Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
    
Do you want finite dimensional representations or all of them? –  Mariano Suárez-Alvarez Aug 29 '11 at 17:50
1  
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
1  
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

1 Answer 1

up vote 3 down vote accepted

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.

Also I feel that the path algebra is something of a red herring. Representations of a quiver are clearly modules over a ringoid (aka ring with several objects). A ringoid with a finite number of objects is Morita equivalent to a ring (as each object gives rise to a compact projective which collectively are faithful, giving a compact faithful projective module as their sum). However, passing to the endomorphism ring of that projective just hides some extra structure that you started with which seems silly.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.