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Given two path algebras $A$ and $B$, for example, A=: $1\to2$ B=: $1\to2\to3$, is the tensor product of A and B over a field $k$ a path algebra? if yes, how to represent it by a quiver? also, how to construct $(A,B)$-bimodules?

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1) Paths algebras (over non-trivial quivers) have homological dimension $1$, while the tensor product of two path algebras has homological dimension $\geq 2$, so it is not a path algebra 2) What kind of question is this? There are free bimodules and any other bimodule is a quotient of a free one. –  Fernando Muro May 9 '12 at 10:25
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It can be realized as a quotient of the Cartesian product of path algebras in question. You have to mod out by some commutativity relations.

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How does that work? Specifically, which element in the tensor product would be the image of the element $(a,b)$ of the Cartesian product? The "obvious" choice, $a\otimes b$ won't work, because addition in the Cartesian product ($(a,b)+(c,d)=(a+c,b+d)$) doesn't match addition in the tensor product, so you won't have a homomorphism. (Multiplication by scalars from $k$ doesn't match either unless $k$ is the two-element field.) –  Andreas Blass May 9 '12 at 12:51
    
I think he means take the Cartesian product of the quivers and then put in commutativity relations. For instance, if you take the quiver 1->2 and take the tensor product of this path algebra with itself you get the incidence algebra of the diamond 1 / \ a b \ / 0 The Hasse diagram is the quiver of this algebra, but you have the relation that the two paths to the top are the same. –  Benjamin Steinberg May 9 '12 at 21:12
    
Hmm, my picture of the Hasse diagram of a diamond didn't draw well in the comments. –  Benjamin Steinberg May 9 '12 at 21:13
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What you want is

here.

Basically you take the 1-skeleton of the products of the quivers and you say that certain paths e.g. (edge,vertex)(vertex,edge)=(vertex,edge)(edge,vertex).

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