==== TABLE OF CONTENTS ====
An-Integral-Counterexample-to-Gerstens-Conjecture//arithmetic.tex
An-Integral-Counterexample-to-Gerstens-Conjecture//claimed-vanishing.tex
An-Integral-Counterexample-to-Gerstens-Conjecture//coefficients.tex
An-Integral-Counterexample-to-Gerstens-Conjecture//geometry.tex
An-Integral-Counterexample-to-Gerstens-Conjecture//introduction.tex
An-Integral-Counterexample-to-Gerstens-Conjecture//main-proof.tex
An-Integral-Counterexample-to-Gerstens-Conjecture//main.tex
An-Integral-Counterexample-to-Gerstens-Conjecture//specialization.tex
An-Integral-Counterexample-to-Gerstens-Conjecture//references.bib
===========================

==== BEGIN An-Integral-Counterexample-to-Gerstens-Conjecture//arithmetic.tex ====
\section{An integral cyclotomic class}
\label{sec:arithmetic}

The coefficient exact sequence identifies the obstruction to lifting a
degree-five coefficient class: it is a class in integral $K_4$ killed by
five.  We will show that this obstruction vanishes for one cyclotomic
ring.  The proof first uses localization to detect coefficient $K_4$
geometrically, then uses finite generation to eliminate all five-primary
torsion.  We finish by recording how the chosen integral lift behaves
under the four cyclotomic automorphisms.

Fix a primitive fifth root of unity $\zeta$, and put
\[
 N=\Q(\zeta),\qquad S=\Z[\zeta,1/5],\qquad
 u=\zeta-1,\qquad \Delta=\Gal(N/\Q).
\]
Let $\chi:\Delta\to\F_5^\times$ be the character defined by
$\delta(\zeta)=\zeta^{\chi(\delta)}$.  We use the coefficient convention and product of Section~\ref{sec:coefficients}.

\paragraph{The cyclotomic ring.}
We record the arithmetic facts that will control the localization
sequence.  Let $\Phi_5(X)=X^4+X^3+X^2+X+1$ be the fifth cyclotomic
polynomial.  The polynomial of $u$ is
\[
 \Phi_5(1+T)=T^4+5T^3+10T^2+10T+5,
\]
which is Eisenstein at five.  In particular it is irreducible over
$\Q_5$, so $N$ has exactly one place above five.  The ring $\Z_5[u]$ is
finite free over $\Z_5$; irreducibility makes it a domain, and integrality
gives dimension one.  Its reduction modulo five is $\F_5[T]/(T^4)$,
with unique maximal ideal generated by $T$.  Thus $\Z_5[u]$ is a
Noetherian local domain with maximal ideal $(5,u)$.  The defining equation
gives
\[
 u^4=-5(1+2u+2u^2+u^3).
\]
The factor in parentheses is a unit, since its residue modulo $(5,u)$
is one.  Hence $(5,u)=(u)$, and $\Z_5[u]$ is a discrete valuation ring.
Being integrally closed with fraction field $\Q_5(u)$, it is the integer
ring of that field, with residue field $\F_5$ and ramification index four.

The norm of $u$ is $\Phi_5(1)=5$.  Since
$\Phi_5'(\zeta)=5\zeta^4/(\zeta-1)$, the discriminant of the power
basis $1,\zeta,\zeta^2,\zeta^3$ is
\[
 \operatorname{N}_{N/\Q}\bigl(\Phi_5'(\zeta)\bigr)
 =\frac{5^4}{\operatorname{N}_{N/\Q}(u)}=125.
\]
Let $\mathcal O_N$ be the ring of integers of $N$.  The index of
$\Z[\zeta]$ in $\mathcal O_N$ can therefore have
no prime divisor other than five, by the discriminant-index formula.
The local calculation above shows that this order is already maximal
at five.  Thus the index is one:
\[
 \mathcal O_N=\Z[\zeta],\qquad \operatorname{disc}(N)=125.
\]
There are two complex places and no real places.  Minkowski's ideal-class
bound \cite[Chapter~4, Theorem~4.3]{MilneANT2020} is consequently
\[
 \frac{4!}{4^4}\left(\frac4\pi\right)^2\sqrt{125}<2.
\]
Every ideal class has an integral ideal representative of positive
integral norm at most this bound, hence of norm one.  It follows that
$\Pic(\mathcal O_N)=0$, and localization gives $\Pic(S)=0$.

The norm calculation also shows that $u$ and all its conjugates are
units in $S$.  Every closed point $v$ of $\Spec S$ has a finite
residue field $k_v$ of characteristic different from five.  The image of
$\zeta$ in $k_v$ has order exactly five: its order divides five, and it
cannot be one because $u$ is a unit.  These are precisely the residue
fields to which we will apply the finite-field clause of
Lemma~\ref{lem:bott}.

\paragraph{An equivariant coefficient class.}
The determinant splits the unit homomorphism
$S^\times\to K_1(S)$, so $[\zeta]\in K_1(S)$ has order
five.  The coefficient exact sequence supplies
$\beta'\in K_2(S;5)$ with
$\partial_{\mathrm{coef}}\beta'=[\zeta]$.  This coefficient group is an
$\F_5$-vector space by Lemma~\ref{lem:moore-product}, and $\Delta$ acts
on it by functoriality.  Define
\[
 e_\chi=\frac14\sum_{\delta\in\Delta}\chi(\delta)^{-1}\delta
       \in\F_5[\Delta],\qquad \beta=e_\chi\beta'.
\]
Here $1/4$ is the inverse of $4$ in $\F_5$.  Naturality makes the
coefficient boundary $\Delta$-equivariant, and
$\delta[\zeta]=\chi(\delta)[\zeta]$.  Hence
$\partial_{\mathrm{coef}}\beta=e_\chi[\zeta]=[\zeta]$.
Relabeling the sum gives $\tau e_\chi=\chi(\tau)e_\chi$ for every
$\tau\in\Delta$, so
\begin{equation}
 \label{eq:equivariant-bott}
 \partial_{\mathrm{coef}}\beta=[\zeta],\qquad
 \delta\beta=\chi(\delta)\beta\quad(\delta\in\Delta).
\end{equation}
For any $S$-scheme on which the operation is used, let $b$ denote
left multiplication by the restriction of $\beta$, and put
$b^2=b\circ b$.  Thus $b^2(x)=\beta\cdot(\beta\cdot x)$, with the
indicated parentheses.  The coefficient localization boundaries below
are different from $\partial_{\mathrm{coef}}$; Lemma~\ref{lem:bott-localization}
states that $b$ commutes with the former.

The next lemma is the arithmetic reason that localization will leave
only a single coefficient line.  It concerns reductions of integral
$K_2$-classes, not all of $K_2(N;5)$.

\begin{lemma}[Unramified degree-two classes]
\label{lem:cyclotomic-unramified}
Let $N=\Q(\zeta)$ and $S=\Z[\zeta,1/5]$ as above.  If a class in
$K_2(N)/5$ has zero tame residue in $k_v^\times/k_v^{\times5}$ at every
closed point $v$ of $\Spec S$, then the class is zero.
\end{lemma}

\begin{proof}
Matsumoto's symbol theorem
\cite[\S5, Theorem~5.10 and Corollary~5.11]{Matsumoto1969} gives the
classical Steinberg-symbol presentation; see \cite[III.6.1]{Weibel2013}
for its modern formulation.  The classical-to-Quillen comparison
\cite[IV.1.10.1 and IV.7.2]{Weibel2013} identifies this group with
$K_2(N)$.  The degree-two norm-residue theorem
\cite[Theorem~6.16]{Voevodsky2011} then gives
\[
 K_2(N)/5\simeq H^2(N,\mu_5^{\otimes2}).
\]
Use $\zeta$ to identify one copy of $\mu_5$ with the constant module
$\Z/5$.  The Kummer sequence identifies the resulting group
$H^2(N,\mu_5)$ with $\Br(N)[5]$.  Write $a$ for the Brauer class
corresponding to the class in the lemma.

For a closed point $v$ of $\Spec S$, the residue characteristic is
not five.  The Quillen boundary on $K_2(N)$ is the tame symbol.  Under
norm residue and the chosen twist identification, its reduction modulo
five and the Brauer residue agree up to an overall sign depending on
boundary conventions; in particular, their vanishing is equivalent.
Indeed, write each field element as a power of a uniformizer times a
unit.  Both residues kill unit--unit symbols, and on a symbol consisting
of a unit $\epsilon$ and a uniformizer they give the class of
$\bar\epsilon$, up to the corresponding sign.  Bilinearity and the
Steinberg relations give the general case
\cite[V.6.1.2 and V.(6.6)]{Weibel2013}
\cite[Proposition~7.10]{Harari2023Brauer}.  Thus $a$ has zero Brauer
residue at every finite place $v\nmid5$.

The completion $N_v$ has finite residue field $k_v$.  For its henselian
valuation ring, the kernel of the Brauer residue is $\Br(k_v)=0$
\cite[Proposition~6.7 and Example~6.8(a)]{Harari2023Brauer}.
Consequently $a|_{N_v}=0$ at every $v\nmid5$.  The infinite completions
are complex and have zero Brauer group.  Only the single place above
five remains.  In the global Brauer invariant sequence, the sum of all
local invariants is zero, so the invariant at this last place is zero.
The local invariant is injective there, and the global restriction map
is injective.  These are the local and global Brauer statements in
\cite[IV, Example~2.14(c)--(e), Proposition~4.3 and Remark~4.4(a)]{MilneCFT2020}; see also
\cite[III.6 and VI.8.1.1]{Weibel2013}.  They imply $a=0$, hence the
original class is zero.  No tame-residue calculation at the place above
five is used.
\end{proof}

We can now eliminate the obstruction in integral $K_4$.  The first part
of the proof detects coefficient classes after passage to an algebraic
closure; the last part explains why this also rules out integral
five-primary torsion.

\begin{lemma}[The integral torsion obstruction vanishes]
\label{lem:cyclotomic-k4}
For $S=\Z[\zeta,1/5]$, one has $K_4(S)[5]=0$.
\end{lemma}

\begin{proof}
Because $S$ is a Dedekind ring with $\Pic(S)=0$, every divisor
on $\Spec S$ is principal.  Equivalently, the normalized valuation
map
\[
 N^\times\longrightarrow\bigoplus_{v\in(\Spec S)^{(1)}}\Z,
 \qquad f\longmapsto\bigl(v(f)\bigr)_v,
\]
is surjective.  The sums here and below have finite support.  Quillen
localization and d\'evissage for $S$
\cite[\S7, Proposition~3.2 and paragraph~3.4]{Quillen1973Higher} identify the
boundary from $K_1(N;5)$ with this valuation map modulo five.  We use
the positive normalized valuation convention
\cite[Example~V.6.1.2 and (V.6.6)]{Weibel2013}.  Thus
\[
 K_1(N;5)\longrightarrow\bigoplus_vK_0(k_v;5)
\]
is surjective.

The number field $N$ has $5$-cohomological dimension two
\cite[VI.\S7 and VI.8.2]{Weibel2013}.  The odd-degree clause of
Lemma~\ref{lem:bott} therefore makes
$b^2:K_1(N;5)\to K_5(N;5)$ an isomorphism.  For every $v$, the finite
field $k_v$ contains the primitive image of $\zeta$, so the finite-field
clause makes $b^2:K_0(k_v;5)\to K_4(k_v;5)$ an isomorphism as well.
The same restricted class $\beta$ acts on every term.  Its compatibility
with localization in Lemma~\ref{lem:bott-localization} carries the last
surjection to a surjection
\[
 K_5(N;5)\longrightarrow\bigoplus_vK_4(k_v;5).
\]
In the exact localization segment
\[
 K_5(N;5)\longrightarrow\bigoplus_vK_4(k_v;5)
 \longrightarrow K_4(S;5)\longrightarrow K_4(N;5),
\]
the middle arrow is consequently zero.  We obtain an injection
\begin{equation}
 \label{eq:k4-restriction}
 K_4(S;5)\hookrightarrow K_4(N;5).
\end{equation}

We next determine where this injected group can lie.  Take
$x\in K_4(S;5)$ and write $x_N$ for its image.  The even-degree
clause of Lemma~\ref{lem:bott} gives
$b:K_2(N;5)\xrightarrow{\sim}K_4(N;5)$, so $x_N=b(z)$ for some
$z\in K_2(N;5)$.  The localization residues of $x_N$ in
$K_3(k_v;5)$ vanish because $x$ extends over $S$.  Write
$\partial_v$ for the localization boundary at $v$.  Compatibility with
$b$ gives
\[
 0=\partial_v(x_N)=b\bigl(\partial_v z\bigr).
\]
Here $\partial_v z\in K_1(k_v;5)$, and
$b:K_1(k_v;5)\to K_3(k_v;5)$ is injective by the finite-field clause.
Thus every localization residue of $z$ is zero.

On the other hand,
$\partial_{\mathrm{coef}}z\in K_1(N)[5]=N^\times[5]=\langle\zeta\rangle$.
Choose $m\in\F_5$ such that this boundary equals $m[\zeta]$, and consider
the corrected class $z-m\beta_N$, where $\beta_N$ is the restriction of
$\beta$.  Equation~\eqref{eq:equivariant-bott} makes its coefficient
boundary zero.  Its localization residues are still zero, since $\beta$ extends
over $S$.  Exactness of the coefficient sequence identifies this
corrected class with an element of $K_2(N)/5$.  Naturality of reduction
with localization identifies its coefficient residues with the tame
residues of that element.  Lemma~\ref{lem:cyclotomic-unramified} therefore makes
the corrected class zero.  We have proved
\[
 z=m\beta_N,\qquad x_N=m\,b(\beta_N).
\]

Let $\overline N$ be an algebraic closure of $N$.  By the closed-field
clause of Lemma~\ref{lem:bott}, the restriction of $b(\beta_N)$ is
the nonzero square of a geometric Bott generator in
$K_4(\overline N;5)$.  Restriction is therefore injective on the
$\F_5$-line spanned by $b(\beta_N)$.  Together with
\eqref{eq:k4-restriction}, this proves
\begin{equation}
 \label{eq:k4-geometric-injection}
 K_4(S;5)\hookrightarrow K_4(\overline N;5).
\end{equation}

It remains to pass from coefficient detection to integral torsion.
The same closed-field clause gives $K_5(\overline N;5)=0$.  The
coefficient exact sequence then gives $K_4(\overline N)[5]=0$.
In fact $K_4(\overline N)$ has no nonzero element of any finite
five-power order: a suitable multiple of such an element would have
order five.

Set $G=K_4(S)$, and write $\rho$ for coefficient reduction at each
ring.  Naturality gives the commutative square
\[
 \begin{array}{ccc}
 G &\xrightarrow{\rho}& K_4(S;5)\\
 \big\downarrow && \big\downarrow\\
 K_4(\overline N)&\xrightarrow{\rho}&K_4(\overline N;5).
 \end{array}
\]
The right vertical arrow is injective by
\eqref{eq:k4-geometric-injection}.  If $g\in G$ has finite five-power
order, its integral image in $K_4(\overline N)$ is zero.  The square
therefore gives $\rho(g)=0$.  Since $G/5$ injects into
$K_4(S;5)$, this says $g\in5G$.  This conclusion holds for every
element of five-power order, not just for elements killed by five.

Finite generation now rules out a nonzero divisible primary subgroup.
Quillen's theorem makes $K_4(\mathcal O_N)$ finitely generated
\cite[\S1, Theorem~1 and Remark~(2)]{Quillen1973Finite}; see also
\cite[IV.6.9]{Weibel2013}.  Integral localization at the single place
above five has the segment
\[
 0=K_4(\F_5)\longrightarrow K_4(\mathcal O_N)
 \longrightarrow G\longrightarrow K_3(\F_5)=\Z/24.
\]
The two finite-field values are Quillen's integral computation
\cite{Quillen1972FiniteFields}; see \cite[IV.1.13]{Weibel2013}.
Thus $G$ is finitely generated.  Write
$G\simeq\Z^r\oplus P\oplus T$, where $P$ is its finite five-primary
torsion subgroup and $T$ is finite of order prime to five.  We proved
$P\subseteq5G$.  If $p=5g\in P$, the free component of $g$ is zero,
and its $T$-component is zero because multiplication by five is an
automorphism of $T$.  Hence $g\in P$, and $P\subseteq5P$.  The reverse
inclusion is automatic.  Iterating $P=5P$ to the exponent of the finite
group $P$ gives $P=0$.  In particular $K_4(S)[5]=0$.
\end{proof}

The coefficient obstruction has now been removed.  We choose the
degree-five class whose two Bott factors produce the square of the
cyclotomic character.

The cyclotomic-unit/Bott pattern has a methodological predecessor in
Soul\'e's coefficient and tower constructions
\cite[\S\S4.2--4.4, pp.~239--240]{Soule1987}.  The fixed integral
$K_5(S)$-lift below is a separate deduction from
$K_4(S)[5]=0$, proved here.

\begin{proposition}[Integral lift and conjugates]
\label{prop:integral-lift}
Let $u=\zeta-1\in S^\times$, and let $b$ be multiplication by the
class $\beta$ in \eqref{eq:equivariant-bott}.  There exists
$c\in K_5(S)$ such that
\begin{equation}
 \label{eq:integral-c}
 c\bmod5=b^2[u]_5\quad\text{in }K_5(S;5).
\end{equation}
For every $\delta\in\Delta$,
\begin{equation}
 \label{eq:conjugate-c}
 (\delta c)\bmod5=\chi(\delta)^2b^2[\delta u]_5
       \quad\text{in }K_5(S;5).
\end{equation}
For any fixed homomorphism $\iota:S\to H$ to a field, let
$\operatorname{res}_\iota$ denote extension of scalars.  Restriction of
these equalities gives the corresponding identities in $K_5(H;5)$.
In the second identity this means
\[
 \operatorname{res}_\iota(\delta c)\bmod5
 =\chi(\delta)^2 b_\iota^2[\iota(\delta u)]_5,
\]
where $b_\iota$ is multiplication by $\operatorname{res}_\iota(\beta)$; no action of
$\delta$ on $H$ is required.
\end{proposition}

\begin{proof}
Lemma~\ref{lem:cyclotomic-k4} and the coefficient exact sequence give
\[
 0\longrightarrow K_5(S)/5\longrightarrow K_5(S;5)
 \longrightarrow K_4(S)[5]=0.
\]
Consequently $K_5(S)\to K_5(S;5)$ is surjective, and
$b^2[u]_5$ has an integral lift $c$.  Applying $\delta$ after reduction,
then using naturality and bilinearity of the parenthesized product, gives
\[
 \begin{aligned}
 (\delta c)\bmod5
 &=\delta\bigl(\beta\cdot(\beta\cdot[u]_5)\bigr)\\
 &=(\delta\beta)\cdot\bigl((\delta\beta)\cdot[\delta u]_5\bigr)\\
 &=\chi(\delta)^2\beta\cdot(\beta\cdot[\delta u]_5).
 \end{aligned}
\]
This proves \eqref{eq:conjugate-c}; functoriality gives its restriction
along $\iota$.  The construction made no equivariant choice of the
integral lift $c$.
\end{proof}
==== END An-Integral-Counterexample-to-Gerstens-Conjecture//arithmetic.tex ====

==== BEGIN An-Integral-Counterexample-to-Gerstens-Conjecture//claimed-vanishing.tex ====
\section{A claimed vanishing theorem for principal divisors}
\label{app:claimed-vanishing}

Mochizuki's Theorem~5 in version~8 of
\cite{Mochizuki2020} asserts that, for a commutative Noetherian local
ring $B$ and a nonzerodivisor $g$, the inclusion from finitely generated
projective $B/gB$-modules to finitely generated $B$-modules of projective
dimension at most one induces zero on $K$-theory. This statement reaches
the exact support in the present paper: Lemma~\ref{lem:local-ring} gives
$B=A$, $g=Y_0x-X_0y$, and $A/gA=\OO$, with $g$ a regular parameter.
Composing the claimed zero map with the inclusion into coherent
$A$-modules would annihilate $i_*\widetilde c$.

We give a counterexample to the functor assertion in
\cite[Lemma~13(III)]{Mochizuki2020}, which is used in that proof.
The counterexample already has both $B$ and $B/gB$ regular. It concerns
the supplied proof of the vanishing assertion; a failure of this
construction alone does not disprove the assertion itself. The proof
of Theorem~\ref{thm:main} above is independent of it.

\subsection*{The complexes and the proposed operation}

Let $B$ be a regular local ring and let $g$ be a nonzero regular
parameter. In particular $g$ is a nonzerodivisor in the maximal ideal,
and $1-g$ is a unit. The exact category $\mathcal C$ used in the cited
argument consists of two-term complexes of finite free $B$-modules,
in homological degrees one and zero, whose differential is injective
and whose cokernel is killed by $g$. Exact sequences are degreewise
exact. We use only the objects
\[
 (n,m)_B=
 \left[B^n\oplus B^m
 \xrightarrow{\operatorname{diag}(gI_n,I_m)}B^n\oplus B^m\right]
 \quad(n,m\geq0)
\]
and split exact sequences between them. Thus no classification of
other objects of $\mathcal C$ is needed for the example.

For $X=(1,1)_B$, put $D=\operatorname{diag}(g,1)$. Following the
block convention in \cite[Conventions~11, equations~(15)--(16)]{Mochizuki2020},
an encoded endomorphism
$M=\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)$ has
actual matrices
\begin{equation}\label{eq:mochizuki-chain-matrices}
 M_1=\begin{pmatrix}a&b\\gc&d\end{pmatrix},
 \qquad
 M_0=\begin{pmatrix}a&gb\\c&d\end{pmatrix}.
\end{equation}
These satisfy $DM_1=M_0D$. Conversely, the chain-map condition, together
with the fact that $g$ is a nonzerodivisor, gives this presentation.
Multiplying the actual matrices shows that composition of the encoded
matrices is
\begin{equation}\label{eq:mochizuki-matrix-composition}
 \begin{pmatrix}a'&b'\\c'&d'\end{pmatrix}
 \star
 \begin{pmatrix}a&b\\c&d\end{pmatrix}
 =
 \begin{pmatrix}
 a'a+gb'c&a'b+b'd\\
 c'a+d'c&gc'b+d'd
 \end{pmatrix}.
\end{equation}
Thus $M'\star M$ represents $M'\circ M$. Upper triangular means
$c=0$, and lower triangular means $b=0$; these terms refer to the
encoded matrices.

Write $\operatorname{Typ}(h)=[B\xrightarrow{h}B]$ for $h\in B$.
The assignments $\mu'_1,\mu'_2$ preceding Definition~12 in
\cite{Mochizuki2020} replace $(n,m)_B$ by
$\operatorname{Typ}(g)^{\oplus n}$ or
$\operatorname{Typ}(1)^{\oplus n}$, respectively. On a morphism they
retain its upper-left block in both chain degrees. On all morphisms
this is not a functor, because the upper-left entry of
\eqref{eq:mochizuki-matrix-composition} contains $gb'c$.
The cited argument therefore restricts the diagrams and their maps.

Here is the part of that restriction needed below. An $S_2$-object in
an exact category is an admissible short exact sequence
$X_{01}\rightarrowtail X_{02}\twoheadrightarrow X_{12}$, together
with the zero objects on its diagonal; morphisms are commuting maps
of these sequences. Definition~12 of \cite{Mochizuki2020} requires
upper triangular structure arrows and allows lower triangular
componentwise isomorphisms between the diagrams. Lemma~13(III) asserts
that $\mu'_1,\mu'_2$ induce functors on this modified $S$-construction.
The following example violates the naturality required of an image
morphism already in degree two.

\begin{lemma}\label{lem:triangular-naturality-failure}
For every regular local ring $B$ with a nonzero regular parameter $g$,
the upper-left-block assignments just described fail to send all
lower triangular isomorphisms between upper triangular $S_2$-objects
to morphisms of the target diagrams.
\end{lemma}

\begin{proof}
Take the encoded endomorphisms of $X=(1,1)_B$
\[
 U=\begin{pmatrix}1&1\\0&1\end{pmatrix},\qquad
 U'=\begin{pmatrix}1-g&1\\0&1\end{pmatrix},\qquad
 L=\begin{pmatrix}1&0\\1&1\end{pmatrix},\qquad
 L'=\begin{pmatrix}1&0\\1&1-g\end{pmatrix}.
\]
Their actual chain matrices are
\[
\begin{array}{c@{\qquad}c}
 U_1=\begin{pmatrix}1&1\\0&1\end{pmatrix}&
 U_0=\begin{pmatrix}1&g\\0&1\end{pmatrix}\\[6pt]
 U'_1=\begin{pmatrix}1-g&1\\0&1\end{pmatrix}&
 U'_0=\begin{pmatrix}1-g&g\\0&1\end{pmatrix}\\[6pt]
 L_1=\begin{pmatrix}1&0\\g&1\end{pmatrix}&
 L_0=\begin{pmatrix}1&0\\1&1\end{pmatrix}\\[6pt]
 L'_1=\begin{pmatrix}1&0\\g&1-g\end{pmatrix}&
 L'_0=\begin{pmatrix}1&0\\1&1-g\end{pmatrix}.
\end{array}
\]
In each degree the determinants of $U,U',L,L'$ are, respectively,
$1,1-g,1,1-g$. All are units, so these are chain isomorphisms.
The first two are upper triangular and the last two lower triangular.
A direct calculation with \eqref{eq:mochizuki-matrix-composition} gives
\[
 L'\star U=\begin{pmatrix}1&1\\1&1\end{pmatrix}=U'\star L.
\]
Equivalently, both degree-one composites are
$\left(\begin{smallmatrix}1&1\\g&1\end{smallmatrix}\right)$ and both
degree-zero composites are
$\left(\begin{smallmatrix}1&g\\1&1\end{smallmatrix}\right)$.
Hence the diagram
\[
\begin{tikzcd}[column sep=large]
 X\arrow[r,"U"]\arrow[d,"L"'] &
 X\arrow[r]\arrow[d,"L'"] &0\arrow[d,equal]\\
 X\arrow[r,"U'"'] &X\arrow[r]&0
\end{tikzcd}
\]
commutes. Each row is split exact since its first arrow is an
isomorphism. Besides that first arrow, all structure arrows of the
corresponding $S_2$-object are zero or identity maps and are therefore
upper triangular. The vertical maps $(L,L',\operatorname{id}_0)$
are lower triangular isomorphisms. Thus the diagram is a morphism in
the proposed source category.

Both upper-left entries of $L,L'$ are $1$, whereas those of $U,U'$ are
$1,1-g$. Either assignment would therefore send the left square to
\[
\begin{tikzcd}[column sep=large,row sep=large]
 \operatorname{Typ}(h)\arrow[r,"1"]\arrow[d,"1"']&
 \operatorname{Typ}(h)\arrow[d,"1"]\\
 \operatorname{Typ}(h)\arrow[r,"1-g"']&\operatorname{Typ}(h),
\end{tikzcd}
\qquad h=g\text{ or }h=1.
\]
Its two composites are $1$ and $1-g$ in both degrees. They are unequal
because $g\ne0$. The proposed image is therefore not a morphism of
$S_2$-diagrams.
\end{proof}

The calculation identifies the precise missing condition: preservation
of upper--upper and lower--lower compositions does not imply preservation
of mixed upper--lower commuting squares. In version~8 the encoded
composition and assignments appear on page~9, Lemma~13(III) on page~10,
and the subsequent additivity argument on page~11. That argument uses
the asserted functors in Lemma~13(IV) and in the proof of assertion
$(\beta)$. Although the two projected composites can be chain homotopic,
the target category requires strictly commuting natural transformations.
The later homotopy comparison does not define replacement functors on
this square. A construction that supplies the required higher
compatibilities and additivity would be an additional argument.
==== END An-Integral-Counterexample-to-Gerstens-Conjecture//claimed-vanishing.tex ====

==== BEGIN An-Integral-Counterexample-to-Gerstens-Conjecture//coefficients.tex ====
\section{Coefficients and Bott multiplication}
\label{sec:coefficients}

Our first goal is to control multiplication by a chosen coefficient
class, including its interaction with localization. These operations
have two uses. In the arithmetic construction they carry surjectivity
of a valuation map to the degree needed to remove the integral lifting
obstruction. In the geometric argument they translate a relation among
supported coefficient classes into ordinary tame symbols.

We write $G(X)$ for the $K$-theory spectrum of coherent modules on a
Noetherian scheme $X$.  For the regular schemes used below, the canonical
resolution equivalence identifies $K(X)$ with $G(X)$: these schemes have
finite dimension and are affine or quasiprojective over an affine base.
We use Quillen
localization and d\'evissage, including localization by a multiplicative
set and the resulting residue maps
\cite[\S4, Theorem~3 and Corollary~2; \S5, Theorems~4--5;
\S7, Proposition~3.2 and paragraph~3.4]{Quillen1973Higher}.

Let $M=\mathbb S/5$ be the cofiber of multiplication by $5$ on the sphere
spectrum.  Our coefficient convention is
\[
 K_i(X;5)=\pi_i(K(X)\wedge M),\qquad
 G_i(X;5)=\pi_i(G(X)\wedge M).
\]
For $i\geq1$ there is a natural exact sequence
\begin{equation}
\label{eq:coefficient-sequence}
 0\longrightarrow K_i(X)/5\longrightarrow K_i(X;5)
 \xrightarrow{\partial_{\mathrm{coef}}}K_{i-1}(X)[5]
 \longrightarrow0,
\end{equation}
and likewise for $G$. Here $A/5=A/5A$ and $A[5]=\ker(5:A\to A)$ for any abelian group $A$. If $H$ is a field, the identities $K_1(H)=H^\times$ and $K_0(H)=\Z$ give
\begin{equation}
\label{eq:low-coefficients}
 K_1(H;5)=H^\times/H^{\times5},\qquad K_0(H;5)=\mathbb Z/5.
\end{equation}
For a unit $v$, the notation $[v]_5$ denotes its image in $K_1(-;5)$.
We use additive notation in $K$-groups, including groups of unit classes.
For a scheme $X$, the notation $X^{(j)}$ denotes its points of codimension $j$.

We make the coefficient product explicit so that localization acts on
classes with nonzero coefficient boundary as well as on reductions of
integral classes. At the prime five this agrees with the product in
\cite[Proposition~16.1 and Theorem~16.2]{FriedlanderSuslin2002}.

\begin{lemma}[Coefficient products]
\label{lem:moore-product}
The spectrum $M$ has a unique unital pairing
$\mu:M\wedge M\to M$ in the homotopy category.  Multiplication by $5$
on $M$ is nullhomotopic, so all the coefficient groups above are
$\mathbb F_5$-vector spaces.  Tensor product, followed by $\mu$, defines
natural bilinear products in coefficient $K$-theory and the corresponding
module actions on coefficient $G$-theory.
\end{lemma}

\begin{proof}
The cofiber sequence for $M$, together with
$\pi_1\mathbb S=\pi_2\mathbb S=\mathbb Z/2$, gives
$\pi_1M=\pi_2M=0$.  Applying maps into $M$ to this cofiber sequence shows
that restriction along the unit $\eta:\mathbb S\to M$ is injective on
\[
 [M,M]\longrightarrow[\mathbb S,M]=\mathbb Z/5.
\]
Thus $5\operatorname{id}_M=0$.  The cofiber sequence
\[
 M\xrightarrow{5}M\xrightarrow{\eta\wedge1}M\wedge M
 \longrightarrow\Sigma M
\]
therefore permits an extension of $\operatorname{id}_M$ to a pairing
$\mu$.  Its restriction along $1\wedge\eta$ is also the identity:
these endomorphisms of $M$ agree after restriction along $\eta$.
Moreover, $[\Sigma M,M]=0$, by the same cofiber sequence and the two
vanishing homotopy groups.  The extension is consequently unique.
Smashing $5\operatorname{id}_M=0$ with any spectrum proves the assertion
about coefficients.  The products are induced by the usual tensor
pairings of $K$-theory spectra and this fixed pairing on $M$.
\end{proof}

For a base ring $R$ and a class $\beta\in K_2(R;5)$, let
\[
 b: G_i(X;5)\longrightarrow G_{i+2}(X;5)
\]
be left multiplication by the restriction of $\beta$, whenever $X$ is
an $R$-scheme.  We use the same notation for $K$-groups, and write
$b^2=b\circ b$.  In particular $b^2(x)=\beta\cdot(\beta\cdot x)$, with these parentheses. We use this fixed two-factor pairing without assuming an associative coefficient ring structure.

\begin{lemma}[Compatibility with localization]
\label{lem:bott-localization}
The operation $b$ commutes with the maps and boundary homomorphisms in
the $G$-theory localization sequences over $R$, including localization
at a generic point and arbitrary multiplicative localization in the Noetherian affine situations below. Under d\'evissage its action on a residue-field
term is multiplication by the restricted class $\beta$.  It is also
compatible with finite coherent pushforward and flat pullback over $R$.
\end{lemma}

\begin{proof}
Tensoring with a finitely generated projective $R$-module is exact on
coherent modules and preserves every specified support.  Tensor product
therefore gives compatible pairings by $K(R)$ on the supported, whole,
and localized categories.  These pairings respect the null composite
in the localization diagram and hence its cofiber sequence.  On a
d\'evissage category, tensoring is tensoring with the restriction to the
residue field.  The projection formula for a finite map and the usual
flat base-change identity are identities of these tensor functors.
For each term $B$ of this localization triangle, use the pairing
\[
 (K(R)\wedge M)\wedge(B\wedge M)
 \longrightarrow K(R)\wedge B\wedge M\wedge M
 \longrightarrow B\wedge M,
\]
where the last arrow uses the integral tensor action and the fixed
pairing on $M$. These pairings commute with all arrows of the triangle.
Precomposing with the representative
$\beta:\mathbb S^2\to K(R)\wedge M$ gives the asserted diagrams for
$b$; no integral lift of $\beta$ is required.  Its degree is even, so moving it across a boundary contributes
no sign. Passage to a generic point follows by filtered localization. These assertions concern localization boundaries; the coefficient connecting map $\partial_{\mathrm{coef}}$ is a different map.
\end{proof}

We will use the following precise instances of field Bott periodicity.
They concern multiplication by the specified coefficient class, rather
than an unspecified isomorphism between the groups.

\begin{lemma}[The field Bott maps]
\label{lem:bott}
Let $H$ be a perfect field of characteristic different from $5$
containing a primitive fifth root $\zeta_H$.  Suppose that
$\beta_H\in K_2(H;5)$ satisfies
$\partial_{\mathrm{coef}}\beta_H=[\zeta_H]$, and let $b$ denote its
multiplication operation.
\begin{enumerate}
 \item If $\operatorname{cd}_5H\leq3$, then
 \[
 b:K_{2r}(H;5)\xrightarrow{\ \sim\ }K_{2r+2}(H;5)
 \qquad(r\geq1).
 \]
 \item If $\operatorname{cd}_5H\leq2$, then
 \[
 b:K_{2r+1}(H;5)\xrightarrow{\ \sim\ }K_{2r+3}(H;5)
 \qquad(r\geq0).
 \]
 \item If $H$ is finite, then $b:K_i(H;5)\to K_{i+2}(H;5)$
 is an isomorphism for every $i\geq0$.
\end{enumerate}
Over an algebraically closed field of characteristic different from
$5$, the group in degree $2j\geq0$ is generated by $b^j(1)$, and the
odd-degree groups vanish. Here $1\in K_0(H;5)$ is the unit class,
$b^0(1)=1$, and $b^j$ denotes iterated left multiplication.
\end{lemma}

The arithmetic and geometric arguments use the following instances of
the lemma. For number fields and finite extensions of $\mathbb Q_5$,
the $5$-cohomological dimension is two; see
\cite[VI.\S7 and VI.8.2]{Weibel2013}. For a one-variable function field
over a finite extension of $\mathbb Q_5$, the cohomological-dimension
bound \cite[Lemma~59.95.4, Tag~\texttt{0F0T}]{StacksProject} gives
$5$-cohomological dimension at most $2+1=3$. In particular, the actual
multiplication maps are
\begin{equation}
\label{eq:bott-maps-used}
 \begin{aligned}
  b^2 &:K_1(H;5)\xrightarrow{\ \sim\ }K_5(H;5)
    &&\text{for the number and local fields used below},\\
  b^2 &:K_2(J;5)\xrightarrow{\ \sim\ }K_6(J;5)
    &&\text{for the function field of the generic curve}.
 \end{aligned}
\end{equation}
In each case the primitive root and the Bott class are obtained by
restriction from the indicated base. For a finite field $H$ satisfying
the lemma's hypotheses, its finite-field clause also gives
$b^2:K_0(H;5)\xrightarrow{\sim}K_4(H;5)$. In the arithmetic construction
this will raise the target of the valuation homomorphism from degree
zero to degree four.

\begin{proof}[Proof of Lemma~\ref{lem:bott}]
We use the corrected, strongly convergent motivic spectral sequence
with its finite-coefficient multiplicative structure:
\begin{equation}
\label{eq:motivic-ss}
 E_2^{p,q}=H^{p-q}(H,\mathbb Z/5(-q))
 \Longrightarrow K_{-p-q}(H;5).
\end{equation}
This form of the spectral sequence is recorded in
\cite[VI.4.2, Addendum~VI.4.2.1, and VI.4.4]{Weibel2013}.
For the corrected construction, see
\cite[Introduction and Theorem~6.1]{Suslin2003}
(the locators use the \emph{Trudy} pagination).
The product is the usual $K$-theory product on the abutment and the
motivic cup product on $E_2$
\cite[Proposition~16.1 and Theorem~16.2]{FriedlanderSuslin2002}.
The Moore-space comultiplication used there corresponds by stable
duality to a unital pairing on $M$.  Its coefficient product agrees with
ours by the uniqueness in Lemma~\ref{lem:moore-product}.

Write an $E_2$-term as $H^s(H,\mathbb Z/5(j))$, of total $K$-degree
$2j-s$.  Terms of negative weight vanish. For $j\geq0$, the defining
motivic complex
$C_*\mathbb Z_{\mathrm{tr}}(\mathbb G_m^{\wedge j})[-j]$
is concentrated in cohomological degrees at most $j$. Global sections
on the Nisnevich site of a field are exact, since every covering of
the field point has a splitting, and derived reduction modulo five does not
increase this upper degree. Thus the field motivic group vanishes
when $s>j$. For $s\leq j$, the norm-residue and
Beilinson--Lichtenbaum comparison theorems identify it with
\begin{equation}
\label{eq:motivic-etale}
 H^s_{\mathrm{et}}(H,\mu_5^{\otimes j})
\end{equation}
\cite[Theorems~6.16--6.17]{Voevodsky2011}.
These are the Rost--Voevodsky norm-residue theorem and its motivic
comparison consequence, with the completion of the norm-residue
argument in \cite{Weibel2009}; the earlier motivic formulation is
\cite{SuslinVoevodsky2000}.
In particular the terms of negative cohomological degree vanish.
Under the dimension bounds below, only finitely many terms contribute
to each total degree, so the resulting filtrations are finite.
In these coordinates a differential $d_\rho$, for $\rho\geq2$, goes
from $(s,j)$ to $(s+2\rho-1,j+\rho-1)$.

We first derive the algebraically closed case from this same spectral
sequence. If $\overline H$ is an algebraic closure of $H$, only the
terms with $s=0$ survive. In each degree $2j\geq0$ there is exactly one
term, $H^0(\overline H,\mu_5^{\otimes j})\cong\F_5$, and in odd
degree there is none. There are no possible differentials. Thus the
edge map is an isomorphism in every even degree, and
\[
 K_{2j}(\overline H;5)\cong\F_5,\qquad
 K_{2j+1}(\overline H;5)=0\qquad(j\geq0).
\]
The coefficient sequence surjects
$K_2(\overline H;5)$ onto $\overline H^\times[5]=\mu_5(\overline H)$.
Both groups have dimension one, so this boundary is an isomorphism.
Consequently the restriction of $\beta_H$ is nonzero, and so is its
edge image. Multiplicativity now shows that each $b^j(1)$ over
$\overline H$, computed by iterated left multiplication, has nonzero
edge image and generates the corresponding even group. This proves
the last assertion of the lemma without a
separate rigidity input.

By naturality, the edge image
$\alpha\in H^0(H,\mu_5)$ of $\beta_H$ is nonzero: restriction on this
$H^0$ group is injective, and its geometric image is the nonzero class
just computed. The edge image of $b^r(1)$ is
$\alpha^{\otimes r}$. Since $H$ contains $\mu_5$, this generates the
one-dimensional group $H^0(H,\mu_5^{\otimes r})$. Hence this whole
line consists of permanent cycles for every $r\geq1$. Only
$\alpha\ne0$ is needed; no choice of an identification of $\alpha$
with the specified primitive root is required.

Suppose first that $\operatorname{cd}_5H\leq3$.  In total degree
$2r$, the only possible terms are
\[
 H^0(H,\mu_5^{\otimes r}),\qquad
 H^2(H,\mu_5^{\otimes(r+1)}).
\]
There is no incoming differential to either term.  The only possible
outgoing differential from the first is $d_2$ to an $H^3$-term, and
it vanishes by permanence of the Bott power.  Every outgoing
differential from the second has cohomological degree at least $5$
and is zero.  We obtain a two-step filtration, expressed as
\begin{equation}
\label{eq:even-field-filtration}
 0\longrightarrow H^2(H,\mu_5^{\otimes(r+1)})
 \longrightarrow K_{2r}(H;5)
 \longrightarrow H^0(H,\mu_5^{\otimes r})
 \longrightarrow0.
\end{equation}
Multiplication by $\beta_H$ preserves these filtrations and acts on
both graded pieces by tensoring with $\alpha$.  Each such map is an
isomorphism, proving (1).

If $\operatorname{cd}_5H\leq2$, the sole term in total degree
$2r+1$ is $H^1(H,\mu_5^{\otimes(r+1)})$, with no possible nonzero
differential.  Multiplication by $\beta_H$ again acts by tensoring
with $\alpha$.  This proves (2).  For a finite field, the same
calculation applies in every positive degree since its
$5$-cohomological dimension is one.  In degree zero, the class $1$
generates $K_0(H;5)$ and $b(1)=\beta_H$ generates
$K_2(H;5)=H^0(H,\mu_5)$.  This proves (3).
\end{proof}
==== END An-Integral-Counterexample-to-Gerstens-Conjecture//coefficients.tex ====

==== BEGIN An-Integral-Counterexample-to-Gerstens-Conjecture//geometry.tex ====
\section{An elliptic quotient of the Fermat quartic}
\label{sec:finite-geometry}

The specialization criterion asks for a divisor-class homomorphism
that kills the omitted points but detects a weighted sum of the
chosen specializations. We construct it from a degree-two quotient
of the Fermat quartic. The square of the cyclotomic character takes
values $\pm1$; the quotient will turn precisely these signs into
elliptic negation.

Set
\[
    k_0=\F_{5^4},\qquad k=\F_{5^8}.
\]
Over $\F_5$, let $D$ be the projective plane curve
\[
    X^4+Y^4=Z^4,
\]
and let $E$ be the smooth projective model of
\[
    w^2=s^4+1.
\]
The partial derivatives show that $D$ is smooth, including after any
extension of the ground field. It is geometrically integral: distinct
projective plane components would intersect and produce a singular point.
The polynomial $s^4+1$ has four simple roots. At infinity the coordinates
$q=1/s$ and $r=w/s^2$ give $r^2=1+q^4$, with two unramified points
$r=\pm1$ above $q=0$. Thus the double cover $E\to\mathbf P^1_s$ has
exactly four branch points; the degree-two Hurwitz formula gives
genus one.

There is a finite morphism of degree two
\begin{equation}
\label{eq:elliptic-quotient}
    h:D\longrightarrow E,
    \qquad s=\frac XY,\qquad w=\frac{Z^2}{Y^2}.
\end{equation}
This is the familiar passage from a diagonal quartic to a genus-one
quotient by a sign involution; related quotient constructions appear
in \cite[\S1, equations~(1.1)--(1.3)]{FlynnWetherell2001}.
We verify the particular map needed here. On $Y\ne0$, write $v=Z/Y$. The extension of function fields is
obtained by adjoining $v$ with $v^2=w$; the nontrivial involution
$v\mapsto-v$ proves that its degree is two. The resulting rational map
extends to the smooth projective curves and is finite. It maps the four
points of $D$ on $Z=0$ to the four branch points of $E\to\mathbf P^1$.

The eighth roots of unity lie in $k_0$, since $8\mid(5^4-1)$.
Thus all four branch points are $k_0$-rational. Choose one, denoted by $o$,
as the origin of $E_{k_0}$. Every fiber of $E\to\mathbf P^1$ is linearly
equivalent to $2(o)$, so the involution $w\mapsto-w$ is negation for this
group law, using the identification of a degree-zero divisor class
with its point on a genus-one curve with chosen origin
\cite[Proposition~III.3.4]{Silverman2009}. In particular, every branch point represents a point of
$E[2](k_0)$.

We now construct the promised homomorphism on divisor classes. Write
$D_k$ and $E_k$ for the base changes to $k$, and continue to denote the
quotient map by $h:D_k\to E_k$. The origin $o\in E(k_0)$ defines the
group law on $E_k$. Put
\[
 Z_\infty=D_k\cap\{Z=0\}.
\]
Define
\begin{equation}\label{eq:elliptic-detector}
 \begin{split}
 \psi:\Pic(D_k)&\longrightarrow E(k)/5E(k),\\
 [D']&\longmapsto
 [h_*D'-\deg(D')\,(o)]\pmod{5E(k)}.
 \end{split}
\end{equation}
Every line bundle on the smooth curve $D_k$ has a nonzero rational
section and hence a representing divisor $D'$. On a closed point $r$,
finite divisor pushforward is
$h_*(r)=[\kappa(r):\kappa(h(r))](h(r))$. Its degree is
$[\kappa(r):\kappa(h(r))][\kappa(h(r)):k]=[\kappa(r):k]$,
so it preserves degree over $k$.
It takes principal divisors to principal divisors by the norm formula
\cite[Tag \texttt{02RT}]{StacksProject}. The identification
$\Pic^0(E_k)\cong E(k)$ sends $[(R)-(o)]$ to $R$
\cite[Proposition~III.3.4 and Remark~III.3.5.1]{Silverman2009}.
Thus \eqref{eq:elliptic-detector} is a well-defined homomorphism on the
entire Picard group, including classes represented by nonrational closed
points. For a $k$-rational point $r$ of $D_k$, it gives
$\psi([r])=h(r)$ modulo $5E(k)$.

Each point of $Z_\infty$ is $k$-rational and maps to a branch point of
$E_k\to\mathbb P^1_k$. Its degree-zero class is two-torsion, and hence
zero modulo five. We have therefore proved
\begin{equation}\label{eq:detector-at-infinity}
 \psi([r])=0\qquad(r\in Z_\infty).
\end{equation}
It remains to find an orbit whose weighted image under $\psi$ is nonzero.

For $a\in\F_5^\times$, define an automorphism of $D$ by
\[
    g_a(X:Y:Z)=(aX:aY:Z).
\]
We will use the following orbit, with coefficients interpreted in the
group $E(k)/5E(k)$.

\begin{lemma}
\label{lem:orbit}
There are points $Q\in E(k_0)$ and $P\in D(k)$ such that
\[
    Q\notin5E(k),\qquad h(P)=Q,\qquad Z(P)\ne0,
\]
and
\begin{equation}
\label{eq:weighted-orbit}
 \begin{aligned}
    \sum_{a\in\F_5^\times}a^2\psi([g_aP])
      &=\sum_{a\in\F_5^\times}a^2h(g_aP)\\
      &=4Q\ne0\quad\text{in }E(k)/5E(k).
 \end{aligned}
\end{equation}
\end{lemma}

\begin{proof}
The underlying curve $E$ has four points over $\F_5$. At $s=0$ there are two affine
points, while at each nonzero $s$ the value $s^4+1=2$ is a nonsquare.
There are also two rational points at infinity, since the leading
coefficient of the quartic is a square. Choosing any rational point as
origin over $\F_5$, the Frobenius polynomial is therefore
$T^2-2T+5$, and its roots compute extension-field point counts
\cite[Theorem~V.2.3.1]{Silverman2009}. If $a_m$ is the sum of the $m$th powers of its roots, then
\[
    a_0=a_1=2,\qquad a_m=2a_{m-1}-5a_{m-2}.
\]
It follows that $a_4=-14$ and
\[
    \#E(k_0)=5^4+1-a_4=640.
\]
This count is independent of the choice of origin. Hence, with origin
$o$, the finite group $E(k_0)$ has a point $Q\notin5E(k_0)$.

Such a point remains nondivisible by $5$ over $k$. In fact, if $Q=5R$
with $R\in E(k)$, the trace for the quadratic extension $k/k_0$, using
the group law defined over $k_0$, would give
\[
    2Q=5\bigl(R+\operatorname{Frob}_{k_0}(R)\bigr)
        \quad\text{in }E(k_0).
\]
Multiplication by $2$ is invertible on $E(k_0)/5E(k_0)$, a contradiction.
Every branch point is $2$-torsion and therefore divisible by $5$, so
$Q$ is not a branch point. A fiber of the degree-two morphism $h$ has a
point over an extension of $k_0$ of degree at most two. Since $k$ is its
quadratic extension, we may choose $P\in D(k)$ over $Q$. The description
of $h$ on $Z=0$ gives $Z(P)\ne0$.

Finally, on the dense chart $Y\ne0$ the morphism $h\circ g_a$
fixes $s$ and multiplies $w$ by $a^{-2}$. It therefore agrees with
$h$ when $a^2=1$ and with $[-1]\circ h$ when $a^2=-1$. These are
identities of morphisms on all of $D$: they extend from the dense
chart because $D$ is integral and the target is separated. This
also covers the possibility that $Q$ is a point at infinity. Thus
\[
    h(g_aP)=
    \begin{cases}
        Q,&a^2=1,\\
        -Q,&a^2=-1.
    \end{cases}
\]
Each weighted summand in \eqref{eq:weighted-orbit} equals $Q$ modulo
$5$. Their sum is $4Q$, which is nonzero because multiplication by $4$
is invertible on $E(k)/5E(k)$.
\end{proof}

We fix $Q,P$ as in the lemma. The homomorphism $\psi$ kills
$Z_\infty$ by \eqref{eq:detector-at-infinity} and detects the weighted
orbit of $P$ by \eqref{eq:weighted-orbit}. The next section realizes
these four specializations as the conjugates of one horizontal divisor.
Curves and morphisms in residue characteristic will henceforth be
viewed over $k$.

\section{The regular local ring and its split divisor}
\label{sec:local-geometry}

We next realize the four-point orbit as the specialization of a split
horizontal divisor. The construction has three parts: choose a
cyclotomic uniformizer with controlled conjugates, map the ramified
regular local ring onto the cyclotomic valuation ring, and split this divisor
by a flat scalar extension.

Let $H_0/\Q_5$ be the unramified extension of degree eight, and let $V$
be its ring of integers, with residue field identified with $k$.
Retain the cyclotomic notation $\zeta$, $u=\zeta-1$,
$\Delta=\Gal(\Q(\zeta)/\Q)$, and
$\chi:\Delta\xrightarrow{\sim}\F_5^\times$.
Choose an embedding of $\Q(\zeta)$ into an algebraic closure of $H_0$,
and put
\[
    L=H_0(\zeta),\qquad \OO=\text{the ring of integers of }L.
\]
The translated cyclotomic equation
\[
    u^4+5u^3+10u^2+10u+5=0
\]
is Eisenstein over $V$. Consequently $L/H_0$ is totally ramified of
degree four, $u$ is a uniformizer of $\OO$, and the residue field is
$k$. Restriction identifies $\Gal(L/H_0)$ with $\Delta$.

There is a uniformizer $t\in\OO$ satisfying
\begin{equation}
\label{eq:cyclotomic-uniformizer}
    t^4=-5,\qquad \frac tu\equiv1\pmod u,\qquad
    \frac{\delta(t)}t\in\mu_4(V),\qquad
    \overline{\frac{\delta(t)}t}=\chi(\delta).
\end{equation}
To construct it, the cyclotomic equation gives
\[
    \frac{u^4}{-5}=1+2u+2u^2+u^3.
\]
The inverse of this unit has a fourth root congruent to $1$ modulo $u$
by Hensel's lemma; multiply $u$ by that root. The quotient
$\delta(t)/t$ is a fourth root of unity and has residue $\chi(\delta)$,
because $\delta(u)/u$ has that residue. All fourth roots of unity lie
in $V$, again by Hensel's lemma.

We also have
\begin{equation}
\label{eq:monogenic-dvr}
    V[t]=\OO.
\end{equation}
For completeness, $T^4+5$ is Eisenstein, so $V[t]$ is finite free over
$V$ and has fraction field $L$. Reduction modulo $5$ shows that it is
local, with maximal ideal $(5,t)=(t)$. Its dimension is one, hence it
is a DVR. It is therefore integrally closed in $L$, and every element
of $\OO$, being integral over $V$, belongs to $V[t]$.

Write $P=(\bar X_0:\bar Y_0:1)$. Hensel's lemma supplies lifts
$X_0,Y_0\in V$ with
\begin{equation}
\label{eq:lifted-point}
    X_0^4+Y_0^4=1.
\end{equation}
Indeed, at least one of $\bar X_0,\bar Y_0$ is nonzero, and the
corresponding partial derivative is a unit.

\begin{lemma}
\label{lem:local-ring}
The ring
\begin{equation}
\label{eq:counterexample-ring}
    A=\bigl(V[x,y]/(5+x^4+y^4)\bigr)_{(5,x,y)}
\end{equation}
is a regular local domain of dimension two. The assignment
\begin{equation}
\label{eq:closed-divisor-map}
    A\longrightarrow\OO,\qquad
    x\longmapsto tX_0,\qquad y\longmapsto tY_0
\end{equation}
is a surjective local homomorphism with nonzero height-one kernel.
\end{lemma}

\begin{proof}
The regular local ring $V[x,y]_{(5,x,y)}$ has dimension three. The
equation $5+x^4+y^4$ is congruent to $5$ modulo the square of its
maximal ideal, so it is part of a regular system of parameters.
Its quotient is regular local of dimension two and is a domain
\cite[Tags \texttt{00NQ} and \texttt{00NP}]{StacksProject}.

Equations \eqref{eq:cyclotomic-uniformizer} and
\eqref{eq:lifted-point} show that the assignment respects the defining
equation. Every denominator inverted in \eqref{eq:counterexample-ring}
has nonzero constant residue at $x=y=0$, hence maps to a unit of
$\OO$. Thus the homomorphism is defined and local. At least one of
$X_0,Y_0$ is a unit of $V$, so its image contains $t$. Surjectivity
follows from \eqref{eq:monogenic-dvr}.

The element $Y_0x-X_0y$ belongs to the kernel and has nonzero image in
$\mathfrak m_A/\mathfrak m_A^2$. In particular the kernel is nonzero.
It is a nonmaximal prime, since the quotient $\OO$ is a DVR, and
therefore has height one in the two-dimensional local domain $A$.
Since $Y_0x-X_0y$ is a regular parameter, it generates a height-one
prime and hence generates this kernel.
\end{proof}

We have obtained the closed divisor that will support the integral
class. To detect its pushforward, we now identify the generic curve
on which this support separates into conjugate points.

Let
\[
    \Dcal=\Proj\bigl(\OO[X,Y,Z]/(X^4+Y^4-Z^4)\bigr),
    \qquad C=\Dcal_L,
    \qquad T=\Spec(A\otimes_V L).
\]
The model $\Dcal$ is projective. Over an algebraic closure of either
$L$ or $k$, the three partial derivatives $4X^3,4Y^3,-4Z^3$ cannot
vanish at a projective point. Thus each geometric fiber is smooth of
dimension one. Each is also integral: smoothness makes it reduced,
while two distinct plane components would intersect and create a
singularity. The same Jacobian calculation on the relative affine charts,
where $4$ is invertible, proves smoothness over $\OO$
\cite[Tag \texttt{00TE}]{StacksProject}. Hence $\Dcal$ is regular.
Flatness over the DVR excludes a component supported in the special
fiber, and the generic fiber is integral; therefore $\Dcal$ is an
integral surface with special fiber $D_k$.

\begin{lemma}
\label{lem:splitting}
The scheme $T$ is regular, integral, Noetherian, and one-dimensional,
and its morphism to $\Spec A$ is flat. The substitutions
\[
    X/Z=x/t,\qquad Y/Z=y/t
\]
identify $T$ with the spectrum of a localization of the coordinate
ring of the chart $Z\ne0$ in $C$.

The base change to $T$ of the closed immersion
$\Spec\OO\to\Spec A$ in \eqref{eq:closed-divisor-map} is the disjoint
union of four $L$-rational closed points $p_\delta$, indexed by
$\delta\in\Delta$. Their coordinates are
\begin{equation}
\label{eq:split-points}
    p_\delta:
    \quad (X/Z,Y/Z)=
    \left(\frac{\delta(t)}tX_0,
          \frac{\delta(t)}tY_0\right).
\end{equation}
Their specializations on $\Dcal$ are $g_{\chi(\delta)}P$.
The factor indexed by $\delta$ is obtained from $\OO$ by the map
$a\mapsto\delta(a)$ into $L$.
\end{lemma}

\begin{proof}
The $V$-module $L$ is flat, since $V\to H_0$ is a localization and
$L$ is a vector space over $H_0$. The substitution $x=tX$, $y=tY$
transforms $5+x^4+y^4=0$ into $X^4+Y^4=1$. Tensoring and localizing
therefore give the asserted description of $T$. Its coordinate ring
is a localization of the coordinate ring of a smooth geometrically
integral affine curve over $L$; it is regular, integral, and
Noetherian, of dimension at most one.

Since $5$ is invertible in the second tensor factor, the coordinate
ring of the base-changed divisor is
\begin{equation}
\label{eq:divisor-tensor-splitting}
    \OO\otimes_V L
      =L\otimes_{H_0}L
      \xrightarrow{\sim}\prod_{\delta\in\Delta}L,
    \qquad a\otimes b\longmapsto(\delta(a)b)_\delta.
\end{equation}
Here $\OO[1/5]=L$, and the last isomorphism is the usual splitting
of a finite Galois extension. The images of $x$ and $y$ on the
$\delta$-factor are $\delta(t)X_0$ and $\delta(t)Y_0$, giving
\eqref{eq:split-points}. Every inverted denominator evaluates to a
unit of $\OO$, so these factors indeed define points of $T$.
They are distinct: the ratios $\delta(t)/t$ are the four different
fourth roots of unity, and at least one of $X_0,Y_0$ is a unit.
Thus $T$ has closed points and has dimension one.

The coordinates in \eqref{eq:split-points} are integral over $\OO$
and satisfy the Fermat equation, so they define sections of $\Dcal$.
Reducing \eqref{eq:cyclotomic-uniformizer} gives their stated
specializations. The last assertion follows directly from the
factor maps in \eqref{eq:divisor-tensor-splitting}.
\end{proof}
==== END An-Integral-Counterexample-to-Gerstens-Conjecture//geometry.tex ====

==== BEGIN An-Integral-Counterexample-to-Gerstens-Conjecture//introduction.tex ====
\section{Introduction}
\label{sec:introduction}

For a Noetherian regular local domain $A$ with fraction field $F$,
Gersten's conjecture asserts exactness of the canonical augmented residue
complex
\begin{equation}\label{eq:gersten}
 0\longrightarrow K_n(A)\longrightarrow K_n(F)
 \longrightarrow\bigoplus_{\operatorname{ht}\mathfrak p=1}
 K_{n-1}(\kappa(\mathfrak p))\longrightarrow\cdots .
\end{equation}
Here $n\geq0$, the groups are integral Quillen $K$-groups, and the
residues come from localization and the codimension filtration.
The first assertion is generic injectivity: a class that vanishes after
passage to the fraction field must already vanish on $A$.
We construct a counterexample to this assertion.

\begin{theorem}\label{thm:main}
Let $V$ be the ring of integers in the unramified extension of $\Q_5$ of
degree eight, and put
\[
 A=\bigl(V[x,y]/(5+x^4+y^4)\bigr)_{(5,x,y)}.
\]
Then $A$ is a Noetherian regular local domain of dimension two and
\[
 \ker\bigl(K_5(A)\longrightarrow K_5(\Frac A)\bigr)\ne0.
\]
Consequently the unrestricted integral Gersten conjecture has a negative
answer.
\end{theorem}

The maximal ideal of $A$ is $(x,y)$, and $5\in(x,y)^4$.
Its special fiber has dimension one and embedding dimension two at the
closed point. Thus the regular total ring is ramified and is not smooth
over $V$. This distinction is central both to the example and to its
relation with positive results on Gersten's conjecture.

\paragraph{History and related results.}
Gersten's questions about transfer and local higher $K$-theory
\cite{Gersten1973Problems}, together with his work on
exact sequences \cite{Gersten1973Exact}, place the conjecture in the
study of localization and residue maps. Quillen formulated it for
arbitrary regular local rings and proved it for
semilocalizations of finite-type algebras over a field at regular primes
\cite[\S7, Conjecture~5.10 and Theorem~5.11]{Quillen1973Higher}.
Panin established the full equicharacteristic case
\cite[Theorem~A]{Panin2003}.

In mixed characteristic, Gillet and Levine's relative theorem reduces
the smooth local case to the associated discrete valuation ring case
\cite{GilletLevine1987}. Geisser and Levine prove finite-coefficient
Gersten exactness for semilocal rings essentially smooth over a discrete
valuation ring, for every finite coefficient order, including powers of
the residue characteristic \cite[Theorem~8.2]{GeisserLevine2000}; their
text credits B.~Kahn for pointing out this consequence. Skalit proved
finite-coefficient Gersten exactness for arbitrary unramified regular
local rings, again for every finite coefficient order
\cite[Corollary~5.7]{Skalit2017}. In mixed characteristic $(0,p)$,
unramified means $p\notin\mathfrak m^2$; our ring is excluded since
$5\in\mathfrak m^4$. Version~1 of
Druzhinin's preprint states a result for essentially smooth local
Henselian schemes over a discrete valuation ring
\cite[Theorem~4.5]{Druzhinin2025}. The smoothness hypotheses exclude the
structure map $V\to A$ of Theorem~\ref{thm:main}.

Feld's finite-coefficient example is a close methodological predecessor
\cite[Theorem~1.1 and Corollary~5.1]{Feld2026}. It gives a degree-two
augmentation kernel modulo three at a ramified regular local ring whose
integral $K_2$ map is injective. Its supported pushforward, character
averaging, cyclotomic scalar extension, conjugate splitting and
divisor-class detection also organize our construction. In that example,
an integral lift of the coefficient kernel has nonzero but
three-divisible generic image, as follows from integral injectivity and
the coefficient exact sequence. An integral generic kernel therefore
requires an additional argument. Here a vanishing result for integral
$K_4$ permits a cyclotomic coefficient class to lift to integral $K_5$.
Two Bott factors contribute the square of the cyclotomic character,
whose signs are realized by an elliptic quotient of the Fermat quartic.

Our companion paper, \emph{An integral degree-three Gersten counterexample}
\cite[Theorem~1.1]{OpenAIDegreeThree2026}, constructs an integral
$K_3$ kernel at a different ramified ring: the local ring at the special
point $[0:0:1]$ of the cyclotomic quintic model
$(X-Z)^5-XY^4+Y^5+\zeta Z^5=0$ over the integers of $\Q_5(\zeta)$,
where $\zeta$ is a primitive fifth root of unity.
Its proof uses a Steinberg relation and a norm valuation test on coherent
$K$-theory \cite[Lemma~4.2, Proposition~3.1, and \S5]{OpenAIDegreeThree2026}.
Here the global arithmetic lift, smooth Picard specialization criterion
and quartic construction are proved directly. Neither proof uses results
of the other paper, and neither determines a smallest degree of failure.

Our proof detects the reduction modulo five of an integral class already
killed over the fraction field. It therefore also gives generic
noninjectivity with coefficients $\Z/5$ for the same ring. This detection
does not determine the order of the integral class or whether its
rationalization is nonzero. The conclusion concerns the unrestricted
integral conjecture and does not settle its rational-coefficient variant.

\paragraph{Construction and proof guide.}
A class pushed forward from a closed divisor of $\Spec A$ vanishes over
$\Frac A$. We must construct such a class integrally and prove that its
pushforward is nonzero. These tasks account for the two branches of the
proof.

For the integral construction, put $S=\Z[\zeta,1/5]$.
Section~\ref{sec:coefficients} supplies coefficient multiplication and
its exact field isomorphisms at the low degrees used here.
Section~\ref{sec:arithmetic} chooses $\beta\in K_2(S;5)$, writes $b$
for left multiplication by $\beta$, and constructs
\[
 c\in K_5(S),\qquad c\bmod5=b^2[\zeta-1]_5,
\]
where $b^2(x)=\beta\cdot(\beta\cdot x)$ and $[v]_5$ is the coefficient
$K_1$-class of a unit. The obstruction
is $K_4(S)[5]$, the subgroup killed by five. Brauer reciprocity, geometric
coefficient detection and finite generation prove that this subgroup
vanishes. Conjugating the reduction of $c$ multiplies its two Bott
factors by the square of the cyclotomic character.

Write $H_0=\Frac V$, $L=H_0(\zeta)$, and let $\OO$ be the valuation
ring of $L$. Integral localization supplies the following route:
\[
 c\in K_5(S)\longmapsto c_L\in K_5(L)
 \longleftarrow\widetilde c\in K_5(\OO).
\]
The leftward arrow is restriction from $\OO$, and $\widetilde c$ is a
chosen preimage; its existence uses integral finite-field $K_4$ vanishing.
Section~\ref{sec:local-geometry} constructs a surjection $A\to\OO$.
Its closed immersion $i$ then gives the integral class
$\gamma=i_*\widetilde c\in K_5(A)$, with zero generic restriction.

For detection, flat extension from $V$ to $L$ splits this divisor into
four rational points on the generic Fermat quartic $X^4+Y^4=Z^4$.
Section~\ref{sec:specialization} proves a general criterion on an arbitrary
localization of a smooth affine curve. A hypothetical relation among
supported coefficient classes untwists, by two Bott operations, to
ordinary integral tame symbols. On a smooth proper arithmetic surface,
the coniveau identity $d_1^2=0$ makes their horizontal specialization a
principal divisor. A homomorphism from the entire special-fiber Picard
group can therefore test the proposed relation.

Section~\ref{sec:finite-geometry} constructs this test. The Fermat quartic
maps with degree two to the genus-one curve $w^2=s^4+1$. With a branch
point as origin, the four chosen specializations map to $Q$ or $-Q$,
with the same signs as the character square. Their weighted sum is
$4Q\ne0$ modulo five. Section~\ref{sec:main-proof} verifies all hypotheses
of the Picard criterion and proves that the coefficient reduction of
$\gamma$ remains nonzero after flat pullback. This detects the already
integral class $\gamma$.

Appendix~\ref{app:claimed-vanishing} treats a separate issue in
Mochizuki's version~8 preprint \cite{Mochizuki2020}. Its Theorem~5 would
annihilate the principal-divisor pushforward used here. We prove an
explicit failure of the strict functor assertion in Lemma~13(III), for
every nonzero regular parameter. That calculation concerns the supplied
proof construction, and by itself does not refute its theorem statement.
No argument in the kernel construction depends on the appendix.
==== END An-Integral-Counterexample-to-Gerstens-Conjecture//introduction.tex ====

==== BEGIN An-Integral-Counterexample-to-Gerstens-Conjecture//main-proof.tex ====
\section{The nonzero integral kernel class}
\label{sec:main-proof}

The proof now has all three inputs: the integral cyclotomic class,
the divisor $\Spec\OO\subset\Spec A$ and its four split points, and
the specialization criterion. We first push forward integrally to
obtain a class killed by generic restriction. We then detect its
coefficient reduction after flat pullback by the elliptic orbit.

\begin{proof}[Proof of Theorem~\ref{thm:main}]
Let $c\in K_5(S)$ be the class from Proposition~\ref{prop:integral-lift},
and write $c_L\in K_5(L)$ for its restriction along $S\to L$.
There is no map $S\to\OO$ in this construction, since $5$ is invertible
in $S$. Instead, integral localization for
the DVR $\OO$ gives
\[
 K_5(\OO)\longrightarrow K_5(L)\longrightarrow K_4(k)=0,
\]
where the last equality is Quillen's computation for finite fields
\cite{Quillen1972FiniteFields}; see also \cite[IV.1.13]{Weibel2013}.
The residue field here has characteristic five; the equality is the
integral finite-field calculation. Choose an integral lift
$\widetilde c\in K_5(\OO)$. The closed immersion
$i:\Spec\OO\hookrightarrow\Spec A$ of Lemma~\ref{lem:local-ring} defines
\[
 \gamma=i_*\widetilde c\in G_5(A)=K_5(A).
\]
The equality uses the canonical resolution isomorphism for the regular
ring $A$. Restriction to $\operatorname{Frac}A$ annihilates every coherent
module supported on $\Spec\OO$. Thus $\gamma$ is in the kernel of the
canonical augmentation, integrally.

To show that it is nonzero, make the flat base change $V\to L$. The square
\[
\begin{tikzcd}[column sep=large,row sep=large]
 \displaystyle\coprod_{\delta\in\Delta}\Spec L
       \arrow[r,"\coprod p_\delta"] \arrow[d]
       & T \arrow[d] \\
 \Spec\OO \arrow[r,"i"] & \Spec A
\end{tikzcd}
\]
is Cartesian by Lemma~\ref{lem:splitting}. To check flat base change with the point indices, put $B_T=A\otimes_VL$.
For every finitely generated $\OO$-module $M$, scalar extension gives
natural isomorphisms of $B_T$-modules
\begin{equation}\label{eq:module-base-change}
 \begin{aligned}
 B_T\otimes_A i_*M
 &\simeq(\OO\otimes_VL)\otimes_{\OO}M\\
 &\simeq\bigoplus_{\delta\in\Delta}L\otimes_{\OO,\delta}M.
 \end{aligned}
\end{equation}
The $\delta$-summand is a $B_T$-module through the quotient defining
$p_\delta$. These are exact functors: $B_T$ is flat over $A$, and
each map $\OO\to L$ is a localization followed by an automorphism.
Thus \eqref{eq:module-base-change} proves the required identity on
integral coherent $K$-theory.

Write $\iota:S\to L$ for the fixed embedding. Since $\widetilde c$
restricts to $c_L=\operatorname{res}_\iota c$, its image on the
$\delta$-factor is
\[
 \operatorname{res}_{\delta:\OO\to L}\widetilde c
 =\operatorname{res}_{\iota\circ\delta}c
 =\operatorname{res}_{\iota}(\delta c).
\]
The equality uses the specified identification
$\Delta=\Gal(L/H_0)$ and requires no equivariance of the chosen DVR
preimage. Consequently, still integrally,
\[
 \gamma|_T=\sum_{\delta\in\Delta}(p_\delta)_*
 \operatorname{res}_\iota(\delta c).
\]
Naturality of coefficient reduction and
Proposition~\ref{prop:integral-lift} now give
\begin{equation}\label{eq:gamma-split}
 \overline{\gamma|_T}
 =\sum_{\delta\in\Delta}(p_\delta)_*
       \bigl(\chi(\delta)^2b^2[\delta u]_5\bigr)
       \quad\text{in }G_5(T;5).
\end{equation}
Here $b$ uses the fixed embedding $S\to L$, and the character factors
come from Proposition~\ref{prop:integral-lift}. No choice of the integral
lift $\widetilde c$ affects this formula.

Apply Proposition~\ref{prop:specialization} to the model $\Dcal/\OO$,
to $U=T$, and to the set $Z_\infty$ and homomorphism $\psi$ constructed
in Section~\ref{sec:finite-geometry}. The required
specialization condition follows directly from the definition of $A$:
let $q$ be any closed point of $C$ and let $\OO_q$ be the valuation
ring of its finite residue-field extension $\kappa(q)/L$. If $q$
specializes in the chart $Z\ne0$, the proper extension
$\Spec\OO_q\to\Dcal$ lies in that chart, since an open set containing
the closed point of a DVR contains its generic point. Hence $X/Z$ and
$Y/Z$ belong to $\OO_q$. The original coordinates
$x=tX/Z$ and $y=tY/Z$ have positive valuation. Every denominator inverted
outside $(5,x,y)$ has nonzero constant residue and hence is a unit at $q$.
Thus $q$ belongs to $T$.

All the hypotheses of Proposition~\ref{prop:specialization} are now
verified: the smooth model and localized chart come from
Section~\ref{sec:local-geometry}, the retained-point condition holds
for every closed point, and \eqref{eq:detector-at-infinity} says that
$\psi$ kills the entire set $Z_\infty$.
The points $p_\delta$ specialize to $g_{\chi(\delta)}P$ and are
$L$-rational. Every $\delta u$ is a uniformizer, so
$v_L(\delta u)=1$. The sum tested by
Proposition~\ref{prop:specialization} for~\eqref{eq:gamma-split} is therefore
\[
 \sum_{\delta\in\Delta}\chi(\delta)^2
        \psi([g_{\chi(\delta)}P])
 =4Q\ne0\quad\text{in }E(k)/5E(k),
\]
by Lemma~\ref{lem:orbit}. It follows that
$\overline{\gamma|_T}\ne0$. Hence $\gamma\ne0$ in the original integral
group, as required.
\end{proof}

\begin{remark}
Although the divisor $i$ is principal, its pushforward need not vanish on
all higher $K$-groups. The self-intersection formula gives $i^*i_*=0$ here;
the projection formula gives $i_*i^*=0$. Neither formula asserts $i_*=0$.
In particular, the proof does not assume that $\widetilde c$ is the
restriction of a class on $A$.
\end{remark}


The positions in the signed orbit matter for this detector. Forgetting
them and retaining only the four normalized valuations, all equal to one,
gives the scalar sum $1+4+4+1=0$ in $\F_5$. The Picard test retains the
points and instead gives $4Q$. This calculation describes that particular
scalar forgetting; it makes no assertion about all scalar functionals.
==== END An-Integral-Counterexample-to-Gerstens-Conjecture//main-proof.tex ====

==== BEGIN An-Integral-Counterexample-to-Gerstens-Conjecture//main.tex ====
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\hypersetup{pdftitle={An integral counterexample to Gersten's conjecture},pdfauthor={OpenAI},pdfsubject={Integral algebraic K-theory of a ramified regular local ring}}

\newtheorem{theorem}{Theorem}[section]
\newtheorem{proposition}[theorem]{Proposition}
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\title{An integral counterexample to Gersten's conjecture}
\author{OpenAI}
\date{September 25, 2026}

\begin{document}
\maketitle
\begin{center}\small\nolinkurl{An-Integral-Counterexample-to-Gerstens-Conjecture-September-25-2026}\end{center}
\begin{abstract}
We construct a two-dimensional ramified regular local ring $A$ of mixed
characteristic $(0,5)$ for which $K_5(A)\to K_5(\Frac A)$ has a nonzero
kernel. This disproves Gersten's conjecture in its unrestricted integral
form.
\end{abstract}

\input{sections/introduction}
\input{sections/coefficients}
\input{sections/arithmetic}
\input{sections/specialization}
\input{sections/geometry}
\input{sections/main-proof}

\appendix
\input{sections/claimed-vanishing}

\bibliographystyle{amsplain}
\bibliography{references}
\end{document}
==== END An-Integral-Counterexample-to-Gerstens-Conjecture//main.tex ====

==== BEGIN An-Integral-Counterexample-to-Gerstens-Conjecture//specialization.tex ====
\section{A specialization criterion for supported classes}
\label{sec:specialization}

Our goal is to detect a sum of coefficient classes pushed forward from
closed points of a curve.  A vanishing sum would be a localization
boundary.  We first use the two Bott maps of Lemma~\ref{lem:bott} to
express that boundary by ordinary tame symbols.  The divisor boundary on
a smooth model then imposes a relation on the special fiber, where a
homomorphism from its Picard group can detect the contradiction.

\begin{proposition}[Picard specialization]
\label{prop:specialization}
Let $L$ be a finite extension of $\Q_5$ containing $\mu_5$, with valuation
ring $R$ and residue field $k$.  Let $\mathcal X\to\Spec R$ be a smooth
projective morphism of relative dimension one with geometrically integral
fibers
\[
 C=\mathcal X_L,\qquad D=\mathcal X_k.
\]
For a closed point $q$ of $C$, let $r(q)$ be its specialization on $D$:
it is the image of the closed point under the unique extension of
$\Spec\kappa(q)\to\mathcal X$ to the valuation ring of $\kappa(q)$.
For a closed point $r$ of $D$, write $[r]\in\Pic(D)$ for the class of the
divisor $(r)$.  Let $m\geq0$ be an integer, and suppose that:
\begin{enumerate}
\item $U$ is the spectrum of an arbitrary localization of the coordinate
ring of a nonempty affine open of $C$, and $p_1,\ldots,p_m$ are distinct
$L$-rational closed points of $U$;
\item $Z$ is a finite set of closed points of $D$, and every closed point
$q$ of $C$ with $r(q)\notin Z$ belongs to $U$;
\item $B$ is an abelian group killed by $5$, and
$\psi:\Pic(D)\to B$ is a homomorphism satisfying $\psi([r])=0$ for every
$r\in Z$.
\end{enumerate}
Put $r_i=r(p_i)\in D(k)$.  Choose $\beta_L\in K_2(L;5)$ with
$\partial_{\mathrm{coef}}\beta_L=[\zeta]$ for a primitive fifth root of
unity $\zeta$, and let $b$ be multiplication by its restrictions, with
$b^2=b\circ b$.  Let $v_L$ be the normalized valuation of $L$.
For $a_i\in L^\times$ and $e_i\in\F_5$, the condition
\begin{equation}\label{eq:specialization-test}
 \sum_{i=1}^m e_i v_L(a_i)\,\psi([r_i])\ne0\quad\text{in }B
\end{equation}
implies
\begin{equation}\label{eq:specialization-conclusion}
 \sum_{i=1}^m (p_i)_*\bigl(e_i b^2[a_i]_5\bigr)
       \ne0\quad\text{in }G_5(U;5).
\end{equation}
Here $[a_i]_5$ is the class of $a_i$ in
$K_1(L;5)=L^\times/L^{\times5}$, and the scalars in
\eqref{eq:specialization-test} act through the natural $\F_5$-module
structure on $B$.
\end{proposition}

\begin{proof}
The sum in \eqref{eq:specialization-test} is zero when $m=0$, so its
nonvanishing implies $m>0$.  Suppose, for a contradiction, that the sum in
\eqref{eq:specialization-conclusion} is zero.  We first turn this assumed
relation into prescribed ordinary tame symbols on the generic curve.

\smallskip\noindent\emph{Localization and Bott untwisting.}
Put $J=L(C)$ and $A_U=\Gamma(U,\mathcal O_U)$.  The ring $A_U$ is a
localization of the coordinate ring of a smooth affine curve.  It is
therefore a Noetherian regular domain of dimension at most one, with
fraction field $J$.  Since one of the $p_i$ is retained, its dimension is
one.  Every finitely generated torsion $A_U$-module is killed by a nonzero
element; the corresponding quotient of $A_U$ has dimension at most zero
and is Noetherian, hence Artinian.  Such a module has finite length, with simple
factors the residue fields of the retained closed points.  This argument
does not require $U$ to be semilocal or the multiplicative set to be finite.

Quillen localization for the torsion category, followed by d\'evissage and
the coefficient convention in Section~\ref{sec:coefficients}, gives the
exact sequence
\begin{equation}\label{eq:curve-localization}
 K_6(J;5)\longrightarrow
 \bigoplus_{q\in U^{(1)}}K_5(\kappa(q);5)
 \longrightarrow G_5(U;5).
\end{equation}
The second arrow is the sum of coherent pushforwards from the points; the
first is the vector of localization residues.  These are ordinary
localization and d\'evissage for a multiplicative set
\cite[\S4, Theorem~3 and Corollary~2; \S5, Theorems~4--5;
\S7, Proposition~3.2 and paragraph~3.4]{Quillen1973Higher}.
The index set of this direct sum can be infinite, but each vector has finite
support.  Every $q\in U^{(1)}$ is a closed point of
$C$, so $\kappa(q)$ is a finite extension of $L$.

Exactness now supplies $\xi\in K_6(J;5)$ whose residue at $p_i$ is
$e_i b^2[a_i]_5$ and whose other residues in $U$ are zero.  The field $J$
has $\operatorname{cd}_5J\leq3$, and each $\kappa(q)$ has
$\operatorname{cd}_5\kappa(q)=2$.  All these fields are perfect and contain
the same primitive root by restriction from $L$.  Thus Lemma~\ref{lem:bott}
applies to the actual multiplication operations used here: it makes
\[
 \begin{aligned}
 b^2 &:K_2(J;5)\longrightarrow K_6(J;5)
     &&\text{surjective},\\
 b^2 &:K_1(\kappa(q);5)\longrightarrow K_5(\kappa(q);5)
     &&\text{injective}.
 \end{aligned}
\]
Choose $\xi_0\in K_2(J;5)$ with $\xi=b^2\xi_0$.  Write
$\partial_q^{(5)}$ for a coefficient localization residue.  The
compatibility in Lemma~\ref{lem:bott-localization} gives
$\partial_q^{(5)}(b^2\xi_0)=b^2\partial_q^{(5)}\xi_0$, with no sign
because $b$ has even degree.  Injectivity on the residue fields then yields
\begin{equation}\label{eq:untwisted-residues}
 \partial_q^{(5)}\xi_0=
 \begin{cases}
 e_i[a_i]_5,&q=p_i,\\
 0,&q\in U^{(1)}\setminus\{p_1,\ldots,p_m\}.
 \end{cases}
\end{equation}

There is still a coefficient obstruction to representing $\xi_0$
integrally.  Its coefficient boundary belongs to
\[
 K_1(J)[5]=\mu_5(J)=\langle\zeta\rangle.
\]
The last equality holds because $L$ already contains all fifth roots of
unity.  Hence there is $\lambda\in\F_5$ with
$\partial_{\mathrm{coef}}\xi_0=\lambda[\zeta]$.  Let $\beta_U$ and
$\beta_J$ be the restrictions of $\beta_L$.  Naturality gives
$\partial_{\mathrm{coef}}\beta_J=[\zeta]$, while all localization residues
of $\beta_J$ on $U$ vanish because it extends as $\beta_U$.  This uses an
extension over the $L$-scheme $U$, not an extension over $R$.  The class
\[
 \eta=\xi_0-\lambda\beta_J\in K_2(J;5)
\]
therefore has zero coefficient boundary and the same residues as $\xi_0$.
The coefficient exact sequence in Section~\ref{sec:coefficients} supplies
an integral representative $z\in K_2(J)$ of $\eta$.  Write
$\partial_q z\in K_1(\kappa(q))=\kappa(q)^\times$ for its ordinary tame
symbol.  Reduction of localization residues commutes with coefficient
reduction, so \eqref{eq:untwisted-residues} becomes
\begin{equation}\label{eq:integral-tame-symbols}
 [\partial_q z]_5=
 \begin{cases}
 e_i[a_i]_5,&q=p_i,\\
 0,&q\in U^{(1)}\setminus\{p_1,\ldots,p_m\}.
 \end{cases}
\end{equation}
In particular, $\partial_q z$ is a fifth power in $\kappa(q)^\times$ at
every retained point other than the $p_i$.  We have obtained an ordinary
$K_2$-class with the required control of its tame symbols.  It remains to
derive the relation that these symbols must satisfy on $D$.

\smallskip\noindent\emph{The horizontal divisor.}
For each closed point $q$ of $C$, let $R_q$ be the valuation ring of the
unique extension of the valuation of $L$ to $\kappa(q)$, let $k_q$ be its
residue field, and let $v_q$ be its normalized valuation.  The extension
$\kappa(q)/L$ is finite, so $R_q$ is a DVR finite over $R$.  Properness
gives the extension
\[
 j_q:\Spec R_q\longrightarrow\mathcal X
\]
used to define $r(q)$.  This map is finite.  Indeed, its graph in
$\Spec R_q\times_R\mathcal X$ is a closed immersion because
$\mathcal X/R$ is separated, and the projection from this fiber product
to $\mathcal X$ is finite, being the base change of $\Spec R_q\to\Spec R$.
Set
\[
 f_q=[k_q:\kappa(r(q))].
\]

Smoothness over the regular DVR $R$ makes $\mathcal X$ a regular surface,
and its function field is $J$.  The codimension filtration for coherent
$K$-theory gives the following row of the coniveau complex:
\begin{equation}\label{eq:surface-complex}
 K_2(J)\longrightarrow
 \bigoplus_{s\in\mathcal X^{(1)}}\kappa(s)^\times
 \longrightarrow
 \bigoplus_{r\in\mathcal X^{(2)}}\Z.
\end{equation}
The assertion needed here is only that the composite is zero: the first
differential $d_1$ of the coniveau spectral sequence satisfies $d_1^2=0$
\cite[\S7, Theorem~5.4 and equation~(5.5)]{Quillen1973Higher}.
There is exactly one vertical point in $\mathcal X^{(1)}$, the generic
point of the integral curve $D$.  Its residue
$\partial_Dz\in k(D)^\times$ contributes the principal divisor
$\operatorname{div}_D(\partial_Dz)$.  The other codimension-one points are
the closed points $q$ of $C$.

We compute their contributions without assuming that a horizontal closure
is normal.  On $\Spec R_q$, the localization boundary
\[
 K_1(\kappa(q))\longrightarrow K_0(k_q)=\Z
\]
is the normalized valuation $v_q$: a unit has boundary zero, and a
uniformizer $\pi_q$ has boundary $[R_q/\pi_qR_q]=[k_q]$
\cite[V.6.1.2 and V.(6.6)]{Weibel2013}.  Functoriality of coherent
localization for the finite map $j_q$ identifies the horizontal boundary
on $\mathcal X$ with the pushforward of this DVR boundary
\cite[\S7, paragraphs~2.7--2.8, Proposition~3.2 and paragraph~3.4]{Quillen1973Higher}.
At the generic point this pushforward is the identity on $\kappa(q)$.
At the closed point it sends $[k_q]$ to
$[k_q:\kappa(r(q))]\,[\kappa(r(q))]$.  Thus the contribution at $r(q)$ is
exactly $f_qv_q(\partial_qz)$.  There is no additional ramification
factor: the boundary already uses the normalized valuation on
$\kappa(q)$, and the closed-point pushforward contributes only the residue
degree $f_q$.

Define the horizontal divisor
\begin{equation}\label{eq:horizontal-divisor}
 H(z)=\sum_{q\in C^{(1)}}f_qv_q(\partial_qz)\,(r(q)).
\end{equation}
The sum is finite because the first arrow of \eqref{eq:surface-complex}
takes values in a direct sum.  All codimension-two points of the proper
surface lie on $D$.  The identity $d_1^2z=0$ therefore says, as an equality
of divisors on $D$,
\[
 H(z)+\operatorname{div}_D(\partial_Dz)=0.
\]
Consequently
\begin{equation}\label{eq:horizontal-pic-zero}
 [H(z)]=0\quad\text{in }\Pic(D).
\end{equation}

\smallskip\noindent\emph{The Picard test.}
Apply $\psi$ to \eqref{eq:horizontal-pic-zero}.  If $r(q)\in Z$, the
corresponding summand vanishes because $\psi([r(q)])=0$.  If
$r(q)\notin Z$, the hypothesis on $U$ retains $q$.  For such a point other
than the $p_i$, \eqref{eq:integral-tame-symbols} makes
$v_q(\partial_qz)$ divisible by five, so its contribution also vanishes in
$B$.  Finally, $\kappa(p_i)=L$, $k_{p_i}=k=\kappa(r_i)$, and hence
$f_{p_i}=1$ and $v_{p_i}=v_L$.  The remaining congruence in
\eqref{eq:integral-tame-symbols} gives
\[
 v_{p_i}(\partial_{p_i}z)\equiv e_i v_L(a_i)\pmod5.
\]
The identity $\psi([H(z)])=0$ is therefore
\[
 \sum_{i=1}^m e_i v_L(a_i)\,\psi([r_i])=0\quad\text{in }B,
\]
contrary to \eqref{eq:specialization-test}.  This proves
\eqref{eq:specialization-conclusion}.
\end{proof}

\begin{remark}[No Gersten exactness assumption]
\label{rem:no-gersten}
The exactness in \eqref{eq:curve-localization} comes from the torsion-category
localization above, valid for the arbitrary localization $U$.
On the mixed-characteristic surface we use only $d_1^2=0$ in
\eqref{eq:surface-complex}; no Gersten injectivity or exactness is assumed.
\end{remark}
==== END An-Integral-Counterexample-to-Gerstens-Conjecture//specialization.tex ====

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==== END An-Integral-Counterexample-to-Gerstens-Conjecture//references.bib ====