Lemma 10.30.1. Let R \subset S be an inclusion of domains. Assume that R \to S is of finite type. There exists a nonzero f \in R, and a nonzero g \in S such that R_ f \to S_{fg} is of finite presentation.
Proof. By induction on the number of generators of S over R. During the proof we may replace R by R_ f and S by S_ f for some nonzero f \in R.
Suppose that S is generated by a single element over R. Then S = R[x]/\mathfrak q for some prime ideal \mathfrak q \subset R[x]. If \mathfrak q = (0) there is nothing to prove. If \mathfrak q \not= (0), then let h \in \mathfrak q be a nonzero element with minimal degree in x. Write h = f x^ d + a_{d - 1} x^{d - 1} + \ldots + a_0 with a_ i \in R and f \not= 0. After inverting f in R and S we may assume that h is monic. We obtain a surjective R-algebra map R[x]/(h) \to S. We have R[x]/(h) = R \oplus Rx \oplus \ldots \oplus Rx^{d - 1} as an R-module and by minimality of d we see that R[x]/(h) maps injectively into S. Thus R[x]/(h) \cong S is finitely presented over R.
Suppose that S is generated by n > 1 elements over R. Say x_1, \ldots , x_ n \in S generate S. Denote S' \subset S the subring generated by x_1, \ldots , x_{n-1}. By induction hypothesis we see that there exist f\in R and g \in S' nonzero such that R_ f \to S'_{fg} is of finite presentation. Next we apply the induction hypothesis to S'_{fg} \to S_{fg} to see that there exist f' \in S'_{fg} and g' \in S_{fg} such that S'_{fgf'} \to S_{fgf'g'} is of finite presentation. We leave it to the reader to conclude. \square
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