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\title{Reduction of the Gorenstein symmetry conjecture to finite rings}
\author{Qiyue Tang}
\date{}
\begin{document}
\maketitle
\begingroup
\renewcommand{\thefootnote}{}
\footnotetext{\textit{2020 Mathematics Subject Classification.}
Primary 16E10; Secondary 16E05, 16G10.\newline
\textit{Keywords.} Gorenstein symmetry conjecture, self-injective dimension,
Artin algebra, finite ring, specialization.}
\endgroup
\begin{abstract}
We prove that the Gorenstein symmetry conjecture for Artin algebras
is equivalent to its restriction to finite rings of prime-power
cardinality, preserving the finite self-injective dimension of a
counterexample. For finite-dimensional algebras over fields, the
conjecture reduces to finite prime fields. In characteristic zero,
a counterexample yields counterexamples of one fixed dimension over
every prime field outside a finite set.
\end{abstract}

\begingroup
\color{blue}
\section*{AI use and scholarly credit}
ChatGPT and Codex generated and revised the research arguments and
performed AI proof review. Codex prepared the manuscript and checked
the proof dependencies and references.

By posting this manuscript on this platform, I do not seek to assert ownership of the results or claim priority. I welcome anyone interested in preparing a finished account or checking the arguments; the eventual scholarly credit should go to those who undertake that work.
\par
\endgroup


\section{Introduction}

All algebras in this paper are nonzero, associative, and unital.
Coefficient rings are commutative and act centrally; ring maps preserve
the identity.
An \emph{Artin algebra} is an algebra finite as a module over a
commutative Artinian coefficient ring. A \emph{finite ring} has a
finite underlying set. For an Artin algebra \(A\), write
\({}_A A\) and \(A_A\) for its left and right regular modules,
and \(\id\) for injective dimension in the corresponding category
of all unital modules. The Gorenstein symmetry conjecture asserts
\begin{equation}\label{eq:gsc}
 \id({}_A A)<\infty\quad\Longleftrightarrow\quad
 \id(A_A)<\infty.
\end{equation}
When both dimensions are finite, they are equal.
The argument in Section~\ref{sec:certificate} applies to Artin
algebras as explained in Section~\ref{sec:dvr}.

The question belongs to the study of left--right symmetry in the
homological dimensions of Artinian rings. Hoshino's 1994 paper
\cite[Introduction]{Hoshino} formulates it in terms of the equality
of the two self-injective dimensions and recalls Zaks's result that
they agree whenever both are finite. It also records the affirmative
cases of self-injective dimension at most one and of finite
representation type. The conjecture appears as Conjecture~(13) in
Auslander, Reiten, and Smal\o{}'s list of homological conjectures
\cite[pp.~409--410]{ARS}. That list places it alongside the finitistic
dimension conjecture and records the implication from the latter.
Thus the finiteness of one self-injective dimension, rather than the
comparison of two already finite dimensions, is the central issue.

Subsequent work has studied the conjecture through decompositions
and equivalences of module and derived categories. Qin and Han
\cite[Section~5]{HQ} reduce it to \(2\)-derived-simple algebras:
these are finite-dimensional algebras whose derived categories admit
no nontrivial \(2\)-recollement by derived categories of algebras.
Here a \(2\)-recollement is a decomposition equipped with the
successive adjoint functors specified in \cite[arXiv version, Definition~2, p.~3]{HQ}.
Their reduction provides a smaller class on which to test the
conjecture. Zhang \cite[arXiv version, Example~2.2(3) and
Theorem~3.2]{Zhang} proves invariance under excellent extensions;
finite separable scalar extensions are among the examples.

More recently, Han, Qin, and Wang \cite[Theorem~3]{HQW} prove
invariance under an essentially surjective eventually homological
isomorphism between bounded derived categories of finite-dimensional
algebras. Essential surjectivity means that every object of the
target is isomorphic to an image. The eventual homological condition
requires the functor-induced maps on morphisms between shifted
objects to be isomorphisms in all sufficiently large degrees,
with the bound allowed to depend on the pair of objects
\cite[Definition~4]{HQW}. We use the version of 17 October 2026 for this statement.
These results describe categorical operations under which the
conjecture can be transferred.

A different reduction concerns the coefficient field. For the
finitistic dimension conjecture, Igusa \cite[Theorem~5.1]{Igusa}
compares algebras over the algebraic closure of a prime field with
algebras over finite fields of the same characteristic, keeping a
bound on Loewy length. In the Gorenstein symmetry problem, finite
homological data can also retain the failure of a finiteness
condition. Indeed, a finite projective resolution bounds one
self-injective dimension, and one suitable nonzero Ext group forces
the other to be infinite. This permits specialization of a
counterexample while preserving its finite self-injective dimension.

For a general Artin algebra, the coefficient ring may have mixed
characteristic, so field specialization alone does not cover the
problem. We treat this case using finite-length modules over a
Cohen ring and obtain the following reduction.

\begin{theorem}\label{thm:artin}
The following statements are equivalent.
\begin{enumerate}
\item Every Artin algebra satisfies the Gorenstein symmetry conjecture.
\item Every finite ring satisfies the Gorenstein symmetry conjecture.
\item Every finite ring of prime-power cardinality satisfies the
Gorenstein symmetry conjecture.
\end{enumerate}
More precisely, if an Artin algebra \(A\) satisfies
\(\id({}_A A)=d<\infty\) and \(\id(A_A)=\infty\), then a finite
ring \(A'\) of prime-power cardinality satisfies
\(\id({}_{A'}A')=d\) and \(\id(A'_{A'})=\infty\).
\end{theorem}

For a local center of residue characteristic \(p>0\), the finite
ring can be chosen with cardinality a power of \(p\) and the same
characteristic as the original algebra (Lemma~\ref{lem:local-center}).
A coefficient field or a Cohen ring is constructed in
Lemma~\ref{lem:artin-coefficients}. In the latter case, Smith matrices encode lengths in finite
homological complexes, and successive square-zero lifts realize the
required finite data over a finite coefficient ring
(Sections~\ref{sec:dvr}--\ref{sec:artin}).

For algebras over fields, the reduction takes the following form.

For a prime \(p\), let \(\Fp=\Z/p\Z\) be the field with \(p\) elements.
\begin{theorem}\label{thm:main}
The following statements are equivalent.
\begin{enumerate}
\item Every finite-dimensional algebra over every field satisfies
the Gorenstein symmetry conjecture.
\item Every finite-dimensional algebra over every finite field satisfies
the Gorenstein symmetry conjecture.
\item For every prime number \(p\), every finite-dimensional
\(\Fp\)-algebra satisfies the Gorenstein symmetry conjecture.
\end{enumerate}
For a fixed prime \(p\), the conjecture for all fields of characteristic
\(p\) is equivalent to the conjecture for \(\Fp\).
\end{theorem}

The proof records a counterexample by finitely many matrices.
A projective resolution of the right dual regular module gives a
finite upper bound \(d\ge0\) on the left self-injective dimension. A nonzero
Ext group in degree \(d+1\) for the left dual regular module forces
the right self-injective dimension to be infinite. We construct a
finite free coefficient model preserving both conditions and the
exact value \(d\) at every field-valued fiber
(Theorem~\ref{thm:model}). In characteristic zero this gives the
following arithmetic refinement.

For a positive integer \(N\), the ring
\(\Z[1/N]=\{a/N^j:a\in\Z,\ j\ge0\}\) is viewed as a subring of \(\Q\).
\begin{theorem}\label{thm:charzero}
Suppose that \(k\) has characteristic zero and that a finite-dimensional
\(k\)-algebra \(A\) satisfies
\[
 \id({}_A A)=d<\infty,\qquad \id(A_A)=\infty.
\]
There exist positive integers \(m,N\) and an associative unital
\(\Z[1/N]\)-algebra \(\mathcal C\), free of rank \(m\), with the
following property. Write
\(\mathcal C_{\Q}=\mathcal C\otimes_{\Z[1/N]}\Q\) and, for a prime
\(p\nmid N\), \(\mathcal C_p=\mathcal C\otimes_{\Z[1/N]}\Fp\).
Then
\[
\begin{aligned}
 \id({}_{\mathcal C_{\Q}}\mathcal C_{\Q})&=d,
 &\id((\mathcal C_{\Q})_{\mathcal C_{\Q}})&=\infty,\\
 \id({}_{\mathcal C_p}\mathcal C_p)&=d,
 &\id((\mathcal C_p)_{\mathcal C_p})&=\infty
 \quad(p\nmid N).
\end{aligned}
\]
In particular, every \(\mathcal C_p\) has the same dimension \(m\)
over \(\Fp\).
\end{theorem}

The integer \(m\) can differ from \(n=\dim_k A\), because restriction
of scalars from a number field increases vector-space dimension.
There is also a version retaining \(n\), with finite extensions of
\(\Fp\) of uniformly bounded degree
(Proposition~\ref{prop:bounded}). Theorems~\ref{thm:main}
and~\ref{thm:charzero} are proved in Sections~\ref{sec:fields}
and~\ref{sec:zero}, respectively.

\section{Finite homological certificates}\label{sec:certificate}

In Sections~\ref{sec:certificate}--\ref{sec:algorithm}, all modules
used in constructions are finite-dimensional over the indicated field. We write \(\pd_A M\) and \(\Ext_A^j(M,S)\) for
left \(A\)-modules; right modules are regarded as left modules over
the opposite algebra \(A^\op\), whose multiplication is reversed.
Projective and injective dimensions are the least lengths of projective
and injective resolutions, respectively, or infinity if no finite
resolution exists. We assign both dimensions zero to the zero module.
For \(j\ge0\),
\(\Ext_A^j(M,S)\) is the degree-\(j\) cohomology of
\(\Hom_A(P_\bullet,S)\) for a projective resolution \(P_\bullet\) of \(M\);
in particular \(\Ext_A^0(M,S)=\Hom_A(M,S)\).

For a finite-dimensional vector space \(V\), set \(D_kV=\Hom_k(V,k)\).
For a left module \(X\), its dual is a right module with
\((\lambda a)(x)=\lambda(ax)\); for a right module \(X\), its dual
is a left module with \((a\lambda)(x)=\lambda(xa)\), where
\(a\in A\), \(x\in X\), and \(\lambda\in D_kX\).
The dual of the regular module is an \(A\)-bimodule with action
\begin{equation}\label{eq:dual}
 (a\lambda b)(x)=\lambda(bxa)
 \qquad(a,b,x\in A,\ \lambda\in D_kA).
\end{equation}
For any left \(A\)-module \(X\), evaluation at \(1\in A\) gives
\[
 \Hom_A(X,D_kA)\cong\Hom_k(X,k),\qquad
 f\longmapsto\bigl(x\longmapsto f(x)(1)\bigr).
\]
The inverse sends a linear functional \(\ell:X\to k\) to the map
\(x\mapsto(a\mapsto\ell(ax))\). This map is \(A\)-linear by
\eqref{eq:dual}. Since vector-space duality is exact, \(D_kA\) is
injective in the category of all left modules. The right-handed
statement follows by replacing \(A\) by \(A^\op\).

Every finite left module \(X\) embeds into a finite direct sum of
copies of \(D_kA\): dualize a finite free right-module surjection
onto \(D_kX\) and identify \(D_kD_kX\) with \(X\) by evaluation.
Repeating this construction on the finite cokernels gives an
injective resolution with finite terms. A finite injective module
is a direct summand of such a finite sum, and its dual is therefore
projective. Conversely, the dual of a finite projective module is
a direct summand of a finite sum of copies of \(D_kA\), hence injective.
If \(\id_A X\le d\) in the full module category, dimension shifting
along this resolution makes its \(d\)-th cokernel injective:
its first Ext against any module is \(\Ext_A^{d+1}(-,X)=0\).
Here the zeroth cokernel means \(X\) itself. Dualizing the resulting
finite resolution, and conversely dualizing finite projective
resolutions, proves
\begin{equation}\label{eq:dimensions}
 \id({}_A A)=\pd_{A^\op}(D_kA),\qquad
 \id(A_A)=\pd_A(D_kA).
\end{equation}
Finite projective resolutions can be taken with finite terms:
successive finite free surjections have finite kernels, and the
kernel at the projective-dimension bound is projective by dimension
shifting.

\begin{lemma}\label{lem:top}
If a nonzero finite-dimensional left \(A\)-module \(M\) has
\(\pd_A M=e<\infty\), then \(\Ext_A^e(M,A)\ne0\).
\end{lemma}
\begin{proof}
For \(e=0\), choose a positive integer \(t\) and embed the nonzero
projective module \(M\) as a direct summand of \(A^t\).
At least one coordinate map \(M\to A\) is nonzero.
For \(e>0\), take a projective resolution of length \(e\) with finite
terms, and write \(u:P_e\to P_{e-1}\) for its first map.
Choose a positive integer \(t\) and \(A\)-linear maps \(\iota:P_e\to A^t\) and
\(\pi:A^t\to P_e\) with \(\pi\iota=1\).
If \(\Ext_A^e(M,A)=0\), every coordinate of \(\iota\) extends across
\(u\). The extensions form a map \(g:P_{e-1}\to A^t\) satisfying
\(gu=\iota\). Thus \((\pi g)u=1\), so \(u\) is a split monomorphism.
Write \(P_{e-1}=u(P_e)\oplus Q\). The next differential vanishes on
\(u(P_e)\), and its restriction to the projective summand \(Q\) is
injective onto the same image, by exactness. Replacing the first two
terms by \(Q\) gives a resolution of length \(e-1\), a contradiction.
\end{proof}

\begin{lemma}\label{lem:gap}
Let \(d\ge0\) be an integer. If \(\id({}_A A)\le d\), every
finite-dimensional left \(A\)-module of finite projective dimension
has projective dimension at most \(d\). Consequently,
\[
 \id({}_A A)\le d<\pd_A(D_kA)
 \quad\Longrightarrow\quad \id(A_A)=\infty.
\]
\end{lemma}
\begin{proof}
For a nonzero module of finite projective dimension \(e\),
Lemma~\ref{lem:top} gives a nonzero \(\Ext_A^e(M,A)\).
The assumed injective dimension forces \(e\le d\).
Apply this to \(D_kA\) if its left projective dimension is finite,
and then use \eqref{eq:dimensions}.
\end{proof}

In particular, if both regular injective dimensions are finite,
Lemma~\ref{lem:gap} and its opposite-algebra version bound each by the
other. This proves the equality used in the introduction.

\begin{lemma}\label{lem:witness}
Let \(d\ge0\) be an integer. If \(M\) is a finite-dimensional left
\(A\)-module and \(\pd_A M>d\), there is a
simple left \(A\)-module \(S\) with
\(\Ext_A^{d+1}(M,S)\ne0\). One can choose \(\dim_k S\le\dim_k A\).
\end{lemma}
\begin{proof}
Starting from \(\Omega^0M=M\), choose finite free surjections
\(F_j\to\Omega^jM\) and define \(\Omega^{j+1}M\) to be their kernels,
for \(0\le j<d\). Thus
\(0\to\Omega^{j+1}M\to F_j\to\Omega^jM\to0\) is exact.
For any left \(A\)-module \(T\), the long exact Ext sequence gives
\[
 \Ext_A^i(\Omega^{j+1}M,T)
 \cong\Ext_A^{i+1}(\Omega^jM,T)\qquad(i\ge1)
\]
because \(F_j\) is projective.
The module \(\Omega^dM\) is nonprojective, since otherwise this
sequence would give a projective resolution of \(M\) of length \(d\).
Choose a finite free surjection \(F\to\Omega^dM\) with kernel \(T\).
It does not split, so
\(\Ext_A^1(\Omega^dM,T)\ne0\): in the long exact sequence obtained
by applying \(\Hom_A(\Omega^dM,-)\), the connecting map sends the
identity to zero exactly when it lifts to \(F\).
Iterating the displayed isomorphism gives
\(\Ext_A^{d+1}(M,T)\ne0\), also for \(d=0\).

The finite module \(T\) has a composition series, obtained by taking
maximal proper submodules until the dimension reaches zero.
If \(\Ext_A^{d+1}(M,S)=0\) for every simple factor \(S\) of this
series, the long exact sequence in the second variable, applied
successively to its short exact sequences, would give
\(\Ext_A^{d+1}(M,T)=0\). Some simple factor therefore has the
required nonvanishing. For any nonzero \(s\in S\), the map
\(A\to S\), \(a\mapsto as\), is surjective, proving the dimension bound.
\end{proof}

For an integer \(d\ge0\), an algebra with left self-injective dimension at most
\(d\) and right self-injective dimension infinite is characterized by
two finite pieces of homological data:
\begin{equation}\label{eq:certificate}
\begin{gathered}
 0\longrightarrow P_d\longrightarrow\cdots\longrightarrow P_0
 \longrightarrow (D_kA)_A\longrightarrow0
 \quad\text{with each \(P_i\) projective},\\
 \Ext_A^{d+1}(D_kA,S)\ne0
 \quad\text{for some finite-dimensional left module \(S\)}.
\end{gathered}
\end{equation}
Existence follows from \eqref{eq:dimensions} and
Lemma~\ref{lem:witness}; sufficiency follows from Lemma~\ref{lem:gap}.
To preserve an exact finite value \(d>0\), we will also retain
\begin{equation}\label{eq:exact}
 \Ext_{A^\op}^{d}(D_kA,A)\ne0,
\end{equation}
which follows from Lemma~\ref{lem:top} applied over \(A^\op\).

\section{Specialization of the certificates}\label{sec:model}

We first describe the finite linear algebra used to preserve the Ext
conditions. For finite-dimensional left modules \(M,S\), the
unnormalized bar resolution has terms
\[
 B_j=A\otimes_k A^{\otimes_k j}\otimes_k M\quad(j\ge0),
 \qquad B_{-1}=M.
\]
The initial factor carries the left \(A\)-action, and
\(A^{\otimes_k0}=k\). The augmentation
\(\partial_0:B_0\to M\) is \(a_0\otimes m\mapsto a_0m\).
For \(j\ge1\), \(a_0,\ldots,a_j\in A\), and \(m\in M\), the differential
\(\partial_j:B_j\to B_{j-1}\) is
\begin{align*}
 \partial_j(a_0\otimes\cdots\otimes a_j\otimes m)
 &=\sum_{i=0}^{j-1}(-1)^i
 a_0\otimes\cdots\otimes a_i a_{i+1}\otimes\cdots
 \otimes a_j\otimes m\\
 &\quad+(-1)^j a_0\otimes\cdots\otimes a_{j-1}\otimes a_jm.
\end{align*}
For \(j\ge2\), the terms of \(\partial_{j-1}\partial_j\) pair
according to two original positions \(i<\ell\), where
\(0\le i<\ell\le j\). Multiplying first at \(\ell\) and then at
\(i\) has sign \((-1)^{i+\ell}\); multiplying first at \(i\) and
then at \(\ell-1\) has sign \((-1)^{i+\ell-1}\). The tensors agree
by associativity, including the last module action, so they cancel.
At the augmentation,
\(\partial_0\partial_1(a_0\otimes a_1\otimes m)
=(a_0a_1)m-a_0(a_1m)=0\).

Define \(k\)-linear maps \(h_j:B_j\to B_{j+1}\) by
\(h_j(x)=1\otimes x\), for \(j\ge-1\). For \(j\ge0\), the first
term of \(\partial_{j+1}(1\otimes x)\) is \(x\). Each remaining
term is \(1\) tensored with a term of \(\partial_jx\), with its sign
reversed. Hence
\[
 \partial_{j+1}h_j(x)=x-h_{j-1}\partial_j(x),\qquad
 \partial_0h_{-1}(m)=m.
\]
These identities give exactness: if \(\partial_jx=0\), then
\(x=\partial_{j+1}h_j(x)\), and the augmentation is surjective.
Each \(B_j\) is a finite free left \(A\)-module, with a basis coming
from a \(k\)-basis of \(A^{\otimes_k j}\otimes_k M\).
Thus \(B_\bullet\) is a projective resolution.

Applying \(\Hom_A(-,S)\) identifies its cochains with
\[
 C^j=\Hom_k(A^{\otimes_k j}\otimes_k M,S)\quad(j\ge0).
\]
The identification sends an \(A\)-linear map to its restriction to
\(1\otimes A^{\otimes_k j}\otimes_k M\); its inverse sends \(c\) to
\(a_0\otimes v\mapsto a_0c(v)\).
For \(c\in C^j\), its differential \(\delta^j:C^j\to C^{j+1}\) is
\begin{align}\label{eq:bar}
 (\delta^jc)(a_1,\ldots,a_{j+1},m)
 &=a_1c(a_2,\ldots,a_{j+1},m)\notag\\
 &\quad+\sum_{i=1}^j(-1)^i
 c(a_1,\ldots,a_i a_{i+1},\ldots,a_{j+1},m)\notag\\
 &\quad+(-1)^{j+1}c(a_1,\ldots,a_j,a_{j+1}m).
\end{align}
For \(j=0\) the sum is empty, so
\((\delta^0c)(a,m)=ac(m)-c(am)\).
For an integer \(q\ge0\), we use
\(H^q(C^\bullet)=\ker\delta^q/\im\delta^{q-1}\), with
\(\delta^{-1}=0\). It equals \(\Ext_A^q(M,S)\).
In fixed bases, all these matrices have entries given by integer
polynomials in the multiplication and action coefficients.

\begin{lemma}\label{lem:nonzero}
Let \(q\ge1\) be an integer, let \(R\) be a subring of a field \(k\),
and let \(C^{q-1}_R\to C^q_R\to C^{q+1}_R\) be a complex of finite free
\(R\)-modules. If its degree-\(q\) cohomology after
tensoring with \(k\) is nonzero, the same holds after tensoring with
any field \(F\) along a unital map \(R\to F\).
\end{lemma}
\begin{proof}
Let \(r,s\) be the ranks over \(k\) of the incoming and outgoing
differentials. Every minor of size \(r+1\), respectively \(s+1\),
is zero in \(k\), and therefore zero in \(R\).
The corresponding ranks over \(F\) are at most \(r,s\).
Here \(\rank_R C^q_R\) is the number of elements in a free basis
of the middle term. The complex identity remains valid over \(F\),
so the incoming image lies in the outgoing kernel. Rank--nullity gives
\begin{align*}
 \dim_F H^q(C^\bullet_R\otimes_R F)
 &\ge \rank_R C^q_R-r-s\\
 &=\dim_k H^q(C^\bullet_R\otimes_R k)>0.\qedhere
\end{align*}
\end{proof}

\begin{theorem}\label{thm:model}
Let \(A\) be an \(n\)-dimensional algebra over a field \(k\), and suppose
\[
 \id({}_A A)=d<\infty,\qquad \id(A_A)=\infty.
\]
There exist a subring \(R\subset k\), finitely generated as a unital
ring, and an \(R\)-algebra \(A_R\), free of rank \(n\), such that
\(A_R\otimes_R k\cong A\) and, for every unital map \(R\to F\) to a
field, the fiber \(A_F=A_R\otimes_R F\) satisfies
\[
 \dim_F A_F=n,\qquad
 \id({}_{A_F}A_F)=d,\qquad \id((A_F)_{A_F})=\infty.
\]
\end{theorem}
\begin{proof}
Fix a basis \(a_1=1,a_2,\ldots,a_n\) of \(A\).
Choose the resolution and the module \(S\) in
\eqref{eq:certificate}, and fix vector-space bases for the \(P_i\)
and \(S\), and use the dual basis to \(a_1,\ldots,a_n\) for
\(D_kA\). For every \(0\le i\le d\), choose a nonnegative integer
\(t_i\) and right \(A\)-linear maps
\begin{equation}\label{eq:split}
 P_i\xrightarrow{\iota_i}A^{t_i}
 \xrightarrow{\pi_i}P_i,\qquad \pi_i\iota_i=1_{P_i}.
\end{equation}

Regard the augmented resolution as a finite complex of vector spaces.
Write \(P_{-1}=D_kA\) and set \(P_i=0\) for \(i<-1\) or \(i>d\).
Let \(\partial_i:P_i\to P_{i-1}\)
be its maps for \(0\le i\le d\), and set
\(\partial_{-1}=0\) and \(\partial_{d+1}=0\).
Its exactness gives \(k\)-linear maps
\(h_i:P_i\to P_{i+1}\) for \(-1\le i<d\) such that
\begin{equation}\label{eq:contraction}
 \partial_{i+1}h_i+h_{i-1}\partial_i=1_{P_i}
 \qquad(-1\le i\le d).
\end{equation}
The maps outside the specified range are zero. To construct \(h_i\),
choose a complement to \(\ker\partial_i\) in \(P_i\).
Exactness identifies this complement, by \(\partial_i\), with
\(\ker\partial_{i-1}\). Define \(h_{i-1}\) on that kernel by the
inverse and set it to zero on the chosen complement. On each of the
two summands of \(P_i\), one of the two terms in
\eqref{eq:contraction} is the identity and the other is zero.
At the endpoints this reads
\(\partial_0h_{-1}=1_{D_kA}\) and
\(h_{d-1}\partial_d=1_{P_d}\).

Let \(R\) be the unital subring of \(k\) generated by the following
finite set of entries: the multiplication table of \(A\), the action
matrices of the \(P_i\) and \(S\), and the matrices of
\(\partial_i,\iota_i,\pi_i,h_i\).
On the free \(R\)-module with basis \(a_1,\ldots,a_n\), define
\(A_R\) using this multiplication table.
Replace each chosen vector space by the free \(R\)-module on its
basis, with actions and maps given by the selected matrices.
Associativity, unit laws, module laws, module linearity of
\(\partial_i,\iota_i,\pi_i\),
and the equations \(\partial_i\partial_{i+1}=0\),
\eqref{eq:split}, and \eqref{eq:contraction} are polynomial
identities in these entries with integer coefficients.
They hold over \(R\), since \(R\hookrightarrow k\) is injective.
The first basis vector of \(A_R\) is its identity, and every field
fiber is nonzero of dimension \(n\).

Define \(D_RA_R=\Hom_R(A_R,R)\) with the action
\eqref{eq:dual}. Its action matrices come from the multiplication table.
The finite free dual basis gives, for every \(R\to F\), an isomorphism
of \(A_F\)-bimodules
\begin{equation}\label{eq:dualbase}
 (D_RA_R)\otimes_R F\cong D_F(A_R\otimes_R F).
\end{equation}
This isomorphism sends \(\lambda\otimes u\) to
\(a\otimes v\mapsto uv\lambda(a)\), where \(\lambda(a)\) is mapped
from \(R\) to \(F\). The dual bases show that it is bijective, and
\eqref{eq:dual} shows compatibility with both actions.
Use \(D_RA_R\) as the augmentation term of the model.
Equations \eqref{eq:split} make each specialized \(P_i\) a direct
summand of a finite free right \(A_F\)-module.
Equations \eqref{eq:contraction} make the specialized augmented
complex exact as a complex of \(F\)-vector spaces, hence also as a
complex of \(A_F\)-modules. Therefore
\begin{equation}\label{eq:upperfiber}
 \pd_{A_F^\op}(D_FA_F)\le d.
\end{equation}

Let \(S_R\) be the free \(R\)-module with the fixed basis and action
matrices of \(S\). Apply the bar construction over \(R\) to the
left \(A_R\)-modules \(D_RA_R,S_R\).
Each cochain term is finite free over \(R\).
After tensoring with any field, it is the bar cochain complex of the
corresponding fiber modules, by the fixed bases and
\eqref{eq:dualbase}. Lemma~\ref{lem:nonzero} in degree \(d+1\) gives
\[
 \Ext_{A_F}^{d+1}(D_FA_F,S_R\otimes_R F)\ne0.
\]
Together with \eqref{eq:upperfiber}, Lemma~\ref{lem:gap} implies
\(\id((A_F)_{A_F})=\infty\).

If \(d>0\), apply the bar construction over \(A_R^\op\) with source
\(D_RA_R\) and target \(A_R\). Equation~\eqref{eq:exact} and
Lemma~\ref{lem:nonzero} imply
\(\Ext_{A_F^\op}^{d}(D_FA_F,A_F)\ne0\).
Thus \eqref{eq:upperfiber} is an equality.
If \(d=0\), it is already an equality because the dual module is
nonzero. Finally \eqref{eq:dimensions} gives the asserted left
injective dimension.
\end{proof}

\section{Finite fields and prime fields}\label{sec:fields}

The first assertion below is the weak Nullstellensatz of
Zariski \cite[Section~1, assertion H3]{Zariski}.
An element is integral over a commutative ring if it satisfies a
monic polynomial with coefficients in that ring.

\begin{lemma}\label{lem:residue}
A field finitely generated as an algebra over a field is a finite
field extension. A field finitely generated as a unital ring is finite.
\end{lemma}
\begin{proof}
First, if a field is integral over a subring, that subring is a field.
Indeed, for a nonzero element \(x\) of the subring, multiply a monic
relation of degree \(e\ge1\) for \(x^{-1}\) by \(x^{e-1}\).
The resulting equation expresses \(x^{-1}\) in the subring.

Let \(k\) be a field and \(E=k[c_1,\ldots,c_t]\) a field generated by
\(t\ge0\) elements as a \(k\)-algebra. After reordering the generators,
let \(c_1,\ldots,c_r\) be a maximal algebraically independent subset
over \(k\), where \(0\le r\le t\). Here
\(k(c_1,\ldots,c_r)\) is the fraction field of the polynomial ring
\(k[c_1,\ldots,c_r]\); both rings mean \(k\) when \(r=0\).
Each remaining generator is algebraic over this fraction field:
a polynomial dependence on \(c_1,\ldots,c_r,c_i\) gives a nonzero
polynomial in \(c_i\), since the first \(r\) elements are independent.
Clearing the denominators of
their monic equations gives a nonzero
\(f\in k[c_1,\ldots,c_r]\) such that these generators are integral
over \(k[c_1,\ldots,c_r,1/f]\).
If these equations have degrees \(e_i\ge1\), for \(r<i\le t\),
their monomials \(\prod_{i=r+1}^{t}c_i^{a_i}\), with
\(0\le a_i<e_i\), span the resulting algebra over that subring.
An empty product is \(1\). Thus
\(E=k[c_1,\ldots,c_r,1/f][c_{r+1},\ldots,c_t]\) is a finite
module over that subring. To check integrality, take module generators
\(u_1=1,u_2,\ldots,u_s\), where \(s\ge1\), and \(x\in E\).
Write \(xu_i=\sum_{j=1}^s b_{ij}u_j\), with
\(b_{ij}\in k[c_1,\ldots,c_r,1/f]\).
Let \(B=(b_{ij})_{1\le i,j\le s}\) and let \(I_s\) be the identity matrix.
Then \((xI_s-B)(u_1,\ldots,u_s)^{\mathsf T}=0\), where the superscript
denotes transpose. Multiplication by the adjugate gives
\(\det(xI_s-B)u_i=0\) for every \(i\).
Since \(u_1=1\), the monic polynomial \(\det(TI_s-B)\) in the
indeterminate \(T\) vanishes at \(x\).
The first paragraph therefore makes
\(k[c_1,\ldots,c_r,1/f]\) a field.

If \(r>0\), choose a monic irreducible polynomial
\(g\in k[c_1]\) which does not divide \(f\).
Such a polynomial exists: \(k[c_1]\) has infinitely many monic
irreducibles, since a product of any finite list plus \(1\) has
an irreducible factor outside the list; only finitely many can
divide a nonzero coefficient of \(f\) when it is written as a
polynomial in \(c_2,\ldots,c_r\).
The ideal generated by \(g\) in \(k[c_1,\ldots,c_r]\) is prime,
because its quotient is a polynomial ring over the field
\(k[c_1]/(g)\). If \(g\) were invertible after adjoining \(1/f\),
then \(gh=f^j\) for some \(h\in k[c_1,\ldots,c_r]\) and \(j\ge0\),
contrary to primality and \(g\nmid f\).
Thus \(r=0\). The finitely many algebraic generators now give
\([E:k]<\infty\).

For the second assertion, in positive characteristic \(p\) the field is a
finite extension of \(\Fp\). In characteristic zero, write such a
field as \(E=\Z[c_1,\ldots,c_t]\). It is finitely generated over
\(\Q\), hence a number field. Choose \(N\ge1\) so that the
coefficients of the monic minimal polynomials of all the \(c_i\)
lie in \(\Z[1/N]\). Then
\(E=\Z[1/N][c_1,\ldots,c_t]\) is integral over \(\Z[1/N]\).
The first paragraph would make \(\Z[1/N]\) a field, whereas a prime
not dividing \(N\) is not invertible in this ring.
\end{proof}

Thus every maximal-ideal residue field of a nonzero finitely
generated unital commutative ring is finite.

\begin{proposition}\label{prop:finite}
Suppose \(\dim_k A=n\), \(\id({}_A A)=d<\infty\), and
\(\id(A_A)=\infty\). There is an \(n\)-dimensional algebra \(B\)
over a finite field \(F\) such that
\[
 \id({}_B B)=d,\qquad \id(B_B)=\infty.
\]
If \(k\) has positive characteristic \(p\), \(F\) can be chosen
of characteristic \(p\).
\end{proposition}
\begin{proof}
Take \(R,A_R\) from Theorem~\ref{thm:model} and a maximal ideal
\(\mathfrak m\) of \(R\).
The ring \(R\) is a nonzero finitely generated \(\Z\)-algebra,
so \(F=R/\mathfrak m\) is finite. The fiber
\(B=A_R\otimes_R F\) has the asserted dimensions.
In characteristic \(p\), \(R\) is an \(\Fp\)-algebra, and its
residue fields have characteristic \(p\).
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:main}]
Statement (1) implies (2) and (3).
If (1) fails, replace a counterexample by its opposite algebra if
necessary to put its finite injective dimension on the left.
Proposition~\ref{prop:finite} then produces a counterexample over a
finite field, proving that (2) implies (1).

If \(F/\Fp\) is a finite field extension and \(B\) is a finite-dimensional
\(F\)-algebra, regard \(B\) as an \(\Fp\)-algebra by restricting
its central scalar action. The underlying unital ring, its regular
modules, and their injective dimensions are unchanged, whereas
\[
 \dim_{\Fp}B=[F:\Fp]\dim_F B.
\]
Thus (3) implies (2). For a fixed \(p\), use the
characteristic-preserving clause of Proposition~\ref{prop:finite}
and the same restriction of scalars. This proves the last assertion.
\end{proof}

\section{Characteristic zero}
\label{sec:zero}

\begin{lemma}\label{lem:number}
Under the hypotheses of Theorem~\ref{thm:charzero}, with
\(n=\dim_k A\), there is a finite field extension \(L/\Q\) and an
\(n\)-dimensional \(L\)-algebra \(B\) with
\(\id({}_B B)=d\) and \(\id(B_B)=\infty\).
The same ring, viewed as a \(\Q\)-algebra, has dimension \(n[L:\Q]\)
and the same two injective dimensions.
\end{lemma}
\begin{proof}
The coefficient ring \(R\) from Theorem~\ref{thm:model} has
characteristic zero. Its localization \(R\otimes_{\Z}\Q\) is nonzero
and finitely generated over \(\Q\). A maximal ideal gives a residue
field \(L\) finite over \(\Q\) by the weak Nullstellensatz.
Apply Theorem~\ref{thm:model} to \(R\to L\).
Restriction of scalars to \(\Q\) changes dimension by the factor
\([L:\Q]\) and leaves the ring and its regular modules unchanged.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:charzero}]
Let \(C\) be the \(\Q\)-algebra obtained in Lemma~\ref{lem:number}
and put \(m=\dim_{\Q}C\).
Apply the construction of Theorem~\ref{thm:model} anew to \(C\),
using \(D_{\Q}C\). This gives a finite collection of rational matrix
entries for its multiplication, modules, resolution, splitting maps,
and vector-space contraction.
Choose a positive integer \(N\) divisible by every denominator.
All these entries lie in \(\Z[1/N]\).

The same matrices define a free rank-\(m\) algebra \(\mathcal C\)
over \(\Z[1/N]\). The splitting and contraction identities hold in
this ring, since it embeds in \(\Q\). All minors above the ranks of
the relevant bar differentials also vanish in this ring.
The proof of Theorem~\ref{thm:model} therefore applies to each map
\(\Z[1/N]\to\Fp\), for \(p\nmid N\), preserving both the finite value
\(d\) and the infinite opposite dimension.
The rational fiber recovers \(C\).
\end{proof}

The original dimension can instead be retained over finite extensions
of the prime fields. The next statement also bounds their degrees.

\begin{proposition}\label{prop:bounded}
Under the hypotheses of Theorem~\ref{thm:charzero}, let \(n=\dim_k A\).
There exist positive integers \(r,N\) such that for every prime
\(p\nmid N\) there are a finite extension \(F/\Fp\) with
\([F:\Fp]\le r\) and an \(n\)-dimensional \(F\)-algebra \(B_p\)
satisfying
\[
 \id({}_{B_p}B_p)=d,\qquad \id((B_p)_{B_p})=\infty.
\]
\end{proposition}
\begin{proof}
Take \(L,B\) as in Lemma~\ref{lem:number} and set \(r=[L:\Q]\).
Select all the finite matrix data for \(B\) used in the proof of
Theorem~\ref{thm:model}. Choose a \(\Q\)-basis
\(b_1=1,b_2,\ldots,b_r\) of \(L\).
There is a positive integer \(N\) such that the multiplication
coefficients of this basis and the coordinates of every selected
matrix entry belong to \(\Z[1/N]\).
Consequently
\[
 O=\bigoplus_{i=1}^r\Z[1/N]b_i\ \subset L
\]
is a subring containing every selected entry and free of rank \(r\)
over \(\Z[1/N]\). The identities and vanishing minors used in
Theorem~\ref{thm:model} hold in \(O\).

For \(p\nmid N\), the commutative algebra \(O/pO\) is nonzero of
dimension \(r\) over \(\Fp\). A maximal ideal of this finite algebra
has residue field \(F\) with \([F:\Fp]\le r\).
Specializing the selected \(O\)-model of \(B\) along \(O\to F\)
gives the required \(n\)-dimensional algebra.
\end{proof}

\begin{corollary}\label{cor:infinite}
If the Gorenstein symmetry conjecture holds for every finite-dimensional
\(\Fp\)-algebra for infinitely many primes \(p\), then it holds for
every finite-dimensional algebra over every field of characteristic zero.
\end{corollary}
\begin{proof}
A characteristic-zero counterexample, after taking the opposite
algebra if necessary, would give a counterexample over every \(\Fp\)
with \(p\nmid N\), by Theorem~\ref{thm:charzero}.
An infinite set of primes contains one not dividing \(N\).
\end{proof}

\section{Finite detection over fields}\label{sec:algorithm}

The field reduction also has a computability consequence. A
semidecision procedure for failure means an algorithm which halts
with a counterexample if the conjecture fails, and never outputs a
counterexample if it holds. It need not halt in the latter case.

\begin{proposition}\label{prop:algorithm}
Failure of the Gorenstein symmetry conjecture for finite-dimensional
algebras over arbitrary fields is semidecidable.
\end{proposition}
\begin{proof}
Enumerate primes \(p\), positive integers \(n\), and nonnegative
integers \(d\), in such a way that every triple is eventually reached.
For each triple, enumerate the finitely many multiplication tables
on \(\Fp^n\) with the first basis vector prescribed as identity.
Associativity and the unit laws are finite equations, so their
validity is decidable.

For each resulting algebra \(A\), form the two actions on \(D_{\Fp}A\)
using \eqref{eq:dual}. For either side, compute \(d\) successive kernels
of finite free surjections. More explicitly, for a finite left module
\(M\) with vector-space basis \(m_1,\ldots,m_t\), use
\(A^t\to M\), \((a_i)\mapsto\sum_i a_im_i\).
Its kernel and its action matrices are obtained by linear algebra.
Use the right-handed version for right modules.
Set \(\Omega^0M=M\), and let \(\Omega^{j+1}M\) denote the kernel
at the next step, for \(0\le j<d\).

For a left module, \(\pd_A M\le d\) is equivalent to
projectivity of \(\Omega^dM\).
Indeed dimension shifting identifies
\(\Ext_A^1(\Omega^dM,X)\) with \(\Ext_A^{d+1}(M,X)\);
vanishing for all \(X\) characterizes projectivity, and the converse
follows by truncating the resolution.
To test a finite module \(K\) for projectivity, take the above finite
free surjection \(q:A^{\dim_{\Fp}K}\to K\).
It splits precisely when there is an \(A\)-linear map
\(s:K\to A^{\dim_{\Fp}K}\)
with \(qs=1_K\). Both module linearity and this last equation are
finite linear equations in the entries of \(s\), decidable over
\(\Fp\). For the zero module the zero section suffices.
Apply the same test over \(A^\op\) for right modules.

Halt if exactly one of the two dual modules has projective dimension
at most \(d\). Lemma~\ref{lem:gap}, on the appropriate side, certifies
that the other injective dimension is infinite.
If a counterexample exists over any field, Theorem~\ref{thm:main}
provides one over a prime finite field; the enumeration reaches its
multiplication table and a suitable \(d\).
\end{proof}

\section{Finite-length modules over a discrete valuation ring}\label{sec:dvr}

Fix a prime \(p\), a discrete valuation ring \(O\) of characteristic
zero with maximal ideal \(pO\), and an integer \(N\ge1\).
Write \(T=O/(p^N)\). For an \(O\)-module of finite length, \(\ell_O\)
denotes its composition length; its simple factors are copies of
\(O/pO\). A finitely generated \(T\)-module has a decomposition
\begin{equation}\label{eq:cyclic}
 L=\bigoplus_{j=1}^{r}O/(p^{e_j}),\qquad 1\le e_j\le N.
\end{equation}
The empty sum represents the zero module, and
\(\ell_O(L)=\sum_{j=1}^r e_j\).
These are the elementary divisors over the principal ideal domain
\(O\). They will allow us to retain lengths after changing coefficients.

\begin{lemma}\label{lem:chain-dual}
The functor \(D_T=\Hom_T(-,T)\) is an exact duality on finitely
generated \(T\)-modules. If \(A\) is a nonzero \(T\)-algebra finite
as a \(T\)-module, then
\[
 \id({}_A A)=\pd_{A^\op}(D_TA),\qquad
 \id(A_A)=\pd_A(D_TA).
\]
\end{lemma}
\begin{proof}
Every ideal of \(T\) is \(p^aT\) for \(0\le a\le N\).
A \(T\)-linear map \(p^aT\to T\) sends \(p^a\) to an element
annihilated by \(p^{N-a}\), hence to \(p^ab\) for some \(b\in T\).
It extends to multiplication by \(b\) on \(T\). Baer's criterion
therefore makes \(T\) injective over itself, so \(D_T\) is exact
on all \(T\)-modules.

For elements \(x=(x_j)\) and \(y=(y_j)\) of the module in
\eqref{eq:cyclic}, consider the pairing
\begin{equation}\label{eq:chain-pairing}
 \langle x,y\rangle=\sum_{j=1}^r p^{N-e_j}x_jy_j\pmod{p^N}
\end{equation}
It identifies a second copy of \(L\) with \(D_TL\).
Indeed the image of \(1\) under a homomorphism
\(O/(p^e)\to T\) is an arbitrary element of
\(p^{N-e}T\), which is uniquely \(p^{N-e}y\) for
\(y\in O/(p^e)\). This also proves that evaluation
\(L\to D_TD_TL\) is an isomorphism. Thus \(D_T\) is a duality.

Give \(D_TA\) the bimodule action in \eqref{eq:dual}, with values
in \(T\). For every left \(A\)-module \(X\), the same evaluation
and inverse maps as in Section~\ref{sec:certificate} give
\(\Hom_A(X,D_TA)\cong\Hom_T(X,T)\). Hence \(D_TA\) is
injective in the full module category. Dualizing a finite free
surjection onto \(D_TX\) embeds each finitely generated \(X\)
into a finite sum of copies of \(D_TA\). The cokernel is again
finitely generated, since \(A\) is finite over the Artinian ring
\(T\). Iteration gives an injective resolution with finitely
generated terms. Finitely generated injectives split off such sums,
so their duals are projective; duals of finitely generated projectives
are injective. If the injective dimension is at most \(d\), dimension
shifting makes the \(d\)-th cokernel injective, as its first Ext
against every module vanishes. Dualizing the truncated resolution
and, conversely, dualizing finite projective resolutions proves both
asserted equalities, including infinite values.
\end{proof}

The proofs of Lemmas~\ref{lem:top}--\ref{lem:witness} also apply to
finitely generated modules over an Artin algebra: finite free
surjections have finitely generated kernels, projectives split off
finite free modules, and composition series have finite length.
In Lemma~\ref{lem:witness}, the last dimension bound is specific to
field algebras. The Ext nonvanishing assertion uses only the
composition series. Consequently, for a finite \(T\)-algebra,
\begin{equation}\label{eq:chain-gap}
 \id({}_A A)\le d<\pd_A(D_TA)
 \quad\Longrightarrow\quad \id(A_A)=\infty.
\end{equation}
Here and below a finite \(T\)-algebra means finite as a module over
\(T\); its underlying set can be infinite.

\subsection{Matrices and lengths}

Choose decompositions \eqref{eq:cyclic} for each module in a finite
collection. An \(O\)-linear map
\[
 f:\bigoplus_{j=1}^{s}O/(p^{e_j})
   \longrightarrow\bigoplus_{i=1}^{r}O/(p^{b_i})
\]
is represented by a matrix \(F=(F_{ij})\) over \(O\), with entries
in row \(i\) read modulo \(p^{b_i}\). It is well defined precisely
when
\begin{equation}\label{eq:well-defined}
 p^{e_j}F_{ij}=p^{b_i}c_{ij}\qquad(1\le i\le r,\ 1\le j\le s)
\end{equation}
for suitable \(c_{ij}\in O\). Likewise an equality of maps can be
expressed by requiring each row of their matrix difference to be a
multiple of its target relation \(p^{b_i}\). Recording the quotients
turns these conditions into polynomial equalities over \(O\).

The matrix from \(O^{r+s}\) to \(O^r\) obtained by adjoining the
target relations to \(F\) is
\begin{equation}\label{eq:smith-presentation}
 Q_f=[\operatorname{diag}(p^{b_1},\ldots,p^{b_r})\mid F].
\end{equation}
Its cokernel is the cokernel of \(f\). The first \(r\) columns have
full rank over the fraction field of \(O\). Smith reduction therefore
gives invertible matrices \(U\) of size \(r\) and \(V\) of size
\(r+s\) such that
\begin{equation}\label{eq:smith}
 UQ_fV=[\operatorname{diag}(p^{a_1},\ldots,p^{a_r})\mid0],
 \qquad 0\le a_i\le N.
\end{equation}
The bound follows because the cokernel is killed by \(p^N\).
One obtains this diagonal form by choosing a nonzero entry of least
valuation, moving it to the first position, and clearing its row and
column: it divides every entry over a valuation ring. Repeat on the
remaining submatrix, and absorb unit factors by multiplying rows by
units. The resulting row and column operations are invertible.

Record \(U,V\), their inverses, and \eqref{eq:smith}.
Suppose that these coefficients and identities are realized over
\(T'=O'/(p^N)\), where \(O'\) is another discrete valuation ring
with uniformizer \(p\). Replacing each cyclic summand by
\(O'/(p^{e_j})\) defines the new modules and maps. Then
\begin{equation}\label{eq:lengths}
 \ell_{O'}(\operatorname{coker}f')=\sum_i a_i,\qquad
 \ell_{O'}(\im f')=\sum_i b_i-\sum_i a_i.
\end{equation}
Indeed the recorded inverses give invertible row and column operations
over \(T'\), and the cokernel of multiplication by \(p^{a_i}\) on
\(T'\) is \(O'/(p^{a_i})\). At \(a_i=0\) this summand is zero;
at \(a_i=N\) it is \(T'\). If the target is zero, both lengths
are zero and the matrices are empty.

For consecutive maps \(L_1\xrightarrow f L_2\xrightarrow g L_3\)
with \(gf=0\), length additivity gives
\begin{equation}\label{eq:homology-length}
 \ell_O(\ker g/\im f)
 =\ell_O(L_2)-\ell_O(\im g)-\ell_O(\im f).
\end{equation}
Thus the identities and Smith matrices for the two maps retain the
length of this cohomology after realization over \(T'\).
In particular they retain exactness and nonzero cohomology.
An initial zero map tests injectivity; a final zero map tests
surjectivity.

\subsection{Algebras, duals, and Hom complexes}

A multiplication on a module \eqref{eq:cyclic} is determined by the
products of its chosen cyclic generators. Well-definedness in each
argument, associativity, and the identity element are finitely many
congruences in these coefficients. Module actions, commuting left
and right actions, and linearity of specified maps are expressed in
the same way. Include witnesses as in \eqref{eq:well-defined} for
every congruence. These equations then remain valid after realization
over \(T'\).

Use the cyclic decomposition of \(A\) and \eqref{eq:chain-pairing}
for \(D_TA\). Record its two actions together with the equations
\[
 \langle a\lambda b,x\rangle=\langle\lambda,bxa\rangle
 \qquad(a,b,x\in A,\ \lambda\in D_TA).
\]
It suffices to impose these equations on the cyclic generators.
The pairing remains perfect over \(T'\), by the cyclic calculation
in Lemma~\ref{lem:chain-dual}. The recorded actions consequently
identify the resulting module with \(D_{T'}A'\) as a bimodule.

For a finitely generated projective left module \(P_i\), choose
an integer \(m_i\ge0\) and \(A\)-linear maps
\[
 P_i\xrightarrow{\iota_i} A^{m_i}
     \xrightarrow{\pi_i}P_i,\qquad \pi_i\iota_i=1.
\]
Their equations preserve projectivity. To preserve Ext we also
describe the actual Hom terms. Put \(E_i=\iota_i\pi_i\), an
idempotent endomorphism of \(A^{m_i}\). For any left module \(X\),
identify \(\Hom_A(A^{m_i},X)\) with \(X^{m_i}\) by evaluation
on the free generators. For an \(A\)-linear map \(F\) between free
modules, write \(F^*\) for precomposition on these Hom modules.
Explicitly, take integers \(a,b\ge0\), and let \(e_1,\ldots,e_b\)
be the target free generators of \(F:A^a\to A^b\).
Write its value on the \(j\)-th source generator as
\(\sum_{i=1}^b f_{ij}e_i\), with \(f_{ij}\in A\). Then
\[
 F^*:X^b\longrightarrow X^a,\qquad
 (F^*(x))_j=\sum_{i=1}^b f_{ij}x_i\quad(1\le j\le a).
\]
Multiplication denotes the given left action on \(X\).

The map \(h\mapsto h\pi_i\) identifies
\(\Hom_A(P_i,X)\) with \(\im E_i^*\subset X^{m_i}\);
its inverse is restriction along \(\iota_i\).
Choose a cyclic presentation \(H_i\) for this image and maps
\begin{equation}\label{eq:hom-split}
 H_i\xrightarrow{u_i}X^{m_i}\xrightarrow{v_i}H_i,\qquad
 v_iu_i=1,\quad u_iv_i=E_i^*.
\end{equation}
If \(\partial_i:P_i\to P_{i-1}\) is a differential, define
the map between the corresponding free modules by
\[
 F_i=\iota_{i-1}\partial_i\pi_i:
 A^{m_i}\longrightarrow A^{m_{i-1}}.
\]
The Hom differential \(\delta_i:H_{i-1}\to H_i\) is
\begin{equation}\label{eq:hom-differential}
 \delta_i=v_iF_i^*u_{i-1},\qquad
 u_i\delta_i=F_i^*u_{i-1}.
\end{equation}
For verification, if \(h:P_{i-1}\to X\), then
\[
 (h\pi_{i-1})F_i=h\partial_i\pi_i.
\]
Moreover \(F_iE_i=F_i\), so \(F_i^*u_{i-1}\) lies in
\(\im E_i^*\), proving the second identity in
\eqref{eq:hom-differential}.
Record \eqref{eq:hom-split}--\eqref{eq:hom-differential} as well
as the Smith data for the required differentials. They identify the
realized complex with the Hom complex over \(A'\) and preserve its
specified cohomology lengths. For right modules use \(A^\op\).

\section{Coefficient rings and finite specialization}\label{sec:smooth}

The auxiliary rings and algebras in
Lemmas~\ref{lem:field-lifting}--\ref{lem:smooth-realization}
are commutative. For a prime \(p\), write
\(\Z_{(p)}=\{a/b\in\Q:a,b\in\Z,\ p\nmid b\}\).
An algebra \(S\) over a ring \(R\) is \emph{formally smooth}
if every \(R\)-algebra map \(S\to B/J\), for an \(R\)-algebra
\(B\) and a nilpotent ideal \(J\), lifts to \(B\).
It suffices to treat square-zero ideals: apply that case successively
to \(B/J^{i+1}\to B/J^i\), for \(i\ge1\).
The coefficient-ring and lifting constructions used here belong to
classical commutative algebra; see \cite[Sections 10.160 and 16.5]{Stacks}.

\begin{lemma}\label{lem:field-lifting}
Every field is formally smooth over its prime field. Every finitely
generated field extension of \(\Fp\) has a separating transcendence
basis: it is finite separable over a purely transcendental subfield.
\end{lemma}
\begin{proof}
First consider a field \(k\) of characteristic \(p\). Write
\(k^{p^e}=\{a^{p^e}:a\in k\}\) for \(e\ge1\).
A \(p\)-basis is a subset \(\mathcal B\subset k\) whose monomials
\(\prod_{b\in\mathcal B}b^{i_b}\), with finite support and
\(0\le i_b<p\), form a basis of \(k\) over \(k^p\).
To obtain one, take a maximal subset with these monomials linearly
independent over \(k^p\). Such a subset exists by the maximality
principle, since every linear relation involves finitely many elements.
The span of its monomials is the field \(k^p(\mathcal B)\): each
finite part is a finite-dimensional domain over \(k^p\), hence a
field. If \(a\notin k^p(\mathcal B)\), its minimal polynomial
there is \(X^p-a^p\), of degree \(p\), so adjoining \(a\)
preserves independence, a contradiction. Here the degree is \(p\)
because \(X^p-a^p\) has a single root in an algebraic closure and
is reducible only if that root belongs to the coefficient field.
Iterating this basis expansion shows that the same monomials with
\(0\le i_b<p^e\) form a basis over \(k^{p^e}\): write each
exponent in base \(p\) and apply the Frobenius map
\(a\mapsto a^p\) to the preceding
expansion. Independence follows at each step by the original basis.

Let \(k\to B/J\) be a homomorphism of \(\Fp\)-algebras, with
\(J\) nilpotent. Choose \(e\ge1\) with \(J^{p^e}=0\).
For \(a\in k\), the \(p^e\)-th power of any lift of its image
is independent of the lift. Addition and multiplication commute
with these powers, so this defines a homomorphism
\(k^{p^e}\to B\). Choose arbitrary lifts in \(B\) of the
images of the elements of \(\mathcal B\). Their \(p^e\)-th powers
have the prescribed values. The basis just proved gives the presentation
\[
 k\cong k^{p^e}[X_b\mid b\in\mathcal B]/
       (X_b^{p^e}-b^{p^e}\mid b\in\mathcal B).
\]
Indeed, reduction by these relations gives precisely the basis monomials.
Sending the generators
to the chosen lifts therefore defines the desired map \(k\to B\).

Now let \(k\) have characteristic zero and let \(k\to B/J\)
be a homomorphism of \(\Q\)-algebras with \(J\) nilpotent.
Choose a transcendence basis of \(k/\Q\).
Arbitrary lifts of its elements define a map from its polynomial
ring; every nonzero polynomial maps to a unit modulo \(J\), hence
to a unit in \(B\), so this extends to the rational function field.
The remaining extension is algebraic separable. Over each finite
subextension, a primitive element has a separable minimal polynomial.
A lift \(u\) of its image can be corrected to a root: for a
square-zero kernel replace \(u\) by \(u-f(u)/f'(u)\), where
\(f\) is its minimal polynomial, evaluated using the lifted coefficient map, and \(f'\) is its formal derivative. The derivative is a unit, and
Taylor expansion proves the assertion. Iteration handles a nilpotent
kernel. The root lifting a given root is unique, since the difference
of two such roots times a unit is zero. Consequently the maps on
finite subextensions agree on overlaps and give a map on \(k\).

Finally let \(L/\Fp\) be finitely generated, of transcendence
degree \(r\ge0\). Choose any transcendence basis
\(u_1,\ldots,u_r\). Frobenius identifies the finite degrees
\([L:\Fp(u_1,\ldots,u_r)]\) and
\([L^p:\Fp(u_1^p,\ldots,u_r^p)]\). The tower formula gives
\[
 [L:L^p]\,[L^p:\Fp(u_1^p,\ldots,u_r^p)]
 =p^r[L:\Fp(u_1,\ldots,u_r)],
\]
so \([L:L^p]=p^r\). Thus a \(p\)-basis of \(L\) consists of
\(r\) elements \(t_1,\ldots,t_r\). They are algebraically
independent over \(\Fp\). Indeed, a nonzero polynomial relation
of least total degree can be grouped by its exponents modulo \(p\) as
\[
 \sum_{0\le i_1,\ldots,i_r<p}
 t_1^{i_1}\cdots t_r^{i_r}
 g_{i_1,\ldots,i_r}(t_1,\ldots,t_r)^p=0,
\]
where the \(g_{i_1,\ldots,i_r}\) are polynomials in
\(\Fp[X_1,\ldots,X_r]\), at least one of them nonzero.
Each nonzero such polynomial has smaller total degree than the original
relation. Independence over \(L^p\) forces all their values to vanish.
A nonzero constant cannot vanish; any other nonzero polynomial gives a
relation of smaller degree. Both alternatives are contradictions.
Set \(F=\Fp(t_1,\ldots,t_r)\). The finite extension \(L/F\)
satisfies \(L=FL^p\). Here \(FL^{p^e}\) denotes the subfield of
\(L\) generated by \(F\) and \(L^{p^e}\), for \(e\ge1\).
Taking powers and adjoining \(F\) inductively gives
\(L=FL^{p^e}\) for every \(e\ge1\).
For some \(e\), all \(p^e\)-th powers of a finite set of field
generators are separable over \(F\): remove the successive
\(p\)-powers in their irreducible polynomials. Their powers
generate \(FL^{p^e}\), which is therefore separable over \(F\).
Since this field is \(L\), the basis is separating. The empty
basis means \(L\) is finite separable over \(\Fp\).
\end{proof}

\begin{lemma}\label{lem:flat-lifting}
Let \(R\to S\) be flat, and let \(I\subset R\) satisfy
\(I^2=0\). If \(S/IS\) is formally smooth over \(R/I\),
then \(S\) is formally smooth over \(R\).
\end{lemma}
\begin{proof}
Choose a polynomial algebra \(P=R[x_t\mid t\in T]\) surjecting
onto \(S\), and denote its kernel by \(J\). The index set \(T\)
may be infinite; every polynomial uses finitely many variables.
Let \(E=\bigoplus_{t\in T}S\,dx_t\) be the free \(S\)-module
with indicated basis. Formal differentiation gives an \(S\)-linear
map
\[
 d:J/J^2\longrightarrow E,\qquad
 f\bmod J^2\longmapsto
 \sum_{t\in T}(\partial f/\partial x_t\bmod J)\,dx_t.
\]
It is well defined by the product rule.

We first verify that \(S\) is formally smooth exactly when \(d\)
has an \(S\)-linear left inverse. If \(S\) is formally smooth,
lift the identity to a section \(S\to P/J^2\). The difference
between \(P\to P/J^2\) and its composite through this section
is an \(R\)-derivation with values in \(J/J^2\). It sends
\(f\in J\) to \(f\bmod J^2\); its values on the variables
define a left inverse of \(d\).
Conversely, given a left inverse \(q\) and a map \(S\to B/K\)
with \(K^2=0\), lift all variables to obtain \(P\to B\).
The images of relations define an \(S\)-linear map
\(\eta:J/J^2\to K\), where \(S\) acts on \(K\) through
\(S\to B/K\). Replace the image of each \(x_t\) by that
image minus \(\eta(q(dx_t))\). Taylor expansion modulo \(K^2\)
shows that the new image of every \(f\in J\) is
\(\eta(f\bmod J^2)-\eta(q(d(f\bmod J^2)))=0\). The corrected map factors through
\(S\), proving the criterion.

Flatness of \(S\) gives \(J\cap IP=IJ\): tensor \(0\to J\to P\to S\to0\) with \(R/I\);
its left exactness follows from flatness of \(S\), and its
kernel identification is precisely this equality. Consequently the
relation module for \(P/IP\to S/IS\) is
\(J/(IJ+J^2)\). The criterion gives a left inverse to the reduction
of \(d\) modulo \(I\). Lift its values on the free basis of \(E\)
to obtain an \(S\)-linear map \(q_0:E\to J/J^2\).
The endomorphism \(\Delta=1-q_0d\) of \(J/J^2\) has image in
\(I(J/J^2)\), so \(\Delta^2=0\). Hence
\[
 ((1+\Delta)q_0)d=(1+\Delta)(1-\Delta)=1.
\]
The criterion applied again proves the assertion.
\end{proof}

A \emph{Cohen ring} for a field \(k\) of characteristic \(p>0\)
is a complete discrete valuation ring of characteristic zero with
maximal ideal generated by \(p\) and residue field \(k\).

\begin{lemma}\label{lem:cohen-exists}
Every field of positive characteristic has a Cohen ring.
\end{lemma}
\begin{proof}
Let the field be \(k\) of characteristic \(p\). Begin with
\(\Z_p=\varprojlim_{m\ge1}\Z/p^m\Z\), the ring of \(p\)-adic
integers. Its nonzero elements are a power of \(p\) times a unit,
as seen from the first nonzero digit; it is a complete discrete
valuation ring with residue field \(\Fp\).
We construct a flat local \(\Z_p\)-algebra \(C\) whose maximal
ideal is \(pC\) and whose residue field is \(k\).
Well-order a set of generators of \(k/\Fp\). Suppose a local
algebra with residue field \(K\subset k\) has been constructed.
For a next generator transcendental over \(K\), take a polynomial
algebra in one variable \(t\) and localize at the prime ideal generated
by \(p\); its residue field is \(K(t)\). For an algebraic
generator, lift its monic minimal polynomial to the current ring
and take the quotient by this lift. The quotient is finite free,
and is local with maximal ideal generated by \(p\): it is integral
over the old local ring and its quotient modulo \(p\) is the
required field. At a limit stage take the union, or equivalently
the direct limit, of the preceding rings. These extensions are
flat local maps. They are injective because a flat local map is
faithfully flat. Thus the construction preserves flatness over
\(\Z_p\), and gives the asserted \(C\). In particular,
multiplication by \(p\) is injective on \(C\).

Set \(O=\varprojlim_{m\ge1} C/p^mC\). For each \(n\ge1\),
projection onto \(C/p^nC\) is surjective: a representative in \(C\)
gives a compatible sequence with the prescribed component. Its kernel is \(p^nO\): if a compatible sequence
vanishes modulo \(p^n\), divide its component modulo \(p^{m+n}\)
by \(p^n\) to obtain a unique class modulo \(p^m\). Injectivity
of multiplication by \(p\) on \(C\) ensures uniqueness and
compatibility. Thus \(O/p^nO=C/p^nC\), and \(O\) is complete
and separated for powers of \(p\).
If \(px=0\) in \(O\), the component of \(x\) modulo
\(p^{m+1}\) belongs to \(p^mC/p^{m+1}C\), again by cancellation
in \(C\). Its component modulo \(p^m\) is zero for every \(m\),
so \(x=0\).

An element of \(O\) nonzero modulo \(p\) is a unit: lift its
inverse modulo \(p\) and invert the remaining factor \(1-py\)
by the convergent series \(\sum_{j\ge0}p^jy^j\).
Every nonzero element therefore has the form \(p^a u\), for a
unique integer \(a\ge0\) and a unit \(u\). Separation makes
\(a\) finite. Since \(p\) is a nonzerodivisor, these factorizations
show that a product of nonzero elements is nonzero. A nonzero ideal is generated by an element of
least such exponent, so every nonzero ideal is a power of \(pO\).
It follows that \(O\) is a discrete valuation ring. It has
characteristic zero because \(p\) is a nonzerodivisor and every
integer prime to \(p\) is a unit. Its residue field is \(k\).
\end{proof}

\begin{lemma}\label{lem:artin-coefficients}
Let \(R\) be a commutative Artinian local ring with residue field
\(k\). Equal characteristic means \(\operatorname{char}R=
\operatorname{char}k\); otherwise the characteristics are unequal.
In equal characteristic \(R\) contains a copy of \(k\)
and is finite-dimensional over it. In unequal characteristic,
there is a Cohen ring \(O\) with residue field \(k\) and a
homomorphism \(O\to R\) inducing the identity on residue fields;
\(R\) is finite as an \(O\)-module.
\end{lemma}
\begin{proof}
Let \(\mathfrak m\) be the maximal ideal and choose \(h\ge1\)
with \(\mathfrak m^h=0\). In equal characteristic the prime field
embeds in \(R\). Starting from the identity of \(k=R/\mathfrak m\),
Lemma~\ref{lem:field-lifting} successively lifts it to
\(k\to R/\mathfrak m^j\), for \(2\le j\le h\).
Each kernel is square-zero, since \(2(j-1)\ge j\). The last
map is injective and is a section of the residue map.

In unequal characteristic, \(k\) has characteristic \(p>0\).
Choose \(O\) by Lemma~\ref{lem:cohen-exists}. For every \(j\ge1\),
\(O/p^jO\) is flat over \(\Z/p^j\Z\): \(O\) is torsion-free,
hence flat, over the principal ideal domain \(\Z_{(p)}\), and
passage to these quotients is base change. Lemmas
\ref{lem:field-lifting} and~\ref{lem:flat-lifting}, applied
successively with the square-zero ideal \((p^{j-1})\), show that
\(O/p^jO\) is formally smooth over \(\Z/p^j\Z\).
Since \(p\in\mathfrak m\), the ring \(R/\mathfrak m^j\) is
an algebra over \(\Z/p^j\Z\). Starting from the residue-field
isomorphism, lift the compatible maps
\(O/p^jO\to R/\mathfrak m^j\) one step at a time. At step \(j\)
use formal smoothness of \(O/p^jO\) and compose the previous map
with \(O/p^jO\to O/p^{j-1}O\). At \(j=h\) this gives \(O\to R\).

In either case, denote the coefficient field or ring by \(C\).
Choose generators \(x_1,\ldots,x_s\) of \(\mathfrak m\).
Every element of \(R\) is a coefficient from \(C\) plus an
element of \(\mathfrak m\). Expand the coefficients of the latter
in the same way, and repeat \(h\) times. The final remainder
belongs to \(\mathfrak m^h=0\). Hence the monomials
\(x_1^{a_1}\cdots x_s^{a_s}\) with \(a_i\ge0\) and
\(\sum_i a_i<h\) span \(R\) over \(C\). If \(s=0\), the
single monomial is \(1\). This proves finiteness.
\end{proof}

For a ring \(R\), a \emph{standard smooth presentation} means
\[
 S=R[x_1,\ldots,x_m]/(f_1,\ldots,f_c),\qquad 0\le c\le m,
\]
where \(f_i\) are polynomials and the determinant of
\((\partial f_i/\partial x_j)_{1\le i,j\le c}\) is a unit in
\(S\). The derivatives are formal polynomial derivatives; for
\(c=0\) the determinant is \(1\). An algebra with such a
presentation will be called standard smooth.

\begin{lemma}\label{lem:standard-lifting}
A standard smooth algebra is formally smooth. Polynomial extensions
and localizations at one element of a standard smooth algebra are
standard smooth. A standard smooth algebra over \(R/I\), for a
nilpotent ideal \(I\subset R\), is the reduction of a standard
smooth \(R\)-algebra.
\end{lemma}
\begin{proof}
For formal smoothness, first let the kernel \(J\) of a target
surjection satisfy \(J^2=0\). Choose lifts of the variable images.
The values of the \(f_i\) lie in \(J\). The displayed derivative
matrix is invertible at these lifts, because its determinant is a
unit modulo \(J\). Correct the first \(c\) lifted variables by
the negative of the inverse matrix times this vector of errors,
and keep the other variables fixed. Taylor expansion has no terms
of order two in the corrections, so all \(f_i\) now vanish.
Successive square-zero lifts handle a nilpotent kernel.

Adjoining polynomial variables leaves the same invertible minor.
For a localization \(S_g\), represent \(g\) by a polynomial and
adjoin a variable \(y\) with relation \(gy-1\). Use the original
\(c\) variables and \(y\) for the new minor. Its determinant is
the old determinant times \(g\), which is a unit in \(S_g\).
Reordering these variables first gives the required presentation.

For the last assertion, lift the defining polynomials from \(R/I\)
to \(R\), take their quotient algebra, and invert the determinant
of the lifted minor. Its reduction is the original algebra, since
the original determinant was already a unit. Adjoining an inverse
by the preceding construction makes this a standard smooth
presentation over \(R\).
\end{proof}

\begin{lemma}\label{lem:finite-level-smooth}
Let \(p\) be prime and \(n\ge1\). Let \(\Lambda\) be a flat
commutative local \(\Z/p^n\Z\)-algebra such that
\(\Lambda/p\Lambda\) is a field. Every homomorphism from a
finitely presented commutative \(\Z/p^n\Z\)-algebra to
\(\Lambda\) factors through a standard smooth \(\Z/p^n\Z\)-algebra.
\end{lemma}
\begin{proof}
We prove the assertion by induction on \(n\). For \(n=1\), the
images of finitely many algebra generators lie in a finitely
generated subfield \(L\) of \(\Lambda\). By Lemma~\ref{lem:field-lifting}, there are algebraically independent
elements \(t_1,\ldots,t_r\), with \(r\ge0\), such that
\(L/\Fp(t_1,\ldots,t_r)\) is finite separable.
The empty list gives \(\Fp(t_1,\ldots,t_r)=\Fp\).
Choose a primitive element \(u\) and its monic minimal polynomial.
After inverting a nonzero polynomial in the \(t_i\), its
coefficients lie in the resulting polynomial algebra. Adjoin \(u\)
and invert the derivative of its minimal polynomial. The quotient embeds in \(L\): division by the monic polynomial
reduces any element to degree below its degree, and linear
independence over \(\Fp(t_1,\ldots,t_r)\) makes the kernel zero.
The invertible derivative gives a standard smooth subalgebra of \(L\). Inverting finitely many additional
nonzero elements includes all the specified images. Their relations
hold in this subalgebra because it embeds in \(L\). Thus the given
map factors through it.

Suppose \(n\ge2\), and put \(R=\Z/p^n\Z\) and
\(I=p^{n-1}R\), so \(I^2=0\). The quotient
\(\Lambda/I\Lambda\) is flat over \(R/I\), by base change,
and has the same residue field. We first show that every map
\(A\to\Lambda\), with \(A\) finitely presented over \(R\),
factors as
\[
 A\longrightarrow B/J\longrightarrow\Lambda,
 \qquad B\text{ standard smooth over }R,\quad J\subseteq IB.
\]
All ideals appearing in finite type \(R\)-algebras are finitely
generated, since \(R\) is Noetherian.

By induction, \(A/IA\to\Lambda/I\Lambda\) factors through a
standard smooth \(R/I\)-algebra \(\bar B\). Choose nonnegative integers \(s,t\) and a finite presentation
\[
 \bar B=(A/IA)[z_1,\ldots,z_s]/(g_1,\ldots,g_t).
\]
Lift the \(g_j\) to \(A[z_1,\ldots,z_s]\), and lift the images
of the \(z_i\) to \(\Lambda\). The evaluated relations have the
form \(p^{n-1}\mu_j\), with \(\mu_j\in\Lambda\). Consequently
\[
 A_1=A[z_1,\ldots,z_s,v_1,\ldots,v_t]/
       (g_j-p^{n-1}v_j\mid 1\le j\le t)
\]
maps to \(\Lambda\), with \(v_j\mapsto\mu_j\), and
\(A_1/IA_1=\bar B[v_1,\ldots,v_t]\).
Lemma~\ref{lem:standard-lifting} lifts this last standard smooth
algebra to a standard smooth \(R\)-algebra \(B\). Formal smoothness gives a map
\(B\to A_1\) inducing the chosen isomorphism modulo \(I\).
It is surjective: \(A_1=\operatorname{im}(B)+IA_1\), and
substitution of this equality into its last term, using \(I^2=0\),
gives \(A_1=\operatorname{im}(B)\). Its kernel \(J\) is
contained in \(IB\), as required.

We now remove the relations in \(J\). Let
\(\varphi:B\to\Lambda\) be the resulting map, and first consider
one generator \(h=p^{n-1}b\) of \(J\), with \(b\in B\).
Flatness identifies the kernel of multiplication by \(p^{n-1}\)
on \(\Lambda\) with \(p\Lambda\), since that kernel on \(R\)
is \(pR\). Hence \(\varphi(b)=p\lambda\) for some
\(\lambda\in\Lambda\). The algebra
\(B[x]/(b-px)\) maps to \(\Lambda\) by \(x\mapsto\lambda\).
Apply the preceding construction to factor this map through
\(B'/J'\), where \(B'\) is standard smooth and \(J'\subseteq IB'\).
Formal smoothness lifts \(B\to B'/J'\) to
\(\alpha:B\to B'\). The composite with \(B'\to\Lambda\)
is exactly \(\varphi\), because the latter map factors through
\(B'/J'\). If \(\xi\in B'\) lifts the image of \(x\), then
\(\alpha(b)-p\xi\in J'\). Therefore
\[
 \alpha(h)=p^{n-1}\alpha(b)
          \in p^nB'+p^{n-1}J'=0.
\]
The last equality uses \(p^n=0\) and \(J'\subseteq p^{n-1}B'\).
Apply this procedure successively to the finitely many generators
of \(J\). Previously killed generators remain zero, and the
remaining images still belong to \(I\) times the current standard smooth
algebra. The final map kills \(J\), so it factors through \(B/J\).
Composing with \(A\to B/J\) completes the induction.
\end{proof}

\begin{lemma}\label{lem:smooth-realization}
Let \(O\) be a Cohen ring of residue characteristic \(p\), and fix
\(N\ge1\). Every finite collection of elements of \(O\), together
with finitely many polynomial equalities over \(\Z_{(p)}\) that
they satisfy, can be realized in \(O'/(p^N)\) for a Cohen ring
\(O'\) with finite residue field.
\end{lemma}
\begin{proof}
The module \(O\) is torsion-free over the principal ideal domain
\(\Z_{(p)}\), hence flat. Base change shows that \(O/(p^N)\)
is flat over \(\Z/p^N\Z\). Introduce one variable for each
specified element and impose the specified polynomial equalities,
with coefficients reduced modulo \(p^N\). The resulting finitely
presented \(\Z/p^N\Z\)-algebra maps to \(O/(p^N)\).
Lemma~\ref{lem:finite-level-smooth} factors this map through a
standard smooth \(\Z/p^N\Z\)-algebra \(S\).

The map \(S\to O/pO\) shows that \(S/pS\ne0\).
It is a finitely generated \(\Fp\)-algebra, so a maximal ideal
has finite residue field \(F\), by Lemma~\ref{lem:residue}.
Choose a Cohen ring \(O'\) with residue field \(F\), using
Lemma~\ref{lem:cohen-exists}.
Formal smoothness over \(\Z/p^N\Z\) lifts \(S\to F\)
successively to \(S\to O'/(p^j)\), for \(2\le j\le N\).
The kernel of \(O'/(p^j)\to O'/(p^{j-1})\) is square-zero,
since \(2(j-1)\ge j\). For \(N=1\), the map to \(F\) itself
is the required map. The last lift realizes all the specified
equalities. The ring \(O'/(p^N)\) is finite: its successive
quotients \(p^jO'/p^{j+1}O'\), for \(0\le j<N\), are
one-dimensional vector spaces over \(F\).
\end{proof}

\begin{proposition}\label{prop:mixed}
Let \(O\) be a Cohen ring of residue characteristic \(p\), and let
\(A\) be a nonzero unital \(O\)-algebra of finite length as an
\(O\)-module. Suppose
\[
 \id({}_A A)=d<\infty,\qquad \id(A_A)=\infty.
\]
There is a finite ring \(A'\) whose cardinality is a power of \(p\)
such that
\[
 \id({}_{A'}A')=d,\qquad \id(A'_{A'})=\infty,\qquad
 \operatorname{char}A'=\operatorname{char}A.
\]
\end{proposition}
\begin{proof}
Let \(N\ge1\) be the least integer with \(p^NA=0\).
Such an integer exists by finite length. Then
\(\operatorname{char}A=p^N\), and the cyclic decomposition of
\(A\) contains a summand \(O/(p^N)\). Put \(T=O/(p^N)\).
Every finitely generated \(A\)-module has finite length over \(O\).

By Lemma~\ref{lem:chain-dual}, choose a right projective resolution
\[
 0\longrightarrow P_d\longrightarrow\cdots\longrightarrow P_0
 \longrightarrow (D_TA)_A\longrightarrow0
\]
with finitely generated terms. Choose the projective splittings from
Section~\ref{sec:dvr}, and Smith matrices for all its maps, including
the augmentation. Together with zero compositions, their lengths
encode exactness at every term, including injectivity at \(P_d\).
If \(d>0\), Lemma~\ref{lem:top} over \(A^\op\) gives
\[
 \Ext_{A^\op}^d(D_TA,A)\ne0.
\]
Record the Hom terms, differentials, and their Smith matrices that
compute this group from the same resolution; the outgoing map at
degree \(d\) is zero.

On the left, \(\pd_A(D_TA)=\infty\). The argument of
Lemma~\ref{lem:witness} gives a finitely generated left module
\(X\) such that \(\Ext_A^{d+1}(D_TA,X)\ne0\).
Take a projective resolution of \({}_A(D_TA)\) through degree
\(d+2\), with finitely generated terms. Retain exactness at the
augmentation and degrees \(0,\ldots,d+1\), using Smith matrices.
Such an exact prefix computes Ext in degree \(d+1\): it extends
to a projective resolution by taking finite free covers of its
leftmost kernel. Record the Hom terms in degrees
\(d,d+1,d+2\), their two differentials, and Smith matrices.

Choose cyclic decompositions for all these modules. Record the
multiplication, unit, actions, projective splittings, dual pairing,
and Hom identities as described in Section~\ref{sec:dvr}.
Choose representatives in \(O\) for their coefficients. Include
the division witnesses for every congruence and the inverse matrices
in every Smith identity. This is a finite set of elements satisfying
finitely many polynomial equalities over \(\Z_{(p)}\).

Apply Lemma~\ref{lem:smooth-realization}. Set \(T'=O'/(p^N)\)
for its resulting Cohen ring. Replacing each \(O/(p^e)\) by
\(O'/(p^e)\) and using the realized multiplication and unit gives
an associative unital algebra \(A'\). If the exponents in \(A\)
are \(e_1,\ldots,e_r\), then
\[
 |A'|=|O'/pO'|^{\sum_{j=1}^r e_j}.
\]
The dual pairing identifies the realized dual module with
\(D_{T'}A'\). Projective splittings and
\eqref{eq:homology-length} give a right projective resolution of
this module of length \(d\). Hence
\(\id({}_{A'}A')\le d\).

Equations \eqref{eq:hom-split}--\eqref{eq:hom-differential}
identify the realized Hom complexes with those computing Ext
over \(A'\). Let \(X'\) be the realized left module. Their
cohomology lengths are unchanged. Therefore
\[
 \Ext_{A'}^{d+1}(D_{T'}A',X')\ne0,
\]
Equation
\eqref{eq:chain-gap} gives \(\id(A'_{A'})=\infty\).
For \(d>0\), the nonzero degree-\(d\) right Ext group also
survives, proving \(\id({}_{A'}A')=d\). For \(d=0\), the
upper bound already gives equality.

The summand with exponent \(N\) gives \(p^{N-1}A'\ne0\),
while \(p^NA'=0\). Since \(A'\) is unital, these statements
give \(\operatorname{char}A'=p^N\).
\end{proof}

\section{Artin algebras and finite rings}\label{sec:artin}

An Artin algebra is finite over its center, which is a commutative
Artinian ring. To see this from the definition, choose a commutative
Artinian coefficient ring \(R\) over which \(A\) is finite.
The center \(Z(A)\) is an \(R\)-submodule of \(A\), hence has
finite \(R\)-length and satisfies the descending chain condition
on ideals. The same finite generating set for \(A\) over \(R\)
generates it over \(Z(A)\).

\begin{lemma}\label{lem:local-center}
Suppose \(A\) is an Artin algebra whose center is local, with
\(\id({}_A A)=d<\infty\) and \(\id(A_A)=\infty\).
There exists a finite ring \(A'\) of prime-power cardinality
with these same two self-injective dimensions.
If the residue field of \(Z(A)\) has characteristic \(p>0\),
the cardinality of \(A'\) can be chosen to be a power of \(p\),
and \(\operatorname{char}A'=\operatorname{char}A\).
\end{lemma}
\begin{proof}
Let \(R=Z(A)\), let \(\mathfrak m\) be its maximal ideal, and
put \(k=R/\mathfrak m\). Lemma~\ref{lem:artin-coefficients}
supplies a coefficient field \(\Lambda=k\) in equal characteristic
or a Cohen ring \(\Lambda\) in unequal characteristic, with \(R\)
finite over \(\Lambda\). Since \(A\) is finite over \(R\), it
is finite over \(\Lambda\).
In equal characteristic, \(A\) is a finite-dimensional
\(k\)-algebra, and Proposition~\ref{prop:finite} applies.
If \(k\) has characteristic \(p\), its finite residue-field
output has the same characteristic \(p\).

In unequal characteristic the residue characteristic is \(p>0\).
The image of \(p\) lies in \(\mathfrak m\), so some \(p^N\)
kills \(A\). The finite \(\Lambda\)-module \(A\) is therefore
of finite length. Proposition~\ref{prop:mixed} gives the result
and preserves its characteristic.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:artin}]
Decompose the commutative Artinian ring \(Z(A)\) into its local
factors. Its central orthogonal idempotents give
\(A=A_1\times\cdots\times A_s\), with each \(Z(A_i)\) local.
Every module decomposes by these idempotents; Hom and exact
sequences decompose componentwise. Applying this to injective
resolutions and to their components gives
\[
 \id({}_A A)=\max_i\id({}_{A_i}A_i),\qquad
 \id(A_A)=\max_i\id((A_i)_{A_i}).
\]
If the first is \(d<\infty\) and the second is infinite, choose
a factor \(A_i\) with infinite right self-injective dimension.
Its left dimension is an integer \(d_i\le d\).
Lemma~\ref{lem:local-center} produces a finite ring \(A_i'\) of
cardinality a power of a prime \(p\), with left dimension \(d_i\)
and infinite right dimension.

To retain \(d\), consider the finite-dimensional \(\Fp\)-algebra
\(B_d=\Fp Q_d/J^2\), where \(Q_d\) is the directed path
\(1\to2\to\cdots\to d+1\), \(\Fp Q_d\) is its path algebra
(products concatenate paths from left to right), and \(J\) is
the ideal generated by arrows. For \(d=0\) this means \(B_0=\Fp\).
The ideal \(\operatorname{rad}B_d\), generated by the images of
the arrows, has square zero and quotient \(\Fp^{d+1}\).
It is the Jacobson radical: every nilpotent ideal annihilates
simple modules, and the quotient is a product of fields.
Let \(e_j\) be the trivial path at vertex \(j\), and let
\(S_j=e_jB_d/e_j\operatorname{rad}B_d\) be the corresponding
simple right module, for \(1\le j\le d+1\).
For \(j\le d\), the module \(e_j\operatorname{rad}B_d\) is
\(S_{j+1}\), whereas
\(e_{d+1}B_d=S_{d+1}\). Thus
\[
 0\longrightarrow e_{d+1}B_d\longrightarrow\cdots
 \longrightarrow e_jB_d\longrightarrow S_j\longrightarrow0
\]
is a projective resolution of length \(d+1-j\).
The map from \(e_{t+1}B_d\) to \(e_tB_d\) sends \(e_{t+1}\)
to the arrow \(t\to t+1\); its image is
\(e_t\operatorname{rad}B_d\), and its kernel is
\(e_{t+1}\operatorname{rad}B_d\). The augmentation is the
quotient map. For \(j=d+1\) it is an isomorphism.
Its differentials have image in the radicals of their targets;
applying \(\Hom_{B_d^\op}(-,S_{d+1})\) makes the differentials
zero and leaves a nonzero term in degree \(d+1-j\).
Hence \(\pd_{B_d^\op}S_j=d+1-j\).
Every finitely generated right module has a composition series, so
the long exact Ext sequence bounds its projective dimension by
\(d\). Reversing the path gives the identical bound and a simple
module of projective dimension \(d\) on the left.
In particular the two dual regular modules have projective
dimension at most \(d\), so both self-injective dimensions are
at most \(d\). Lemma~\ref{lem:top}, applied to the simple module
of projective dimension \(d\) on each side, gives the reverse
bounds. Thus both self-injective dimensions of \(B_d\) equal \(d\).

The ring \(A'=A_i'\times B_d\) has cardinality a power of \(p\),
left self-injective dimension \(d\), and infinite right
self-injective dimension. This proves the quantitative assertion.
Taking opposite rings handles the other orientation of a failure.
Every finite ring is an Artin algebra over its finite prime subring.
Thus (1) implies (2), and (2) implies (3). The construction just
given shows that failure of (1) implies failure of (3), completing
the equivalence.
\end{proof}

\Needspace{6\baselineskip}
\begin{corollary}\label{cor:artin-algorithm}
Failure of the Gorenstein symmetry conjecture for Artin algebras
is semidecidable.
\end{corollary}
\begin{proof}
Enumerate primes \(p\), positive integers \(r\), and nonnegative
integers \(d\), so that every triple is eventually reached.
On a set of \(p^r\) elements, enumerate addition
and multiplication tables and distinguished zero and identity
elements. The ring axioms and \(0\ne1\) are finite tests.
For each ring \(A\) obtained, its characteristic is \(p^N\)
for some \(N\ge1\). Set \(T=\Z/p^N\Z\) and compute
\(D_TA=\Hom_T(A,T)\) by enumerating maps of finite sets and
testing additivity and scalar linearity. Its two actions are
given by \eqref{eq:dual}.

For any finite module \(M\), a free surjection is obtained by
taking one free generator for every element of \(M\) and sending
it to that element. Its kernel is a finite set and can be computed
exactly. Iterate \(d\) times on each dual regular module.
The last kernel is projective precisely when its finite free
surjection has a module-linear section; enumeration of all maps
between the two finite sets decides this. At \(d=0\), test the
dual module itself. At a zero kernel, the zero section suffices.
Thus both inequalities \(\pd(D_TA)\le d\) are decidable.

Halt when exactly one holds. Lemma~\ref{lem:chain-dual} and
\eqref{eq:chain-gap} give the finite and infinite sides on that
same ring. If an Artin counterexample exists,
Theorem~\ref{thm:artin} supplies a finite ring of prime-power
cardinality with the same orientation, and the enumeration reaches
its tables and a suitable \(d\).
\end{proof}

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\bigskip
\begingroup\small
\noindent\textsc{Zhili College, Tsinghua University, Beijing 100084, China}\par
\noindent\textit{Email address:}\enspace
\mbox{\href{mailto:tangqy24@mails.tsinghua.edu.cn}{\texttt{tangqy24@mails.tsinghua.edu.cn}}}\par
\endgroup
\end{document}