Let $A$ be a be a finite dimensional, associative, and unital $\mathbb{C}$-algebra. Let $\mathcal{A}$ be the category of finitely generated $A$-modules. Since $A$ is an Artinian ring, there are only finitely many non-isomorphic simple $A$-modules in $\mathcal{A}$. For our purposes assume that there are only two non-isomorphic simple $A$-modules, $\left\{S_{1}, S_{2}\right\}$.

Recall that the Grothendieck group $K_{0}(\mathcal{A})$ of $\mathcal{A}$ is the free abelian group generated by isomorphism classes $[M]$ of $A$-modules satisfying the relation that $[M_{2}] = [M_{1}] + [M_{3}]$ whenever there exists a short exact sequence \begin{equation*} 0 \rightarrow M_{1} \rightarrow M_{2} \rightarrow M_{3} \rightarrow 0 \end{equation*} We also have that $\left\{[S_{1}], [S_{2}]\right\}$ is a basis of $K_{0}(\mathcal{A})$. Also recall that exact functors induce group homomorphisms on Grothendieck groups by \begin{equation*} [F] : K_{0}(\mathcal{C}) \rightarrow K_{0}(\mathcal{D})\\ [F] : [X] \mapsto [FX] \end{equation*} where $F$ is an exact functor between essentially small abelian categories $\mathcal{C}$ and $\mathcal{D}$.

Given this setup, does there exist an exact endofunctor \begin{equation*} F: \mathcal{A} \rightarrow \mathcal{A} \end{equation*} such that \begin{align*} [F]\left([S_{1}]\right) &= [FS_{1}] = [S_{1}]\\ [F]\left([S_{2}]\right) &= [FS_{2}] = [S_{1}] - [S_{2}] \end{align*}

Is there an endofunctor such that $$[F]([S_1]) = [FS_1] = -[S_1]$$

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    $\begingroup$ No. Look at the composition series of $F(S_2)$. $\endgroup$ Apr 6, 2019 at 22:19
  • $\begingroup$ @QiaochuYuan Is there an exact endofunctor such that $[F]([S_1]) = [FS_1] = -[S_1]$? $\endgroup$
    – Anfänger
    Apr 7, 2019 at 14:38


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