(a) The ideal of $A$ generated by the family $(x_\lambda)$ is equal to $m$.

(b) The family $(x_\lambda)$ is a system of generators of the $A$*-algebra* $A$.

(c) For all $i \geq 0$, the $A$*-module* $A_i$ is generated by the elements of the form $\prod_\lambda x_\lambda^{n_\lambda}$ which are of degree $i$ in $A$.

Clearly conditions (b) and (c) are equivalent. If they hold, every element of m is of the form $f((x_i))$ where $f$ is a polynomial of $A_0[X_\lambda]_{\lambda \in L}$ with no constant term; then $m = \sum_{\lambda \in L} A x_\lambda$, which proves that (c) implies (a). Conversely, suppose that condition (a) holds. Let $A' = A_0[x_\lambda]_{\lambda \in L}$ be the sub-A,-algebra of A generated by the family $(x_\lambda)$ and let us show that $A' = A$. For this, it is sufficient to show that $A_i \subset A$ for all $i \geq 0$. We proceed by induction on $i$, the property being obvious for $i = 0$. Then let $y \in A$, with $i \geq 1$. Since $y \in m$, there exists a family $(a_\lambda)_{\lambda \in L}$ of elements of A of finite support such that $y = \sum_\lambda a_\lambda x_\lambda$ and we may assume that each of the $a_\lambda$ is homogeneous of degree $i - \deg(x_\lambda)$ (by replacing it if need be by its homogeneous component of that degree); as $\deg(x_\lambda) > 0$, the induction hypothesis shows that $a_\lambda \in A'$ for all $\lambda \in L$, whence $y \in A'$ and $A_i \subset A'$, which proves that (a) implies (b).

(i) The following conditions are equivalent: (a) The ideal $m$ is a finitely generated A-module. (b) The ring $A$ is a finitely generated A,-module.

(ii) Suppose that the conditions in (i) hold and let $M = \bigoplus_{i \in \mathbf{Z}} M_i$ be a finitely generated graded A-module. Then, for all $i \in \mathbf{Z}$, $M_i$ is a finitely A,-module and there exists $i_0$ such that $M_i = \{0\}$ for $i < i_0$.

(i) If a family $(y_\mu)$ of elements of A is a system of generators of the A-module m (resp. of the A,-module A), so is the family consisting of the homogeneous components of the $y_\mu$; the equivalence of conditions (a) and (b) then follows from Proposition 1.

(ii) We may suppose that A is generated (as an A,-algebra) by homogeneous elements $a_i$ ($1 < i < r$) of degree $> 1$ and M is generated (as an A-module) by homogeneous elements $x_j$ ($1 < j < s$); let $h_i = \deg(a_i), k_j = \deg(x_j)$. Clearly M, consists of the linear combinations with coefficients in $A_0$ of the elements $a_1^{\alpha_1} a_2^{\alpha_2} \ldots a_r^{\alpha_r} x_j$ such that the $\alpha_i$ are integers $\geq 0$ satisfying the relation $k_j + \sum_{i=1}^r \alpha_i h_i = n$; for each $n$ there is only a finite number of families $(\alpha_i)_{1 \leq i \leq r}$ satisfying these conditions, since $h_i \geq 1$ for all $i$; we conclude that $M_n$ is a finitely generated A,-module and moreover clearly $M_n = \{0\}$ when $n < \inf(k_j)$.

(i) $A^{(d)}$ is a finitely generated $A$-module. (ii) $M^{(d, k)}$ is a finitely generated $A^{(d)}$-module.

Let us show that $A$ is a finitely generated $A^{(d)}$-*module*. Let $(a_i)_{1 \leqslant i \leqslant s}$ be a system of generators of the $A$-algebra $A$ consisting of homogeneous elements. The elements of $A$ (finite in number) of the form $a_1^{\alpha_1} a_2^{\alpha_2} \ldots a_s^{\alpha_s}$ such that $0 \leqslant \alpha_i \leqslant d$ for $1 \leqslant i \leqslant s$ constitute a system of generators of the $A^{(d)}$-module $A$; for every system of integers $n_i \geqslant 0$ $(1 \leqslant i \leqslant s)$, there are positive integers $q_i, r_i$ such that $n_i = q_i d + r_i$ where $r_i < d$ $(1 \leqslant i \leqslant s)$; then

which proves our assertion, for every homogeneous element $x \in A$ satisfies $x^d \in A^{(d)}$. Then, if $M$ is a finitely generated $A$-module, it is also a finitely generated $A^{(d)}$-module; as $M$ is the direct sum of the $M^{(d, k)}$ $(0 < k < d - 1)$, each of the $M^{(d, k)}$ is a finitely generated $A^{(d)}$-module, which proves (ii).

Let us apply the above to the graded $A$-module $m = \bigoplus_{i \geqslant 1} A_i$, which is finitely generated by the Corollary to Proposition 1 of no. 2; it is seen that $m^{(d)}$ is a finitely generated $A^{(d)}$-module; therefore (no. 2, Corollary to Proposition 1) $A^{(d)}$ is a finitely generated $A$-algebra.

Let $n \geqslant n_0$ and $x \in M_n$. Since the $y_\lambda$ generate $M$, there exists a family $(a_\lambda)_{\lambda \in L}$ of elements of $A$ of finite support such that $x = \sum_\lambda a_\lambda y_\lambda$; we may further suppose that each $a_\lambda$ is homogeneous and of degree $n + 1 - \deg(y_\lambda)$ (replacing it if need be by its homogeneous component of that degree). As

$A = A_0[A_1]$ and $\deg(a_\lambda) > 0$, each $a_\lambda$ is a sum of elements of the form $bb'$ where $b \in A_0, b' \in A$, whence $x \in A_1 M_n$. Then $M_{n+1} = A_1 M_n$, whence $M_{n+k} = A_k M_n$ by induction on $k$.

(i) For $n \geq n_0$ and $k \geq 0$, $S_{n+k} = S_k \cdot S_n$. (ii) For $d \geq n_0$, $S^{(d)} = S_0[S_d]$.

By Lemma 1 there exists an integer $n_0 \geq 0$ such that, for $n \geq n_0$ and $k \geq 0$, $S_{n+k} = A_k S_n$, whence a fortiori $S_{n+k} = S_k S_n$, which establishes (i). Then, for $d \geq n_0$ and $m > 0$, $S_{m d} = (S_d)^m$ as follows by induction on $m$ applying (i); this establishes (ii).

Let $(x_j)_{1 \leq j \leq s}$ be a system of homogeneous generators of the $R_0$-algebra $R$, whose degrees are $\geq 1$. Let $h_j = \deg(x_j)$, let $q$ be a common multiple of the $h_j$ and let us write $q_j = q/h_j$ for $1 \leq j \leq s$; the elements $x_j^{q_j}$ are then all of degree $q$. Let $B$ be the sub-$R_0$-algebra of $R$ generated by the $x_j^{q_j}$; it is a graded subalgebra of $R$ and $B_i = 0$ if $i$ is not a multiple of $q$. Let $A$ (resp. $S$) be the graded ring whose underlying ring is $B$ (resp. $R^{(q)}$) and whose graduation consists of the $A_i = B_{iq}$ (resp. $S_i = R_{iq}$). Then $A = A_0[A_1]$ by definition of $B$. Consider the elements of $R$ (finite in number) of the form $x_1^{\alpha_1} x_2^{\alpha_2} \ldots x_s^{\alpha_s}$, where $0 \leq \alpha_j \leq q_j$ and $\alpha_1 h_1 + \ldots + \alpha_s h_s \equiv 0$ (mod. $q$); let us show that they generate the $B$-module $R^{(q)}$. It is sufficient to prove that every element of $R^{(q)}$ of the form $x_1^{n_1} x_2^{n_2} \ldots x_s^{n_s}$ is a $B$-linear combination of the above elements. Now, there exist positive integers $k_j, r_j$ such that $n_j = k_j q_j + r_j$ where $r_j < q_j$ ($1 \leq j \leq s$); then

and by hypothesis $\sum_{j=1}^s n_j h_j \equiv 0$ (mod. $q$), hence $\sum_{j=1}^s r_j h_j \equiv 0$ (mod. $q$); as the $x_j^{q_j}$ belong to $B$ by definition, this proves our assertion. As $S$ is a finitely generated $A$-module, Lemma 2 can be applied: there exists $n_0$ such that for $d > n_0$, $S^{(d)} = S_0[S_d]$ and hence $R^{(qd)} = R_0[R_{qd}]$ for $d \geq n_0$. The proposition follows with $e = q n_0$.

The condition is obviously necessary. Conversely, if it is fulfilled, then in the graded ring $A/p = \bigoplus_{i \geq 0} A_i/p_i$ the product of two homogeneous elements $\neq 0$ is $\neq 0$ and hence $A/p$ is an integral domain (Algebra, Chapter II, § 11, no. 4, Proposition 7).

The condition of the statement is obviously necessary. If $a_n = A_n$ for all $n \geq n_0,$ clearly every prime ideal containing $m$ is a solution to the problem; there may therefore be several prime ideals which solve the problem; however, any two of these ideals are obviously equivalent. Suppose then that there exists a homogeneous element $a \in A_d$ (with $d \geq n_0$) not belonging to $a_d$. Let $p$ be the set of $x \in A$ such that $ax \in a$. Clearly $p$ is an ideal of $A$; as the homogeneous components of $ax$ are the products by $a$ of those of $x$ and $a$ is a graded ideal, $p$ is a graded ideal; moreover, $1 \notin p$ and hence $p \neq A$. To prove that $p$ is prime, it is sufficient to show that, if $x \in A, \ y \in A,$ satisfy $x \notin p$ and $y \notin p,$ then $xy \notin p$ (Proposition 4). Then $ax \notin a_{m+d}, ay \notin a_{n+d},$ whence by hypothesis

we conclude that $axy \notin a_{m+n+d}$ since $xy \notin p$. Finally, if $n \geq n_0$ and $x \in A_n,$ the conditions $x \in a,$ and $ax \in a_{n+d}$ are equivalent by hypothesis and hence $p \cap A_n = a_n$ which completes the proof of the existence of the graded prime ideal $p$ which solves the problem. If also $p'$ is another graded prime ideal of $A$ such that $p' \cap A_n = a_n$ for $n \geq n_0,$ then $a \notin p'$ and $ax \in p'$ for all $x \in p,$ whence $p \subset p'$ since $p'$ is prime. On the other hand, if $x$ is a homogeneous element of degree $n \geq 0$ of $p',$ $ax$ is homogeneous of degree $n + d \geq n_0$ and therefore belongs to $p' \cap A_{n+d} = a_{n+d}$, whence by definition $x \in p$, which shows that $p' \subset p$ and finally $p' = p$.

(i) For every essential graded ideal p of A, $p \cap A^{(d)}$ is an essential graded prime ideal of $A^{(d)}$.

(ii) Conversely, for every essential graded prime ideal $p'$ of $A^{(d)}$, there exists a unique (necessarily essential) graded prime ideal $p$ of A such that $p \cap A^{(d)} = p'$.

(i) If $a \in A_k$ does not belong to $p_k$, $a^{kd}$ does not belong to $p_{kd}$ and hence $p \cap A^{(d)}$ is essential.

(ii) For all $n \geq 0$, the set $p \cap A$, must be equal to the set $a_n$ of $x \in A$, such that $x^d \in p'$. Let us show that $a = \bigoplus_{n \geq 0} a_n$ is a graded prime ideal; as $a_n = p'_n$ when $n$ is a multiple of $d$, since $p'$ is prime, this will prove the uniqueness of $p$. Now, if $x \in a_n$, $y \in a_m$, $(x - y)^{2d}$ is the sum of terms each of which is a product of $x^d$ or $y^d$ by a homogeneous element of degree $nd$ and hence $(x - y)^{2d} \in p'$ and, since $p'$ is prime, $(x - y)^d \in p'$ and therefore $a_n$ is a subgroup of $A$. As $p'$ is an ideal of $A^{(d)}$, $a$ is a graded ideal of $A$; finally, the relation $(xy)^d \in p'$ implies $x^d \in p'$ or $y^d \in p'$, which completes the proof by virtue of Proposition 4.

Let $A$ be a graded commutative ring with positive degrees and $p$ an essential graded prime ideal of $A$. The set $S$ of homogeneous elements of $A$ not belonging to $p$ is multiplicative and the ring of fractions $S^{-1}A$ is therefore graded canonically (Chapter II, § 2, no. 9) (note that there will in general be homogeneous elements $\neq 0$ of negative degree in this graduation). We shall denote by $A_{(p)}$ the subring of $S^{-1}A$ consisting of the homogeneous elements of degree 0, in other words the set of fractions $x/s$, where $x$ and $s$ are homogeneous of the same degree in $A$ and $s \notin p$. Similarly, for every graded $A$-module $M$, $S^{-1}M$ is graded canonically (loc. cit.) and we shall denote by $M_{(p)}$ the subgroup of homogeneous elements of degree 0, which is obviously an $A_{(p)}$-module.

If $S$ is the set of homogeneous elements of $A$ not belonging to $p$ and $S' = S \cap A^{(d)}$, the canonical homomorphism $\phi : {S'}^{-1}M^{(d)} \to S^{-1}M$ is a homogeneous homomorphism of degree 0 and it is injective, for, if $x \in M_{nd}$ satisfies $sx = 0$ for $s \in A_n, s \notin p$, then also $s^dx = 0$ and $s^d \in A_{md}, s^d \notin p'$. It remains to show that the image under $\phi$ of $(M^{(d)})_{(p')}$ is the whole of $M_{(p)}$; but if $x \in M_n, s \in A_n$ and $s \notin p$, then also $x/s = (xs^{d-1})/s^d$ where $xs^{d-1} \in A_{nd}, s^d \in A_n$ and $s^d \notin p'$, whence our assertion.

We argue by induction on $n$, the proposition being trivial for $n = 1$. If there exists an index $j$ such that $\mathfrak{a} \cap m \cap p^{(j)}$ is contained in one of the $p^{(k)}$ of index $k \neq j$, it follows from the induction hypothesis that there is a homogeneous element $z' \in \mathfrak{a} \cap m$ not belonging to any of the $p^{(k)}$ for $k \neq j$ and therefore not belonging to $p^{(j)}$ either and this element solves the problem. Suppose then that for every index $j$, $\mathfrak{a} \cap m \cap p^{(j)}$ is not contained in any of the $p^{(k)}$ of index $k \neq j$; the induction hypothesis therefore implies the existence of a homogeneous element $y_j \in \mathfrak{a} \cap m \cap S_j$ belonging to none of the $S_j$ for $k \# j$; as the $y_j$ are all of degrees $\geq 1$, we may assume by replacing them by suitable powers (since the $p^{(k)}$ are prime) that $y_1$ and $\prod_{j=2}^n y_j$ are of the same degree. Then $z = y_1 + \prod_{j=2}^n y_j$ is homogeneous of degree $\geq 1$ and the same argument as in Chapter II, § 1, no. 1, Proposition 2 shows that $z$ solves the problem.