Proposition 3

Let $R = \bigoplus_{i \geq 0} R_i$, be a graded commutative ring with positive degrees which is a finitely generated $R_0$-algebra. There exists an integer $e \geq 1$ such that $R^{(me)} = R_0[R_{me}]$ for all $m \geq 1$.

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 $e = q n_0$.