# The Jacobson radical of an infinite dimensional algebra

Does any one know the Jacobson radical of the path algebra of the following quiver?

$$\bullet \leftrightarrows \bullet$$

How many simplerepresentations of it are there?

Is there any software that computes the Jacobson radicals of infinite dimensional non-commutative algebras?

-
Do you want finite dimensional representations or all of them? –  Mariano Suárez-Alvarez Aug 29 '11 at 17:50
Finite dimensional simple representations of this algebra are given e.g. in Assem, Simson, Skowronski: Elements of the representation theory of associative algebras Volume 1. Chapter III.4 Exercise 13 –  Julian Kuelshammer Aug 29 '11 at 18:02
(if the underlying field is algebraically closed) –  Julian Kuelshammer Aug 29 '11 at 18:03
Heh. That was where I was heading :) –  Mariano Suárez-Alvarez Aug 29 '11 at 18:10
As far as I can see simple modules are finite dimensional. –  Torsten Ekedahl Aug 29 '11 at 18:51

As there seem to be some differing opinions in the comments as to whether all irreducible representations are finite-dimensional let me give the argument I had in mind. A module over the path algebra is the same thing as a representation of the quiver, which in this case means two vector spaces $U$ and $V$ and linear maps $e\colon U\rightarrow V$ and $f\colon V\rightarrow U$. We introduce also $S:=fe\colon U\rightarrow U$ and $T:=ef\colon V\rightarrow V$. Assume now that $(U,V,e,f)$ is irreducible and pick a non-zero $u\in U$ and let $W$ be the subrepresentation generated by $u$. It is clear that the $U$-part of $W$ is the $k[S]$-submodule generated by $u$ so by irreducibility we have that $U=k[S]u$. Now, do the same argument for $Su$ giving us also that $U=k[S]Su$. In particular there is a polynomial $p(S)$ such that $u=p(S)Su$, i.e., $q(S)u=0$ where $q(S)=p(S)S-1$ which in particular is non-zero so that $U=k[S]u$ is finite dimensional. By symmetry the same argument applies to $V$ so the module is finite dimensional.
Note, that slightly extending this also gives us a classification of the finite dimensional modules. In particular there are enough of them to make the Jacobson radical be equal to $0$.
Addendum: Rather than mixing together several steps it is probably better to divide it up: First show that if $(U,V,e,f)$ is irreducible then $U$ is irreducible as $k[S]$-module and then use the classification of simple $k[S]$-module.
You are right. The situation is more similar to the polynomial ring in one variable. What you do is conceptually reducing the algebra $A$ to $e_1Ae_1$, where $e_1$ is the trivial path for one point. $e_1Ae_1$ is then the polynomial ring in one variable, which has only finite dimensional representations. By general theory the functor $F: A-mod\to e_1Ae_1-mod, M\mapsto e_1 M$ sends irreducibles to irreducibles (or zero, which can be excluded in our case). Your argument then shows that irreducible modules are finite dimensional for the polynomial ring. –  Julian Kuelshammer Sep 2 '11 at 6:56