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Let $R$ be a finite-dimensional local (associative, unital, and not necessarily commutative) algebra over a field $k$ (that is, $R$ has a unique maximal two-sided ideal $\mathfrak M$) such that $\kappa:=R/\mathfrak{M}$ is a finite-dimensional division $k$-algebra.

Assuming that $k$ is a perfect field, does the quotient ring homomorphism $R\to \kappa$ split? In other words, does there always exists a subalgebra $\kappa\subseteq R$ such that $R=\kappa\oplus \mathfrak M$ as $k$ vector spaces?

The motivation comes from the representation theory of finite-dimensional algebras. If $A$ is a finite dimensional $k$-algebra, and if $M$ is an indecomposable left $A$-module of finite type, then $R:={\rm End}_A(M)$ is a finite-dimensional local $k$-algebra (the elements of $\mathfrak M$ are those $f\colon M\to M$ which are not isomorphisms).

When $k$ is algebraically closed, the residue field $\kappa$ is $k$ and the answer is yes. I guess that it should also be the case when $R$ is commutative (using Hensel's lemma).

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The Wedderburn-Malcev «Principal» theorem, which gives you this splitting, is proved in that generality in Pierce's Associative Algebras, if I recall correctly. –  Mariano Suárez-Alvarez Apr 5 '12 at 8:07
    
(Hi! by the way :D ) –  Mariano Suárez-Alvarez Apr 5 '12 at 8:12
    
Sorry to nitpick: the zero ideal in $R = M_2(k)$ is the unique maximal two-sided ideal in $R$, but $R$ isn't a division algebra. –  Konstantin Ardakov Apr 5 '12 at 8:47
    
Mariano: Hi! and thanks for the answer. Also, my apologies for this lack of culture. I should stop working only with algebraically closed fields... –  Patrick Le Meur Apr 5 '12 at 9:19
    
Konstantin: You're right. My question was not formulated properly, I was concentrated on the situation of the indecomposable module. I edited the question accordingly. Thanks. –  Patrick Le Meur Apr 5 '12 at 9:21

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