
Lemma 10.55.2. Let $S$ be a graded ring, which is finitely generated over $S_0$. Then for all sufficiently divisible $d$ the algebra $S^{(d)}$ is generated in degree $1$ over $S_0$.

Proof. Say $S$ is generated by $f_1, \ldots , f_ r \in S$ over $S_0$. After replacing $f_ i$ by their homogeneous parts, we may assume $f_ i$ is homogeneous of degree $d_ i > 0$. Then any element of $S_ n$ is a linear combination with coefficients in $S_0$ of monomials $f_1^{e_1} \ldots f_ r^{e_ r}$ with $\sum e_ i d_ i = n$. Let $m$ be a multiple of $\text{lcm}(d_ i)$. For any $N \geq r$ if

$\sum e_ i d_ i = N m$

then for some $i$ we have $e_ i \geq m/d_ i$ by an elementary argument. Hence every monomial of degree $N m$ is a product of a monomial of degree $m$, namely $f_ i^{m/d_ i}$, and a monomial of degree $(N - 1)m$. It follows that any monomial of degree $nrm$ with $n \geq 2$ is a product of monomials of degree $rm$. Thus $S^{(rm)}$ is generated in degree $1$ over $S_0$. $\square$

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