==== TABLE OF CONTENTS ====
The-Bogomolov-Pop-reconstruction-theorem//curve-recovery.tex
The-Bogomolov-Pop-reconstruction-theorem//introduction.tex
The-Bogomolov-Pop-reconstruction-theorem//local-theory.tex
The-Bogomolov-Pop-reconstruction-theorem//paper.tex
The-Bogomolov-Pop-reconstruction-theorem//pencil-bound.tex
The-Bogomolov-Pop-reconstruction-theorem//preliminaries.tex
The-Bogomolov-Pop-reconstruction-theorem//reconstruction.tex
The-Bogomolov-Pop-reconstruction-theorem//residual-tests.tex
The-Bogomolov-Pop-reconstruction-theorem//references.bib
===========================
==== BEGIN The-Bogomolov-Pop-reconstruction-theorem//curve-recovery.tex ====
\section{Recovering the curve subfields}
\label{sec:curve-recovery}
A \emph{curve subfield} of $K/k$ will mean a subfield $E\subset K$
containing $k$, relatively algebraically closed in $K$, and of
transcendence degree one over $k$. Such a field is finite over $k(t)$
for each $t\in E\setminus k$. We identify its completed multiplicative
group $\widehat E$ with its image in $\widehat K$ by
Lemma~\ref{prelim:completion}.
\begin{theorem}\label{recovery:curves}
The map $\Theta$ gives a bijection between the completed curve
subfields of $K/k$ and those of $L/l$. More precisely, for every curve
subfield $E\subset K$ there is a unique curve subfield $P\subset L$
such that
\[
\Theta(\widehat E)=\widehat P.
\]
\end{theorem}
Theorem~\ref{pencil:uniform} bounds the degrees of the prime divisors
detected by a pencil. We first show that infinitely many occur. Their
incidence family gives containment in a completed curve field over a
finite extension of $L$. Intersections and norms then descend this
containment to $L$, and the inverse correspondence gives equality.
\subsection{The divisors detected by a pencil}
Fix $t\in K\setminus k$, let $E$ be the relative algebraic closure of
$k(t)$ in $K$, and put
\[
h_a=\Theta[t-a],\qquad
\mathcal S=\bigcup_{a\in k}\operatorname{Supp}_X(h_a).
\]
Here $X/l$ is the fixed normal projective model used in
Theorem~\ref{pencil:uniform}. That theorem bounds the degree of every
member of $\mathcal S$ by a constant depending on $t$ and $X$.
\begin{lemma}\label{recovery:residue-powers}
The set $\mathcal S$ is infinite. For every $z\in E^\times$, every
$j\geq1$, and every representative $g\in L^\times$ of
$\Theta[z]\bmod\ell^j$, all but finitely many $D\in\mathcal S$ satisfy
\begin{equation}\label{eq:recovery-residue-powers}
\operatorname{ord}_D(g)=0,
\qquad g|_D\in l(D)^{\times\ell^j}.
\end{equation}
\end{lemma}
\begin{proof}
The classes $[t-a]$ are independent in $k(t)^\times/\ell$, as evaluation
at their distinct zeros shows. The kernel of
$k(t)^\times/\ell\to K^\times/\ell$ is finite dimensional: adjoining
roots for any finite independent subset of that kernel gives an
elementary abelian $\ell$-extension contained in the finite extension
$E/k(t)$. Thus their images span an infinite-dimensional subspace of
$K^\times/\ell$, and the same is true of the $h_a$ in
$\widehat L/\ell=L^\times/\ell$.
The kernel of divisor evaluation
\[
L^\times/\ell\longrightarrow
\bigoplus_{D\in X^{(1)}}\mathbb F_\ell
\]
is finite dimensional. Indeed, take a smooth dense open $U\subset X$.
A function whose divisor on $U$ is divisible by $\ell$ defines a
$\mu_\ell$-torsor on $U$: its divisor divided by $\ell$ is Cartier.
The kernel therefore embeds in
$H^1_{\mathrm{\acute et}}(U,\mu_\ell)$, which is finite dimensional
by Deligne's finiteness theorem
\cite[Th\'eor\`emes de finitude, Theorem 1.1]{DeligneSGA4half}.
If $\mathcal S$ were finite, both the image and the kernel
of divisor evaluation on the pencil span would be finite dimensional,
a contradiction.
For $D\in\mathcal S$, let $w_D$ be its divisorial valuation and let $v$
be the quasi-prime divisor of $K/k$ supplied by
Proposition~\ref{local:transport}. Its inertia detects some $[t-a]$.
Consequently $vE/vk$ is nonzero inside the rank-one group $vK/vk$.
The valuation transcendence-degree inequality applied to $E/k$ gives
$\operatorname{trdeg}(Ev/kv)=0$, so $Ev=kv$ because $kv$ is
algebraically closed.
If $\operatorname{ord}_D(\Theta[z])=0$, inertia evaluation implies
$v(z)\in vk$. Multiplying $z$ first by a constant to give it value zero,
and then by a constant with the reciprocal residue, makes it a
principal unit. Thus $D_v$ kills $[z]$, and $D_{w_D}$ kills $\Theta[z]$.
By Corollary~\ref{pencil:finite}, the excluded $D$ form a finite set.
Discard also the finitely many prime divisors in the ordinary divisor
of $g$. For every remaining $D$, $g$ is a unit at $D$, and its residue
has zero evaluation modulo $\ell^j$ against
$D_{w_D}/I_{w_D}=W_{l(D)}$. Finite-level Kummer duality gives
\eqref{eq:recovery-residue-powers}.
\end{proof}
\subsection{One incidence curve for every Kummer test}
The following geometric lemma permits an arbitrary collection of tests.
In particular, neither the constants nor the collection must be
countable or uncountable.
\begin{lemma}\label{recovery:incidence}
Let $l$ be algebraically closed, let $X/l$ be a normal integral projective
variety of dimension $d\geq2$, and let $\mathcal S$ be an infinite set
of prime divisors of bounded degree in a fixed projective embedding.
Let $\mathcal A$ be any collection of pairs $(g,q)$, where
$g\in l(X)^\times$ and $q\geq1$ is prime to $\operatorname{char}l$.
Suppose that, for each pair separately, all but finitely many
$D\in\mathcal S$ satisfy
\[
\operatorname{ord}_D(g)=0,
\qquad g|_D\in l(D)^{\times q}.
\]
There exist a finite field extension $M/l(X)$ and a curve field
$P_0/l$ contained in $M$ such that $M/P_0$ is regular and
\[
g\in(M\overline{P_0})^{\times q}
\quad\text{for every }(g,q)\in\mathcal A.
\]
The same fields $M$ and $P_0$ work for the entire collection.
\end{lemma}
\begin{proof}
\emph{Place the divisors in a bounded family.}
Integral subvarieties of fixed dimension and bounded degree in
$\mathbb P^N_l$ have only finitely many Hilbert polynomials
\cite[Section~2 and Lemma~2.4]{GrothHilbert1961}; see also
\cite[Proposition~5.3]{Tanaka2023}.
The Hilbert points of $\mathcal S$ therefore lie in a finite union of
projective Hilbert schemes of $X$, with their universal flat families
\cite[Theorem~3.2 and the Hilbert specialization after Proposition~3.8]{GrothHilbert1961}.
Their closure has a positive-dimensional
irreducible component on which they are dense. Give it its reduced
structure and shrink to an integral variety $T$ such that the universal
family
\[
Z\subset T\times X\longrightarrow T
\]
has geometrically integral divisor fibers. This is possible by
constructibility of geometric integrality
\cite[Tags 0579, 055B]{Stacks}; the family is flat by construction.
\emph{Fix one incidence curve.}
Choose an integral locally closed curve $T'\subset T$ now, before
making any of the tests in $\mathcal A$, and put $Z'=Z\times_T T'$.
Both $Z$ and $Z'$ are integral. Indeed, flatness over an integral base
excludes vertical components and embeds affine coordinate rings in
their generic-fiber localizations, while the generic fibers are
geometrically integral. Both incidence maps to $X$ are dominant.
Otherwise a proper closed subset of $X$ would contain infinitely many
distinct prime-divisor fibers; each would have to be one of its
finitely many components of dimension $d-1$. Distinct parameter points
give distinct divisors because they are distinct Hilbert points.
Set
\[
M=l(Z'),\qquad P_0=l(T').
\]
Dominance embeds $l(X)$ into $M$. Since $\dim Z'=d$, this is a finite
extension. Geometric integrality of the generic fiber of $Z'\to T'$
means that $M/P_0$ is regular.
Figure~\ref{fig:incidence} records the two projections that determine
these fields.
\begin{figure}[htbp]
\centering
\begin{tikzcd}[column sep=large,row sep=large]
Z' \arrow[r] \arrow[d] & Z \arrow[r,"\operatorname{pr}_X"] \arrow[d]
& X \\
T' \arrow[r,hook] & T &
\end{tikzcd}
\caption{The left square is Cartesian. The composite $Z'\to X$ is
dominant and generically finite, giving $l(X)\subset M=l(Z')$;
the vertical map gives $P_0=l(T')\subset M$. The curve $T'$ is fixed
before the tests $(g,q)$. Auxiliary normal covers used to split a test
may depend on that test, but $M$ and $P_0$ do not.}
\label{fig:incidence}
\end{figure}
\emph{Extend each splitting to the fixed curve.}
It remains to prove all the power assertions without changing $T'$.
Fix one pair $(g,q)$. Let $U\subset X$ be the unit locus of $g$, and let
\[
Y=Z^{\mathrm{sm}/T}\cap\operatorname{pr}_X^{-1}(U).
\]
Then $Y\to T$ is smooth. Its fibers over the generic points of $T$
and of $T'$ are nonempty and geometrically integral: geometric
integrality supplies a dense smooth locus, and dominance of the two
incidence maps supplies the unit condition. Over $Y$ consider the
degree-$q$ finite \'etale cover
\[
C_g=\operatorname{Spec}_Y\mathcal O_Y[u]/(u^q-g).
\]
On every tested fiber satisfying the hypothesis and meeting $Y$, a
rational $q$-th root extends to a regular unit by normality of the
smooth fiber. Thus the cover splits into $q$ copies on those fibers.
The number of geometric irreducible components is constant on a dense
open of $T$ \cite[Tag 055A]{Stacks}. That number must be $q$, by the
dense set of tested fibers. A degree-$q$ finite \'etale cover of a
normal integral scheme with $q$ components has degree one on each component.
It follows that the cover is split on the geometric generic fiber
over $T$.
This last dense open need not meet $T'$. To pass to the fixed curve,
descend the generic splitting to a finite extension of $l(T)$ and
normalize $T$ in it, obtaining a finite surjective morphism
$\widetilde T\to T$. The scheme
$\widetilde Y=Y\times_T\widetilde T$ is normal because it is smooth
over a normal base \cite[Tag 034F]{Stacks}. It is integral because its
generic fiber is geometrically integral and it is flat over the
integral base. The pulled-back cover is generically split and finite
\'etale over this normal integral scheme, hence split everywhere:
each component is finite birational over the normal target and is an
isomorphism \cite[Lemma 58.11.2, Tag 0BQL]{Stacks}.
Surjectivity of $\widetilde T\to T$ provides a lift of a geometric
point above the generic point of $T'$. Restricting the split cover to
that point proves that $g$ has a $q$-th root in
$M\overline{P_0}$. The auxiliary finite extension may depend on
$(g,q)$; the already chosen fields $M$ and $P_0$ do not.
\end{proof}
\subsection{Kummer descent and intersections}
We next turn geometric splitting into membership in a completed curve
field. Two elementary facts keep track of the finite extension
introduced by the incidence construction.
\begin{lemma}\label{recovery:kummer-descent}
Let $M/P$ be a regular field extension, let $q$ be prime to the
characteristic, and suppose $\mu_q\subset P$. If
$g\in M^\times$ becomes a $q$-th power in $M\overline P$, then its
class in $M^\times/M^{\times q}$ comes from a unique class in
$P^\times/P^{\times q}$.
\end{lemma}
\begin{proof}
A $q$-th root $r$ of $g$ is separable over $M$, so it already belongs to
$MP^{\mathrm{sep}}$: the remaining constant extension is purely
inseparable. Choose a finite Galois extension $P'/P$ with $r\in MP'$.
Regularity identifies
$\operatorname{Gal}(MP'/M)$ with $\operatorname{Gal}(P'/P)$.
The ratios $\sigma(r)/r\in\mu_q$ form a character of this group.
By Hilbert's Theorem~90, there is $b\in P'^\times$ with these same
ratios. Then $r/b\in M$ and $b^q\in P$, giving
$g=b^q(r/b)^q$. Finally, if an element of $P$ has a $q$-th root in
$M$, that root is algebraic over $P$ and therefore belongs to $P$.
This proves uniqueness.
\end{proof}
\begin{lemma}\label{recovery:intersection}
Let $N/l$ be a function field over algebraically closed constants of
characteristic different from $\ell$. If $P$ and $Q$ are distinct
curve subfields of $N/l$, then
$\widehat P\cap\widehat Q\subset\widehat N$ has finite
$\Zell$-rank.
\end{lemma}
\begin{proof}
The compositum $PQ$ has transcendence degree two over $l$; otherwise
relative algebraic closedness in $N$ would force $P=Q$. Let $C_P$ and
$C_Q$ be the smooth projective curves of these fields. The product is
integral, and its function field identifies with $PQ$: the rational
map defined by the two subfields has two-dimensional, hence dense,
image in $C_P\times C_Q$.
For a point $x\in C_P(l)$, take the valuation of the divisor
$\{x\}\times C_Q$. Extend it by a Gauss valuation over a transcendence
basis for $N/PQ$, then prolong it across the remaining finite
extension. Its normalized order $w$ on $N$ is zero on $Q^\times$ and
restricts to $e\operatorname{ord}_x$ on $P^\times$, for some integer
$e>0$. This construction allows inseparable extensions and all
ramification indices.
An element of $\widehat P\cap\widehat Q$ has $w$-order zero by its
$Q$-representation. Its $x$-order is therefore zero, since
multiplication by $e$ is injective on $\Zell$. This holds for every
$x$. By Lemma~\ref{prelim:curve-kernel}, the intersection embeds in
the finite-rank Tate module of $\operatorname{Pic}^0(C_P)$.
\end{proof}
\begin{proof}[Proof of Theorem~\ref{recovery:curves}]
Let $E\subset K$ be a curve subfield, and choose $t\in E\setminus k$.
Relative algebraic closedness makes $E$
the relative algebraic closure of $k(t)$ in $K$. Apply
Lemmas~\ref{recovery:residue-powers} and \ref{recovery:incidence} to
one representative of each $\Theta[z]\bmod\ell^j$, for
$z\in E^\times$ and $j\geq1$. We obtain fields $L\subset M$ and
$P_0\subset M$ as in the incidence lemma.
Put $H=\Theta(\widehat E)$. We claim that its image in $\widehat M$
lies in $\widehat P_0$. At level $\ell^j$, every element of
$\widehat E$ is represented by an actual $z\in E^\times$.
Lemma~\ref{recovery:kummer-descent} puts its image in
$P_0^\times/P_0^{\times\ell^j}$. These classes are unique because
$P_0$ is relatively algebraically closed in $M$, and their uniqueness
makes them compatible as $j$ varies. Taking inverse limits proves
the claim. Completion maps across $M/L$ are injective by
Lemma~\ref{prelim:completion}.
Choose a finite normal extension $N/L$ containing $M$, allowing an
inseparable part, and let $P_N$ be the relative algebraic closure of
$P_0$ in $N$. It is a curve subfield of $N/l$. The image of $H$ lies
in $\widehat P_N$. Every $\sigma\in\operatorname{Aut}_L(N)$ fixes
$H$, so $H$ also lies in $\widehat{\sigma(P_N)}$.
It has infinite rank: $\widehat E/\ell$ has infinite dimension, as
the pencil argument in Lemma~\ref{recovery:residue-powers} shows.
Indeed, lifts of any finite mod-$\ell$ independent set are
$\Zell$-independent: a nonzero relation, divided by the smallest
$\ell$-power in its coefficients, would reduce to a nonzero
mod-$\ell$ relation. Torsion freeness justifies this division.
Lemma~\ref{recovery:intersection} therefore forces
$\sigma(P_N)=P_N$ for every $\sigma$.
Set $P_L=P_N\cap L$. This field has transcendence degree one over
$l$. Indeed, for $G=\operatorname{Aut}_L(N)$, the field $P_N^G$ has
transcendence degree one and lies in $N^G$. In characteristic zero
$N^G=L$. In characteristic $p>0$, $N^G/L$ is finite purely inseparable,
so a common $p^a$-th power sends $P_N^G$ into $P_L$. Moreover $P_L$
is relatively algebraically closed in $L$: an element of $L$
algebraic over $P_L$ is algebraic over $P_N$, hence lies in $P_N$.
Taking the relative algebraic closure of any nonconstant rational
subfield shows that $P_L/l$ is finitely generated.
The norm $N_{N/L}$ sends $P_N^\times$ into $P_L^\times$. In fact,
if $s$ is the inseparable degree of $N/L$, then
\[
N_{N/L}(x)=\left(\prod_{\sigma\in G}\sigma(x)\right)^s;
\]
the right side lies in $P_N$ because $P_N$ is $G$-stable, and the norm lies in
$L$. On completions, norm composed with inclusion is multiplication
by $[N:L]$. We conclude that
$[N:L]H\subset\widehat P_L$. The latter is saturated in
$\widehat L$ by Lemma~\ref{prelim:completion}, so
$H\subset\widehat P_L$.
Here the prime-to-$\ell$ factor of $[N:L]$ is a unit of $\Zell$,
and saturation removes its remaining $\ell$-power factor.
Apply the same argument to $\Theta^{-1}$. It gives a curve subfield
$E'\subset K$ with
\[
\widehat E\subset\Theta^{-1}(\widehat P_L)
\subset\widehat E'.
\]
The infinite-rank intersection forces $E'=E$ by
Lemma~\ref{recovery:intersection}. Both inclusions are therefore
equalities. That lemma also proves uniqueness. Reversing $K$ and $L$
proves the asserted bijection.
\end{proof}
\subsection{Divisorial valuations and rational quotients}
We can now distinguish valuations trivial on the constants. Throughout
this subsection, a \emph{divisorial valuation} means such a
quasi-prime divisor, with its discrete value group normalized to
$\mathbb Z$.
The following criterion adapts Topaz's Lemma~A.7
\cite[Lemma~A.7]{TopazTorelli2025} to completed multiplicative groups.
The proof below verifies the needed hypotheses in this setting.
\begin{proposition}\label{recovery:divisorial}
A quasi-prime divisor $v$ of $K/k$ is divisorial if and only if
\begin{equation}\label{eq:recovery-divisorial}
D_v^\perp\cap\widehat E=0
\end{equation}
for some curve subfield $E$ of $K/k$. Consequently $\varphi$
preserves divisorial inertia and decomposition groups in both
directions.
\end{proposition}
\begin{proof}
Suppose $v$ is trivial on $k$. Its residue field has transcendence
degree $d-1\geq1$. Choose $t$ with transcendental residue, and let
$E$ be the relative algebraic closure of $k(t)$ in $K$. The valuation
is trivial on $k(t)$ and hence on its algebraic extension $E$.
Thus $E$ embeds into $Kv$. Its relative algebraic closure in $Kv$
is finite over $E$, so the induced completion map is injective:
factor it through that finite extension and then a relatively
algebraically closed inclusion. Residual evaluation by
$D_v/I_v=W_{Kv}$ proves \eqref{eq:recovery-divisorial}.
Conversely, suppose $v|k$ is nontrivial, and let $E$ be any curve
subfield. Choose $x\in E\setminus k$, replacing it by $x^{-1}$ if
necessary so that $v(x)\geq0$, and choose $c\in k$ with $v(c)>0$.
Then $1+cx$ is a nonconstant principal unit. Its class in
$\widehat E$ is nonzero and is killed by $D_v$, so
\eqref{eq:recovery-divisorial} fails. The criterion is invariant
under $\Theta$ by Theorem~\ref{recovery:curves} and the local
correspondence.
\end{proof}
For a curve subfield $E\subset K$, restriction gives a surjection
\[
\rho_E:W_K\longrightarrow W_E.
\]
This is the continuous surjection of Lemma~\ref{prelim:completion}.
\begin{proposition}\label{recovery:rational}
The point inertia lines of every curve subfield $E$ are precisely
the saturations of the nonzero images $\rho_E(I_v)$ for divisorial
valuations $v$ of $K/k$. The correspondence of
Theorem~\ref{recovery:curves} preserves these lines and the genus.
In particular, it matches exactly the relatively algebraically closed
rational subfields.
If $\Theta(\widehat E)=\widehat P$, the induced continuous
isomorphism $\varphi_E:W_P\to W_E$ makes the diagram
\begin{equation}\label{eq:recovery-quotient-diagram}
\begin{tikzcd}[column sep=large]
W_L \arrow[r,"\varphi"] \arrow[d,two heads,"\rho_P"']
& W_K \arrow[d,two heads,"\rho_E"] \\
W_P \arrow[r,"\varphi_E"'] & W_E
\end{tikzcd}
\end{equation}
commute and matches the point inertia lines on its targets.
For rational $E$ and $P$, these are the geometric rational quotient
diagrams, with their point decomposition data.
\end{proposition}
\begin{proof}
A divisorial valuation restricted nontrivially to $E$ is an integer
multiple of a point valuation on its smooth projective curve: its
value group is a nonzero subgroup of $\mathbb Z$, and it is trivial
on $k$. Conversely, extend any point valuation of $E$ by a Gauss
valuation over a transcendence basis for $K/E$, then prolong it over
the remaining finite extension. The resulting valuation is
divisorial on $K/k$. Its restriction is $e\operatorname{ord}_x$ for
some $e>0$. Saturation of the nonzero image recovers
$\Zell\operatorname{ord}_x$, including when $\ell$ divides $e$.
The dual of $\Theta|_{\widehat E}$ defines $\varphi_E$.
Evaluation against elements of $\widehat E$ proves the commutativity
of \eqref{eq:recovery-quotient-diagram}. Proposition~\ref{recovery:divisorial}
and the preceding description of point lines show that $\varphi_E$
matches them. Their generators may differ by units, which do not
change the common kernel of point-order evaluations modulo $\ell$.
Its dimension is twice the genus by Lemma~\ref{prelim:curve-kernel}.
Hence genus is preserved. A smooth projective curve over an
algebraically closed field has genus zero exactly when its function
field is rational.
If $k(t)$ is relatively algebraically closed in $K$, the extension
$K/k(t)$ is regular. Only separability needs explanation. In
characteristic $p>0$, relative algebraic closedness implies
$t\notin K^p$. Since $k$ is perfect, $t$ can be included in a
separating transcendence basis for $K/k$, proving that $K/k(t)$ is
separably generated. Characteristic zero is immediate. Thus these
are exactly the rational subfields defined by general elements.
Finally, the residue field at a point of a rational curve is the
algebraically closed constant field, whose character group is zero.
Point decomposition therefore equals point inertia, and the diagram
also carries the required point decomposition data.
\end{proof}
The curve correspondence has consequently recovered both the true
divisorial valuations and all geometric rational quotients from the
original datum. The final reconstruction theorem can now be applied
with these intrinsically determined objects.
==== END The-Bogomolov-Pop-reconstruction-theorem//curve-recovery.tex ====
==== BEGIN The-Bogomolov-Pop-reconstruction-theorem//introduction.tex ====
\section{The reconstruction problem}
\label{sec:introduction}
Birational anabelian geometry asks how much of a field can be read from its
Galois groups. Over algebraically closed constants, the arithmetic Galois
action disappears. Bogomolov's program proposes that a particularly small
quotient still remembers every function field of dimension at least two:
the quotient in which commutators are central. We prove the Isom form of
this assertion, including surfaces and the prime $2$.
\subsection{The exact statement}
Fix a prime $\ell$. A function field $F/\kappa$ in this paper is a finitely
generated extension of an algebraically closed field $\kappa$.
For $\operatorname{char}F\ne\ell$, let $G_F^{(\ell)}$ be the maximal
pro-$\ell$ quotient of the absolute Galois group, and set
\[
\Pi_F^a=G_F^{(\ell)}/[G_F^{(\ell)},G_F^{(\ell)}],\qquad
\Pi_F^c=G_F^{(\ell)}/[[G_F^{(\ell)},G_F^{(\ell)}],G_F^{(\ell)}].
\]
All commutator subgroups are closed. The kernel
$\Delta_F=\ker(\Pi_F^c\to\Pi_F^a)$ is central. Our commutator convention is
$[x,y]=x^{-1}y^{-1}xy$. Lifting elements of $\Pi_F^a$ to $\Pi_F^c$ therefore
defines a continuous alternating $\Zell$-bilinear map
\[
[\ ,\ ]_F:\Pi_F^a\times\Pi_F^a\longrightarrow\Delta_F.
\]
An isomorphism $\varphi:\Pi_L^a\to\Pi_K^a$ is
\emph{bracket-compatible} if it is a continuous $\Zell$-module isomorphism
and there is a continuous $\Zell$-module isomorphism
$\psi:\Delta_L\to\Delta_K$ such that
\[
\psi([\sigma,\tau]_L)=[\varphi\sigma,\varphi\tau]_K
\quad(\sigma,\tau\in\Pi_L^a).
\]
Write $\Isom^c(\Pi_L^a,\Pi_K^a)$ for this set. The units $\Zell^\times$
act by multiplication on $\varphi$; the corresponding map on the central
kernel can be multiplied by the square of the same unit.
Write $F^i$ for the perfect closure of $F$; in characteristic zero this is
$F$ itself. Let $\Isom^i(K,L)$ consist of the field isomorphisms
$\alpha:K^i\to L^i$ satisfying $\alpha(k)=l$.
If the common characteristic is $p>0$, put
\[
\Isom_F^i(K,L)=\Isom^i(K,L)/
\bigl(\alpha\sim\Frob_L^n\circ\alpha,\ n\in\mathbb Z\bigr).
\]
Frobenius is an automorphism of a perfect field, so negative integers are
allowed. In characteristic zero take no quotient. If the
characteristics differ, set $\Isom_F^i(K,L)=\varnothing$.
Field isomorphisms induce contravariant Galois isomorphisms. Passing to the
above quotients gives the canonical map in the following theorem.
\begin{theorem}[Bogomolov--Pop reconstruction]\label{thm:main}
Let $k$ and $l$ be arbitrary algebraically closed fields of characteristics
different from $\ell$, and let $K/k$ and $L/l$ be function fields with
$\trdeg(K/k),\trdeg(L/l)\ge2$. Then
\[
\Phi_{K,L}:\Isom_F^i(K,L)\longrightarrow
\Isom^c(\Pi_L^a,\Pi_K^a)/\Zell^\times,
\qquad [\alpha]\longmapsto[\alpha^*],
\]
is a bijection.
\end{theorem}
Thus the theorem resolves the pro-$\ell$ abelian-by-central
Bogomolov--Pop reconstruction conjecture positively. Neither equality of
characteristics nor equality of relative transcendence degrees is assumed.
Both are forced by a bracket-compatible isomorphism. The input contains
no distinguished valuations, inertia groups, or curve quotients.
\subsection{Previous work and the proof}
Bogomolov introduced the abelian-by-central approach to birational
reconstruction \cite{Bogomolov1991}. The local approach detects
valuations through commuting elements and their associated multiplicative
characters \cite{BT2002,TopazCL2017}.
Bogomolov and Tschinkel proved the surface case over algebraic closures
of finite fields \cite{BT2008}, followed by higher-dimensional
reconstruction \cite{BT2011}; Pop established a corresponding
birational reconstruction theorem \cite{PopBI2012}.
The precise Isom formulation, including its scalar and Frobenius
ambiguities, is recorded in
\cite[Introduction, Conjecture~1]{TopazRat2016}.
Pop's global reconstruction
theorem recovers a function field from its total decomposition graph
together with suitable rational quotients
\cite{PopGraphs2012,PopBI2012}. His later reconstruction theorem with
divisorial inertia applies in relative transcendence degree greater than
two \cite[Theorem~1.1]{Pop2018}; its additional inertia datum must itself
be recovered to solve the conjecture from the group alone.
The local theory of commuting pairs supplies quasi-divisorial valuations,
which can be nontrivial on the constants
\cite{Pop2018,Topaz2019}. Pop's minimized decomposition theory also
recognizes certain residual curves over algebraic closures of finite
fields, with Topaz's appendix treating the required residue
characteristic $\ell$ case \cite{PopMin2015}.
Related reconstruction results use additional structure.
Silberstein studies the full absolute Galois group under a
prime-field-closure hypothesis \cite{Silberstein2013}.
Topaz's integral cohomological reconstruction uses a specified
arithmetic Galois action in transcendence degree at least three
\cite[Theorems~A and~B]{Topaz2019}; his Torelli-type theorem uses
mixed Hodge structures and cup products over constants embedded in
$\mathbb C$ \cite{TopazTorelli2025}.
These results concern different input data. Here only the original
pro-$\ell$ abelian-by-central datum is supplied, and neither the
constant field nor geometric inertia is distinguished in advance.
Our task is to recover the missing geometric information uniformly over
the constants. Write $\widehat F=\varprojlim_e F^\times/F^{\times\ell^e}$
for the completed multiplicative group and $[f]$ for the class of a
nonzero function. Choose Kummer identifications, and dualize a
bracket-compatible isomorphism to an isomorphism
$\Theta:\widehat K\to\widehat L$ of completed multiplicative groups.
The central step proves that, for fixed $t\in K\setminus k$, the entire
pencil
\[
h_a=\Theta[t-a],\qquad a\in k,
\]
has uniformly bounded divisor-support degree on a fixed projective model
of $L$. This is stronger than finiteness of each individual support.
Bogomolov--Tschinkel already used bounded-genus curves and Hurwitz
estimates to prove finiteness of $\ell$-adic supports
\cite[Section~6, proof of Proposition~6.1]{BT2011}.
Here the bound is uniform over a whole pencil and is obtained by finite
tests over arbitrary constants. A finite list of Kummer classes
is specialized and restricted to a smooth residual curve of bounded
genus, preserving all relations at the chosen finite level and the
relevant divisor supports. Five fixed members of the pencil then bound
the degree of a separable map to the projective line by Riemann--Hurwitz.
The bound also controls every additional member, at every sufficiently
high $\ell$-power level. The use of five anchors works for $\ell=2$.
The pencil detects infinitely many prime divisors, and the uniform
degree bound places them in a family of finite type. Their incidence
with the model produces a curve subfield
after a finite field extension. A normal-base-change argument allows us
to use one parameter curve for every Kummer class and every exponent;
no uncountability assumption enters. Norms and the finite-rank
intersection of distinct completed curve fields descend this subfield
to $L$. Applying the same argument to $\Theta^{-1}$ gives a bijection
of the completed relatively algebraically closed curve subfields.
These subfields distinguish true divisorial valuations from those that
also value the constants. They also recover point inertia and genus,
and hence all relatively algebraically closed rational subfields.
Their quotient diagrams supply exactly the input to Pop's global
reconstruction theorem. The new geometric steps are the uniform
pencil bound and the incidence argument that recovers curve subfields;
the remaining steps use the local valuation theory and Pop's
reconstruction theorem described above.
The proof follows this order. Section~\ref{sec:preliminaries} fixes the
Kummer topology and the relation between the two formulations of the
group input. Section~\ref{local:section} supplies the local correspondence.
Sections~\ref{sec:tests} and~\ref{sec:pencil} prove the residual test and
uniform support bound. Section~\ref{sec:curve-recovery} recovers curve
subfields and rational quotients. Section~\ref{sec:reconstruction}
checks compatibility on the decomposition graphs before applying Pop's
theorem and completing the Isom statement and its ambiguity checks.
==== END The-Bogomolov-Pop-reconstruction-theorem//introduction.tex ====
==== BEGIN The-Bogomolov-Pop-reconstruction-theorem//local-theory.tex ====
\section{Recovering the local data}
\label{local:section}
The commutator detects a relation between multiplicative characters.
That relation recovers quasi-prime divisors, including those that value
constants nontrivially. At this stage their residue fields may have
characteristic~$\ell$. We therefore work with character groups throughout
and use a Galois interpretation only at the top field.
For the commuting-pair approach to valuation detection and its
development via rigid elements, see
\cite{Bogomolov1991,BT2002,TopazCL2017}.
We use the character-space formulation below to include the required
residue fields.
\subsection{Commutators and alternating characters}
For any field $F$, put
\[
\mathcal G_F=W_F\otimes_{\Zell}\Qell.
\]
A pair $f,g\in\mathcal G_F$ is \emph{alternating} if
\begin{equation}\label{local:alternating}
f(x)g(1-x)=f(1-x)g(x)\qquad(x\in F\setminus\{0,1\}).
\end{equation}
A subspace is alternating if each pair of its elements is alternating.
Every character in $\mathcal G_F$ kills $-1$.
\begin{lemma}\label{local:commuting}
Let $F/\kappa$ be a function field over an algebraically closed field
of characteristic different from~$\ell$. Under Kummer duality $W_F=\Pi_F^a$, two elements of $W_F$
are alternating if and only if their commutator in $\Delta_F$ is zero.
This assertion holds also for $\ell=2$.
\end{lemma}
\begin{proof}
We give the finite-coefficient argument, keeping track of the
infinite rank of $W_F$. Write $G=G_F^{(\ell)}$, $A=G^{\mathrm{ab}}$
and $R=\mathbb Z/\ell^e$, with trivial action. Kummer duality makes
$A$ a product of copies of~$\Zell$.
First, a class in $H^2_{\mathrm{cont}}(A,R)$ is determined by its
commutator form. Indeed, a class with zero commutator defines
an abelian extension of $A$ by $R$. Such an extension splits
continuously: the Pontryagin dual of $A$ is a direct sum of copies of
$\Qell/\Zell$, hence is divisible and injective. Conversely every
continuous alternating bilinear form $A\times A\to R$ is the
commutator form of a cup-product class. To see this, continuity and
compactness give an open subgroup in the radical of the form, so it
depends on finitely many coordinates. In coordinate characters $x_i$
it is a sum of the forms
\[
x_i(a)x_j(b)-x_j(a)x_i(b),\qquad i<j,
\]
which come from the cocycles $x_i\cup x_j$.
Consequently cup products generate $H^2_{\mathrm{cont}}(A,R)$.
This proof uses no division by~$2$. In particular, even when $\ell=2$,
a diagonal cup has zero commutator and therefore vanishes. No extra
power or Bockstein class survives on the torsion-free group~$A$.
Next, inflation
\[
H^2_{\mathrm{cont}}(G,R)\longrightarrow H^2(G_F,R)
\]
is injective. Let $F(\ell)$ be the maximal Galois pro-$\ell$ extension
of $F$. An element of $F(\ell)^\times$ lies in a finite Galois
$\ell$-extension $F'/F$. Adjoin to $F'$ the $\ell^e$-th roots of all
its $F$-conjugates. Since the roots of unity lie in $F$, the resulting
field is again a finite Galois $\ell$-extension of $F$. Thus
$F(\ell)^\times$ is $\ell^e$-divisible. Kummer theory gives
$H^1(G_{F(\ell)},R)=0$, and the five-term sequence proves the claimed
injectivity.
Fix compatible identifications of the roots of unity with the
coefficient groups. The degree-two norm-residue theorem
\cite{MS1982} identifies the kernel of the tensor cup map
\[
(F^\times/F^{\times\ell^e})\otimes_R
(F^\times/F^{\times\ell^e})\longrightarrow H^2(G_F,R)
\]
with the subgroup generated by the Steinberg tensors
$[x]\otimes[1-x]$. Since the tensor cup map onto
$H^2_{\mathrm{cont}}(A,R)$ is surjective, the kernel of inflation from
$A$ to $G$ is generated by the Steinberg cups $[x]\cup[1-x]$.
Finally, put $N=[G,G]$. The five-term sequence for $N\subset G$
identifies this same kernel with
\[
\operatorname{Hom}_{\mathrm{cont}}(N/[N,G],R)
=\operatorname{Hom}_{\mathrm{cont}}(\Delta_F,R).
\]
The commutator form of the transgression of a character $\chi$ is
$\chi([\ ,\ ]_F)$, up to the uniform sign determined by the
transgression convention. Its value at $f,g$ therefore vanishes for
all~$\chi$ precisely when
\[
f(x)g(1-x)-f(1-x)g(x)=0\pmod{\ell^e}
\]
for every $x\ne0,1$. Continuous characters to finite cyclic
$\ell$-groups separate points of $\Delta_F$. Letting $e$ vary proves
the assertion.
\end{proof}
It follows that a bracket-compatible isomorphism preserves alternating
pairs on $W_F$, and hence on $\mathcal G_F$: multiply rationalized
characters by powers of~$\ell$ and use
\eqref{local:alternating}. The latter relation makes sense on residue
fields of every characteristic, without invoking their Galois groups.
\subsection{Character groups of valuations}
We regard equivalent valuations as the same valuation. Write $v\leq w$
if $v$ is a coarsening of $w$. For a valuation $v$ of $F$, let
$\mathcal O_v$, $\mathfrak m_v$, $Fv$ and $vF$ denote its valuation
ring, maximal ideal, residue field and value group. Set
\[
U_v=\mathcal O_v^\times,\qquad U_v^1=1+\mathfrak m_v,
\qquad I_v=U_v^\perp\subset D_v=(U_v^1)^\perp\subset W_F,
\]
and write $\mathcal I_v=I_v\otimes\Qell$ and
$\mathcal D_v=D_v\otimes\Qell$.
These are the \emph{minimized} inertia and decomposition groups.
When both $F$ and $Fv$ have characteristic different from~$\ell$,
they agree with the usual abelian inertia and decomposition groups.
Their definitions above impose no characteristic restriction.
A subset of $\mathcal G_F$ is \emph{valuative} if it is contained in
$\mathcal I_v$ for some valuation~$v$.
We use the following local character theorem
\cite[Fact~2.1, Lemma~2.5 and Corollary~2.7]{Topaz2019}:
\begin{enumerate}
\item Every valuative subset $\Sigma$ has a unique coarsest
valuation $v_\Sigma$ with $\Sigma\subset\mathcal I_{v_\Sigma}$.
It coarsens every valuation with this property.
\item In an alternating subspace $S$, the valuative elements form
a valuative subspace $\Sigma$ of codimension at most one, and
$S\subset\mathcal D_{v_\Sigma}$.
\item If $\Sigma$ is valuative and a character $f$ alternates with
every member of $\Sigma$, then $f\in\mathcal D_{v_\Sigma}$.
\end{enumerate}
These statements use $\operatorname{Hom}(F^\times,\Zell)\otimes\Qell$
and hold for arbitrary fields~$F$. The condition on the value at $-1$
in the local character theorem is automatic here. This is the form
needed below when $\operatorname{char}(Fv)=\ell$.
Let now $F/\kappa$ be a function field over an algebraically closed
field, of transcendence degree~$s$. A \emph{quasi-prime divisor} is a
valuation minimal among those satisfying
\[
vF/v\kappa\cong\mathbb Z,
\qquad \operatorname{trdeg}(Fv/\kappa v)=s-1.
\]
A quasi-prime $r$-divisor is a composition of $r$ successive
quasi-prime divisors of the successive residue function fields. It
satisfies
\begin{equation}\label{local:rank}
vF/v\kappa\cong\mathbb Z^r,
\qquad \operatorname{trdeg}(Fv/\kappa v)=s-r.
\end{equation}
The residue constants $\kappa v$ are algebraically closed.
We recall the finite-generation point implicit in
\eqref{local:rank}. The valuation transcendence-degree inequality is
\begin{equation}\label{local:abhyankar}
\operatorname{rrank}(vF/v\kappa)
+\operatorname{trdeg}(Fv/\kappa v)\leq s,
\end{equation}
where $\operatorname{rrank}$ is dimension after tensoring with
$\mathbb Q$. At equality, choose relative value-independent elements
and elements with algebraically independent residues. Together they
form a transcendence basis, and $F$ is finite over the resulting
rational field. The finite-extension valuation inequality then shows
that the relative value group is finitely generated and the residue
extension is finitely generated. Since $v\kappa$ is divisible,
$vF/v\kappa$ is torsion free, hence is a finitely generated free
abelian group. Applying this argument successively gives
\eqref{local:rank}.
\begin{lemma}\label{local:residue}
For a quasi-prime $r$-divisor $v$ of $F/\kappa$, restriction to units
induces a canonical topological isomorphism
\[
D_v/I_v\cong W_{Fv}.
\]
If $f,g\in\mathcal D_v$, they are alternating if and only if their
residual characters in $\mathcal G_{Fv}$ are alternating.
\end{lemma}
\begin{proof}
There is an exact sequence of abelian groups
\[
0\longrightarrow (Fv)^\times/(\kappa v)^\times
\longrightarrow F^\times/(\kappa^\times U_v^1)
\longrightarrow vF/v\kappa\longrightarrow0.
\]
Its final term is free, so applying $\operatorname{Hom}(-,\Zell)$
gives a surjection $D_v\to W_{Fv}$ with kernel $I_v$.
This map is continuous for pointwise convergence; compactness makes
the induced bijection on the quotient a topological isomorphism.
For the second assertion, use \eqref{local:alternating}. If $v(x)>0$,
then $1-x$ is a principal unit. If $v(x)<0$, the identity
$1-x=-x(1-x^{-1})$ makes the characters of $1-x$ equal those of~$x$.
If $v(x)=0$ and $v(1-x)>0$, then $x$ is a principal unit. In each
case the alternating determinant is zero. In the remaining case
$x$ and $1-x$ are units with residues different from $0,1$, and the
determinant is exactly the residual determinant. Conversely every
residual element different from $0,1$ has such a lift.
\end{proof}
In particular, a unit representing a class modulo $\ell^e$ evaluates
on $D_v/I_v$ exactly as its residue does on $W_{Fv}$. This observation
will allow finite Kummer tests to pass through a valuation.
\subsection{An intrinsic description of quasi-prime divisors}
The next argument follows the character-space proof of
\cite[Fact~3.2 and Theorem~3.3]{Topaz2019}. We include it because we
need it also in residue characteristic~$\ell$, whereas that section
of the cited paper is stated in a cohomological setting with
characteristic different from~$\ell$.
\begin{lemma}\label{local:dimension}
For $F/\kappa$ as above, in any characteristic, $s$ is the maximum
dimension of an alternating subspace of $\mathcal G_F$.
If $S$ has dimension $s$, its valuative subspace is
$\mathcal I_u$ for its associated valuation $u$, and $u$ attains
equality in \eqref{local:abhyankar}.
\end{lemma}
\begin{proof}
For every valuation $u$, characters kill the divisible constant group,
so
\[
\dim_{\Qell}\mathcal I_u
\leq\operatorname{rrank}(uF/u\kappa).
\]
Let $S$ be alternating and let $\Sigma$ be its valuative subspace,
with associated valuation~$u$. If $S=\Sigma$, the asserted dimension
bound follows from \eqref{local:abhyankar}.
Otherwise $\Sigma$ has codimension one in~$S$.
A character in $S\setminus\Sigma\subset\mathcal D_u$ has nonzero
residual character. Thus $Fu\ne\kappa u$, and
$\operatorname{trdeg}(Fu/\kappa u)\geq1$. Consequently
\[
\dim S\leq\operatorname{rrank}(uF/u\kappa)+1\leq s.
\]
This also bounds arbitrary alternating subspaces, by applying the
argument to their finite-dimensional subspaces.
There is a full discrete flag at a smooth closed point on a model
of $F/\kappa$. Its inertia space has dimension~$s$ and is
alternating by the valuation inequality for $x$ and $1-x$.
Thus the maximum equals~$s$.
If $\dim S=s$ and $S=\Sigma$, all inequalities
\[
s\leq\dim\mathcal I_u
\leq\operatorname{rrank}(uF/u\kappa)\leq s
\]
are equalities. If $S\ne\Sigma$, all inequalities
\[
s-1\leq\dim\mathcal I_u
\leq\operatorname{rrank}(uF/u\kappa)
\leq s-\operatorname{trdeg}(Fu/\kappa u)\leq s-1
\]
are equalities. In both cases $\Sigma=\mathcal I_u$ and $u$ has
no transcendence defect.
\end{proof}
\begin{lemma}\label{local:quasiprime}
Assume $s\geq2$. A line $H\subset\mathcal G_F$ is the inertia
space of a quasi-prime divisor if and only if there are two
$s$-dimensional alternating subspaces $S_1,S_2$ with
$S_1\cap S_2=H$. The valuation is unique. If $0\ne h\in H$, its
decomposition space is
\[
\mathcal D_v=\{f\in\mathcal G_F:(h,f)\text{ is alternating}\}.
\]
\end{lemma}
\begin{proof}
Suppose first that $v$ is quasi-prime. The residue function field
$Fv/\kappa v$ has two independent discrete valuations of full rank
$s-1$, trivial on the constants. For example, choose two independent
prime divisors on a rational subfield on a transcendence basis,
complete each to a flag, and prolong the flags to the finite
extension~$Fv$. Independent valuation approximation
\cite[Theorem~1.2]{ADF2020} makes their inertia spaces disjoint.
Composing with $v$ gives two alternating inertia spaces of dimension
$s$ whose intersection is $\mathcal I_v$.
Conversely, let $S_1,S_2$ be as stated. By
Lemma~\ref{local:dimension}, their valuative parts are
$\Sigma_i=\mathcal I_{u_i}$, where each $u_i$ has no transcendence
defect and $\dim\Sigma_i\geq s-1$.
We first show that $H$ is valuative. If, say, $u_1\leq u_2$, then
\[
\Sigma_1\subset\Sigma_2,
\qquad \Sigma_1\subset S_1\cap S_2=H.
\]
Since $\dim\Sigma_1\geq1$, this proves the claim.
If the valuations are incomparable, let $w$ be their finest common
coarsening. Applying independent valuation approximation on $Fw$
shows that the principal-unit groups of $u_1$ and $u_2$ generate
$U_w$. Indeed, approximate a unit $a$ by $b$ for the first residual
valuation and approximate $1$ by $b$ for the second. Then
$a=(a/b)b$ is a product of the two required principal units.
A character in $H$ annihilates both principal-unit groups,
and hence belongs to $\mathcal I_w$. Again $H$ is valuative.
In either case,
\[
H=\Sigma_1\cap\Sigma_2.
\]
Let $v=v_H$ be its coarsest associated valuation. It coarsens both
$u_i$, and
\[
H\subset\mathcal I_v
\subset\mathcal I_{u_1}\cap\mathcal I_{u_2}=H.
\]
A coarsening of a valuation without transcendence defect also has
no transcendence defect: relative rational ranks add under
composition, and the two valuation inequalities sum to the equality
for the finer valuation. Thus $v$ has no transcendence defect.
Its relative value group is finitely generated free, and its rank
is $\dim\mathcal I_v=1$. Hence $vF/v\kappa\cong\mathbb Z$.
A coarsening with the same relative rank would have the same inertia
space, contradicting the defining minimality of~$v_H$.
Therefore $v$ is quasi-prime, and the coarsest-valuation description
also proves uniqueness.
Every member of $\mathcal D_v$ alternates with $h$, by the calculation
in Lemma~\ref{local:residue}. For the converse apply the third part
of the local character theorem to the valuative line~$H$.
\end{proof}
The integral groups are recovered without an index ambiguity:
\[
I_v=\mathcal I_v\cap W_F,
\qquad D_v=\mathcal D_v\cap W_F.
\]
Indeed these are annihilators with values in the torsion-free group
$\Zell$. In particular they are saturated submodules.
\begin{proposition}\label{local:transport}
Let $K/k$ and $L/l$ be function fields over algebraically closed
fields of characteristic different from~$\ell$, each of
transcendence degree at least two. Let
$\varphi:W_L\to W_K$ be a continuous bracket-compatible
$\Zell$-module isomorphism. Then their transcendence degrees agree,
say they are~$d$. For every $1\leq r<d$, the map $\varphi$ carries
exactly the pairs $I_w\subset D_w$ of quasi-prime $r$-divisors of
$L/l$ onto the corresponding pairs of $K/k$.
For matched pairs it induces a continuous isomorphism
\[
D_w/I_w\cong W_{Lw}\ \longrightarrow\
D_v/I_v\cong W_{Kv}
\]
that preserves alternating pairs.
\end{proposition}
\begin{proof}
Lemma~\ref{local:commuting} makes the alternating relation intrinsic
to the bracket. Lemma~\ref{local:dimension} then recovers~$d$, and
Lemma~\ref{local:quasiprime} recovers the rationalized rank-one
pairs. Intersecting with the integral modules recovers the exact
pairs. Lemma~\ref{local:residue} supplies the residual character
spaces and their alternating relation. If $u$ is a valuation of $Fv$,
the inverse images of $I_u$ and $D_u$ under $D_v\to W_{Fv}$ are
$I_{u\circ v}$ and $D_{u\circ v}$, respectively, directly from the
unit and principal-unit definitions. Apply the same
characteristic-free recognition argument to each residue function
field. Continuing while its transcendence degree is at least two
recovers precisely the successive quasi-prime divisors through
rank~$d-1$. The construction applies to $\varphi^{-1}$ as well.
\end{proof}
\subsection{Recognizing the finite-constant residual curves}
The remaining local input recognizes certain residual curves and all
their point inertia groups. It does not yet distinguish true
divisors from quasi-prime divisors on the original function field.
We first describe the topological test. A compact abelian
pro-$\ell$ group $G$ with a family $(T_i)$ of procyclic subgroups is
\emph{complete-curve-like} if one can choose generators $\tau_i$ so
that $\tau_i$ tends to zero outside finite subsets, the resulting map
\[
\prod_i\Zell\longrightarrow G,
\qquad(a_i)\longmapsto\sum_i a_i\tau_i
\]
has kernel the diagonal copy of $\Zell$, and its cokernel is a
finitely generated $\Zell$-module. The diagonal kernel expresses
the single relation among the point inertias of a complete curve.
This property is invariant under continuous module isomorphisms.
For this criterion, let $F/\kappa$ be a function field over an
algebraically closed field of characteristic different from $\ell$;
its residue fields may still have characteristic $\ell$.
Form the closed subset
\[
\mathscr I_F=
\overline{\bigcup_{u\text{ quasi-prime}} I_u}\subset W_F.
\]
This is the closure of a \emph{union}, not the closed subgroup
generated by that union. For a quasi-prime $(s-1)$-divisor $v$, let
\begin{equation}\label{local:residual-set}
\mathscr R_v=
\operatorname{im}\bigl(\mathscr I_F\cap D_v\longrightarrow
D_v/I_v\bigr),
\end{equation}
and consider all maximal procyclic subgroups of $D_v/I_v$
contained in $\mathscr R_v$.
Pop's residual-curve criterion
\cite[Section~4, Theorem~4.2(i)(a)]{PopMin2015} applies when
\begin{enumerate}
\item $vF$ has no nonzero $\ell$-divisible convex subgroup;
\item there is a subfield $\kappa_1\subset F$ with
$\kappa_1v=\kappa v$ and
$\operatorname{trdeg}(F/\kappa_1)
=\operatorname{trdeg}(Fv/\kappa_1v)=1$.
\end{enumerate}
It states that the group with the family just described is
complete-curve-like exactly when $\kappa v$ is an algebraic closure
of a finite field. In that case the family is precisely all point
inertia subgroups of the smooth projective curve with function
field~$Fv$. The criterion allows $\operatorname{char}(Fv)=\ell$;
this is the purpose of the minimized groups and of Topaz's Appendix
to~\cite{PopMin2015}.
\begin{lemma}\label{local:criterion-hypotheses}
Every quasi-prime $(s-1)$-divisor $v$ of $F/\kappa$ satisfies the two
hypotheses of the residual-curve criterion.
\end{lemma}
\begin{proof}
For a quasi-prime divisor $u$, an $\ell$-divisible convex subgroup
$H\subset uF$ has zero image in $uF/u\kappa\cong\mathbb Z$.
Thus $H\subset u\kappa$. Coarsening by $H$ preserves the relative
rank one. Since $u$ attains equality in the valuation inequality,
its coarsening does also, so its residual transcendence degree is
still $s-1$. Minimality forces $H=0$.
Now consider a composite quasi-prime divisor. The value group of
its last residual step is a nonzero convex subgroup $\Gamma$ of
the full value group and has the rank-one property just proved.
If $H$ were a nonzero $\ell$-divisible convex subgroup of the full
group, convexity would make $H\cap\Gamma$ $\ell$-divisible: an
$\ell$-th part of a positive element remains between zero and that
element. Convex subgroups are comparable, so $H\cap\Gamma$ is
nonzero. This contradicts the rank-one case.
For the second hypothesis choose $x_1,\ldots,x_{s-1}$ whose values
are rationally independent modulo $v\kappa$, and put
$\kappa_1=\kappa(x_1,\ldots,x_{s-1})$.
Relative value independence implies algebraic independence. In a
polynomial in the $x_i$, distinct monomials have different values
modulo $v\kappa$, and there is a unique term of least value.
A quotient of polynomials with value zero consequently has residue
in $\kappa v$: equality of the two least values forces the same
monomial, leaving a quotient of constants. It follows that
$\kappa_1v=\kappa v$. The two asserted transcendence degrees are
then one by \eqref{local:rank}.
\end{proof}
\begin{proposition}\label{local:curves}
In the setting of Proposition~\ref{local:transport}, let
$w$ and $v$ be matched quasi-prime $(d-1)$-divisors.
Then $lw$ is algebraic over a finite field if and only if $kv$ is.
When this holds, the induced isomorphism $W_{Lw}\to W_{Kv}$ carries
the full family of point inertia subgroups of the respective smooth
projective curves onto each other. Corresponding normalized order
characters differ by units of~$\Zell$.
\end{proposition}
\begin{proof}
Proposition~\ref{local:transport} recovers every group used in
\eqref{local:residual-set}. A continuous isomorphism preserves the
closure of the union, its intersection with decomposition, its
quotient image and the maximal procyclic subgroups in that image.
Lemma~\ref{local:criterion-hypotheses} verifies the hypotheses of
Pop's criterion on both sides. The complete-curve-like property
therefore holds simultaneously, and the criterion identifies the
entire point families when it does. The resulting isomorphism
identifies each full procyclic subgroup with its counterpart, so it
sends a generator to a generator, differing only by a $\Zell$-unit.
\end{proof}
Only after separately excluding residue characteristic~$\ell$ will
we use these point lines to compare genera. For the present purpose,
Proposition~\ref{local:curves} provides exact point-order vanishing
at every finite level, with no assumption on the cardinality or
transcendence degree of the original constant fields.
==== END The-Bogomolov-Pop-reconstruction-theorem//local-theory.tex ====
==== BEGIN The-Bogomolov-Pop-reconstruction-theorem//paper.tex ====
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\DeclareMathOperator{\ord}{ord}
\DeclareMathOperator{\Pic}{Pic}
\DeclareMathOperator{\supp}{supp}
\DeclareMathOperator{\Spec}{Spec}
\DeclareMathOperator{\Frob}{Frob}
\title{The Bogomolov--Pop reconstruction theorem}
\author{OpenAI}
\date{September 23, 2026}
\begin{document}
\maketitle
\begin{abstract}
We prove the Bogomolov--Pop reconstruction conjecture for function fields of
transcendence degree at least two over arbitrary algebraically closed fields
of characteristic different from $\ell$.
The pro-$\ell$ abelian-by-central datum determines the perfect closure and its
constant field, with precisely the Frobenius and $\ell$-adic unit ambiguities.
\end{abstract}
\tableofcontents
\input{introduction}
\input{preliminaries}
\input{local-theory}
\input{residual-tests}
\input{pencil-bound}
\input{curve-recovery}
\input{reconstruction}
\bibliographystyle{alpha}
\bibliography{references}
\end{document}
==== END The-Bogomolov-Pop-reconstruction-theorem//paper.tex ====
==== BEGIN The-Bogomolov-Pop-reconstruction-theorem//pencil-bound.tex ====
\section{A uniform bound for pencils}\label{sec:pencil}
The preceding construction tests a finite collection of functions at a
finite Kummer level. We now combine such tests with five fixed members
of a pencil. Riemann--Hurwitz bounds the degree of every residual
pencil, which in turn bounds the support of the entire original pencil.
Bounded-genus curve sections and Riemann--Hurwitz already underlie the
finite-support argument of Bogomolov and Tschinkel
\cite[Section~6, proof of Proposition~6.1]{BT2011}.
Here the fixed anchors give one bound for every member of the pencil,
and the finite residual tests allow arbitrary algebraically closed constants.
Keep the model $X$ of the preceding section and the Kummer isomorphism
$\Theta:\widehat K\to\widehat L$.
For $h\in\widehat L$ and $e\ge1$, define
\[
S_e(h)=\{D\subset X\text{ prime divisor}:
\operatorname{ord}_D(h)\not\equiv0\pmod{\ell^e}\},
\qquad
\operatorname{Supp}_X(h)=\{D:\operatorname{ord}_D(h)\ne0\}.
\]
The order maps extend continuously to $\widehat L$.
Each $S_e(h)$ is finite, because $h$ modulo $\ell^e$ is represented by an
ordinary rational function. The exact support need not be finite a
priori. The sets $S_e(h)$ increase with $e$, and their union is
$\operatorname{Supp}_X(h)$.
\begin{theorem}[Uniform pencil support]\label{pencil:uniform}
For every $t\in K\setminus k$, there is a finite constant $B_t$ such that
\[
\sum_{D\in\operatorname{Supp}_X(\Theta[t-a])}\deg D\le B_t
\qquad\text{for every }a\in k.
\]
In particular, all these supports are finite, with a bound independent
of the member of the pencil.
\end{theorem}
\begin{proof}
Choose five distinct anchors $a_1,\ldots,a_5\in k$, including $0$.
In characteristic zero include $1$ and $\ell$ as well.
Put $h_a=\Theta[t-a]$ and
\[
N_i=\sum_{D\in S_1(h_{a_i})}\deg D\quad(1\le i\le5),
\qquad c_\ell=5(1-1/\ell)-2=3-5/\ell.
\]
The $N_i$ are finite before any exact-support assertion is known.
Moreover $c_\ell\ge1/2$, so the constant
\begin{equation}\label{pencil:constant}
B=\max\left\{1,\frac{2G_X-2+\sum_{i=1}^5N_i}{c_\ell}\right\}
\end{equation}
is finite and depends only on $X$, $\Theta$, $t$ and the fixed anchors.
We prove the theorem with $B_t=2B$.
Fix one additional $a\in k$, and list the distinct elements among the
five anchors and $a$ as $a_j$. Write $u_j=t-a_j$.
For all sufficiently large $e$, the classes $[u_j]$ modulo $q=\ell^e$
are nonzero and pairwise distinct and generate a noncyclic subgroup.
Indeed the functions $t-a_j$ are independent modulo constants: a
multiplicative relation would give a constant rational function of the
transcendental element $t$ and hence have all exponents zero.
Their images in the free abelian group $K^\times/k^\times$ are
therefore independent. Integral evaluations detect nonzero elements
and nonzero two-by-two determinants, which remain nonzero modulo a
sufficiently large power of $\ell$.
Equivalently, one can prolong two distinct point valuations of $k(t)$
to $K$ to obtain a diagonal evaluation matrix with nonzero integer
entries. The necessary lower bound on $e$ may depend on $a$.
Choose representatives $g_j\in L^\times$ of $h_{a_j}$ modulo $q$,
and apply Lemma~\ref{tests:residual}.
It gives $w$, a residual curve $C$, and the same noncyclic subgroup with
distinct nonzero generators in $(Lw)^\times/(Lw)^{\times q}$.
Evaluation on $D_w/I_w=W_{Lw}$ detects these classes.
By Proposition~\ref{local:transport}, there is a corresponding quasi-prime
$(d-1)$-divisor $v$ of $K/k$, and the evaluations of the $u_j$ on $D_v$
modulo $q$ are likewise nonzero, distinct and noncyclic.
\emph{Normalize the residual pencil.}
We first show that
\begin{equation}\label{pencil:no-relative-value}
v(k(t))=vk.
\end{equation}
Otherwise $v(k(t))/vk$ is a nonzero subgroup of the finitely generated
free group $vK/vk$. The valuation transcendence-degree inequality for
$k(t)/k$ makes it cyclic and gives $k(t)v=kv$.
Every function of value in $vk$ then becomes a principal unit after
multiplication by constants. Since $D_v$ kills constants and principal
units, evaluation on $D_v$ factors through this cyclic relative value
group. Its image modulo $q$ is cyclic, a contradiction.
All the $v(u_j)$ have a common value $\gamma\in vk$.
For if $v(u_j)<v(u_h)$, then
$u_j/(a_h-a_j)=1+u_h/(a_h-a_j)$ is a principal unit, forcing the
evaluation of $u_j$ to vanish.
Also $v(a_j-a_h)=\gamma$ whenever $j\ne h$:
a strictly larger value would make $u_j/u_h$ a principal unit and
force two evaluations to coincide.
Choose $b\in k^\times$ with $v(b)=\gamma$, and fix one listed anchor
$a_0$. Set
\[
\tau_0=\overline{(t-a_0)/b},\qquad
\beta_j=\overline{(a_j-a_0)/b}.
\]
These expressions are defined, the $\beta_j$ are distinct elements of
$kv$, and
\begin{equation}\label{pencil:residual-linear}
\overline{u_j/b}=\tau_0-\beta_j.
\end{equation}
The element $\tau_0$ is nonconstant: a constant residue would make
$(t-a_0)/b$ a constant times a principal unit and give zero evaluation.
The residue characteristic is different from $\ell$.
This is automatic if $k$ has positive characteristic.
In characteristic zero, the anchor differences $1-0$ and $\ell-0$
have the same value, so $v(\ell)=v(1)=0$.
Thus $\ell$ is nonzero in $kv$ in that case as well.
\emph{Compare the two residual curves.}
The constants $lw$ are algebraic over a finite field.
Proposition~\ref{local:curves} therefore gives the same assertion for $kv$
and identifies the full families of point inertia lines in
$W_{Lw}$ and $W_{Kv}$.
Let $C'/kv$ be the smooth projective curve with function field $Kv$.
By Lemma~\ref{prelim:curve-kernel}, the mod-$\ell$ kernels of the point
order maps have dimensions $2g(C)$ and $2g(C')$.
The residual isomorphism identifies these kernels, and hence
\begin{equation}\label{pencil:genus}
g(C')=g(C)\le G_X.
\end{equation}
Generators of matched point inertia lines differ by units of $\Zell$.
Consequently the number of nonzero point orders modulo $\ell$ or $q$
is preserved by the comparison.
Because constants have zero Kummer class, evaluation identifies the
classes of the functions in \eqref{pencil:residual-linear} with those
of $\overline g_j$ modulo $q$.
\emph{Use the five anchors to bound the degree.}
Write $p=\operatorname{char}(kv)>0$, using the finite-constant
recognition above.
Remove inseparability by writing $\tau_0=\tau^{p^r}$ with $r\ge0$
maximal. Such an $r$ is finite, as a nonzero integer point order of
$\tau_0$ cannot be divisible by arbitrarily high powers of $p$.
The function $\tau$ gives a separable morphism
$\tau:C'\to\mathbb P^1$ of some degree $n\ge1$.
Let $\beta'_j$ be the $p^r$-th root of $\beta_j$ in $kv$.
These constants are distinct, and
\[
\tau_0-\beta_j=(\tau-\beta'_j)^{p^r}.
\]
Since $p\ne\ell$, this power does not affect whether any point order
is zero modulo $\ell$ or $q$.
Over the value $\beta'_i$ corresponding to anchor $i$, at most $N_i$
points have ramification index not divisible by $\ell$.
Indeed such a point contributes a nonzero order modulo $\ell$ to
$\tau_0-\beta_i$; transport to $C$ and
\eqref{tests:support} give the bound $N_i$.
Every other point of that fiber has ramification index at least
$\ell$, so there are at most $n/\ell$ of them.
If $r_i$ denotes the total number of points in the fiber, then
\[
r_i\le N_i+n/\ell,\qquad
\sum_{P\mid\beta'_i}(e_P-1)=n-r_i
\ge(1-1/\ell)n-N_i.
\]
For a separable morphism of smooth curves, the different exponent at
$P$ is at least $e_P-1$, including in the presence of wild ramification.
The five fibers are disjoint. Riemann--Hurwitz therefore yields
\[
2g(C')-2+2n
\ge\sum_{i=1}^5\sum_{P\mid\beta'_i}(e_P-1)
\ge5(1-1/\ell)n-\sum_{i=1}^5N_i.
\]
Together with \eqref{pencil:genus} this gives $c_\ell n\le
2G_X-2+\sum_iN_i$, and hence $n\le B$.
This argument uses the different inequality, not the tame
Riemann--Hurwitz formula; see \cite[Section~53.12, Tag~0C1B]{Stacks}.
\emph{Pass from finite tests to exact supports.}
For every tested $j$, all zeros and poles of $\tau-\beta'_j$ lie in two
fibers, over $\beta'_j$ and infinity. There are at most $2n$ such
points. The number of its nonzero point orders modulo $q$ is therefore
at most $2B$.
Transport this bound to $C$ and apply \eqref{tests:support} to obtain
\begin{equation}\label{pencil:finite-level-bound}
\sum_{D\in S_e(h_a)}\deg D\le2B.
\end{equation}
The right side is independent of $a$, $e$, the finite descent data and
the residual test. For each fixed $a$, the inequality holds for every
sufficiently large $e$.
Finally $S_e(h_a)$ increases to $\operatorname{Supp}_X(h_a)$.
If the exact support had total degree greater than $2B$, a finite
subset would already have degree greater than $2B$; that subset would
lie in $S_e(h_a)$ for all sufficiently large $e$, contradicting
\eqref{pencil:finite-level-bound}.
Every prime divisor has positive integer degree, so the exact support
is finite. This proves the asserted uniform bound.
\end{proof}
\begin{corollary}\label{pencil:finite}
For every $z\in K^\times$, the class $\Theta[z]$ has finite support on
$X$.
\end{corollary}
\begin{proof}
If $z\notin k$, apply Theorem~\ref{pencil:uniform} to $t=z$ and $a=0$.
If $z\in k^\times$, its Kummer class is zero because $k^\times$ is
divisible.
\end{proof}
==== END The-Bogomolov-Pop-reconstruction-theorem//pencil-bound.tex ====
==== BEGIN The-Bogomolov-Pop-reconstruction-theorem//preliminaries.tex ====
\section{Kummer duality and the group input}\label{sec:preliminaries}
We first give the topological form of Kummer duality needed for selected
subfields and residue fields. In this section the residue characteristic
is allowed to equal $\ell$ unless explicitly excluded.
For a function field $F/\kappa$ over algebraically closed constants, put
\[
W_F=\Hom(F^\times,\Zell),\qquad
\widehat F=\varprojlim_{e\ge1} F^\times/F^{\times\ell^e}.
\]
Equip $W_F$ with pointwise convergence and $\widehat F$ with its
$\ell$-adic topology. We use additive notation for these modules and
write $[f]$ for the class of $f\in F^\times$. In particular
$[fg]=[f]+[g]$. All characters vanish on $\kappa^\times$. We abbreviate
$F^\times/F^{\times q}$ to $F^\times/q$. An annihilator $B^\perp$
of $B\subset F^\times$ or $B\subset\widehat F$ is taken in $W_F$;
for $B\subset W_F$ it is taken in $\widehat F$, using evaluation.
\begin{lemma}[Continuous evaluation]\label{prelim:duality}
The group $F^\times/\kappa^\times$ is free abelian. Consequently
$W_F$ is a product of copies of $\Zell$, and evaluation gives
\[
\widehat F=\Hom_{\mathrm{cont}}(W_F,\Zell),\qquad
\widehat F/\ell^e\widehat F=F^\times/F^{\times\ell^e}.
\]
The module $\widehat F$ is torsion free, and evaluations on $W_F$
detect every finite-level class.
\end{lemma}
\begin{proof}
Choose a normal integral projective model of $F/\kappa$. A rational
function with zero Weil divisor on this model is a global unit and hence
a constant. Thus the divisor map embeds $F^\times/\kappa^\times$ in
the free abelian group of prime divisors; a subgroup of a free abelian
group is free.
For a basis indexed by a set $J$, the character group is $\Zell^J$.
Modulo $\ell^e$, a continuous character on this compact product depends
on finitely many coordinates. Compatible such characters are therefore
families $(c_j)_{j\in J}$ in $\Zell$ with only finitely many
$c_j\notin\ell^e\Zell$ for each $e$. This is exactly the completion
of $\bigoplus_{j\in J}\mathbb Z$. Reduction modulo $\ell^e$ and
coordinate evaluations prove the assertions. Divisibility of
$\kappa^\times$ removes the constants at every level.
\end{proof}
When $\operatorname{char}F\ne\ell$, a choice of a compatible system of
$\ell$-power roots of unity identifies the Tate module $\Zell(1)$ with
$\Zell$. Kummer theory then identifies $\Pi_F^a$ with $W_F$.
An isomorphism $\varphi:\Pi_L^a\to\Pi_K^a$ has evaluation dual
\begin{equation}\label{eq:theta}
\Theta:\widehat K\longrightarrow\widehat L,
\qquad \Theta(h)(\sigma)=h(\varphi\sigma).
\end{equation}
Changing the Tate identifications multiplies this map by one unit.
We fix the identifications for the proof.
\begin{lemma}[Subfields and finite extensions]\label{prelim:completion}
Let $E\subset F$ be function fields over the same algebraically closed
field $\kappa$.
\begin{enumerate}
\item If $E$ is relatively algebraically closed in $F$, then
$\widehat E\to\widehat F$ is injective, its image is closed and
$\ell$-saturated, and restriction $W_F\to W_E$ is a continuous
surjection. Here $\ell$-saturated means that $\ell h\in\widehat E$
for $h\in\widehat F$ implies $h\in\widehat E$.
\item If $F/E$ is finite, its completion map is injective, and
inclusion followed by norm on completions is multiplication by
$[F:E]$.
\end{enumerate}
\end{lemma}
\begin{proof}
In the first case, a root in $F$ of an element of $E$ is algebraic over
$E$, so injectivity holds at every finite level. Alternatively, on a
normal integral projective $E$-model of $F$, the divisor map embeds
$F^\times/E^\times$ into a free abelian group: its kernel consists of
global units, which are algebraic over $E$ and therefore lie in $E$.
Thus $F^\times/E^\times$ is free abelian, and
\[
0\longrightarrow E^\times/\kappa^\times
\longrightarrow F^\times/\kappa^\times
\longrightarrow F^\times/E^\times\longrightarrow0
\]
splits as a sequence of abelian groups. Completion preserves this
splitting. The two completed free factors are torsion free, proving
closedness and saturation. Every character on $E^\times$ extends along
the splitting, proving the asserted surjection; continuity follows from
the pointwise topologies.
For a finite extension, the field norm is defined also in the inseparable
case and sends an element of $E$ to its $[F:E]$-th power. It induces
the stated composite on completions. Torsion freeness from
Lemma~\ref{prelim:duality} proves injectivity.
\end{proof}
The next observation lets us apply Pop's reconstruction theorems to
the stated group input.
\begin{lemma}[Lifting bracket-compatible isomorphisms]
\label{prelim:central-lift}
Every bracket-compatible isomorphism in Theorem~\ref{thm:main} is the
abelianization of a continuous isomorphism $\Pi_L^c\to\Pi_K^c$.
This holds also for $\ell=2$.
\end{lemma}
\begin{proof}
Use the given maps to identify the bases of the two central extensions
with one abelian pro-$\ell$ group $A$ and their kernels with one group
$B$. Let $E_1,E_2$ denote these extensions. In the fiber product
$P=E_1\times_A E_2$, quotient by the diagonal central subgroup
$\{(b,b):b\in B\}$. The resulting group $Q$ is abelian: the
commutator of two pairs has identical components by bracket
compatibility. It fits into an exact sequence of abelian pro-$\ell$
groups
\[
0\longrightarrow B\longrightarrow Q\longrightarrow A\longrightarrow0.
\]
By Lemma~\ref{prelim:duality}, $A$ is a product of copies of $\Zell$.
Such products are projective in the category of abelian pro-$\ell$
groups. Indeed Pontryagin duality changes $A$ into a direct sum of
copies of $\mathbb Q_\ell/\Zell$, a divisible, hence injective,
discrete torsion group. Consequently $Q\to A$ has a continuous
section.
The inverse image of this section in $P$ maps isomorphically to each
$E_i$. For fixed first component, the second component can be adjusted
by a unique element of $B$ to land over the section; its uniqueness
also proves that each projection has trivial kernel. These are
continuous bijections between compact Hausdorff groups, hence
isomorphisms. The resulting map $E_1\to E_2$ induces the prescribed
base and kernel maps. The proof uses no division by $2$ and leaves
no additional power datum to specify.
\end{proof}
\subsection{Orders on a curve}
For a smooth projective integral curve $C/\kappa$ with function field
$E$, its closed points are $\kappa$-rational. Each normalized order
$\ord_P:E^\times\to\mathbb Z$ extends to $\widehat E\to\Zell$.
\begin{lemma}[Curve kernels]\label{prelim:curve-kernel}
If $\operatorname{char}\kappa\ne\ell$, then
\[
\ker\left(E^\times/E^{\times\ell}
\xrightarrow{(\ord_P)_P}\bigoplus_{P\in C}\mathbb Z/\ell\right)
\simeq\Pic(C)[\ell].
\]
Its dimension over $\Fell$ is $2g(C)$. The kernel of all exact point
orders on $\widehat E$ is the Tate module $T_\ell\Pic(C)$ and has
$\Zell$-rank $2g(C)$.
\end{lemma}
\begin{proof}
The same argument works with $q=\ell^e$. If
$\operatorname{div}(f)=qD$, send $[f]$ to the divisor class of $D$.
Changing $f$ by a $q$-th power does not change this class. Every
$q$-torsion divisor class occurs. The kernel consists of functions
$cu^q$; the constant $c\in\kappa^\times$ is a $q$-th power. This
proves the finite-level identification. The $q$-torsion of the
Jacobian over an algebraically closed field of characteristic prime to
$q$ is $(\mathbb Z/q)^{2g(C)}$
\cite[Tag~0C1Z, Lemma~53.17.1]{Stacks}. Under reduction from $\ell^{e+1}$
to $\ell^e$, the divisor class is multiplied by $\ell$. Taking the
inverse limit gives the exact-order kernel and its rank.
\end{proof}
In particular, an isomorphism of the $W$-groups carrying the family of
point inertia lines bijectively onto the corresponding family preserves
the genus. A generator
of each line can change by a unit; this changes neither vanishing nor
nonvanishing of any finite-level order. We will use this observation
only after proving that the residue characteristics differ from $\ell$.
==== END The-Bogomolov-Pop-reconstruction-theorem//preliminaries.tex ====
==== BEGIN The-Bogomolov-Pop-reconstruction-theorem//reconstruction.tex ====
\section{From the recovered quotients to the field}
\label{sec:reconstruction}
We have recovered the geometric data needed for the last step. We now
specify that data and Pop's reconstruction theorems, so that their
application uses exactly the original group input.
For a function field $F/\kappa$, a \emph{divisorial flag} is a sequence
of valuations obtained by taking a prime divisor trivial on $\kappa$,
then a prime divisor on its residue field, and continuing in this way.
The empty flag is included. The \emph{total decomposition graph}
records, at each such flag, the abelian pro-$\ell$ group of the residue
field, and at each next divisorial valuation its inertia and
decomposition subgroups and the quotient map to the next residue group.
Thus the graph retains the group maps as well as the flags. At a
one-variable field over algebraically closed constants, its terminal
data are precisely the point inertia lines: the residual groups at
closed points vanish, so point decomposition equals point inertia.
We use the following two results of Pop.
\begin{proposition}[Pop's reconstruction theorems]
\label{reconstruction:pop}
Let $F/\kappa$ and $F'/\kappa'$ be function fields of transcendence
degree greater than one over algebraically closed fields of
characteristic different from $\ell$.
\begin{enumerate}
\item From $\Pi_F^c\to\Pi_F^a$ and the union of its true divisorial
inertia groups, one recovers the total decomposition graph, functorially
under class-two isomorphisms whose abelianizations preserve this union.
\item Suppose an isomorphism between the total decomposition graphs of
$F$ and $F'$ is compatible with all their geometric rational quotient
diagrams. Then its map on the top abelian groups is induced, up to one
unit in $\Zell$, by an isomorphism of perfect closures in the opposite
direction carrying the constant fields onto each other. That field
isomorphism is unique modulo Frobenius twists in positive characteristic,
and unique in characteristic zero.
\end{enumerate}
In (2), a geometric rational quotient is restriction to
$\kappa(u)\subset F$ with $F/\kappa(u)$ regular, together with the
point inertia and decomposition data on its target. Compatibility
means that restriction commutes with the graph isomorphisms and the
induced isomorphisms of these targets.
\end{proposition}
\begin{proof}[Reference and hypotheses]
Part (1) is \cite[Proposition~2.4]{Pop2018}; the surrounding section
assumes transcendence degree greater than one. Part (2) is the Isom
statement of \cite[Theorem~2.1(2)]{PopBI2012}, restating
\cite[Main Theorem]{PopGraphs2012}. The stated theorem requires
Bertini-type families of rational quotients; all geometric rational
quotients form such a family. More explicitly, for algebraically
independent $x,t$ with $x$ separating, the general functions
\[
ax+t,\qquad \frac{t}{a'x+a},\qquad
\frac{a''t+a'x+a+1}{t+a'x+a}
\]
in \cite[Fact/Definition~43]{PopGraphs2012} define relatively
algebraically closed rational subfields for the prescribed cofinite
choices of constants. Every one of these subfields is among the
quotients in (2). Thus its hypotheses imply precisely the required
Bertini compatibility. No restriction to finite-field closures or to
dimension greater than two occurs in this Isom statement.
\end{proof}
We emphasize the distinction between the two citations. The later
\cite[Theorem~1.1]{Pop2018} reconstructs rational quotients from
divisorial inertia under a dimension-greater-than-two hypothesis.
Here those quotients have already been recovered by
Section~\ref{sec:curve-recovery}, so that theorem is unnecessary.
\begin{proof}[Proof of Theorem~\ref{thm:main}]
A field isomorphism $\alpha:K^i\to L^i$ induces a contravariant
isomorphism of absolute Galois groups, well defined up to an inner
automorphism. Purely inseparable extension does not change these
groups. Abelianization removes the inner ambiguity, and the induced
class-two map preserves the commutator. Thus it gives a member of
$\Isom^c(\Pi_L^a,\Pi_K^a)$. In positive characteristic, absolute
Frobenius on algebraic closures commutes with field automorphisms;
composition with any integer Frobenius power therefore gives the same
allowed orbit. This proves well-definedness of $\Phi_{K,L}$.
Conversely, let $\varphi:\Pi_L^a\to\Pi_K^a$ be bracket-compatible.
Lemma~\ref{prelim:central-lift} lifts it to a class-two isomorphism.
After choosing Tate identifications, let $\Theta$ be its dual
\eqref{eq:theta}. Proposition~\ref{local:transport} recovers the common
relative transcendence degree and the quasi-divisorial pairs.
The finite residual tests and the uniform support bound prove the
completed curve-subfield correspondence in
Theorem~\ref{recovery:curves}. The recognition statements in
Section~\ref{sec:curve-recovery} then show that $\varphi$ preserves true
divisorial inertia and every relatively algebraically closed rational
quotient, including its point lines.
Proposition~\ref{recovery:rational} gives the actual commuting
quotient diagrams in \eqref{eq:recovery-quotient-diagram}.
Their vertical maps are surjective, and their target isomorphisms
$\varphi_E$ match all point inertia lines. For rational targets these
are the entire point decomposition data. Thus the same fixed
$\varphi$ is compatible with every geometric rational quotient;
no independent rescaling of the targets is needed.
Part (1) of Proposition~\ref{reconstruction:pop} gives the total
decomposition-graph isomorphism. The quotient squares also commute on
these graphs. Indeed, restriction along an actual regular inclusion
$E=k(t)\subset K$ is a geometric rational quotient morphism
\cite[Section~2, Embeddings and Restrictions]{PopBI2012}.
At a flag vertex its map is obtained by restricting to the decomposition
subgroup and passing to the inertia quotient
\cite[Definition/Remark~24 and Definition~25]{PopGraphs2012}.
If the flag restricts trivially to $E$, its inertia is killed, and the
top square descends to the residual square. Otherwise its nonzero
inertia image lies in a unique point line of $E$, where decomposition
equals inertia and the residue group is trivial. The same description
applies to $P\subset L$. Local images can have finite index: saturation
identifies the point line, but the actual subgroup images and maps are
retained. Applying these restriction and quotient operations successively
along each flag therefore proves
compatibility for the same fixed $\varphi$ and $\varphi_E$.
Part (2), applied to the recovered rational quotients,
gives an isomorphism $\alpha:K^i\to L^i$ with $\alpha(k)=l$ and
$[\varphi]=[\alpha^*]$. Choices of Tate identifications are absorbed
by the one allowed global unit. This proves surjectivity. It also
proves that different characteristics cannot give a bracket-compatible
isomorphism.
Finally, two field isomorphisms with the same Galois-side orbit give
the same compatible graph and quotient data after one global unit
adjustment. The uniqueness assertion in
Proposition~\ref{reconstruction:pop} identifies them modulo Frobenius
in positive characteristic, and identifies them outright in
characteristic zero. On perfect closures the Frobenius powers are
exactly the integer powers appearing in the definition of
$\Isom_F^i(K,L)$. This proves injectivity and completes the theorem.
\end{proof}
==== END The-Bogomolov-Pop-reconstruction-theorem//reconstruction.tex ====
==== BEGIN The-Bogomolov-Pop-reconstruction-theorem//residual-tests.tex ====
\section{Finite tests on residual curves}\label{sec:residual-tests}\label{sec:tests}
We next construct residual curves that retain any prescribed finite collection
of Kummer relations. The curve can depend on the functions and on the
exponent. Its genus, however, is bounded in terms of one fixed projective
model. This distinction will make the pencil estimate uniform.
Fix a normal integral projective model $X\subset\mathbb P^N_l$ of $L/l$,
and put $d=\operatorname{trdeg}(L/l)\ge2$.
For a prime divisor $D$ on $X$, let $\deg D$ denote its degree in this
embedding. For an integer $m\ge2$ and $g\in L^\times$, set
\[
s_m(g)=\sum_{\substack{D\subset X\text{ prime divisor}\\
\operatorname{ord}_D(g)\not\equiv0\pmod m}}
\deg D.
\]
This is a finite sum.
\begin{lemma}[Finite residual tests]\label{tests:residual}
There is an integer $G_X\ge0$, depending only on the embedded variety $X$,
with the following property. Given $g_1,\ldots,g_r\in L^\times$ and
$q=\ell^e$, $e\ge1$, there is a quasi-prime $(d-1)$-divisor $w$ of $L/l$
such that:
\begin{enumerate}
\item $lw$ is an algebraic closure of a finite field of characteristic
different from $\ell$, and $Lw=lw(C)$ for a smooth projective integral
curve $C/lw$ with $g(C)\le G_X$;
\item $w(g_j)=0$ for every $j$, and, for every
$(n_1,\ldots,n_r)\in(\mathbb Z/q)^r$,
\[
\prod_jg_j^{n_j}\in L^{\times q}
\quad\Longleftrightarrow\quad
\prod_j\overline g_j^{n_j}\in(Lw)^{\times q};
\]
\item for $m=\ell$ and $m=q$,
\begin{equation}\label{tests:support}
\#\{P\in C(lw):\operatorname{ord}_P(\overline g_j)
\not\equiv0\pmod m\}=s_m(g_j)
\qquad(1\le j\le r).
\end{equation}
\end{enumerate}
Here $\overline g_j$ is the residue of $g_j$ at $w$.
\end{lemma}
We use the following form of Bertini irreducibility. Its hypotheses include
the finite covering maps that occur below.
\begin{lemma}[Hyperplane sections of a quasi-finite map]\label{tests:bertini}
Let $V$ be an integral variety over an algebraically closed field, and let
$f:V\to\mathbb P^N$ be quasi-finite, with $\dim V=a\ge2$.
For a general hyperplane $H$, the inverse image $f^{-1}(H)$ is
geometrically irreducible. If $f$ factors as a finite \'etale map to a
smooth locally closed subvariety $U\subset\mathbb P^N$, this inverse image
is also smooth for a general $H$.
\end{lemma}
\begin{proof}
We recall the incidence argument for irreducibility; see also
\cite[Lemma~37.32.3, Tag~0G4F]{Stacks}.
Write $B=(\mathbb P^N)^*$ and
\[
I=\{(x,H)\in V\times B:f(x)\in H\}.
\]
The projection to $V$ is a projective bundle, so $I$ is integral.
In $I\times_BI$, the open set lying over pairs $(x,y)$ with
$f(x)\ne f(y)$ is a projective bundle with fiber $\mathbb P^{N-2}$ over
an irreducible open subset of $V\times V$. It is therefore irreducible
of dimension $2a+N-2$.
The locus of pairs with $f(x)=f(y)$ has dimension at most $a$, since $f$
is quasi-finite; the incidence above it has dimension at most $a+N-1$.
This is strictly smaller than $2a+N-2$.
Every component of the generic hyperplane section has dimension $a-1$:
a hyperplane cuts a nonzero nonunit in an integral variety of finite type
over a field. Consequently every component of its square has dimension
$2a-2$. The equal-image locus cannot supply a component of this generic
square. The generic square is thus irreducible. Equivalently, the
function field of $B$ is separably algebraically closed in that of $I$,
so the generic hyperplane section is geometrically irreducible.
Geometric irreducibility spreads to a nonempty open of $B$
\cite[Lemma~37.27.5, Tag~0559]{Stacks}.
Finally, a general hyperplane section of the smooth embedded variety $U$
is smooth by Bertini, and its \'etale inverse image is smooth.
\end{proof}
\begin{proof}[Proof of Lemma~\ref{tests:residual}]
\emph{Spread the finite data.}
Let $D_1,\ldots,D_s$ be the prime divisors occurring in the divisors of
the $g_j$. There is a closed subset $B_X\subset X$ of codimension at
least two such that $X\setminus B_X$ is smooth, the
$D_i\setminus B_X$ are smooth pairwise disjoint Cartier divisors, and,
near each of them, every $g_j$ is a power of a local equation times a
unit. Off their union the $g_j$ are units.
These assertions follow by deleting the singular loci, the pairwise
divisor intersections and the proper closed exceptional subsets on
which the required local identities fail.
Choose a smooth dense open $U\subset X$ on which all $g_j$ are units.
The simultaneous root cover
\[
V=\operatorname{Spec}_U
\mathcal O_U[T_1,\ldots,T_r]/(T_1^q-g_1,\ldots,T_r^q-g_r)
\longrightarrow U
\]
is finite \'etale of degree $q^r$.
Let $H$ be the subgroup generated by the $[g_j]$ in
$L^\times/L^{\times q}$. Kummer theory shows that each connected
component of $V$ has degree $|H|$ over $U$.
Since $U$ is smooth, these components are integral.
All these data descend to a finitely generated domain $R\subset l$ in
which $\ell$ is invertible. Enlarge $R$ and then localize it so that the
following properties hold in every geometric fiber:
\begin{itemize}
\item the embedded model $X_R$ is projective and flat, with normal
integral fibers and the Hilbert polynomial of $X$;
\item the $D_{i,R}$ are distinct geometrically integral divisors with
their original degrees, and the local divisor identities and
smoothness conditions above hold outside a subset of fiberwise
codimension at least two;
\item the root cover and its disjoint decomposition into components
extend, and every component has geometrically integral fibers
and finite locally free degree $|H|$ over $U_R$.
\end{itemize}
These are finite spreading conditions. Flatness makes the projective
Hilbert polynomials constant. Geometric integrality and normality
spread from the generic fiber after shrinking; the latter uses the
proper flat finite-presentation form of
\cite[Theorem~12.2.4(iv),(viii)]{EGAIV3}.
The equations $g_j=u\pi^n$, the inverses specifying units, the component
idempotents, and the finite locally free ranks are all finite algebraic
data. Spreading them preserves the exact orders and degrees, rather
than just the supports. The dimension bound on the omitted closed set
also persists after shrinking.
For later use include a separating transcendence basis
$x_1,\ldots,x_d$ of $L/l$, a primitive element $y$ over $l(x_1,\ldots,x_d)$,
its monic minimal polynomial, and common birational opens with $X$.
After another localization, the same presentation and birational
identification hold in the geometric fibers, with the reduced polynomial
irreducible of its original degree over the rational function field.
Preserve as well the nonzero reductions of the finitely many
numerators and denominators used for the $g_j$.
This follows either by spreading those common opens and geometric
integrality, or by spreading the absolutely irreducible defining
polynomial after clearing denominators.
\emph{Realize a good fiber as a residue field.}
Choose a closed point $s\in\operatorname{Spec}R$.
Its residue field is finite and has characteristic different from $\ell$.
A valuation ring of $l$ dominating $R_s$ exists
\cite[Lemma~10.50.2, Tag~00IA]{Stacks}.
Its residue field is algebraically closed. If that residue field is
not algebraic over a finite field, give a transcendence basis independent
values in an ordered free abelian group, prolong the resulting valuation
to the residue field, and compose. This yields a valuation $s_0$ of $l$
with the same center $s$ and residue field $\lambda$ algebraic over a
finite field. Residues of algebraically closed fields are algebraically
closed, so $\lambda$ is an algebraic closure of a finite field.
The composition does not change the center because every nonzero
element of a finite field has value zero.
This construction uses a transcendence basis of arbitrary cardinality;
there is no restriction on the size of $l$.
Extend $s_0$ to $l(x_1,\ldots,x_d)$ by the Gauss valuation, whose residue
field is $\lambda(x_1,\ldots,x_d)$ and whose value group is $s_0l$.
Prolong it to $L$. The element $y$ is integral and its residue satisfies
the reduced irreducible polynomial, of degree
$[L:l(x_1,\ldots,x_d)]$. The fundamental valuation inequality therefore
forces the residue degree to be that degree and the value-group index
to be one \cite[Section~15.125, Tag~0ASF]{Stacks}.
We obtain a valuation $s_X$ with
\[
s_XL=s_0l,\qquad Ls_X=\lambda(X_s),
\]
and with the $g_j$ reducing to their specified rational functions on
$X_s$. This also explains why no extra ramification or defect is hidden
in the passage to the good fiber.
\emph{Cut down to a curve.}
Over $\lambda$, choose a general flag of $d-1$ hyperplanes in the fixed
projective embedding. Make the final section $C$ avoid the omitted
codimension-two set and meet the $D_{i,s}$ transversely.
The intersections with different $D_{i,s}$ are disjoint.
The final section is smooth and integral. Apply
Lemma~\ref{tests:bertini} successively to the finite list of covering
components as well as to the reduced underlying base sections; each restricted covering
component remains integral over $U_s\cap C$.
At every step the varieties to which irreducibility is applied have
dimension at least two, including the last surface-to-curve step.
Smoothness of the restricted cover follows from its \'etaleness over the
smooth curve open.
For completeness, intermediate scheme sections need only have
irreducible support and be smooth at the generic points used by the
flag. All those points lie in the original smooth locus.
Choose the hyperplanes also to avoid the finitely many associated
points at each stage, so their equations are non-zero-divisors.
The hyperplane exact sequences give
\[
P_C(n)=\Delta^{d-1}P_X(n),\qquad
\Delta P(n)=P(n)-P(n-1).
\]
As $C$ is smooth integral, its genus is consequently fixed by the
embedded model $X$; denote it by $G_X$.
All conditions imposed on the flag are finitely many nonempty open
conditions. They can be met over the infinite algebraically closed
field $\lambda$, even when it is countable.
Compose $s_X$ with the successive divisorial valuations of this flag.
For the first composition, the relative value group is the innermost
convex copy of $\mathbb Z$, and its residue transcendence degree drops
by one. Every proper coarsening kills this copy of $\mathbb Z$ and
therefore has all its values supplied by constants.
The first composition is thus a quasi-prime divisor. The remaining
flag valuations are prime divisors over the residue constants, giving
a quasi-prime $(d-1)$-divisor $w$.
All $g_j$ are units at every generic point in the flag, so $w(g_j)=0$,
and $Lw=\lambda(C)$.
\emph{Check relations and orders.}
An old relation $\prod_jg_j^{n_j}=b^q$ specializes: since $w(g_j)=0$,
torsion-freeness of the value group gives $w(b)=0$.
Thus the old relation kernel in $(\mathbb Z/q)^r$ is contained in the
new one.
On the other hand, every component of the root cover restricted to
$U_s\cap C$ is integral and has degree $|H|$.
Kummer theory gives the same order $|H|$ for the subgroup generated by
the residue classes. The two finite relation kernels therefore have
the same order, so they are equal.
Finally, $C$ meets $D_{i,s}$ in exactly $\deg D_i$ distinct transverse
points. At each such point the order of $\overline g_j$ equals the
unchanged integer $\operatorname{ord}_{D_i}(g_j)$.
There are no other zeros or poles. This proves
\eqref{tests:support}, in fact for every divisor $m$ of $q$, and
completes the construction.
\end{proof}
==== END The-Bogomolov-Pop-reconstruction-theorem//residual-tests.tex ====
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