5
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Let $A = KQ$, where $Q$ is the quiver $$\begin{array}{ccc} & \alpha & \\ 1 & \rightleftarrows & 2 \\ & \beta& \end{array}$$ are there simple right $A$-modules with dimension $\geq3$?

In generally, how to find all simple modules for the given path algebra, especially the infinitely dimensional case?

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    As xymatrix is not working here, you should probably write down that this is the quiver with two vertices 1 and 2 and two arrows, one going $1\to 2$ and one going $2\to 1$.2012-11-16
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    Do you assume the field to be algebraically closed?2012-11-16
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    Hint: In your example you can use Gauss normal formal to get the identity matrix as one of the matrices and Jordan normal form to get the other matrix to Jordan normal form. From that you easy see "canonical" subrepresentations.2012-11-16
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    As you can see from Julian's answer, and also his answer to your previous question http://math.stackexchange.com/questions/227909/the-semisimple-and-local-properties-of-path-algebras , it is in general difficult to determine simples of path algbera which are infinite dimension. For finite dimensional ones, standard result says they correspond to putting 1-dim vector space on one vertex...But I guess this is not you want to know?2012-11-16
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    I forgot to mention this related mathoverflow question: http://mathoverflow.net/questions/73989/the-jacobson-radical-of-an-infinite-dimensional-algebra2012-11-18

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