Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $S = \sum_{n\ge 0} S_n$ 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 finitely generated algebra over $S_0$?

share|cite|improve this question
Can't you just take the generators of $S$ and divide them by appropriate powers of $f$? – Lalit Jain Nov 9 '12 at 16:22
@LalitJain I have no idea what powers of $f$ I should take. – Makoto Kato Nov 9 '12 at 16:33
I've just found a proof of the title question in the Stacks Project, Lemma 7.55.9, p.324, 2012. – Makoto Kato Nov 10 '12 at 9:32
up vote 3 down vote accepted

The answer is yes.

Assume that $\deg f=d>0$. 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.

share|cite|improve this answer
Could you explain why $S^{(d)}$ is finitely generated over $S_0$? – Makoto Kato Nov 9 '12 at 23:55
Yes, but it's easier to give you a reference: N. Bourbaki, Commutative Algebra, Chapter III, Section 1, Proposition 2. – user26857 Nov 10 '12 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 $k_i (1 ≦ i ≦ r)$ over $S_0$. Let $n > 0$ be an integer. Let $e_i ≧ 0 (1 ≦ i ≦ r)$ be integers such that $\sum_i k_ie_i = dn$. Then the degree $x_1^{e_1}...x_r^{e_r}$ is $dn$. Suppose $e_i > d$. Then $x_1^{e_1}...x_r^{e_r} = x_i^d x_1^{e_1}...x_i^{e_i - d}...x_r^{e_r}$. Hence $(S^{(d)})_+ = \sum_{n>0} (S^{(d)})_n$ is generated as a graded $S^{(d)}$-module by $x_1^{e_1}...x_r^{e_r} (0 \le e_i < d)$. Let $U = \{x_1^{e_1}...x_r^{e_r}\mid 0 \le e_i < d\}$. By the 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 $S_0$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.