements of $H$, with $b_i$ homogeneous of degree $n_i > 0$ and such that $M / \sum_{i=1}^n b_i M$ is an $H$-module of finite length. If one supposes that $H$ is generated by $H_1$ as an $H_0$-algebra and that the residue field of $H_0$ is infinite, one may suppose $n_i = 1$ for $i = 1, ..., d$. By Prop. 4, one has $\dim_{S^{-1}H}(S^{-1}M) = d$ and

Then $(b_1/1, ..., b_d/1)$ is a maximal secant sequence for the $S^{-1}H$-module $S^{-1}M$. Put $a_i = b_i^{(n_1 ... n_d)/n_i}$ for $1 \leq i \leq d$; then the $a_i$ are all of the same degree, and $(a_1/1, ..., a_d/1)$ is a maximal secant sequence for $S^{-1}M$ (§ 3, No. 2, Remark 3).