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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