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%opening
\title{A Note on Flatness}
\author{Simon Felten}

\begin{document}

\maketitle

Let $k$ be a field of characteristic $0$, and let $C$ be a $k$-linear rigid abelian symmetric monoidal category with ind-extension $\mathrm{Ind}(C)$; it is a $k$-linear abelian symmetric monoidal category which admits all filtered colimits. $C \subset \mathrm{Ind}(C)$ is the full subcategory of finite/dualizable/compact objects (Yifei is going to prove this). That $C$ is rigid means that all objects are dualizable.
\begin{itemize}
 \item $C \subset \mathrm{Ind}(C)$ is an abelian subcategory; in particular, the embedding is exact (?).
\end{itemize}

\begin{lemma}
 All objects in $\mathrm{Ind}(C)$ are flat.
\end{lemma}
\begin{proof}
 First suppose $V \in C$. Then we have $\mathrm{Hom}_C(V \otimes S,T) = \mathrm{Hom}_C(S,V^\vee \otimes T)$, so the functor $V \otimes -$ has both a right adjoint and a left adjoint; in particular, it is exact, so $V$ is flat. Every object in $\mathrm{Ind}(C)$ is a filtered colimit of objects in $C$, so it is flat as well.
\end{proof}



\begin{lemma}
 Let $V \in C$. If $\dim(V) \in \mathrm{End}_C(1)$ is invertible, then $V$ is faithful, i.e., $V \otimes -$ is a faithful functor. In this case, $V$ is faithfully flat.
\end{lemma}
\begin{proof}
 Assume $V \otimes X = 0$. Then $V^\vee \otimes V \otimes X = 0$, so 
 $$\dim(V) \otimes \mathrm{id}_X: X \to V^\vee \otimes V \otimes X \to X$$
 is $0$; this contradicts the assumption that $\dim(V)$ is invertible.
\end{proof}

Let $v: M \to 1$ be an epimorphism such that $\dim(M)$ is not a non-positive integer. Then $v^\vee: 1 \to M^\vee$ is a monomorphism. It gives rise to a filtered diagram
$$1 \xrightarrow{v^\vee \cdot} M^\vee \xrightarrow{v^\vee \cdot} S^2(M^\vee) \xrightarrow{v^\vee \cdot} S^3(M^\vee) \quad ... $$
with filtered colimit $B = \mathrm{colim}_n S^n(M^\vee)$.

\begin{lemma}
 $B \in \mathrm{Ind}(C)$ is faithfully flat.
\end{lemma}
\begin{proof}
 It suffices to prove $B$ faithful, i.e., that $B \otimes X = 0$ implies $X = 0$. The morphism $1 \to S^n(M^\vee)$ is dual to the morphism $S^n(M) \to S^n(1) = 1$; the latter is an epimorphism, so its dual is a monomorphism. In particular, the induced map $1 = \mathrm{colim}_n 1 \to \mathrm{colim}_n S^n(M^\vee) = B$ is a monomorphism. Since $- \otimes X$ is exact, we get a monomorphism $X \to B \otimes X$. Thus, if $B \otimes X = 0$, then $X = 0$; indeed, $0$ has no non-zero subobject.
\end{proof}







\end{document}
