**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).
$\square$

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