Let $S = \bigoplus_{n\ge 0} S_n$ be a graded commutative ring. Let $f$ be a homogeneous element of $S$ of degree $> 0$. Let $S_{(f)}$ be the degree $0$ part of the graded ring $S_f$, where $S_f$ is the localization with respect to the multiplicative set $\{1, f, f^2,\dots\}$. Suppose $S$ is finitely generated algebra over $S_0$. Then $S_{(f)}$ is a finitely generated algebra over $S_0$.

  • $\begingroup$ Can't you just take the generators of $S$ and divide them by appropriate powers of $f$? $\endgroup$
    – Lalit Jain
    Nov 9, 2012 at 16:22
  • $\begingroup$ @LalitJain I have no idea what powers of $f$ I should take. $\endgroup$ Nov 9, 2012 at 16:33
  • $\begingroup$ I've just found a proof of the title question in the Stacks Project, Lemma 7.55.9, p.324, 2012. stacks.math.columbia.edu $\endgroup$ Nov 10, 2012 at 9:32

2 Answers 2


Let $\deg f=d>0$. Then $$S_{(f)}=\{\frac{x}{f^m}:\deg x=md, m\ge 0\}=S_0\oplus\frac{1}{f}S_d\oplus\frac{1}{f^2}S_{2d}\oplus\cdots.$$ But one knows that $S$ finitely generated over $S_0$ implies $S^{(d)}=S_0\oplus S_d\oplus S_{2d}\oplus\cdots$ finitely generated over $S_0$. Now the system of generators for $S_{(f)}$ over $S_0$ should be clear.

  • $\begingroup$ Could you explain why $S^{(d)}$ is finitely generated over $S_0$? $\endgroup$ Nov 9, 2012 at 23:55
  • $\begingroup$ Yes, but it's easier to give you a reference: N. Bourbaki, Commutative Algebra, Chapter III, Section 1, Proposition 2. $\endgroup$
    – user26857
    Nov 10, 2012 at 1:16

This is not an answer but it's too long for a comment. I've just came up with the following proof of the assertion that

$S^{(d)}$ is finitely generated over $S_0$.

Suppose $S$ is generated by homogeneous elements $x_i$ of degree $k_i (1 \leqslant i \leqslant r)$ over $S_0$. Let $n > 0$ be an integer. Let $e_i \geqslant 0$ $(1 \leqslant i \leqslant r)$ be integers such that $\sum_i k_ie_i = dn$. Then the degree of $x_1^{e_1}\cdots x_r^{e_r}$ is $dn$. Suppose $e_i \geqslant d$. Then $x_1^{e_1}\cdots x_r^{e_r} = x_i^d x_1^{e_1}\cdots x_i^{e_i - d}\cdots x_r^{e_r}$. Hence $(S^{(d)})_+ = \bigoplus_{n>0} (S^{(d)})_n$ is generated as a graded $S^{(d)}$-module by $x_1^{e_1}\cdots x_r^{e_r} (0 \leqslant e_i < d)$.
Let $U = \{x_1^{e_1}\cdots x_r^{e_r}\mid \sum k_ie_i=md,0 \leqslant e_i \leqslant d\}$. By induction on $n$, it is easy to see that $(S^{(d)})_n$ is generated as an $S_0$-module by finite products of elements of $U$ for every integer $n > 0$. Hence $S^{(d)}$ is generated by $U$ over $S_0$.


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