Proof.
Choose h_1, \ldots , h_ n \in S such that \sum h_ i g_ i = 1.
Proof of (1). For each i choose a finite list of elements x_{i, j} \in S_{g_ i}, j = 1, \ldots , m_ i which generate S_{g_ i} as an R-algebra. Write x_{i, j} = y_{i, j}/g_ i^{n_{i, j}} for some y_{i, j} \in S and some n_{i, j} \ge 0. Consider the R-subalgebra S' \subset S generated by g_1, \ldots , g_ n, h_1, \ldots , h_ n and y_{i, j}, i = 1, \ldots , n, j = 1, \ldots , m_ i. Since localization is exact (Proposition 10.9.12), we see that S'_{g_ i} \to S_{g_ i} is injective. On the other hand, it is surjective by our choice of y_{i, j}. The elements g_1, \ldots , g_ n generate the unit ideal in S' as h_1, \ldots , h_ n \in S'. Thus S' \to S viewed as an S'-module map is an isomorphism by Lemma 10.23.2.
Proof of (2). We already know that S is of finite type. Write S = R[x_1, \ldots , x_ m]/J for some ideal J. For each i choose a lift g'_ i \in R[x_1, \ldots , x_ m] of g_ i and we choose a lift h'_ i \in R[x_1, \ldots , x_ m] of h_ i. Then we see that
S_{g_ i} = R[x_1, \ldots , x_ m, y_ i]/(J_ i + (1 - y_ ig'_ i))
where J_ i is the ideal of R[x_1, \ldots , x_ m, y_ i] generated by J. Small detail omitted. By Lemma 10.6.3 we may choose a finite list of elements f_{i, j} \in J, j = 1, \ldots , m_ i such that the images of f_{i, j} in J_ i and 1 - y_ ig'_ i generate the ideal J_ i + (1 - y_ ig'_ i). Set
S' = R[x_1, \ldots , x_ m]/\left(\sum h'_ ig'_ i - 1, f_{i, j}; i = 1, \ldots , n, j = 1, \ldots , m_ i\right)
There is a surjective R-algebra map S' \to S. The classes of the elements g'_1, \ldots , g'_ n in S' generate the unit ideal and by construction the maps S'_{g'_ i} \to S_{g_ i} are injective. Thus we conclude as in part (1).
\square
Comments (0)
There are also: