# 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 $f$? Nov 9, 2012 at 16:22
• @LalitJain I have no idea what powers of $f$ I should take. Nov 9, 2012 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 Nov 10, 2012 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 $S_0$? Nov 9, 2012 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, 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 $$S_0$$.