Let algebras be finite dimensional over a field $K$ and let $J$ denote the Jacobson radical (this is the intersection of all maximal right ideals) of an algebra. Being hereditary means that the algebra has global dimension at most 1.


For the class of algebras $A$ with $Ext_A^1(J,J) \neq 0$, can we explicitly write down a non-split short exact sequence $0 \rightarrow J \rightarrow W \rightarrow J \rightarrow 0$ with a pretty $W$?

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    $\begingroup$ The condition $\text{Ext}^1_A(J,J)=0$ is equivalent to $\text{injdim}J\leq1$, right? So a counterexample would disprove a conjecture of Marczinzik. $\endgroup$ – Jeremy Rickard Jul 19 '18 at 9:36
  • $\begingroup$ @JeremyRickard Oops you are right... Was too blind to see that $Ext_A^2(A/J,J)=Ext_A^1(J,J)$. I delete the question if you dont mind. $\endgroup$ – Mare Jul 19 '18 at 9:37
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    $\begingroup$ No, I don't mind. $\endgroup$ – Jeremy Rickard Jul 19 '18 at 9:48

Assume $J$ is not projective as a right $A$-module, or equivalently that $A$ is not hereditary. Then $\mbox{Ext}^1_A(J,J) \neq 0$ and it is possible to write down a somewhat explicit non-split extension of direct summands of $J$ as follows. Let $e_1, \ldots, e_n$ be a complete set of pairwise orthogonal primitive idempotents in $A$, and assume $A$ is basic. Since $J = \oplus_i e_iJ$, by assumption $e_iJ$ is not projective for some $i$. If we decompose $e_iJ = P \oplus M$ where $P$ is projective and $M$ has no projective direct summands, then by mapping $M$ onto a simple module $S_j = e_jA/e_jJ$ in its top, we get an epimorphism $f : e_i J \to S_j$ that cannot factor through a projective module. Thus the pull-back of the short exact sequence $$0 \to e_j J \to e_jA \to S_j \to 0$$ along $f: e_i J \to S_j$, yields a non-split short exact sequence $$0 \to e_jJ \to X \to e_iJ \to 0.$$ Moreover, using the snake lemma, the induced map $X \to e_jA$ is onto and thus splits, showing that $X \cong e_j A \oplus \ker(f)$, and $\ker(f)$ is a maximal submodule of $e_i J$. It is not hard to see what the maps are either. The maps $e_jJ \to e_jA$ and $\ker(f) \to e_iJ$ are just the inclusions, while the maps $e_jJ \to \ker(f)$ and $e_jA \to e_iJ$ would typically be induced by multiplication by some arrow from $i$ to $j$ in the quiver of $A$.

We can now add a split short exact sequence to get a non-split extension of $J$ by $J$ of the form $$0 \to J \to e_jA \oplus \ker(f) \oplus (1-e_j)J \oplus (1-e_i)J \to J \to 0.$$

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