Lemma 10.122.11. Let $R$ be a ring. Let $S = R[x]/I$. Let $\mathfrak q \subset S$ be a prime. Assume $R \to S$ is quasi-finite at $\mathfrak q$. Let $S' \subset S$ be the integral closure of $R$ in $S$. Then there exists an element $g \in S'$, $g \not\in \mathfrak q$ such that $S'_ g \cong S_ g$.

**Proof.**
Let $\mathfrak p$ be the image of $\mathfrak q$ in $\mathop{\mathrm{Spec}}(R)$. There exists an $f \in I$, $f = a_ nx^ n + \ldots + a_0$ such that $a_ i \not\in \mathfrak p$ for some $i$. Namely, otherwise the fibre ring $S \otimes _ R \kappa (\mathfrak p)$ would be $\kappa (\mathfrak p)[x]$ and the map would not be quasi-finite at any prime lying over $\mathfrak p$. We conclude there exists a relation $b_ m x^ m + \ldots + b_0 = 0$ with $b_ j \in S'$, $j = 0, \ldots , m$ and $b_ j \not\in \mathfrak q \cap S'$ for some $j$. We prove the lemma by induction on $m$. The base case is $m = 0$ is vacuous (because the statements $b_0 = 0$ and $b_0 \not\in \mathfrak q$ are contradictory).

The case $b_ m \not\in \mathfrak q$. In this case $x$ is integral over $S'_{b_ m}$, in fact $b_ mx \in S'$ by Lemma 10.122.1. Hence the injective map $S'_{b_ m} \to S_{b_ m}$ is also surjective, i.e., an isomorphism as desired.

The case $b_ m \in \mathfrak q$. In this case we have $b_ mx \in S'$ by Lemma 10.122.1. Set $b'_{m - 1} = b_ mx + b_{m - 1}$. Then

Since $b'_{m - 1}$ is congruent to $b_{m - 1}$ modulo $S' \cap \mathfrak q$ we see that it is still the case that one of $b'_{m - 1}, b_{m - 2}, \ldots , b_0$ is not in $S' \cap \mathfrak q$. Thus we win by induction on $m$. $\square$

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