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.


Question:



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> 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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$Ext_A^1(J,J)$ for the Jacobson radical $J$ of an algebra $A$
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
$$0 \to J \to e\_jA \oplus \ker(f) \oplus (1-e\_j)J \oplus (1-e\_i)J \to J \to 0.$$


