As $A_0 A_0 \subset A_0$ by definition, it suffices to prove that $1 \in A_0$. Let $1 = \sum_{\lambda \in \Delta} e_\lambda$ be the decomposition of 1 into its homogeneous components. If $x \in A_\mu$, then $x = x . 1 = \sum_{\lambda \in \Delta} x e_\lambda$; comparing the components of degree $\mu$, (since $\mu + \lambda = \mu$ implies $\lambda = 0$) $x = x e_0$. Since this relation is true for every homogeneous element of $A$, it is true for all $x \in A$; in particular $1 = 1 . e_0 = e_0 \in A_0$.

(a) $N$ is the sum of the family $(N \cap M_\lambda)_{\lambda \in \Delta}$. (b) The homogeneous components of every element of N belong to N. (c) $N$ is generated by homogeneous elements.

Every element of N can be written uniquely as a sum of elements of the $M_\lambda$ and hence it is immediate that (a) and (b) are equivalent and that (a) implies (c). We show that (c) implies (b). Then let $(x_i)_{i \in I}$ be a family of homogeneous generators $\neq 0$ of N and let $\delta(i)$ be the degree of $x_i$. Every element of N can be written as $\sum_i a_i x_i$ with $a_i \in A$; if $a_{i,\lambda}$ is the component of $a_i$ of degree $\lambda$, the conclusion follows from the relation

Remark (1) In the above notation, the relation $\sum_{i \in I} a_i x_i = 0$ is therefore equivalent to the system of relations $\sum_{\mu + \delta(i) = \lambda} a_{i,\mu} x_i = 0$. When $\Delta$ is a group, these relations can be written $\sum_{i \in I} a_{i,\lambda - \delta(i)} x_i = 0$.

When a submodule N of M has the equivalent properties stated in Proposition 2, clearly the $N \cap M_\lambda$ form a graduation compatible with the A-module structure of N, called the graduation induced by that on M; N with this graduation is called a graded submodule of M.

It suffices to apply Corollary 1 noting that an element of M has only a finite number of homogeneous components $\neq 0$.

A graded submodule of $A_s$ (resp. $A_d$) is called a graded left (resp. right) ideal of the graded ring A. For every subring B of $A(B \cap A_\lambda)(B \cap A_\mu) \subset B \cap A_{\lambda+\mu}$; if B is a graded sub-$\mathbf{Z}$-module of A, the graduation induced on B by that on A is therefore compatible with the ring structure on B; B is then called a graded subring of A.

Clearly if N (resp. B) is a graded sub-A-module of M (resp. a graded subring of A), the canonical injection $N \to M$ (resp. $B \to A$) is a graded module homomorphism of degree 0 (resp. a graded ring homomorphism).

If N is a graded submodule of a graded A-module M and $(M_\lambda)_{\lambda \in \Delta}$ the graduation of M, the submodules $(M_\lambda + N)/N$ of $M/N$ form a graduation compatible with the structure of this quotient module. For, if $N_\lambda = M_\lambda \cap N$, $(M_\lambda + N)/N$ is identified with $M_\lambda/N_\lambda$ and it follows from Proposition 2 and § 1, no. 6, formula (26) that $M/N$ is their direct sum. Moreover,

and hence $A_\lambda((M_\mu + N)/N) \subset (M_{\lambda+\mu} + N)/N$, which establishes our assertion.

tion. The graduation ((M_\lambda + N)/N)_{\lambda \in \Delta} is called the quotient graduation of that on M by N and the quotient module M/N with this graduation is called the graded quotient module of M by the graded submodule N; the canonical homomorphism M \to M/N is a graded homomorphism of degree 0 for this graduation.

If b is a graded two-sided ideal of A, the quotient graduation on A/b is compatible with the ring structure on A/b; the ring A/b with this graduation is called the quotient graded ring of A by b; the canonical homomorphism A \to A/b is a homomorphism of graded rings for this graduation.

Assertion (i) follows immediately from the definitions and Proposition 2(c). If x is an element of M such that u(x) = 0 and x = \sum_\lambda x_\lambda is its decomposition into homogeneous components (where x_\lambda is of degree \lambda), then

and u(x_\lambda) is of degree \lambda + \delta; if \delta is regular the relation \lambda + \delta = \mu + \delta implies \lambda = \mu, hence the u(x_\lambda) are the homogeneous components of u(x) and necessarily u(x_\lambda) = 0 for all \lambda \in \Delta, which proves (ii). The bijection v: M/Ker(u) \to Im(u) canonically associated with u is then a graded homomorphism of degree \delta, as follows from the definition of the quotient graduation; whence (iii) when \delta = 0.

It suffices to apply Proposition 3 to u considered as a homomorphism of degree 0 of graded \mathbf{Z}-modules.

If $(\mathbf{N}_i)$ is a family of graded submodules of $M$, property (c) of Proposition 2 shows that the sum of the $\mathbf{N}_i$ is generated by homogeneous elements and property (b) of Proposition 2 proves that the homogeneous components of every element of $\bigcap_i \mathbf{N}_i$ belongs to $\bigcap_i \mathbf{N}_i$; whence (i).

To prove (ii), it suffices to note that $\operatorname{Ann}(x)$ is the kernel of the homomorphism $a \mapsto ax$ of the $A$-module $A_s$ into $M$ and that this homomorphism is graded of degree $\mu$; the conclusion follows from Proposition 3(ii). Finally (iii) is a consequence of (i) and (ii) for the annihilator of a graded submodule $N$ of $M$ is the intersection of the annihilators of the homogeneous elements of $N$, by virtue of Proposition 2.

Suppose that $a$ is of degree $\delta$; let $b = \sum_\lambda b_\lambda$ be an element permutable with $a$, $b_\lambda$ being the homogeneous component of $b$ of degree $\lambda$ for all $\lambda \in \Delta$. Then by hypothesis $\sum_\lambda (ab_\lambda - b_\lambda a) = 0$ and $ab_\lambda - b_\lambda a$ is homogeneous of degree $\lambda + \delta$; as $\delta$ is cancellable, it follows that $ab_\lambda = b_\lambda a$ for all $\lambda$, which proves our assertion.

It is the intersection of the centralizers of the homogeneous elements of $B$.

Remark (3) *A direct system $(A_\alpha, \phi_{\beta\alpha})$ of graded rings of type $\Delta$ (resp. a direct system $(M_\alpha, f_{\beta\alpha})$ of graded $A_\alpha$-modules of type $\Delta$) is a direct system of rings (resp. $A_\alpha$-modules) such that each $A_\alpha$ (resp. $M_\alpha$) is graded of type $\Delta$ and each $\phi_{\beta\alpha}$ (resp. $f_{\beta\alpha}$) is a homomorphism of graded rings (resp. an $A_\alpha$-homomorphism of degree 0 of graded modules). If $(A^\lambda_\alpha)_{\lambda \in \Delta}$ (resp. $(M^\lambda_\alpha)_{\lambda \in \Delta}$) be the graduation of $A_\alpha$ (resp. $M_\alpha$) and we write

it follows from § 6, no. 2, Proposition 5 that $(A^\lambda)$ (resp. $(M^\lambda)$) is a graduation of A (resp. M) and it follows from I, § 10, nos. 3 and 4 that this graduation is compatible with the ring structure on A (resp. the A-module structure on M). The graded ring A (resp. graded A-module M) is called the direct limit of the direct system of graded rings $(A_\alpha, \phi_{\beta\alpha})$ (resp. graded modules $(M_\alpha, f_{\beta\alpha})$). If $\phi_\alpha : A_\alpha \to A$ (resp. $f_\alpha : M_\alpha \to M$) is the canonical mapping, $\phi_\alpha$ (resp. $f_\alpha$) is a homomorphism of graded rings (resp. a homomorphism of degree 0 of graded $A_\alpha$-modules).

Let $x$ be a non-zero element of $M_{\lambda_0}$; suppose that $x \in \mathfrak{I}_0 M$. Then $x = \sum_i a_i x_i$, where the $a_i$ are homogeneous elements $\neq 0$ of $\mathfrak{I}_0$ and the $x_i$ homogeneous elements $\neq 0$ of M with $\deg(x) = \deg(a_i) + \deg(x_i)$ for all $i$ (no. 2). But, as $\deg(a_i) > 0$, $\lambda_0 = \deg(a_i) + \deg(x_i) > \deg(x_i)$, which contradicts the hypothesis.

Suppose $M \neq \{0\}$. Let $\lambda_0$ be a minimal element of the set of degrees of a finite generating system of M consisting of homogeneous elements $\neq 0$; then the hypotheses of Proposition 6 would be fulfilled, which implies a contradiction.

$\mathbf{M}/\mathbf{N}$ is a finitely generated graded $\mathbf{A}$-module and the hypothesis implies that $\mathfrak{J}_0 . (\mathbf{M}/\mathbf{N}) = \mathbf{M}/\mathbf{N}$; hence $\mathbf{M}/\mathbf{N} = 0$.

is surjective, then $u$ is surjective.

$u(\mathbf{M})$ is a graded submodule of $\mathbf{N}$ and the $(\mathbf{A}/\mathfrak{J}_0)$-module

is isomorphic to $(N \otimes_{\mathbf{A}} (\mathbf{A}/\mathfrak{J}_0))/\mathrm{Im}(u \otimes 1)$ (§ 3, no. 6, Proposition 6). The hypothesis therefore implies $(N/u(\mathbf{M})) \otimes_{\mathbf{A}} (\mathbf{A}/\mathfrak{J}_0) = 0$ and hence $\mathbf{N} = u(\mathbf{M})$ by Corollary 1.

We know (no. 2, Remark 3) that there is a graded free $\mathbf{A}$-module $\mathbf{L}$ (of graduation $(L_\lambda)$) and a surjective homomorphism $p : \mathbf{L} \to \mathbf{M}$ of degree 0 such that $p(e_{i\lambda}) = x_{i\lambda}$ for all $(i, \lambda) \in I$ ($(e_{i\lambda})_{(i, \lambda) \in I}$ being a basis of $\mathbf{L}$ consisting of homogeneous elements $e_{i\lambda} \in L_\lambda$). It follows from the above Remark that $p$ is surjective. Consider the graded $\mathbf{A}$-module $\mathbf{R} = \mathrm{Ker}(p)$ and note that $R_\lambda = \{0\}$ for $\lambda < \lambda_0$ by definition; we need to prove that $\mathbf{R} = \{0\}$ and by Proposition 6 it will suffice to show that $\mathfrak{J}_0 R = R$. Consider the commutative diagram (§ 3, no. 6, Proposition 5)

where $j$ is the canonical injection, $a, b, c$ deriving from the canonical injection $\mathfrak{J}_0 \to A$ (§ 3, no. 4, Proposition 4); it must be shown that $a$ is surjective. Note that, as $L$ is free, $b$ is injective (§ 3, no. 7, Corollary 6 to Proposition 7) and $c$ is injective by hypothesis. Then let $t$ be an element of $R$ and $t$ its class in $R / \mathfrak{J}_0 R$; then there is an exact sequence (§ 3, no. 6, Proposition 5 and Corollary 2 to Proposition 6)

where $j$ and $\bar{p}$ derive from $j$ and $p$ when passing to the quotients and $\bar{p}$ is by hypothesis a bijection; then $j(\bar{t}) = 0$, in other words $j(t) \in \mathfrak{J}_0 L$. Then there is an element $z \in \mathfrak{J}_0 \otimes L$ such that $j(t) = b(z)$; as $p(b(z)) = 0$, $c((1 \otimes p)(z)) = 0$ and, as $c$ is injective, $(1 \otimes p)(z) = 0$. In other words, $z$ is the image of an element $t' \in \mathfrak{J}_0 \otimes R$ under $1 \otimes j$ and then $j(a(t')) = b(z) = j(t)$; as $j$ is injective, this implies $t = a(t')$.

We shall show later (Commutative Algebra, II, § 3, no. 2, Proposition 5) how this proposition can be extended to non-graded modules.

If there exists such an order structure on $\Delta$ and if $\lambda > 0$, then $\lambda + \mu > 0$ for all $\mu \geq 0$ and in particular, by induction on the integer $n > 0$, $n.\lambda > 0$, which proves that $\Delta$ is torsion-free (since every element $\neq 0$ of $\Delta$ is either $> 0$ or $< 0$). Conversely, if $\Delta$ is torsion-free, $\Delta$ is a sub-$\mathbf{Z}$-module of a vector $\mathbf{Q}$-space (§ 7, no. 10, Corollary 1 to Proposition 26) which may be assumed of the form $\mathbf{Q}^{(I)}$; if $I$ is given a well-ordering (Set Theory, III, § 2, no. 3, Theorem 1) and $\mathbf{Q}$ its usual ordering, the set $\mathbf{Q}^{(I)}$ with the lexicographical ordering is totally ordered (Set Theory, III, § 2, no. 6); it is immediate that this ordering is compatible with the additive group structure of $\mathbf{Q}^{(I)}$.

Let $\Delta$ be given a total ordering compatible with its group structure (Lemma

1) and let $x = \sum_{\lambda \in \Delta} x_\lambda, y = \sum_{\lambda \in \Delta} y_\lambda$ be two non-zero elements of $A$ ($x_\lambda$ and $y_\lambda$ being homogeneous of degree $\lambda$ for all $\lambda \in \Delta$); let $\alpha$ (resp. $\beta$) be the greatest of the elements $\lambda \in \Delta$ such that $x_\lambda \neq 0$ (resp. $y_\lambda \neq 0$); it is immediate that if $\lambda \neq \alpha$ or $\mu \neq \beta$, either $x_\lambda y_\mu = 0$ or $\deg(x_\lambda y_\mu) < \alpha + \beta$; the homogeneous component of $xy$ of degree $\alpha + \beta$ is therefore $x_\alpha y_\beta$, which is non-zero by hypothesis; whence $xy \neq 0$.