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I'm trying to compute some examples and I'm unable to come up with a following example:

What is(are) the example(s) of an acyclic quiver $Q$ with relations such that the 2-Kronecker quiver is NOT a sub-quiver of $Q$ and the corresponding finite-dimensional algebra $A=\mathbb{K}Q/I$, where $I\subseteq\mathbb{K}Q$ is an admissible ideal of relations of the quiver $Q$ and $\mathbb{K}$ is an algebraically closed field, is a tame, triangular algebra of global dimension at least $3$ and is brick-infinite(i.e., $A-\operatorname{mod}$ has infinitely many bricks, up to isomorphism)?

Edit: triangular means that the quiver $Q$ is acyclic. An element $M\in A-\operatorname{mod}$ is called a brick if $\operatorname{End}_{A}(M)\cong\mathbb{K}$.

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  • $\begingroup$ Cross-posted at MSE: math.stackexchange.com/q/4543412/884739 $\endgroup$
    – It'sMe
    Oct 2, 2022 at 5:46
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    $\begingroup$ Could you define "triangular tame algebra"? I've come across the term "tame triangular algebra", but I wouldn't expect it to be a familiar term, even to many specialists in quiver algebras. And I'm not even sure that's what you mean, anyway. $\endgroup$
    – Jeremy Rickard
    Oct 2, 2022 at 11:16
  • $\begingroup$ @JeremyRickard I mean "tame triangular algebra". I'll make the correction. $\endgroup$
    – It'sMe
    Oct 2, 2022 at 12:02
  • $\begingroup$ Tensor the path algebra of Dynkin type $A_2$ three times and you get an algebra with all properties except that I am not sure whether it is brick-infinite (is it?). $\endgroup$
    – Mare
    Oct 3, 2022 at 10:04
  • $\begingroup$ @Mare What does the quiver look like and what are your relations? $\endgroup$
    – It'sMe
    Oct 3, 2022 at 10:09

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