Let $H$ be an associative $k$-algebra, where $k$ is a field. Let $1 = e + f$ be a decomposition of $1$ into a sum of two orthogonal idempotents.
Question: When is $H \cong \left[ {\begin{array}{cc} eHe & eHf \\ 0 & fHf \ \end{array} } \right]$?
Note that the RHS is indeed an algebra with unit $\left[ {\begin{array}{cc} e & 0 \\ 0 & f \ \end{array} } \right]$. I'd like to write down an iso as follows:
$\varphi: \left[ {\begin{array}{cc} eHe & eHf \\ 0 & fHf \ \end{array} } \right] \to H$, $ \left[ {\begin{array}{cc} eh_1e & eh_2f \\ 0 & fh_3f \ \end{array} } \right] \mapsto eh_1e + eh_2f + fh_2e + fh_3f$
$\psi: H \to \left[ {\begin{array}{cc} eHe & eHf \\ 0 & fHf \ \end{array} } \right]$. $h \mapsto \left[ {\begin{array}{cc} ehe & ehf \\ 0 & fhf \ \end{array} } \right]$
If we assume that $\varphi$ is well-defined (I'm not sure about that since $h_2$ in the expression $eh_2f$ is not unique) and that $\varphi$ and $\psi$ are algebra homomorphisms, then they are inverse to each other and we are done. But I guess it does not work in this generality.
Context: I'm reading the paper "On the structure of modules over wild hereditary algebras"*. There $H$ is assumed to be a path algebra $kQ$ where $Q$ is a connected finite wild quiver without oriented cycles. Moreover, $e$ and $f$ are chosen in such a way that $He$ is the projective cover of the preprojective simple modules and $Hf$ a projective cover of the remaining simple modules. How can we use this to get an isomorphism? I also should point out that they don't say how the iso is defined, so the construction above is just my guess.
- Kerner, O. & Skowroński, A. Manuscripta Math. (2002) 108: 369. doi:10.1007/s002290200268
As was pointed out by the comments of Matthias Klupsch, we only need to show that Hom$(Hf, He) = 0$. I asked the first author of the paper per E-Mail why this holds. Here is an elaboration of his answer:
Basic fact 1: Local modules are indecomposable.
Basic fact 2: If $P(T) \to T$ is the projective cover of the simple module $T$ and $M$ is a module with Hom$(P(T), M) \neq 0$, then $[M : T] \neq 0$, so $T$ is a composition factor of $M$.
Less basic fact 3: In our setting (more general: over finite dimensional connected hereditary $k$-algebras) the following holds: Let $X$ preprojective, $Y$ regular and $Z$ preinjective, then Hom$(Z, X)$ = Hom$(Z, Y)$ = Hom$(Y, X)$ = 0. A reference for this is Chapter 2 of the paper "Representations of wild quivers" by Otto Kerner, where he says that it goes back to the original papers "Representation theory of artin algebras" by M. Auslander and I. Reiten.
For all Lemmas, let $P \to S$ be the projective cover of the preprojective simple module $S$.
Lemma 1: If $X \subseteq$ rad$(P)$, then $P / X$ is also preprojective.
Proof: $P$ is local with unique maximal submodule rad$(P)$. It follows that $P/X$ is local with unique maximal submodule rad$(P)/X$, so by fact 1 $P/X$ is indecomposable. Clearly, there is a canonical non-zero map $P/X \to P/$rad$(P) = S$, so Hom$(P/X, S) \neq 0$. By fact $3$ it follows that also $P/X$ is preprojective. $\square$
Lemma 2: Every composition factor of $P$ is preprojective.
Proof: Let $S'$ be a composition factor of $P$. Then $S' = Y / X$ for submodules $X \subset Y \subseteq P$. It follows that $X \subseteq$ rad$P$, so by Lemma 1, $P/X$ is preprojective. Clearly, Hom$(S', P/X) \neq 0$ (just take the inclusion $S' = Y/X \to P/X$), and so again by fact 3 we get that $S'$ is preprojective. $\square$
Lemma 3: If $P(T) \to T$ is the projective cover of the simple, non-preprojective module T, then Hom$(P(T), P) = 0$.
Proof: If the Hom-space is non-zero, we get that $T$ is a composition factor of $P$ by fact 2, and so by Lemma 2 $T$ is also preprojective, a contradiction. $\square$
Proposition: Every homomorphism $Hf \to He$ is trivial.
Proof: Follows by Lemma 3 by composing $Hf$ and $He$ into direct sums of indecomposable projective modules and using the fact that Hom is additive: Hom$(\bigoplus_i M_i, \bigoplus_j M_j) = \bigoplus_i \bigoplus_j$Hom$(M_i, M_j)$. $\square$