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\fancyhead[L]{\small FPRD Lab}
\fancyhead[R]{\small Semisimplicity under odd symmetry}
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\title{\textbf{Absolutely flat finite algebras in\\
characteristic-zero supertannakian categories}}
\author{FPRD Lab\\\small Research conducted by agentic AI under the direction of Joshua Gay}
\date{Internally audited preprint -- 6 September 2026}

\begin{document}
\maketitle

\begin{abstract}
Coulembier and Sherman ask whether six classical semisimplicity conditions are
equivalent for a finite algebra object in a characteristic-zero
supertannakian category. We give an affirmative answer. The new structural
step is that a finite algebra with no nonzero nilpotent invariant ideal is
absolutely flat. After passage to the connected algebraic supergroup, a
minimal invariant ideal is differentiably simple. Kac's normal-form theorem
then identifies its underlying superalgebra with
$S\shat\Lambda(V)$, where $S$ is a simple associative superalgebra. The
Grassmann factor is intrinsic: it is exactly the supercenter. That center is
a simple commutative algebra object and is therefore absolutely flat by the
commutative theorem of Coulembier and Sherman. Separability of $S$ makes
tensor product over $S\shat\Lambda(V)$ a natural retract of tensor product
over the center, after applying the faithful super fiber functor. This lifts
absolute flatness from the center to the full algebra. A separate internal
$\Ext^1$ argument supplies the remaining implication. The proof is
structural and does not classify supergroups or invariant algebra laws.
\end{abstract}

\begin{center}
\fcolorbox{accent}{soft}{\parbox{0.89\textwidth}{\small
\textbf{Status.} This manuscript has passed a hostile internal audit but has
not received external specialist review. A search through 6 September 2026
found no earlier resolution of the cited August 2026 question; this bounded
search is not a claim of priority.}}
\end{center}

\section{The question and the result}

Throughout, $k$ is an algebraically closed field of characteristic zero,
$\C$ is a supertannakian category over $k$, and $A\in\C$ is a unital
associative algebra object. Algebras are not assumed commutative. Modules and
relative tensor products live in $\Ind\C$.

For an artinian algebra object, Coulembier and Sherman consider the following
conditions \cite[Theorem 3.17]{CS}:
\begin{enumerate}[label=(\arabic*)]
\item $A$ is absolutely flat;
\item $\underline{\Ext}^{1}_{A}(M,-)=0$ for every finitely presented left
  $A$-module $M$;
\item $A$ is a finite product of simple algebra objects;
\item $A$ is semisimple, equivalently its internal Jacobson radical vanishes;
\item the only nilpotent two-sided ideal object in $A$ is zero;
\item the monoidal category of finitely presented $A$-bimodules is rigid.
\end{enumerate}
They prove
\[
 (6)\Longleftrightarrow(1)\Longrightarrow(2),\qquad
 (3)\Longrightarrow(4)\Longleftrightarrow(5),
\]
and, because $A\in\C$, also $(1)\Rightarrow(3)$. They then ask whether all
six conditions are equivalent when $\C$ is supertannakian in characteristic
zero \cite[Question 5.16]{CS}.

\begin{theorem}[Main theorem]\label{thm:main}
Let $\C$ be a supertannakian category over an algebraically closed field of
characteristic zero and let $A\in\C$ be a finite associative algebra object.
Then conditions \emph{(1)--(6)} above are equivalent.
\end{theorem}

The two arrows added here are
\[
  (5)\Longrightarrow(1),\qquad (2)\Longrightarrow(5).
\]
The first occupies Sections \ref{sec:connected}--\ref{sec:flatness}; the
second is proved in Section \ref{sec:ext}.

\section{Reduction to an algebraic supergroup}\label{sec:reduction}

We use the same finite-generation reduction as Coulembier and Sherman
\cite[Section 5.5]{CS}. The tensor subcategory generated by $A$ embeds as a
full tensor subcategory of $\Rep^f G$ for an algebraic supergroup $G$ in
$\sVec$. Exactness, ideals, and relative tensor products are preserved by
this embedding. It is therefore enough to prove the assertion for
\[
  \C=\Rep^fG.
\]
Write $H=G^\circ$. The identity component is a normal connected algebraic
supergroup and $G/H$ is finite etale. Since $k$ is algebraically closed, the
component group is represented by finitely many $k$-points.

The action of $H$ on the underlying finite-dimensional superalgebra of $A$
is by even algebra automorphisms. Its Lie superalgebra
$\mathfrak h=\Lie H$ therefore acts by superderivations:
\[
 x(ab)=x(a)b+(-1)^{|x||a|}a\,x(b).
\]
For a connected algebraic supergroup in characteristic zero, a graded
subspace of a finite rational module is $H$-stable if it is
$\mathfrak h$-stable. One way to see this is through the super
Harish--Chandra-pair equivalence \cite{Ma,Sh}: stability under
$\mathfrak h_{\bar 1}$ handles the odd directions, while stability under
$\mathfrak h_{\bar 0}$ integrates to the connected even group.

\section{Connected reduction and differential simplicity}\label{sec:connected}

Assume condition (5): $A$ has no nonzero nilpotent $G$-stable two-sided
ideal.

\begin{lemma}[Connected reduction]\label{lem:connected}
$A$ has no nonzero nilpotent $H$-stable two-sided ideal.
\end{lemma}

\begin{proof}
Suppose $I$ is such an ideal. Every nilpotent ideal of the underlying
finite-dimensional associative algebra lies in its ordinary Jacobson radical
$J_{\mathrm{ord}}(A)$. Choose representatives $g_1,\ldots,g_r$ of $G/H$ and
put
\[
  I_G=\sum_{j=1}^{r}g_jI.
\]
The ordinary Jacobson radical is characteristic, so every summand lies in
$J_{\mathrm{ord}}(A)$. Thus $I_G$ is a nonzero $G$-stable ideal contained in
the nilpotent ideal $J_{\mathrm{ord}}(A)$, hence is nilpotent. This
contradicts (5).
\end{proof}

Choose a minimal nonzero $H$-stable graded two-sided ideal $I\subseteq A$.
Its square is again $H$-stable. Minimality gives $I^2=0$ or $I^2=I$; the
first alternative contradicts Lemma \ref{lem:connected}. Hence
\[
 I^2=I.
\]

\begin{lemma}[Saturation lemma]\label{lem:saturation}
The possibly nonunital superalgebra $I$ is differentiably simple in Kac's
sense.
\end{lemma}

\begin{proof}
Let $0\ne J\triangleleft I$ be a graded ideal invariant under every
superderivation of $I$. In particular, $J$ is invariant under the
restrictions of all elements of $\mathfrak h$. Connected integration makes
$J$ an $H$-stable subspace.

It is important that $J$ need not itself be an ideal of $A$. Saturate it:
$AJA$ is a nonzero $H$-stable two-sided ideal of $A$ contained in $I$.
Minimality of $I$ therefore gives $AJA=I$. Since $A$ is unital and $I$ is a
two-sided ideal, $IA=AI=I$. Consequently
\[
 I=I^3=I(AJA)I=IJI\subseteq J.
\]
Thus $J=I$. Together with $I^2\ne0$, this is precisely differentiable
simplicity.
\end{proof}

Kac's Proposition 5.1.1 applies to finite-dimensional differentiably simple
superalgebras, not only to unital ones \cite{Kac}. It gives an isomorphism of
underlying associative superalgebras
\begin{equation}\label{eq:kac}
 I\cong S\shat\Lambda(V),
\end{equation}
where $S$ is a finite-dimensional simple associative superalgebra and
$\Lambda(V)$ is a finite Grassmann algebra. The isomorphism is not asserted
to be $H$-equivariant.

In particular, $I$ has an even identity $e$. Since $I$ is a two-sided ideal
of $A$, this identity is central in $A$: for $a\in A$, both $ae$ and $ea$
belong to $I$, and
\[
 ae=eae=ea.
\]
The identity of $I$ is unique, so $e$ is $H$-fixed. Hence
\[
 A=I\times(1-e)A
\]
as $H$-algebras. The second factor inherits Lemma \ref{lem:connected};
induction on dimension yields
\begin{equation}\label{eq:Hdecomp}
 A=B_1\times\cdots\times B_t,
 \qquad B_i\cong S_i\shat\Lambda(V_i),
\end{equation}
where every $B_i$ is $H$-simple. Normality of $H$ makes the finite component
group permute the $B_i$. The product over each component-group orbit is a
simple $G$-algebra object. Thus the same argument already proves
$(5)\Rightarrow(3)$.

\section{The Grassmann factor is the supercenter}\label{sec:center}

For a superalgebra $B$, write
\[
 Z_s(B)=\{z\in B:zb=(-1)^{|z||b|}bz\text{ for homogeneous }b\in B\}.
\]

\begin{lemma}\label{lem:scenter-simple}
If $S$ is a finite-dimensional simple associative superalgebra over $k$,
then $Z_s(S)=k$.
\end{lemma}

\begin{proof}
A nonzero homogeneous supercentral element generates a nonzero graded ideal,
so simplicity forces it to be invertible. An odd supercentral element $z$
satisfies $z^2=-z^2$, hence $z^2=0$ in characteristic zero and cannot be
invertible. The even supercenter is therefore a finite-dimensional
commutative division algebra over the algebraically closed field $k$, and so
equals $k$.
\end{proof}

\begin{lemma}[Canonical Grassmann center]\label{lem:center}
For $B=S\shat\Lambda(V)$ as in \eqref{eq:kac},
\[
 Z_s(B)=1\shat\Lambda(V).
\]
If $B$ is $H$-simple, this supercenter is a simple commutative $H$-algebra
object.
\end{lemma}

\begin{proof}
Expand a homogeneous supercentral element in a homogeneous basis of
$\Lambda(V)$. Supercommuting it with every $s\shat1$ shows coefficient by
coefficient that its coefficients in $S$ lie in $Z_s(S)$. Lemma
\ref{lem:scenter-simple} makes them scalar. Conversely,
$1\shat\Lambda(V)$ is supercentral because the Grassmann algebra is
supercommutative. This proves the identity, including for queer-type $S$.

The supercenter is defined intrinsically and is therefore $H$-stable. If it
had a nonzero proper $H$-stable ideal $L$, then $BL$ would be a nonzero
$H$-stable two-sided ideal of $B$. Under any isomorphism
$B\cong S\shat\Lambda(V)$, one has $BL=S\shat L$, which is proper when $L$
is proper. This contradicts $H$-simplicity.
\end{proof}

The category $\Rep^fH$ admits its symmetric super fiber functor to $\sVec$.
Coulembier and Sherman's commutative theorem therefore applies
\cite[Corollary 6.6]{CS}:
\begin{equation}\label{eq:centerflat}
 C_i:=Z_s(B_i)\quad\text{is absolutely flat in }\Rep^fH.
\end{equation}

\section{Separability lifts flatness}\label{sec:flatness}

We first make the superalgebra input explicit. By the graded
Artin--Wedderburn classification \cite{Wall}, a finite-dimensional simple
associative superalgebra over $k$ is of matrix type $M(p|q)$ or queer type
$Q(n)$.

\begin{proposition}[Explicit separability]\label{prop:separable}
Every finite-dimensional simple associative superalgebra over $k$ is a
separable algebra object in $\sVec$.
\end{proposition}

\begin{proof}
For $M(p|q)$, choose homogeneous matrix units $E_{ij}$ and fix one index
$r$. The element
\[
 e_M=\sum_i E_{ir}\otimes E_{ri}
\]
is even and has product $1$. For every matrix unit $E_{k\ell}$,
\[
 (E_{k\ell}\otimes1)e_M
 =E_{kr}\otimes E_{r\ell}
 =e_M(1\otimes E_{k\ell}).
\]
Thus $b\mapsto be_M=e_Mb$ is an even bimodule section of multiplication.

For $Q(1)=k\langle u\rangle/(u^2-1)$ with $u$ odd,
\[
 e_Q=\tfrac12(1\otimes1+u\otimes u)
\]
is even, has product $1$, and satisfies
$(u\otimes1)e_Q=e_Q(1\otimes u)$. Finally
$Q(n)\cong M_n(k)\shat Q(1)$ with $M_n(k)$ purely even, and tensor products
of separable algebra objects are separable.
\end{proof}

\begin{remark}[A second explicit section]\label{rem:casimir}
The averaged element
$d^{-1}\sum_{i,j}E_{ij}\otimes E_{ji}$ is also an even separability element.
Proposition \ref{prop:separable} uses the fixed-column element because its
bimodule-centrality reduces to a single surviving summand on each side. The
publication controls verify both formulas with the outer bimodule actions.
\end{remark}

If $B=S\shat C$ and $C=1\shat C\subseteq Z_s(B)$, tensoring a separability
section for $S$ with $C$ makes the extension $C\to B$ separable after
forgetting the $H$-action. This section need not be $H$-equivariant. The
following formulation explains why equivariance is unnecessary.

\begin{lemma}[Faithful-forgetful separability transfer]\label{lem:transfer}
Let $C\to B$ be a morphism of finite superalgebras in $\Rep^fH$. Suppose
that, after applying the forgetful functor
$U:\Ind\Rep^fH\to\operatorname{sVEC}$, multiplication
$B\otimes_C B\to B$ has an even $B$-bimodule section. If $C$ is absolutely
flat in $\Rep^fH$, then $B$ is absolutely flat in $\Rep^fH$.
\end{lemma}

\begin{proof}
Let $M$ be any right $B$-module in $\Ind\Rep^fH$. For a left $B$-module
$N$, the separability element $e=\sum x_j\otimes_Cy_j$ gives, after
forgetting equivariance, a natural section of the canonical quotient
\[
 U(M\otimes_C N)\longrightarrow U(M\otimes_B N),
 \qquad
 m\otimes_B n\longmapsto\sum_j mx_j\otimes_Cy_jn.
\]
Well-definedness is exactly the bimodule-centrality of $e$, and the composite
back to $M\otimes_BN$ is the identity because $\sum_jx_jy_j=1$. Hence the
functor $U(M\otimes_B-)$ is a natural retract of $U(M\otimes_C-)$ on the
category of equivariant $B$-modules.

Restriction of a short exact sequence of $B$-modules to $C$ is exact.
Absolute flatness of $C$ makes $M\otimes_C-$ exact on that sequence. A
natural retract of an exact additive functor is exact, so
$U(M\otimes_B-)$ is exact. The exact faithful functor $U$ reflects
exactness; therefore $M$ is flat over $B$. Since $M$ was arbitrary, $B$ is
absolutely flat.
\end{proof}

This lemma deliberately does not assert that internal $C$-flatness implies
ordinary flatness against all nonequivariant $C$-modules. It uses the
forgetful functor only on equivariant short exact sequences, which is the
statement needed here.

Apply Lemma \ref{lem:transfer} to each $C_i\to B_i$ in
\eqref{eq:Hdecomp}, using \eqref{eq:centerflat} and Proposition
\ref{prop:separable}. Every $B_i$ is absolutely flat in $\Rep^fH$, hence so
is their finite product $A$. Restriction
\[
 \Rep^fG\longrightarrow\Rep^fH
\]
is exact and faithful and preserves the coequalizers defining relative
tensor products. Exactness of tensor product over $A$ after restriction
therefore reflects exactness before restriction. Thus $A$ is absolutely flat
in $\Rep^fG$, proving
\[
 \boxed{(5)\Longrightarrow(1).}
\]
The argument applies to arbitrary modules in the ind-category, not only to
finite modules.

\section{The internal Ext arrow}\label{sec:ext}

\begin{lemma}\label{lem:ext}
For a finite algebra object in a supertannakian category,
\[
 (2)\Longrightarrow(5).
\]
\end{lemma}

\begin{proof}
Let $R$ be a nilpotent two-sided ideal object and put $M=A/R$. First, $M$ is
finitely presented. Indeed, multiplication gives an epimorphism
$A\otimes M\twoheadrightarrow M$. Its kernel $K$ lies in $\C$, and the
epimorphism $A\otimes K\twoheadrightarrow K$, followed by the inclusion
$K\hookrightarrow A\otimes M$, gives a finite free presentation of $M$.

Apply condition (2) to
\[
 0\longrightarrow R\longrightarrow A\longrightarrow M\longrightarrow0.
\]
The internal long exact sequence gives an epimorphism
\[
 \underline{\Hom}_A(M,A)\longrightarrow
 \underline{\Hom}_A(M,M).
\]
Let $\omega:\C\to\sVec$ be the super fiber functor. Because $M$ is finite
and dualizable, $\omega$ carries internal Hom to ordinary super Hom; because
it is exact, it also preserves the equalizer imposing $A$-linearity. Thus
the even identity of the underlying $A$-module $M$ lifts to an even
underlying $A$-linear section $s:M\to A$.

Set $e=s(1+R)$. Then
\[
 e\equiv1\pmod R,\qquad e^2=e,\qquad Re=0.
\]
Indeed, $e$ acts as the identity on $1+R$, while elements of $R$ act as zero,
and $s$ is $A$-linear. The element $1-e\in R$ is idempotent. A nilpotent
ideal contains no nonzero idempotent, so $1-e=0$. Hence $e=1$ and
$R=Re=0$.
\end{proof}

\section{Proof of the main theorem}

Coulembier and Sherman prove
\[
 (6)\Longleftrightarrow(1)\Longrightarrow(2),\qquad
 (3)\Longrightarrow(4)\Longleftrightarrow(5),\qquad
 (1)\Longrightarrow(3)
\]
for the finite algebra $A$. Sections \ref{sec:connected}--\ref{sec:flatness}
give $(5)\Rightarrow(1)$, and Lemma \ref{lem:ext} gives
$(2)\Rightarrow(5)$. Every one of the six conditions therefore implies all
the others. This proves Theorem \ref{thm:main}.

\section{Audit record and scope}\label{sec:audit}

The argument was tested at the bridges most likely to hide a false
equivariance or grading step.
\begin{center}
\small
\begin{tabular}{>{\raggedright\arraybackslash}p{0.25\textwidth}>{\raggedright\arraybackslash}p{0.63\textwidth}}
\toprule
Bridge & Audit conclusion \\
\midrule
Connected integration & Reduced to the super Harish--Chandra pair and the
connected ordinary even group; no finite-component invariance is assumed.\\
Kac's scope & Proposition 5.1.1 explicitly allows a nonunital algebra; the
hypothesis $I^2\ne0$ is supplied by nilpotent-ideal exclusion.\\
Differential ideal & The original shortcut was invalid. The corrected
$AJA$ saturation in Lemma \ref{lem:saturation} supplies the missing ambient
ideal.\\
Supercenter & The coefficientwise proof uses only $Z_s(S)=k$ and therefore
covers both matrix and queer types.\\
Separability & Explicit even fixed-column, averaged-matrix, and queer
sections were checked with the outer bimodule actions.\\
Equivariance and Ind scope & The section is used only after the faithful
forgetful functor; the retract argument runs on every equivariant short exact
sequence and on arbitrary ind-modules.\\
Internal $\Ext$ & Finite presentation, parity of the lifted identity, and
the internal-Hom equalizer are written explicitly in Lemma \ref{lem:ext}.\\
\bottomrule
\end{tabular}
\end{center}

As of 6 September 2026, the cited Coulembier--Sherman manuscript is arXiv
version 1, dated 13 August 2026, and still states Question 5.16. Exact-phrase
and structural searches found no later paper containing this argument. Such
a search cannot establish priority. The theorem in this manuscript has no
external review, and any use should retain that qualification until a
specialist has checked the proof.

The decisive external ingredients are clearly separated:
\begin{itemize}
\item Coulembier and Sherman supply the question, the established implication
diagram, the supertannakian reduction, and the commutative absolute-flatness
theorem.
\item Kac supplies the differentiably-simple-superalgebra normal form,
following Block's method.
\item Super Harish--Chandra-pair theory supplies the connected integration
step.
\item The matrix/queer classification supplies explicit separability of the
simple factor.
\end{itemize}
The application of these ingredients to Question 5.16, including the
saturation and faithful-forgetful transfer lemmas, is the contribution proved
here.

\begin{thebibliography}{9}

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R. E. Block,
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\href{https://doi.org/10.2307/1970745}{doi:10.2307/1970745}.

\bibitem{CS}
K. Coulembier and A. Sherman,
\emph{Absolutely flat algebras in tensor categories},
arXiv:2608.13137v1 (13 August 2026).
\href{https://arxiv.org/abs/2608.13137}{arXiv:2608.13137}.

\bibitem{Kac}
V. G. Kac,
\emph{Lie superalgebras},
Adv. Math. 26 (1977), 8--96.
\href{https://doi.org/10.1016/0001-8708(77)90017-2}{doi:10.1016/0001-8708(77)90017-2}.

\bibitem{Ma}
A. Masuoka,
\emph{Harish--Chandra pairs for algebraic affine supergroup schemes over an
arbitrary field},
Transform. Groups 17 (2012), 1085--1121.
\href{https://arxiv.org/abs/1111.2387}{arXiv:1111.2387}.

\bibitem{Sh}
A. Sherman,
\emph{Two geometric proofs of the classification of algebraic supergroups
with semisimple representation theory},
Transform. Groups 28 (2023), 1053--1073.
\href{https://arxiv.org/abs/2012.11317}{arXiv:2012.11317}.

\bibitem{Wall}
C. T. C. Wall,
\emph{Graded Brauer groups},
J. Reine Angew. Math. 213 (1964), 187--199.
\href{https://doi.org/10.1515/crll.1964.213.187}{doi:10.1515/crll.1964.213.187}.

\end{thebibliography}

\input{fprd-afterword.tex}
\end{document}