The Stacks project

Proof. There exist finitely many elements $x_1, \ldots , x_ n \in S$ such that $S$ is finite over the $R$-sub algebra generated by $x_1, \ldots , x_ n$. (For example generators of $S$ over $R$.) We prove the proposition by induction on the minimal such number $n$.

The case $n = 0$ is trivial, because in this case $S' = S$, see Lemma 10.36.3.

The case $n = 1$. We may replace $R$ by its integral closure in $S$ (Lemma 10.122.9 guarantees that $R \to S$ is still quasi-finite at $\mathfrak q$). Thus we may assume $R \subset S$ is integrally closed in $S$, in other words $R = S'$. Consider the map $\varphi : R[x] \to S$, $x \mapsto x_1$. (We will see that $\varphi $ is not injective below.) By assumption $\varphi $ is finite. Hence we are in Situation 10.123.4. Let $J \subset S$ be the “conductor ideal” defined in Situation 10.123.4. Consider the diagram

According to Lemma 10.123.6 the image of $x$ in the quotient $S/\sqrt{J}$ is strongly transcendental over $R/ (R \cap \sqrt{J})$. Hence by Lemma 10.123.10 the ring map $R/ (R \cap \sqrt{J}) \to S/\sqrt{J}$ is not quasi-finite at any prime of $S/\sqrt{J}$. By Lemma 10.122.6 we deduce that $\mathfrak q$ does not lie in $V(J) \subset \mathop{\mathrm{Spec}}(S)$. Thus there exists an element $s \in J$, $s \not\in \mathfrak q$. By definition of $J$ we may write $s = \varphi (f)$ for some polynomial $f \in R[x]$. Let $I = \mathop{\mathrm{Ker}}(\varphi : R[x] \to S)$. Since $\varphi (f) \in J$ we get $(R[x]/I)_ f \cong S_{\varphi (f)}$. Also $s \not\in \mathfrak q$ means that $f \not\in \varphi ^{-1}(\mathfrak q)$. Thus $\varphi ^{-1}(\mathfrak q)$ is a prime of $R[x]/I$ at which $R \to R[x]/I$ is quasi-finite, see Lemma 10.122.5. Note that $R$ is integrally closed in $R[x]/I$ since $R$ is integrally closed in $S$. By Lemma 10.123.11 there exists an element $h \in R$, $h \not\in R \cap \mathfrak q$ such that $R_ h \cong (R[x]/I)_ h$. Thus $(R[x]/I)_{fh} = S_{\varphi (fh)}$ is isomorphic to a principal localization $R_{h'}$ of $R$ for some $h' \in R$, $h' \not\in \mathfrak q$.

The case $n > 1$. Consider the subring $R' \subset S$ which is the integral closure of $R[x_1, \ldots , x_{n-1}]$ in $S$. By Lemma 10.122.9 the extension $S/R'$ is quasi-finite at $\mathfrak q$. Also, note that $S$ is finite over $R'[x_ n]$. By the case $n = 1$ above, there exists a $g' \in R'$, $g' \not\in \mathfrak q$ such that $(R')_{g'} \cong S_{g'}$. At this point we cannot apply induction to $R \to R'$ since $R'$ may not be finite type over $R$. Since $S$ is finitely generated over $R$ we deduce in particular that $(R')_{g'}$ is finitely generated over $R$. Say the elements $g'$, and $y_1/(g')^{n_1}, \ldots , y_ N/(g')^{n_ N}$ with $y_ i \in R'$ generate $(R')_{g'}$ over $R$. Let $R''$ be the $R$-sub algebra of $R'$ generated by $x_1, \ldots , x_{n-1}, y_1, \ldots , y_ N, g'$. This has the property $(R'')_{g'} \cong S_{g'}$. Surjectivity because of how we chose $y_ i$, injectivity because $R'' \subset R'$, and localization is exact. Note that $R''$ is finite over $R[x_1, \ldots , x_{n-1}]$ because of our choice of $R'$, see Lemma 10.36.4. Let $\mathfrak q'' = R'' \cap \mathfrak q$. Since $(R'')_{\mathfrak q''} = S_{\mathfrak q}$ we see that $R \to R''$ is quasi-finite at $\mathfrak q''$, see Lemma 10.122.2. We apply our induction hypothesis to $R \to R''$, $\mathfrak q''$ and $x_1, \ldots , x_{n-1} \in R''$ and we find a subring $R''' \subset R''$ which is integral over $R$ and an element $g'' \in R'''$, $g'' \not\in \mathfrak q''$ such that $(R''')_{g''} \cong (R'')_{g''}$. Write the image of $g'$ in $(R'')_{g''}$ as $g'''/(g'')^ n$ for some $g''' \in R'''$. Set $g = g''g''' \in R'''$. Then it is clear that $g \not\in \mathfrak q$ and $(R''')_ g \cong S_ g$. Since by construction we have $R''' \subset S'$ we also have $S'_ g \cong S_ g$ as desired. $\square$