For a basic hereditary algebra A over algebraic closed field K, prove its Ext-quiver $\Gamma_{A}$ is a full sub-quiver of its AR-quiver $\Delta_{A}$. I have no clue for this.
2026-03-25 04:43:05.1774413785
Should Ext-quiver be a full sub-quiver of its AR-quiver for a basic hereditary algebra A over algebraic closed field K?
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Yes, this is true.
You can check that for arrow $j\to i$ in the Ext-quiver, the corresponding map $P_i\to P_j$ (resp. $I_j\to I_i$) is irreducible. In particular, the full subquiver of the AR-quiver consisting of the vertices corresponding to the projective indecomposables (resp. injective indecomposables) is the $Q^{op}$ (resp. $Q$), where $Q$ is the Ext-quiver of $A$. In fact, since AR-translation $\tau= DTr$ and its inverse $\tau^{-1}=TrD$ is not defined for projectives and injectives respectively, the projective indecomposables (injective indecomposable) lie in the leftmost (resp. rightmost) part of the AR-quiver.