b) The canonical homomorphism of $H$ into $S^{-1}H$ induces for each $n$ an isomorphism $u_n$ of $H_n$ onto $S^{-1}H_{\geq n}/S^{-1}H_{\geq n+1}$; the $u_n$ are the homogeneous components of an isomorphism of graded rings of $H$ onto the graded ring associated with $S^{-1}H$, filtered by the $S^{-1}H_{\geq n}$.

For every ideal $a$ of $H$, let us denote by $a^{\mathrm{gr}}$ the greatest graded ideal contained in $a$; we have $a^{\mathrm{gr}} = \sum_n (a \cap H_n)$.

b) Every maximal element of the set of graded ideals of $H$ distinct from $H$ is a maximal ideal of $H$ which contains $H_{\geq 1}$.

c) Every minimal prime ideal of $H$ is graded.

a) This follows from III, § 1, no. 4, prop. 4.

b) Let $m$ be a graded ideal of $H$ distinct from $H$. Then one has

If $m$ is maximal, then one has $m = m_0 + H_{\geq 1}$, where $m_0$ is a maximal ideal of $H_0$, whence $b$.

c) Let $p$ be a minimal prime ideal of $H$. Since $p^{gr}$ is prime by $a$ and contained in $p$, one has $p = p^{gr}$, whence $c$.

\* Remark 1. --- Let us take up again the notation of example 1 of No. 1. Lemma 2 implies that, if two irreducible subvarieties $Y$ and $Z$ of $\mathbf{C}^{n+1}$ have the same projecting cone and if $Z \subset Y$ and $Z \neq Y$, then $Y$ is the projecting cone of $Z$, and $Z$ is of codimension 1 in $Y$. \*

Replacing $H$ by $H/q^{gr}$, one is reduced to the case where $q^{gr} = \{0\}$. Then $H$ is an integral domain (lemma 1, $a$), $p^{gr} = 0$, and it is a matter of proving that $ht(p) \leq 1$: this will indeed imply that $ht(q) = 0$, hence that $q = \{0\}$. Since $p^{gr} = \{0\}$, one has $p \cap H_n = \{0\}$ for every $n$, and $p$ is disjoint from the multiplicative subset $T = \bigcup (H_n - \{0\})$. The ring $H_p$ is therefore isomorphic to a ring of fractions of $T^{-1}H$, and one has therefore

(§ 1, No. 3, prop. 6 and 7). But, by lemma 4 of V, § 1, No. 8, $T^{-1}H$ is a field or is isomorphic to a ring $K[X, X^{-1}]$, where $K$ is a field; hence one has $\dim(T^{-1}H) \leq 1$ and $ht(p) \leq 1$, which was to be proved.

By Lemma 1, $a$, the ideal $p^{gr}$ is prime and contained in $p$, hence $ht(p^{gr}) \leq ht(p) - 1$. Since the proposition is trivial when $ht(p^{gr}) = + \infty$, one may suppose that $ht(p^{gr}) < + \infty$. Let us prove the inequality $ht(p) \leq ht(p^{gr}) + 1$ by induction on $ht(p^{gr})$. It is enough to prove that, for every prime ideal $q$ contained in $p$ and distinct from $p$, one has $ht(q) \leq ht(p^{gr})$. Let us distinguish two cases according as $q^{gr} \neq p^{gr}$ or $q^{gr} = p^{gr}$. If $q^{gr} \neq p^{gr}$, then one has $ht(q^{gr}) < ht(p^{gr})$; one has

by the induction hypothesis if $q \neq q^{gr}$ and trivially if $q = q^{gr}$; consequently, one has $ht(q) \leq ht(q^{gr}) + 1 \leq ht(p^{gr})$, which was to be proved. If $q^{gr} = p^{gr}$, then one has $q = q^{gr}$ by Lemma 2, hence $ht(q) = ht(q^{gr}) \leq ht(p^{gr})$, whence again the desired conclusion.

a) Every chain of graded prime ideals of H, saturated as a chain of graded prime ideals, is saturated as a chain of prime ideals.

b) For every graded prime ideal p of H, one has htgr(p) = ht(S^{-1}p) = ht(p).

c) One has dimgr(H) = dim(S^{-1}H) = dim(H).

To prove a), it is enough to prove that, if p and q are two distinct graded prime ideals of H such that q ⊂ p and every graded prime ideal contained between q and p is equal either to p or to q, then ht(p/q) = 1. Passing to the quotient by q, one is reduced to the case where q = {0}. It is therefore a matter of proving that, if H is an integral domain, and if p is an ideal of H, minimal among the graded prime ideals ≠ {0}, one has ht(p) = 1. Now let a be a nonzero homogeneous element of p, and let r be a prime ideal of H such that a ∈ r ⊂ p and minimal for these properties (II, § 2, No. 6, Lemma 2). Since r^{gr} is prime (Lemma 1, a)) and nonzero, one has r^{gr} = p, hence p = r. Since H is an integral domain and noetherian, p has height 1 (§ 3, No. 1, Prop. 1), whence a).

Let us prove b). Let p be a graded prime ideal of H. One has

(Prop. 2); let us prove the inequality ht(p) ≤ htgr(p) by induction on htgr(p). If htgr(p) = 0, p is minimal among the graded prime ideals, hence minimal (Lemma 1, c)), and one has ht(p) = 0. Suppose that one has htgr(p) > 0 and let us prove the inequality ht(q) ≤ htgr(p) − 1 for every prime ideal q contained in p and distinct from p. Let us distinguish two cases. If q is graded, one concludes by the induction hypothesis. If q is not graded, then one has q^{gr} ≠ p, hence ht(q^{gr}) ≤ htgr(q^{gr}) by the induction hypothesis, whence ht(q) ≤ htgr(q^{gr}) + 1 by Prop. 3; it remains to prove the inequality htgr(q^{gr}) ≤ htgr(p) − 2; but if one had htgr(q^{gr}) = htgr(p) − 1, the chain q^{gr} ⊂ p would be saturated by a), which it is not since q^{gr} ≠ q ≠ p.

Finally let us prove c). One has dimgr(H) ≤ dim(S^{-1}H) ≤ dim(H) (Prop. 2), and it remains to prove dim(H) ≤ dimgr(H), or again ht(p) ≤ dimgr(H) for every prime ideal p of H. Let therefore p be a prime ideal of H. If p is graded, one has ht(p) = htgr(p) ≤ dimgr(H). If p is not graded, one has ht(p) = htgr(p^{gr}) + 1 by Prop. 3; let m be a maximal graded ideal of H containing p^{gr}; by Lemma 1, b), m is maximal, hence distinct from p^{gr}, and one has htgr(p^{gr}) + 1 ≤ htgr(m) ≤ dimgr(H), whence again ht(p) ≤ dimgr(H). This completes the proof.

One has, for $n \geq n_0$, $M_{\geq n+1} = H_1 \cdot M_{\geq n}$, hence $S^{-1}M_{\geq n+1} = H_1 \cdot S^{-1}M_{\geq n} = S^{-1}H_{\geq 1} \cdot S^{-1}M_{\geq n}$.

Let a be the annihilator of the $H$-module $M$; it is a graded ideal of $H$. Since $M$ is a finitely generated $H$-module, the annihilator of the $S^{-1}H$-module $S^{-1}M$ is the ideal $S^{-1}\alpha$ of $S^{-1}H$. One has $\dim_H(M) = \dim(H/\alpha)$ and $\dim_{S^{-1}H}(S^{-1}M) = \dim(S^{-1}H/S^{-1}\alpha)$. Proposition 4 then follows from Theorem 1, c) of No. 2 applied to the graded ring $H/\alpha$.

The ring $S^{-1}H$ is local and noetherian (No. 1, remark 1), the $S^{-1}H$-module $S^{-1}M$ is nonzero and finitely generated, and of dimension $d = \dim_H(M)$ (prop. 4). Moreover, $S^{-1}H_{\geq 1}$ is an ideal of definition of $S^{-1}H$ (§ 3, No. 2, lemma 2), and $S^{-1}M_{\geq n}$ is an $S^{-1}H_{\geq 1}$-good filtration on $S^{-1}M$ (lemma 3). Finally, one has $\operatorname{long}_{S^{-1}H} S^{-1}M_{\geq n}/S^{-1}M_{\geq n+1} = \operatorname{long}_{H_0}(M_n)$ for all $n$. It is enough therefore to apply Theorems 2 and 3 of § 4 (Nos. 3 and 4).

Applying proposition 5 to the case $M = H = \operatorname{gr}_q(A)$, one obtains the relation

with $d = \dim(\operatorname{gr}_q(A))$ and $Q(1) \neq 0$. One has $d = \dim(A)$ by Theorem 3 of § 4, No. 4, whence the corollary.

a) Let $a_1, ..., a_n$ be elements of $H$, homogeneous of degrees $> 0$, and let $\varphi$ be the homomorphism (of $H_0$-algebras) of $H_0[X_1, ..., X_n]$ into $H$ which maps $X_i$ to $a_i$ for $1 \leq i \leq n$. The $S^{-1}H$-module $S^{-1}M / \sum_{i=1}^n (a_i/1).S^{-1}M$ is of finite length if and only if $\varphi_*(M)$ is a finitely generated module over $H_0[X_1, ..., X_n]$.

b) There exists a family $(a_1, ..., a_d)$ of elements of $H$, all homogeneous of the same degree $> 0$, with $d = \dim_H(M)$, and such that $(a_1/1, ..., a_d/1)$ is a maximal secant sequence for the $S^{-1}H$-module $S^{-1}M$. If moreover $H$ is generated by $H_1$ as an $H_0$-algebra, and if the residue field of $H_0$ is infinite, one can take the $a_i$ of degree 1.

a) Put $N = M / \sum_{i=1}^n a_i M$. We have $\dim_H(N) = \dim_{S^{-1}H}(S^{-1}N)$ by prop. 4. Consequently, the $S^{-1}H$-module $S^{-1}N$ is of finite length if and only if the $H$-module $N$ is of finite length, that is to say if and only if $N$ is a finitely generated module over $H_0$. If $\varphi_*(M)$ is the module over $H_0[X_1, ..., X_n]$ deduced from $M$ by the homomorphism $\varphi : H_0[X_1, ..., X_n] \to M$, we have $N = \varphi_*(M)/\sum_{i=1}^n X_i \cdot \varphi_*(M)$. Consequently (A, II, p. 171, cor. 3 and remark) $\varphi_*(M)$ is a finitely generated module over $H_0[X_1, ..., X_n]$ if and only if $N$ is a finitely generated module over $H_0$. This proves a).

To prove b), we shall first establish a lemma.

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).

By Prop. 6 there exists an $H_0$-homomorphism $\varphi$ of the indicated form which makes $H$ into a finite $H_0[X_1, ..., X_n]$-algebra. By Theorem 1 of § 2, No. 3, one then has

since one has and since $H_0[X_1, ..., X_n]$ is an integral domain, this implies $\mathrm{Ker}\ \varphi = \{0\}$.

The condition is sufficient (§ 4, No. 4, Theorem 3 and § 5, No. 2, Theorem 1). Let us show that it is necessary. Consider the graded ring gr(A) = gr_{m_A}(A); let k be an infinite extension of the field $\kappa_A$, and put $H = k \otimes_{\kappa_A} \mathrm{gr}(A)$. The ring H is of dimension $\geq n$ (Prop. 5 and its corollary); one therefore deduces from Corollary 1 the existence of an injective graded homomorphism of graded k-algebras $\varphi : H_0[X_1, ..., X_n] \to H$. Consequently one has, for every integer $r \geq 0$,

and equality for every $r$ implies the bijectivity of $\varphi$, hence the regularity of A (§ 5, No. 2, Theorem 1).

The equalities

and

then imply the analogous assertions for the function $r \mapsto \mathrm{long}_A(A/m_A^{r+1})$.

Let $m_1, ..., m_n$ be homogeneous elements of M whose images in $M/rM$ generate the $R/rR$-module $M/rM$. Let N be the (graded) sub-$R$-module of M generated by $\{ m_1, ..., m_n \}$. For every $i \in \mathbf{Z}$, we have $M_i = N_i + rM_i$, hence $M_i = N_i$ (II, § 3, No. 2, Prop. 4); consequently we have $M = N$.

Let X be a finite generating set of the B-algebra C. For every $x \in X$, the image of x in $S^{-1}C$ is integral over $S^{-1}B$, and consequently there exist an integer $n(x) \geq 0$, elements $b_1(x), ..., b_{n(x)}(x) \in B$ and an element $f(x) \in S$ such that

Let $f = \prod_{x \in X} f(x)$; the image of every element x of X in $C_f$ is integral over $B_f$, hence $C_f$ is a finite $B_f$-algebra (V, § 1, No. 1, Prop. 4).

Since H is a finitely generated $H_0$-algebra, each $H_i$ is a finitely generated $H_0$-module (III, § 1, No. 2, corollary to Prop. 1) and H is generated as an $H_0$-algebra by $H_0 \oplus H_1 \oplus \cdots \oplus H_r$ for a suitable integer $r \geq 0$. Let $p \in \mathrm{Spec}(H_0)$ and put $\dim(H \otimes_{H_0} \kappa(p)) = n \geq 0$. By Corollary 1 to Prop. 6, there exist elements $a_1, ..., a_n$ of H, all homogeneous of the same degree $d > 0$, such that the $\kappa(p)$-homomorphism $\overline{\varphi} : \kappa(p)[X_1, ..., X_n] \to H \otimes_{H_0} \kappa(p)$ which maps $X_i$ to $a_i \otimes 1$ for $1 \leq i \leq n$, makes $H \otimes_{H_0} \kappa(p)$ into a finite $\kappa(p)[X_1, ..., X_n]$-algebra. Let $\varphi$ denote the $H_0$-homomorphism of $H_0[X_1, ..., X_n] = R$ into H which maps $X_i$ to $a_i$ for $1 \leq i \leq n$. If, for every $m \in \mathbf{Z}$, we put $H'_m = \sum_{(m-1)d < i \leq md} H_i$, we obtain a $\mathbf{Z}$-grading on H, compatible with the structure of R-module given by $\varphi$. Each $H'_m$ is finitely generated over $H_0$. By Lemma 5, $H_p$ is a finitely generated $R_p$-module. By Lemma 6, there therefore exists $f \in H_0 - p$ such that $H_f$ is a finitely generated $R_f$-module. For every $q \in \mathrm{Spec}(H_0)_f$, $H \otimes_{H_0} \kappa(q)$ is a finite $\kappa(q)[X_1, ..., X_n]$-algebra, therefore $\dim(H \otimes_{H_0} \kappa(q)) \leq n$ (§ 2, No. 3, Theorem 1), which completes the proof.

(iii) the $a_i$ are algebraically independent over $H_0$ and generate $H$ as an $H_0$-algebra.

One has $\dim(H) = \dim(S^{-1}H)$ (No. 2, th. 1), and $[H_+/H_+^2 : H_0] = [(S^{-1}H_+)/ (S^{-1}H_+)^2 : H_0]$ (II, § 3, No. 3, prop. 9); after § 5, No. 1, this implies $a$ and the equivalences (i) $\Leftrightarrow$ (ii) in $b$ and $c$. The two implications (iii) $\Rightarrow$ (i) in $b$ and $c$ are trivial. Let us prove the implications (i) $\Rightarrow$ (iii). Assume therefore that one has $\dim(H) = [H_+/H_+^2 : H_0]$ and let $a_1, ..., a_n$ be homogeneous elements of $H$, of degrees $> 0$, whose classes form a basis of $H_+/H_+^2$; consider the homomorphism of graded algebras $\varphi : H_0[X_1, ..., X_n] \to H$ which sends $X_i$ to $a_i$. The ideal $H_+$ of $H$ is generated by the $a_i$ (A, II, p. 171, cor. 2 and remark); consequently the homomorphism $\varphi$ is surjective (III, § 1, No. 2, prop. 1). But one has

(§ 2, No. 4, cor. 1 of th. 3), therefore $\mathrm{Ker}\, \varphi = 0$ since $H_0[X_1, ..., X_n]$ is an integral domain. This implies assertions (iii).

Under the hypotheses of $b$, one says that $H$ is a regular graded $H_0$-algebra, or a polynomial graded $H_0$-algebra. Under the hypotheses of $c$, one says that the $a_i$ form a graded system of coordinates of $H$.