A quasi-finite one-generator quotient is locally its integral closure

We first produce a relation over $S^{\mathrm{\prime }}$ of positive degree with a coefficient outside $\mathfrak{q}$. Suppose that every element of $I$ had all its coefficients in $\mathfrak{p}$, that is $I\subseteq \mathfrak{p}R[x]$. Then by [L8] the quotient $S\mathrm{/}\mathfrak{p}S\cong R[x]\mathrm{/}(I+\mathfrak{p}R[x])=R[x]\mathrm{/}\mathfrak{p}R[x]\cong (R\mathrm{/}\mathfrak{p})[x]$ is a polynomial ring over the domain $R\mathrm{/}\mathfrak{p}$; writing $\bar{\mathfrak{q}}$ for the image of $\mathfrak{q}$ in $S\mathrm{/}\mathfrak{p}S$ and using [L7] with $T=S\setminus \mathfrak{q}$, we get $S_{\mathfrak{q}}\mathrm{/}\mathfrak{p}{S}_{\mathfrak{q}}\cong (S\mathrm{/}\mathfrak{p}S{)}_{\bar{\mathfrak{q}}}$, a localisation of $(R\mathrm{/}\mathfrak{p})[x]$.

Such a localisation is infinite-dimensional over $\kappa (\mathfrak{p})=\mathrm{Frac}(R\mathrm{/}\mathfrak{p})$. Indeed the powers $1,\bar{x},{\bar{x}}^2,\dots$ of the class of $x$ in $(R\mathrm{/}\mathfrak{p})[x]$ are $\kappa (\mathfrak{p})$-linearly independent: a relation ${\sum }_i{c}_i{\bar{x}}^i=0$ with $c_i\in \kappa (\mathfrak{p})$ becomes, after multiplying by a common nonzero denominator $d\in R\mathrm{/}\mathfrak{p}$, a relation ${\sum }_i(d{c}_i){\bar{x}}^i=0$ with coefficients in the domain $R\mathrm{/}\mathfrak{p}$ and not all zero, which by [L9] is a nonzero polynomial of $(R\mathrm{/}\mathfrak{p})[x]$ equal to zero — impossible by [L10] unless all coefficients vanish; and this linear independence is preserved in the localisation because the localisation map is injective on the domain $(R\mathrm{/}\mathfrak{p})[x]$ by [L11].

Therefore $I\subseteq ̸\mathfrak{p}R[x]$: otherwise steps 1.1 and 1.2 would exhibit $S_{\mathfrak{q}}\mathrm{/}\mathfrak{p}{S}_{\mathfrak{q}}$ as an infinite-dimensional $\kappa (\mathfrak{p})$-algebra, contradicting the quasi-finiteness hypothesis through [L1]. So there are $n\ge 0$ and $a_0,\dots ,a_n\in R$ with $f={\sum }_j{a}_j{x}^j\in I$ and $a_j\mathrm{\notin }\mathfrak{p}$ for at least one $j$; applying $\phi$ and writing $b_j=\phi (a_j)$ (the element of the relative integral closure $S^{\mathrm{\prime }}$ given by the image of $R$) gives a relation ${\sum }_{j=0}^n{b}_j\phi (x{)}^j=0$ in $S$ with some $b_j\mathrm{\notin }\mathfrak{q}$, because $a_j\in \mathfrak{p}$ if and only if $\phi (a_j)\in \mathfrak{q}$.

Among all relations ${\sum }_{j=0}^m{b}_j\phi (x{)}^j=0$ in $S$ with all $b_j\in S^{\mathrm{\prime }}$ and at least one $b_j\mathrm{\notin }\mathfrak{q}$, choose one of minimal degree $m$, which exists by step 2.1; then $m\ge 1$, because a relation of degree $0$ reads $b_0=0$ with $b_0\in S^{\mathrm{\prime }}$, and $b_0\mathrm{\notin }\mathfrak{q}$ would be impossible. We show that its leading coefficient satisfies $b_m\mathrm{\notin }\mathfrak{q}$.

Assume first that $b_m\mathrm{\notin }\mathfrak{q}$. By [L5] applied to the relation of step 3.1 with $A=S^{\mathrm{\prime }}$ and $t=\phi (x)$, the element $b_m\phi (x)$ is integral over $S^{\mathrm{\prime }}$; hence the inclusion $S^{\mathrm{\prime }}\to S^{\mathrm{\prime }}[b_m\phi (x)]$ is an integral ring map, and since $S^{\mathrm{\prime }}={\mathrm{Int}}_R(S)$ consists of elements integral over $R$ by [L4], the composite $R\to S^{\mathrm{\prime }}[b_m\phi (x)]$ is integral by [L6]. So $b_m\phi (x)$ is integral over $R$, that is $b_m\phi (x)\in {\mathrm{Int}}_R(S)=S^{\mathrm{\prime }}$.

Assume instead that $b_m\in \mathfrak{q}$. The argument of step 4.1 still gives $b_m\phi (x)\in S^{\mathrm{\prime }}$. Set $b_{m-1}^{\mathrm{\prime }}=b_m\phi (x)+b_{m-1}\in S^{\mathrm{\prime }}$. Then ${\sum }_{j=0}^{m-2}{b}_j\phi (x{)}^j+b_{m-1}^{\mathrm{\prime }}\phi (x{)}^{m-1}=0$ by step 3.1, and $b_{m-1}^{\mathrm{\prime }}-b_{m-1}=b_m\phi (x)\in \mathfrak{q}$ because $b_m\in \mathfrak{q}$; so $b_{m-1}^{\mathrm{\prime }}\mathrm{\notin }\mathfrak{q}$ if $b_{m-1}\mathrm{\notin }\mathfrak{q}$, and otherwise one of $b_{m-2},\dots ,b_0$ lies outside $\mathfrak{q}$ by step 3.1. Either way we have a relation of degree $m-1$ with all coefficients in $S^{\mathrm{\prime }}$ and at least one coefficient outside $\mathfrak{q}$, contradicting the choice of the minimal degree $m$. Hence the case $b_m\in \mathfrak{q}$ cannot occur.

By steps 3.1 and 5.1 the minimal relation of step 3.1 has $b_m\in S^{\mathrm{\prime }}\setminus \mathfrak{q}$, and $b_m\phi (x)\in S^{\mathrm{\prime }}$ by step 4.1. Put $g=b_m$. In the localisation $S_g^{\mathrm{\prime }}$ the element $g$ is a unit by [L3], so $\phi (x)=(b_m\phi (x))\mathrm{/}{b}_m\in S_g^{\mathrm{\prime }}$; since $S$ is generated as an $S^{\mathrm{\prime }}$-algebra by $\phi (x)$ by [L2], the subring $S_g^{\mathrm{\prime }}\subseteq S_g$ already contains a generating set of $S_g$ over $S_g^{\mathrm{\prime }}$ and hence equals $S_g$.

The canonical homomorphism $S_g^{\mathrm{\prime }}\to S_g$ is injective, so it is an isomorphism: if a fraction $a\mathrm{/}{g}^k$ with $a\in S^{\mathrm{\prime }}$ maps to zero in $S_g$, then by [L11] there is $l\ge 0$ with $g^{l}a=0$ in $S$; since $a,g\in S^{\mathrm{\prime }}$, the product $g^{l}a$ lies in $S^{\mathrm{\prime }}$, and [L11] again gives $a\mathrm{/}{g}^k=0$ in $S_g^{\mathrm{\prime }}$.

Steps 2.1, 6.1 and 7.1 produce $g\in S^{\mathrm{\prime }}\setminus \mathfrak{q}$ with $S_g^{\mathrm{\prime }}\to S_g$ an isomorphism, which is the assertion. The argument is choice-free: the only selection is that of a relation of minimal degree in step 3.1, and the minimal-degree argument uses only that degrees are natural numbers. ∎