I'm learning about representation theory of associative algebras. In my studies arrised less technical questions involving Morita theory and quivers that will exposed.

Morita theory give us a way to find a more "economic" algebra for a given algebra such that they have same modular theory. For Artin algebras we have a complete set o idempotents which allow us to construct this basic algebra.

1) There examples of non basic algebras? There are explicit examples of algebras without this complet set of idempotents? That is, an algebra with not only the trivial idempotents, but in wich we can't show that there are only those (could has a infinite number).

2) What is the quiver of a non basic algebra? The book (Assem-Simson-Skowronski) always talk about quiver involving basic and connected algebras, but don't show the issues arrised when consider an algebra without this properties. There exist quiver representation for non connected non basic algebras?

I know that is a lot of questions, but any help/commentarie/references will be really apreciated. Thanks in advance!


1 Answer 1


There are several questions here, and I will try to answer them in order.

First, an example of an algebra which is not basic would be the $k$-algebra $M_2(k)$ of $2\times 2$ matrices over a field $k$. To see that it is not basic, note that as a right module over itself, it is the direct sum of the two submodules $R_1$ and $R_2$ of matrices with zero entries in row $2$ and row $1$, respectively. Since $R_1$ and $R_2$ are isomorphic as right $M_2(k)$-modules, the algebra is not basic.

Second, on the existence of certain sets of idempotents. I am assuming here that you are interested in complete sets of pairwise orthogonal primitive idempotents, that is, collections of elements $e_1, \ldots, e_n$ such that $e_i\cdot e_j$ is $e_i$ if $i=j$ (idempotence) and $0$ otherwise (orthogonality); $\sum_{i=1}^n e_i = 1$ (completeness); and none of the $e_i$ can be written as a sum of two orthogonal non-zero idempotents (primitiveness). If an algebra is artinian, then such a set of idempotents always exists. An algebra which does not admit such a set is $\prod_{\mathbb{N}} k$, the product of a countably infinite number of copies of a field $k$. This algebra admits complete sets of pairwise orthogonal idempotents, but we cannot require that these idempotents are all primitive. Another example would be the $\mathbb{R}$-algebra of continuous functions from $\mathbb{R}$ to $\mathbb{R}$.

Next, you ask about the quiver of a non-basic algebra. This is somewhat ill-defined. Let's review the theory over an algebraically closed field $k$. In this setting, for any basic, finite-dimensional $k$-algebra $A$, there exists a unique quiver $Q$ and a (non-unique) admissible ideal $I$ of $kQ$ such that $A$ is isomorphic to $kQ/I$. Moreover, any algebra of the form $kQ/I$ as above is basic. It follows from these results that if $A$ is not basic, then it cannot be of the form $kQ/I$, with $I$ an admissible ideal of $kQ$.

In the above discussion, the operative word is admissible. For any finite-dimensional $k$-algebra, you can find lots of quivers $Q$ and (not necessarily admissible) ideals $I$ such that $A$ is isomorphic to $kQ/I$. For matrix algebras $M_n(k)$, some such descriptions are given in this mathoverflow question and its answers. Still, for non-basic algebras, it will never be possible to find an admissible ideal, and there will be no good notion of the uniqueness of the quiver.

Finally, if an algebra is basic but not connected, then the description by a quiver with an admissible ideal of relations works fine; all that happens is that the quiver is not connected.

  • $\begingroup$ thank very much for this answer! $\endgroup$ Dec 22, 2020 at 13:20
  • $\begingroup$ @JúlioCésarM.Marques No problem :-) $\endgroup$ Dec 22, 2020 at 13:50

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