**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

\[ \sum \nolimits _{e\deg (f) = \sum e_ i\deg (f_ i)} \lambda _{e_1 \ldots e_ n} f_1^{e_1} \ldots f_ n^{e_ n}/f^ e \]

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

\[ f_1^{e_1} \ldots f_ n^{e_ n}/f^ e = f_ i^{\deg (f)}/f^{\deg (f_ i)} \cdot f_1^{e_1} \ldots f_ i^{e_ i - \deg (f)} \ldots f_ n^{e_ n}/f^{e - \deg (f_ i)} \]

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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