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. $\square$

Comment #4243 by Kazuki Masugi on

"Write $g$" should be "Write $h$".

7th sentense in the proof.

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