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\hypersetup{pdftitle={Graded absolute integral closures and Picard groups},
 pdfsubject={Bounded graded bases and faithfully flat trivialization of invertible modules over polynomial rings}}
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\title[Graded integral closures and Picard groups]{Graded absolute integral closures\\and Picard groups}
\author{}
\date{}
\begin{document}
\raggedbottom
\begin{abstract}
We deduce from Bhatt's graded Cohen--Macaulay and vanishing results that
an integer-graded absolute integral closure of a weighted polynomial ring
over $\Z_p$, modulo $p$, is free with basis degrees smaller than the sum
of the weights. A direct $p$-adic expansion turns this degree bound into
a uniform solution of the cusp relation modulo nilpotents. Consequently
every ring $R$ admits a faithfully flat algebra with seminormal reduction,
and every invertible $R[T]$-module becomes free after a faithfully flat
extension of $R$, answering a question of Drinfeld.
\end{abstract}
\maketitle

\section{The Picard statement}

Vladimir Drinfeld asked whether every invertible $R[T]$-module becomes
trivial after a faithfully flat extension of the coefficient ring $R$;
see \cite[\S3.1, Question following Corollary~3.4]{BMV}. The
$p$-nilpotent case is proved in \cite[Corollary~3.4]{BMV}.
We establish the following stronger statement.

\begin{theorem}\label{thm:main}
For every commutative ring $R$, there is a faithfully flat $R$-algebra $S$
whose reduction is seminormal. In particular,
\[
 \Pic(S)\xrightarrow{\ \sim\ }\Pic(S[T]).
\]
Consequently, every invertible $R[T]$-module becomes free after a faithfully
flat extension of $R$.
\end{theorem}

The trivializing extension cannot in general be chosen ind-fppf. Gabber's
counterexample, recorded by Rosengarten \cite[Remark~2.2.16]{Rosengarten}
and recalled in \cite[\S3.1]{BMV}, gives an invertible $R[T]$-module that
does not become free fppf locally on $R$. The same holds
ind-fppf locally: a trivialization over a filtered colimit of fppf
algebras descends to a finite stage, since invertible modules are
finitely presented \cite[Tag~05LI]{Stacks}.

The proof uses a bounded graded basis theorem over $\Z_p$, separately for
each prime $p$. Only its reduction modulo $p$ is needed; completion is
handled by a single convergent expansion. The passage from seminormality
to the Picard statement is the Traverso--Swan theorem
\cite{Traverso,Swan}, in the form \cite[Theorem~I.3.11]{Weibel}.

For a nonnegatively integer-graded characteristic-zero domain $B$, write
$\grplus B$ for the union of its finite nonnegatively integer-graded domain
extensions inside a fixed algebraic closure of $\Frac(B)$. Degree-zero
extensions are allowed. These rings form a directed system: the Euler
derivation extends uniquely to the algebraic closure, so the gradings
agree on intersections and on composita. In particular,
\begin{equation}\label{eq:invariance}
 \grplus B=\grplus H\qquad
 \text{if }B\subset H\text{ is finite and graded}.
\end{equation}
This follows by transitivity of finiteness and taking composita. All
closures below use compatible fields. We set $M(a)_j=M_{a+j}$ and write
$\widehat M$ for $p$-adic completion when $p$ is fixed. Numbering for \cite{Bhatt}
refers to arXiv:2008.08070v2.

\section{Bounded graded bases}

\begin{theorem}[Consequence of Bhatt's graded results]\label{thm:basis}
Fix a prime $p$, and set
\[
 P=\Z_p[x_1,\ldots,x_r],\qquad r\geq2,\qquad
 \deg(x_i)=w_i\in\Z_{>0},\qquad C=\grplus P.
\]
There is a nonempty homogeneous basis giving an isomorphism
\begin{equation}\label{eq:basis}
 C/pC\simeq\bigoplus_{\lambda\in\Lambda}(P/pP)(-s_\lambda),
 \qquad 0\leq s_\lambda<\sum_i w_i.
\end{equation}
Any homogeneous lifts $u_\lambda\in C$ of this basis induce an isomorphism
of completed $P$-modules
\begin{equation}\label{eq:completed-basis}
 \left(\bigoplus_{\lambda\in\Lambda}P(-s_\lambda)\right)^{\wedge}
 \xrightarrow{\ \sim\ }\widehat C,
 \qquad e_\lambda\longmapsto u_\lambda.
\end{equation}
In particular, reduction of this isomorphism gives compatible homogeneous
bases modulo every power of $p$.
\end{theorem}

\begin{proof}
First suppose $w_i=1$ for all $i$, and write the variables as $y_i$.
We explain the two inputs from Bhatt. Put $V=\Z_p$,
$X_0=\mathbf P^{r-1}_V$, and let $X^+\to X_0$ be an absolute integral
closure, with $\cL$ the pullback of $\cO_{X_0}(1)$. Then
\begin{equation}\label{eq:sections}
 C=\bigoplus_{m\geq0}H^0(X^+,\cL^{\otimes m}).
\end{equation}
For a finite graded domain extension $P\subset H$, the quotient
$H/(y_1,\ldots,y_r)H$ is a finite graded $V$-module, hence is bounded in
degree. Its positive-degree ideal is therefore nilpotent, so the opens
$D_+(y_i)$ cover $Y=\Proj H$. On each chart $y_i$ is an invertible element
of degree one. Homogeneous $P$-module generators $h_j$ of $H$ give generators
$h_j/y_i^{\deg h_j}$ of $(H[y_i^{-1}])_0$ over $(P[y_i^{-1}])_0$.
Thus $\pi:Y\to X_0$ is finite and surjective. The same trivializations give
$\widetilde{H(m)}=\pi^*\cO_{X_0}(m)$ and an injection of $H$ into its
section ring \cite[Tags~01M7 and~01MT]{Stacks}.
Conversely, the section ring of a finite surjective integral cover of $X_0$
is a finite graded $P$-algebra, by the projection formula and
\cite[Lemma~30.14.1(5), Tag~01YS]{Stacks}. These section rings are cofinal.
Their embeddings use the algebraic closure of $K(X_0)$ inside the chosen
closure of $\Frac(P)$ and the generic trivialization $y_1$.
Normalizing the covers preserves finiteness, since $V$ is excellent.
Taking their filtered colimit gives \eqref{eq:sections}; global sections
commute with this limit, as can be checked on the finite standard affine
cover. No generation-in-degree-one hypothesis on $H$ is needed.

Let $\overline V$ be an absolute integral closure of $V$. Replacing
$X_0$ by $X_0\times_V\overline V$ does not change $X^+$. In
\cite[Lemma~6.6 and Notation~6.7]{Bhatt}, the ring in \eqref{eq:sections}
is $R^{+,\mathrm{gr}}$, the integer-degree summand of the rationally graded
ring $R^{+,\mathrm{GR}}$. By \cite[Theorem~6.24 and Corollary~6.25]{Bhatt},
$R^{+,\mathrm{GR}}/p$ is Cohen--Macaulay over the regular ring $P/pP$.
Here regularity of systems of parameters and their powers, followed by
the Koszul direct-limit description of local cohomology, gives
cohomological Cohen--Macaulayness. Thus \cite[Lemma~2.9]{Bhatt} gives
flatness. Projection onto integer degrees is $P$-linear and remains
split modulo $p$, so $C/pC$ is flat. Moreover, \cite[Proposition~6.12]{Bhatt} gives
$H^q(X^+_{p=0},\cL^{\otimes m})=0$ for $q>0$ and $m\geq0$.
The affine-cover \v Cech comparison \cite[Tag~01XD]{Stacks}, applied to
the degree-$m$ part of the localization complex on the $y_i$, identifies
\[
 H^r_{(y_1,\ldots,y_r)}(C/pC)_m
 =H^{r-1}(X^+_{p=0},\cL^{\otimes m}).
\]
Indeed, on $U_i\subset X^+$ above $D_+(y_i)$ one has
$(C[y_i^{-1}])_m=\Gamma(U_i,\cL^{\otimes m})$, and similarly on
intersections. These opens are affine, so their sections commute with
reduction modulo $p$. No such assertion for global sections is assumed. As $r\geq2$,
we have obtained
\begin{equation}\label{eq:inputs}
 C/pC\text{ is }P/pP\text{-flat},\qquad
 \bigl(H^r_{(y_1,\ldots,y_r)}(C/pC)\bigr)_{\geq0}=0.
\end{equation}

A nonnegatively graded flat module $M$ over a positively graded polynomial
ring $Q$ over a field is graded free, without a finite-generation
hypothesis. To see this, lift a homogeneous basis of $M/Q_+M$ to obtain
$F\to M$ with $F$ graded free. Induction on degree makes this map
surjective. If $K$ is its kernel, flatness gives $K/Q_+K=0$, so the same
induction gives $K=0$.

Apply this observation to $C/pC$. A free summand with generator in degree
$s$ contributes the nonzero class $1/(y_1\cdots y_r)$ in degree $s-r$ to
top local cohomology. The vanishing in \eqref{eq:inputs} therefore forces
$s<r$. The basis is nonempty because $P\subset C$ is integral and
$C/pC\ne0$ by lying over.

For arbitrary weights, adjoin $y_i$ with $y_i^{w_i}=x_i$ and
$\deg(y_i)=1$. The ring $P'=V[y_1,\ldots,y_r]$ is finite free over $P$,
and \eqref{eq:invariance} identifies their graded closures. Restrict the
basis over $P'/pP'$ to $P/pP$, using the monomials
$y_1^{a_1}\cdots y_r^{a_r}$ with $0\leq a_i<w_i$. The resulting basis
degrees are at most
\[
 (r-1)+\sum_i(w_i-1)=\sum_iw_i-1,
\]
which proves \eqref{eq:basis}.

Finally, set $F=\bigoplus_\lambda P(-s_\lambda)$. The map $F\to C$
defined by the chosen lifts is an isomorphism after derived reduction
modulo $p$, since both modules are $p$-torsion-free. Its derived completion
is an isomorphism: the cofiber is derived complete and has zero reduction
modulo $p$, so derived Nakayama applies \cite[Tag~0G1U]{Stacks}.
By \cite[Lemma~5.7]{Bhatt}, applied to the one-element sequence $(p)$,
these derived completions are the ordinary completions and remain
$p$-torsion-free. This proves \eqref{eq:completed-basis} and its
reductions modulo $p^n$.
\end{proof}

\begin{corollary}\label{cor:completion}
In Theorem~\ref{thm:basis}, put $W=\sum_iw_i$ and $B=\widehat C$.
Then $B$ is flat over $P$, $B/pB=C/pC$, and
\begin{equation}\label{eq:membership}
 \operatorname{im}(C_{\geq W}\longrightarrow B)
 \subseteq(x_1,\ldots,x_r)B.
\end{equation}
\end{corollary}
\begin{proof}
Since $C$ is $p$-torsion-free and $C/pC$ is flat over $P/pP$, completed
flatness follows from \cite[Lemmas~5.7 and~5.15]{Bhatt}. The identity
$B/pB=C/pC$ is the usual reduction property of derived completion
\cite[\S15.93, Tag~091N]{Stacks}.

By \eqref{eq:basis}, $C/(p,x_1,\ldots,x_r)C$ has no components in
degrees $\geq W$. Given $z_0\in C_m$, $m\geq W$, choose successively
\[
 z_j=\sum_i x_i a_{i,j}+p z_{j+1},\qquad
 a_{i,j}\in C_{m-w_i},\quad z_{j+1}\in C_m.
\]
For every $N\geq1$ the exact identity
\[
 z_0=\sum_i x_i\left(\sum_{j=0}^{N-1}p^j a_{i,j}\right)+p^N z_N
\]
holds in $C$. In $B$, the remainder tends to zero and each coefficient
series converges. Thus $z_0=\sum_i x_i(\sum_{j\geq0}p^j a_{i,j})$,
proving \eqref{eq:membership} without any closedness assertion for the
parameter ideal. Only the degree bound modulo $p$ is used.
\end{proof}

\section{The universal cusp}

Set $A=\Z[a,b]$ and $\Delta=b^2-a^3$.

\begin{proposition}\label{prop:cusp}
There is a faithfully flat $A$-algebra $E$ containing an element $c$ such
that
\begin{equation}\label{eq:universal}
 (c^2-a)^6\in\Delta E,\qquad (c^3-b)^6\in\Delta E.
\end{equation}
\end{proposition}
\begin{proof}
For a prime $p$, put
\[
 A_p=\Z_p[a,b],\qquad P_p=\Z_p[a,t],\qquad
 D_p=\Z_p[a,b,t]/(b^2-a^3-t^6),
\]
with $\deg(a,b,t)=(2,3,1)$. The ring $D_p$ is a domain: $a^3+t^6$
has odd pole order at $a=\infty$ in $\Q_p(t)(a)$ and hence is not a square.
The decompositions $D_p=P_p\oplus bP_p=\bigoplus_{j=0}^5 A_pt^j$ show
that both polynomial inclusions are finite free. Thus \eqref{eq:invariance} gives
\[
 C_p:=\grplus{D_p}=\grplus{A_p}=\grplus{P_p}.
\]
Let $B_p=\widehat{C_p}$. Applied to $P_p$, the degree estimate
\eqref{eq:membership} gives $b\in(a,t)B_p$, since
$\deg b=3=\deg a+\deg t$. Write $b=ac_p+td_p$.
The same corollary makes $B_p$ flat over $P_p$, so multiplication by $a$
is injective on $B_p/tB_p$. In that quotient,
\[
 a^2(c_p^2-a)=b^2-a^3=0.
\]
Hence $c_p^2=a$ and $c_p^3=ac_p=b$ modulo $t$, giving
\begin{equation}\label{eq:errors}
 c_p^2-a,\ c_p^3-b\in tB_p.
\end{equation}

Applied to $A_p$, Corollary~\ref{cor:completion} shows that $B_p$ is
$A_p$-flat and that $B_p/pB_p$ is a nonzero free $A_p/pA_p$-module.
Since $\Z_p$ is flat over $\Z$ \cite[Tag~00MB]{Stacks}, the base change
$A_p=A\otimes_{\Z}\Z_p$ is $A$-flat, and so is every $B_p$. Set
\[
 E=\prod_{p\ \mathrm{prime}}B_p.
\]
Products of flat modules over the Noetherian ring $A$ are flat
\cite[Tag~05CZ]{Stacks}. Every maximal ideal of $A$ has positive residue
characteristic \cite[Tag~00GB]{Stacks}. If this characteristic is $p$,
then $A_p/pA_p=A/pA$ and
\[
 B_p/\mathfrak mB_p=(B_p/pB_p)\otimes_{A/pA}(A/\mathfrak m)\ne0
\]
by freeness of $B_p/pB_p$. Projection to this factor gives
$E/\mathfrak mE\ne0$.
Thus $E$ is faithfully flat. Finally take $c=(c_p)_p$; equations
\eqref{eq:errors} and $t^6=\Delta$ give \eqref{eq:universal} componentwise.
\end{proof}

\section{Seminormal reduction and Picard groups}

\begin{proof}[Proof of Theorem~\ref{thm:main}]
Starting with $R_0=R$, form $R_{i+1}$ by tensoring, over $R_i$, the base
changes of $A\to E$ for every map $A\to R_i$, equivalently for every
pair $(a,b)\in R_i^2$. Infinite tensor products mean filtered colimits
of finite tensor products. All these maps are faithfully flat, as is
$R\to S:=\varinjlim_iR_i$: flatness passes to filtered colimits, and a
finite equation putting $1$ in the extension of a proper ideal would
already occur at a finite stage.

A pair $\bar a,\bar b\in S_{\red}$ with $\bar b^2=\bar a^3$ lifts to
$a,b\in S$ coming from some $R_i$. At the next stage,
\eqref{eq:universal} supplies an element $c$. Choose $N\geq1$ with
$(b^2-a^3)^N=0$ in $S$; then
$(c^2-a)^{6N}=(c^3-b)^{6N}=0$ in $S$. Thus the image of $c$ is the
required root, and $S_{\red}$ is seminormal by the square-and-cube criterion
\cite[Section~I.3]{Weibel}.

The Traverso--Swan theorem \cite[Theorem~1]{Swan}, in the formulation
\cite[Theorem~I.3.11]{Weibel}, now gives $\Pic(S)\simeq\Pic(S[T])$;
this formulation applies to arbitrary rings in terms of their reductions. If $L$ is an invertible $R[T]$-module,
its pullback to $S[T]$ is consequently extended from
$M=(L/TL)\otimes_RS$. Choose a finite principal open cover $D(f_j)$ of
$\Spec S$ trivializing $M$. Then $R'=\prod_j S_{f_j}$ is faithfully flat
over $R$, and $L\otimes_{R[T]}R'[T]\simeq R'[T]$.
\end{proof}

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\end{document}