# If $S$ is a finitely generated graded algebra over $S_0$, $S_{(f)}$ is finitely generated algebra over $S_0$?

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 $$S_0$$.

• 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. stacks.math.columbia.edu – Makoto Kato Nov 10 '12 at 9:32

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.
• Could you explain why $S^{(d)}$ is finitely generated over $S_0$? – Makoto Kato Nov 9 '12 at 23:55
$$S^{(d)}$$ 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 of $$x_1^{e_1}\cdots x_r^{e_r}$$ is $$dn$$. Suppose $$e_i > d$$. 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 $$x_1^{e_1}\cdots x_r^{e_r} (0 \le e_i < d)$$.
Let $$U = \{x_1^{e_1}\cdots x_r^{e_r}\mid 0 \le e_i < d\}$$. 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 $$S_0$$.