Lemma 10.57.9. Let $R$ be a ring. Let $S$ be a graded $R$-algebra. Let $f \in S_{+}$ be homogeneous. Assume that $S$ is of finite type over $R$. Then
the ring $S_{(f)}$ is of finite type over $R$, and
for any finite graded $S$-module $M$ the module $M_{(f)}$ is a finite $S_{(f)}$-module.
Proof. Choose $f_1, \ldots , f_ n \in S$ which generate $S$ as an $R$-algebra. We may assume that each $f_ i$ is homogeneous (by decomposing each $f_ i$ into its homogeneous components). An element of $S_{(f)}$ is a sum of the form
with $\lambda _{e_1 \ldots e_ n} \in R$. Thus $S_{(f)}$ is generated as an $R$-algebra by the $f_1^{e_1} \ldots f_ n^{e_ n} /f^ e$ with the property that $e\deg (f) = \sum e_ i\deg (f_ i)$. If $e_ i \geq \deg (f)$ then we can write this as
Thus we only need the elements $f_ i^{\deg (f)}/f^{\deg (f_ i)}$ as well as the elements $f_1^{e_1} \ldots f_ n^{e_ n} /f^ e$ with $e \deg (f) = \sum e_ i \deg (f_ i)$ and $e_ i < \deg (f)$. This is a finite list and we see that (1) is true.
To see (2) suppose that $M$ is generated by homogeneous elements $x_1, \ldots , x_ m$. Then arguing as above we find that $M_{(f)}$ is generated as an $S_{(f)}$-module by the finite list of elements of the form $f_1^{e_1} \ldots f_ n^{e_ n} x_ j /f^ e$ with $e \deg (f) = \sum e_ i \deg (f_ i) + \deg (x_ j)$ and $e_ i < \deg (f)$. $\square$
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