Algebraic Zariski Main localization at a quasi-finite prime

We assume the Axiom of Choice throughout; it is recorded in [L12] and will be used exactly once below, through [L6] in the nowhere-quasi-finiteness argument. By [L3] the finite type $R$-algebra $S$ has the form $S=R[x_1,\dots ,x_n]$ for some $n\ge 0$ and some $x_1,\dots ,x_n\in S$; for these elements $S$ is module-finite over $R[x_1,\dots ,x_n]=S$, generated as a module over itself by $1$. We prove by induction on $n\ge 0$ the statement $P(n)$: for every unital ring map $A\to B$ of finite type that is quasi-finite at a prime ${\mathfrak{q}}_B\in \mathrm{Spec}(B)$ in the sense of [L1], and every $y_1,\dots ,y_n\in B$ such that $B$ is module-finite over the subalgebra $A[y_1,\dots ,y_n]$ of [L3], there exists $g$ in the relative integral closure $B^{\mathrm{\prime }}$ of the image of $A$ in $B$ with $g\mathrm{\notin }{\mathfrak{q}}_B$ and $B_g^{\mathrm{\prime }}\to B_g$ an isomorphism. The Theorem is the case $A=R$, $B=S$, ${\mathfrak{q}}_B=\mathfrak{q}$ and $y_i=x_i$, and it is exactly the assertion to be proved.

We begin with the case $n=0$ of $P(0)$: here $B$ is module-finite over $A[y_1,\dots ,y_0]=A$, so every element of $B$ is integral over $A$ by [L10]; hence $B\subseteq B^{\mathrm{\prime }}$, while $B^{\mathrm{\prime }}\subseteq B$ by [L2], so that $B^{\mathrm{\prime }}=B$. Take $g=1$: the prime ${\mathfrak{q}}_B$ is proper by [L16], so $1\mathrm{\notin }{\mathfrak{q}}_B$, and $B_1^{\mathrm{\prime }}=B^{\mathrm{\prime }}=B=B_1$ by [L15], an isomorphism of localisations. Thus $P(0)$ holds.

Now suppose $n=1$, write $y=y_1$ and put $T:=B^{\mathrm{\prime }}$, the integral closure of the image of $A$ in $B$, which is an intermediate $A$-subalgebra by [L2]; also $B$ is module-finite over $T[y]$ because $A[y]\subseteq T[y]$. By [L4] the map $T\to B$ is of finite type and quasi-finite at ${\mathfrak{q}}_B$, and $T$ is integrally closed in $B$: if $b\in B$ is integral over $T$, then $b$ is integral over $A$ by [L9], since every element of $T$ is integral over $A$ by [L2], so $b\in B^{\mathrm{\prime }}=T$. Consequently, for $P(1)$ it suffices to prove the following special case. $(†)$ If $R$ is a ring integrally closed in a ring $S$, if $R\to S$ is of finite type and quasi-finite at $\mathfrak{q}\in \mathrm{Spec}(S)$, and if $S$ is module-finite over $R[x]$ for one element $x\in S$, then there is $g\in R\setminus \mathfrak{q}$ with the canonical map $R_g\to S_g$ an isomorphism. Indeed, applying $(†)$ to $T\to B$ with the element $y$ and the prime ${\mathfrak{q}}_B$ produces $g\in T\setminus {\mathfrak{q}}_B=B^{\mathrm{\prime }}\setminus {\mathfrak{q}}_B$ with $T_g\cong B_g$, which is the conclusion of $P(1)$ because $T=B^{\mathrm{\prime }}$.

Now let $n\ge 2$ and assume $P(m)$ for all $m<n$; suppose $S$ is module-finite over $R[x_1,\dots ,x_n]$ and $R\to S$ is of finite type and quasi-finite at $\mathfrak{q}$. Put $R^{\mathrm{\prime }}:={\mathrm{Int}}_{R[x_1,\dots ,x_{n-1}]}(S)$, the integral closure of $R[x_1,\dots ,x_{n-1}]$ in $S$ by [L2], an intermediate $R$-subalgebra containing $x_1,\dots ,x_{n-1}$. By [L4] the map $R^{\mathrm{\prime }}\to S$ is of finite type and quasi-finite at $\mathfrak{q}$, and $S$ is module-finite over $R^{\mathrm{\prime }}[x_n]$ because $R[x_1,\dots ,x_n]\subseteq R^{\mathrm{\prime }}[x_n]$.

We now prove $(†)$, so we assume $R=S^{\mathrm{\prime }}={\mathrm{Int}}_R(S)$, that $R\to S$ is of finite type and quasi-finite at $\mathfrak{q}$, and that $S$ is module-finite over $R[x]$ for some $x\in S$. Let $\phi :R[x]\to S$ be the $R$-algebra homomorphism with $\phi (x)=x$, put $I=\ker \phi$ and $A=\mathrm{Im}(\phi )=R[x]\mathrm{/}I$, so that $R=\phi (R)\subseteq A\subseteq S$; the map $\phi$ is finite, that is, $S$ is a finitely generated $R[x]$-module. Let $J=\{g\in S:gS\subseteq A\}$ be the conductor of $A$ in $S$. Every element of $S$ integral over $R$ lies in $R\subseteq A$ because $R=S^{\mathrm{\prime }}$, so the hypothesis of [L5] is satisfied and $J$ is an ideal of $S$.

The ring $R^{\mathrm{\prime }}$ is integrally closed in $S$: if $s\in S$ is integral over $R^{\mathrm{\prime }}$, then $s$ is integral over $R[x_1,\dots ,x_{n-1}]$ by [L9], because every element of $R^{\mathrm{\prime }}$ is integral over $R[x_1,\dots ,x_{n-1}]$ by [L2]; hence $s\in R^{\mathrm{\prime }}$.

The quotient $\bar{S}:=S\mathrm{/}\sqrt{J}$ is reduced, and so is its subring $\bar{R}:=R\mathrm{/}(R\cap \sqrt{J})$: if $s\in S$ has nilpotent image in $\bar{S}$, say $s^m\in \sqrt{J}$ for some $m\ge 1$, then by [L17] there is $k\ge 1$ with $s^{mk}=(s^m{)}^k\in J$, so $s\in \sqrt{J}$ by [L17] again, that is, the image of $s$ in $\bar{S}$ is zero; thus $\bar{S}$ has no nonzero nilpotent and is reduced by [L19].

The image $\bar{x}$ of $x$ in $\bar{S}$ is strongly transcendental over $\bar{R}$ in the sense of [L8], and $\bar{S}$ is module-finite over $\bar{R}[\bar{x}]$. For the first assertion, let $\bar{u}\in \bar{S}$ and $\bar{P}={\sum }_i{\bar{a}}_i{z}^i\in \bar{R}[z]$ satisfy $\bar{u}\bar{P}(\bar{x})=0$, and lift $\bar{u}$, $\bar{P}$ to $u\in S$, $P={\sum }_i{a}_i{z}^i\in R[z]$; then $u\phi (P)\in \sqrt{J}$, so [L5] gives $u\phi (a_i)\in \sqrt{J}$ for every $i$, which says $\bar{u}{\bar{a}}_i=0$ in $\bar{S}$ for every $i$, exactly the condition of [L8]. Here $\bar{R}\to \bar{S}$ is injective, since an element of $R$ lies in $R\cap \sqrt{J}$ precisely when its image in $\bar{S}$ vanishes. For the second assertion, the image of $R[x]$ in $\bar{S}$ is $\bar{R}[\bar{x}]$ and the images of finitely many $R[x]$-module generators of $S$ generate $\bar{S}$ over that subring, so $\bar{S}$ is module-finite over $\bar{R}[\bar{x}]$ by [L3].

By [L6], whose hypothesis is exactly the combination of steps 4.1 and 5.1, the map $\bar{R}\to \bar{S}$ is of finite type and quasi-finite at no prime of $\bar{S}$. This is the only step of the proof that uses the Axiom of Choice, through the going down and lying over arguments inside [L6].

We deduce that $J\subseteq ̸\mathfrak{q}$. Suppose instead that $\sqrt{J}\subseteq \mathfrak{q}$; then $\bar{\mathfrak{q}}:=\mathfrak{q}\mathrm{/}\sqrt{J}$ is a prime of $\bar{S}$, and [L4] applied to the finite type map $R\to S$ and the ideal $\sqrt{J}\subseteq \mathfrak{q}$ shows that $R\to \bar{S}$ is of finite type and quasi-finite at $\bar{\mathfrak{q}}$. Applying [L4] once more with the intermediate $R$-subalgebra $\bar{R}=\mathrm{Im}(R\to \bar{S})\subseteq \bar{S}$ gives that $\bar{R}\to \bar{S}$ is quasi-finite at $\bar{\mathfrak{q}}$, contradicting step 6.1. Since the primes containing $J$ are precisely those containing $\sqrt{J}$, by [L18], the assumption $J\subseteq \mathfrak{q}$ is therefore impossible, and we may choose $s\in J\setminus \mathfrak{q}$.

Because $J=\{g\in S:gS\subseteq A\}\subseteq A$, the element $s$ lies in $A=\mathrm{Im}(\phi )$, so $s=\phi (f)$ for some $f\in R[x]$; and $f\mathrm{\notin }{\mathfrak{q}}_0:={\phi }^{-1}(\mathfrak{q})$ because $\phi (f)=s\mathrm{\notin }\mathfrak{q}$. Put ${\mathfrak{q}}^{\mathrm{\prime }}:={\mathfrak{q}}_0\mathrm{/}I$. By [L20] the ideal ${\mathfrak{q}}^{\mathrm{\prime }}$ is a prime of $A=R[x]\mathrm{/}I$ with ${\mathfrak{q}}^{\mathrm{\prime }}\cap R={\mathfrak{q}}_0\cap R=\mathfrak{p}$, and ${\mathfrak{q}}^{\mathrm{\prime }}=\mathfrak{q}\cap A$.

The canonical $R$-algebra homomorphism $A_f\to S_s$ is an isomorphism: it is injective as the localisation of the injective map $A\to S$ by [L13], and it is surjective because $sS\subseteq A$. Indeed, for $\sigma \in S$ the element $s\sigma$ lies in $A$, say $s\sigma =\phi (r)$ with $r\in R[x]$, so $\sigma \mathrm{/}1=(s\sigma )\mathrm{/}s=\phi (r)\mathrm{/}s$ is the image of $\phi (r)\mathrm{/}f$ (note $\phi (f)=s$).

The map $R\to A$ is quasi-finite at ${\mathfrak{q}}^{\mathrm{\prime }}$ by [L4]: the ring $A$ is an intermediate $R$-subalgebra of $S$ that is of finite type over $R$ as a quotient of $R[x]$ by [L3]; the element $f$ lies in $A\setminus {\mathfrak{q}}^{\mathrm{\prime }}$ because $\phi (f)=s\mathrm{\notin }\mathfrak{q}$; and step 9.1 says that $A_f=S_f$ as subrings of $S_f$, with ${\mathfrak{q}}^{\mathrm{\prime }}=\mathfrak{q}\cap A$ by step 8.1.

The integral closure of the image of $R$ in $A$ is $R$ itself: every element of $A$ that is integral over $R$ is an element of $S$ integral over $R$, hence lies in $S^{\mathrm{\prime }}=R$ because $R$ is integrally closed in $S$; and $R\subseteq A$. Since $A=R[x]\mathrm{/}I$ and $R\to A$ is quasi-finite at ${\mathfrak{q}}^{\mathrm{\prime }}$ by step 10.1, [L7] provides $h\in R\setminus {\mathfrak{q}}^{\mathrm{\prime }}$ such that the canonical map $R_h\to A_h$ is an isomorphism.

Inside $A_h=R_h$ the element $f\in A$ has the form $f=r\mathrm{/}{h}^k$ for some $r\in R$ and some $k\ge 0$. We claim $r\mathrm{\notin }\mathfrak{q}$. Otherwise $r\in \mathfrak{q}\cap A={\mathfrak{q}}^{\mathrm{\prime }}$; localising $A$ at the prime ${\mathfrak{q}}^{\mathrm{\prime }}$ by [L21], the equation $f{h}^k=r$ would exhibit $f{h}^k$ as an element of the maximal ideal ${\mathfrak{q}}^{\mathrm{\prime }}{A}_{{\mathfrak{q}}^{\mathrm{\prime }}}$, although $f\mathrm{\notin }{\mathfrak{q}}^{\mathrm{\prime }}$ and $h\mathrm{\notin }{\mathfrak{q}}^{\mathrm{\prime }}$ make $f{h}^k$ a unit of $A_{{\mathfrak{q}}^{\mathrm{\prime }}}$.

Set $g:=rh\in R$; then $g\mathrm{\notin }\mathfrak{q}$ because $r\mathrm{\notin }\mathfrak{q}$, $h\mathrm{\notin }\mathfrak{q}$ and $\mathfrak{q}$ is prime by [L16]. We show that the canonical map $R_g\to S_g$ is an isomorphism. First $A_{fh}\cong S_{sh}$: localising the isomorphism $A_f\cong S_s$ of step 9.1 at the common element $h$ gives $(A_f{)}_h\cong (S_s{)}_h$, and $(A_f{)}_h\cong A_{fh}$, $(S_s{)}_h\cong S_{sh}$ by [L14]. Second $A_{fh}\cong R_{rh}=R_g$: by step 11.1 we have $A_h\cong R_h$, so $A_{fh}\cong (R_h{)}_f$ by [L14], and inside $R_h$ the elements $f$ and $r$ differ by the unit $h^{-k}$, so $(R_h{)}_f\cong (R_h{)}_r\cong R_{hr}$ by [L14]. Third $S_{sh}\cong S_{rh}=S_g$: the relation $f{h}^k=r$ holds in $A_h$ and hence in $S_h$, and applying the $R$-algebra map $\phi$ gives $s{h}^k=\phi (f{h}^k)=\phi (r)=r$ there; so $s$ and $r$ also differ by the unit $h^{-k}$ in $S_h$, whence $(S_h{)}_s\cong (S_h{)}_r$, that is $S_{sh}\cong S_{rh}=S_g$ by [L14]. Composing these isomorphisms gives an isomorphism $R_g\to S_g$; every map in the chain is the canonical localisation of one of the inclusions $R\subseteq A\subseteq S$, so the composite is the map induced by the inclusion $R\subseteq S$. This proves $(†)$.

Steps 3.1 to 13.1 prove $(†)$, and step 2.1 reduces $P(1)$ to $(†)$. Hence $P(1)$ holds: for a finite type map $A\to B$ quasi-finite at ${\mathfrak{q}}_B$ with $B$ module-finite over $A[y_1]$, there is $g\in B^{\mathrm{\prime }}\setminus {\mathfrak{q}}_B$ with $B_g^{\mathrm{\prime }}\cong B_g$.

By $(†)$, that is by step 13.1 applied under the hypotheses verified in steps 2.2 and 3.2, there is $g^{\mathrm{\prime }}\in R^{\mathrm{\prime }}\setminus \mathfrak{q}$ with $(R^{\mathrm{\prime }}{)}_{g^{\mathrm{\prime }}}\cong S_{g^{\mathrm{\prime }}}$ via the inclusion.

The localisation $S_{g^{\mathrm{\prime }}}$ is a finitely generated $R$-algebra: $S$ is generated as an $R$-algebra by finitely many elements by [L3], and adjoining $1\mathrm{/}{g}^{\mathrm{\prime }}$ exhibits $S_{g^{\mathrm{\prime }}}$ as generated by those elements together with $1\mathrm{/}{g}^{\mathrm{\prime }}$ by [L3]. Choose $z_1,\dots ,z_M\in S_{g^{\mathrm{\prime }}}$ generating $S_{g^{\mathrm{\prime }}}$ over $R$; by step 14.2 each $z_j$ lies in $(R^{\mathrm{\prime }}{)}_{g^{\mathrm{\prime }}}$, so $z_j=y_j\mathrm{/}(g^{\mathrm{\prime }}{)}^{n_j}$ for some $y_j\in R^{\mathrm{\prime }}$ and $n_j\ge 0$. Put $R^{\mathrm{\prime }\mathrm{\prime }}:=R[x_1,\dots ,x_{n-1},y_1,\dots ,y_M,g^{\mathrm{\prime }}]\subseteq R^{\mathrm{\prime }}$, an $R$-subalgebra that is of finite type over $R$ by [L3]. Then $(R^{\mathrm{\prime }\mathrm{\prime }}{)}_{g^{\mathrm{\prime }}}=S_{g^{\mathrm{\prime }}}$: the inclusion $(R^{\mathrm{\prime }\mathrm{\prime }}{)}_{g^{\mathrm{\prime }}}\subseteq (R^{\mathrm{\prime }}{)}_{g^{\mathrm{\prime }}}=S_{g^{\mathrm{\prime }}}$ is clear, while each $z_j=y_j\mathrm{/}(g^{\mathrm{\prime }}{)}^{n_j}$ lies in $(R^{\mathrm{\prime }\mathrm{\prime }}{)}_{g^{\mathrm{\prime }}}$, so $S_{g^{\mathrm{\prime }}}=R[z_1,\dots ,z_M]\subseteq (R^{\mathrm{\prime }\mathrm{\prime }}{)}_{g^{\mathrm{\prime }}}$.

The algebra $R^{\mathrm{\prime }\mathrm{\prime }}$ is module-finite over $R[x_1,\dots ,x_{n-1}]$: each of $y_1,\dots ,y_M,g^{\mathrm{\prime }}$ lies in $R^{\mathrm{\prime }}$ and is therefore integral over $R[x_1,\dots ,x_{n-1}]$ by [L2], so [L11] applies to $R^{\mathrm{\prime }\mathrm{\prime }}=R[x_1,\dots ,x_{n-1}][y_1,\dots ,y_M,g^{\mathrm{\prime }}]$.

By [L4] applied to the intermediate subalgebra $R^{\mathrm{\prime }\mathrm{\prime }}$, the element $g^{\mathrm{\prime }}\in R^{\mathrm{\prime }\mathrm{\prime }}\setminus \mathfrak{q}$ and the equality $(R^{\mathrm{\prime }\mathrm{\prime }}{)}_{g^{\mathrm{\prime }}}=S_{g^{\mathrm{\prime }}}$ of step 15.1, the map $R\to R^{\mathrm{\prime }\mathrm{\prime }}$ is quasi-finite at ${\mathfrak{q}}^{\mathrm{\prime }\mathrm{\prime }}:=\mathfrak{q}\cap R^{\mathrm{\prime }\mathrm{\prime }}$; it is of finite type by step 15.1. Since $R^{\mathrm{\prime }\mathrm{\prime }}$ is module-finite over $R[x_1,\dots ,x_{n-1}]$ by step 16.1, the induction hypothesis $P(n-1)$ applies to the map $R\to R^{\mathrm{\prime }\mathrm{\prime }}$, the prime ${\mathfrak{q}}^{\mathrm{\prime }\mathrm{\prime }}$ and the elements $x_1,\dots ,x_{n-1}\in R^{\mathrm{\prime }\mathrm{\prime }}$: there is $g^{\mathrm{\prime }\mathrm{\prime }}\in R^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}\setminus {\mathfrak{q}}^{\mathrm{\prime }\mathrm{\prime }}$ with $(R^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}{)}_{g^{\mathrm{\prime }\mathrm{\prime }}}\cong (R^{\mathrm{\prime }\mathrm{\prime }}{)}_{g^{\mathrm{\prime }\mathrm{\prime }}}$ via the inclusion, where $R^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}={\mathrm{Int}}_R(R^{\mathrm{\prime }\mathrm{\prime }})\subseteq S^{\mathrm{\prime }}$ is the integral closure of the image of $R$ in $R^{\mathrm{\prime }\mathrm{\prime }}$.

The image of $g^{\mathrm{\prime }}$ in $(R^{\mathrm{\prime }\mathrm{\prime }}{)}_{g^{\mathrm{\prime }\mathrm{\prime }}}=(R^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}{)}_{g^{\mathrm{\prime }\mathrm{\prime }}}$ has the form $g^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}\mathrm{/}(g^{\mathrm{\prime }\mathrm{\prime }}{)}^m$ for some $g^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}\in R^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}$ and some $m\ge 0$, since the elements of that localisation are fractions with numerator in $R^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}$ by [L21]. If $g^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}\in \mathfrak{q}$, then $g^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}\mathrm{/}(g^{\mathrm{\prime }\mathrm{\prime }}{)}^m$ would lie in the prime $\mathfrak{q}(R^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}{)}_{g^{\mathrm{\prime }\mathrm{\prime }}}$ of $(R^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}{)}_{g^{\mathrm{\prime }\mathrm{\prime }}}$, because $g^{\mathrm{\prime }\mathrm{\prime }}\mathrm{\notin }\mathfrak{q}$; but this element is the image of $g^{\mathrm{\prime }}\in R^{\mathrm{\prime }\mathrm{\prime }}\setminus \mathfrak{q}$ under $R^{\mathrm{\prime }\mathrm{\prime }}\to (R^{\mathrm{\prime }\mathrm{\prime }}{)}_{g^{\mathrm{\prime }\mathrm{\prime }}}$, and $g^{\mathrm{\prime }\mathrm{\prime }}\mathrm{\notin }{\mathfrak{q}}^{\mathrm{\prime }\mathrm{\prime }}=\mathfrak{q}\cap R^{\mathrm{\prime }\mathrm{\prime }}$ forces that image to lie outside $\mathfrak{q}(R^{\mathrm{\prime }\mathrm{\prime }}{)}_{g^{\mathrm{\prime }\mathrm{\prime }}}$. Hence $g^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}\mathrm{\notin }\mathfrak{q}$, and since also $g^{\mathrm{\prime }\mathrm{\prime }}\mathrm{\notin }\mathfrak{q}$ and $\mathfrak{q}$ is prime by [L16], the product $g:=g^{\mathrm{\prime }\mathrm{\prime }}{g}^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}\in R^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}\subseteq S^{\mathrm{\prime }}$ satisfies $g\mathrm{\notin }\mathfrak{q}$.

Finally $(R^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}{)}_g\cong S_g$. By [L14], $(R^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}{)}_g=((R^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}{)}_{g^{\mathrm{\prime }\mathrm{\prime }}}{)}_{g^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}}\cong ((R^{\mathrm{\prime }\mathrm{\prime }}{)}_{g^{\mathrm{\prime }\mathrm{\prime }}}{)}_{g^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}}$, and inside $(R^{\mathrm{\prime }\mathrm{\prime }}{)}_{g^{\mathrm{\prime }\mathrm{\prime }}}$ the element $g^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}$ differs from $g^{\mathrm{\prime }}$ by the unit $(g^{\mathrm{\prime }\mathrm{\prime }}{)}^{-m}$: indeed step 18.1 says that the image of $g^{\mathrm{\prime }}$ equals $g^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}\mathrm{/}(g^{\mathrm{\prime }\mathrm{\prime }}{)}^m$, that is $g^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}=g^{\mathrm{\prime }}(g^{\mathrm{\prime }\mathrm{\prime }}{)}^m$. Hence inverting $g^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}$ is the same as inverting $g^{\mathrm{\prime }}$, and $((R^{\mathrm{\prime }\mathrm{\prime }}{)}_{g^{\mathrm{\prime }\mathrm{\prime }}}{)}_{g^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}}\cong ((R^{\mathrm{\prime }\mathrm{\prime }}{)}_{g^{\mathrm{\prime }\mathrm{\prime }}}{)}_{g^{\mathrm{\prime }}}=(R^{\mathrm{\prime }\mathrm{\prime }}{)}_{g^{\mathrm{\prime }\mathrm{\prime }}{g}^{\mathrm{\prime }}}$ by [L14]. By step 15.1, $(R^{\mathrm{\prime }\mathrm{\prime }}{)}_{g^{\mathrm{\prime }}}=S_{g^{\mathrm{\prime }}}$, so $(R^{\mathrm{\prime }\mathrm{\prime }}{)}_{g^{\mathrm{\prime }\mathrm{\prime }}{g}^{\mathrm{\prime }}}=(S_{g^{\mathrm{\prime }}}{)}_{g^{\mathrm{\prime }\mathrm{\prime }}}=S_{g^{\mathrm{\prime }\mathrm{\prime }}{g}^{\mathrm{\prime }}}$ by [L14]. The same relation $g^{\mathrm{\prime }}=g^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}\mathrm{/}(g^{\mathrm{\prime }\mathrm{\prime }}{)}^m$, read in $S_{g^{\mathrm{\prime }\mathrm{\prime }}}$ through the injective map $R^{\mathrm{\prime }\mathrm{\prime }}\to S$ and its localisation, shows that $g^{\mathrm{\prime }}$ and $g^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}$ differ by a unit of $S_{g^{\mathrm{\prime }\mathrm{\prime }}}$, so also $S_{g^{\mathrm{\prime }\mathrm{\prime }}{g}^{\mathrm{\prime }}}\cong S_{g^{\mathrm{\prime }\mathrm{\prime }}{g}^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}}=S_g$ by [L14]. Thus $(R^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}{)}_g\cong S_g$ via the canonical maps. As $R^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}\subseteq S^{\mathrm{\prime }}\subseteq S$, the inclusion $S^{\mathrm{\prime }}\to S$ becomes an isomorphism after inverting $g$, since $S_g^{\mathrm{\prime }}$ lies between the subrings $(R^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}{)}_g$ and $S_g$, which coincide. This is the conclusion of $P(n)$.

Steps 1.2, 14.1 and 15.1 to 19.1 establish $P(m)$ for every $m\ge 0$ by induction. Applying $P(n)$ to the originally given map $R\to S$, the prime $\mathfrak{q}$ and the elements $x_1,\dots ,x_n$ of step 1.1 produces $g\in S^{\mathrm{\prime }}\setminus \mathfrak{q}$ such that $S^{\mathrm{\prime }}\to S$ induces an isomorphism $S_g^{\mathrm{\prime }}\cong S_g$, which is the assertion of the Theorem. The Axiom of Choice was used only in step 6.1 through [L6]; every other step selected only finitely many elements (the generators $x_i$, the element $s$, the finite lists $y_j$ and $z_j$, the elements $r$ and $g$), so no use of the axiom is hidden elsewhere. ∎