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Noetherian Leavitt path algebras and their regular algebras

Published 5 Nov 2013 in math.RA | (1311.1064v1)

Abstract: In the past, it has been shown that the Leavitt path algebra $L(E)=L_K(E)$ of a graph $E$ over a field $K$ is left and right noetherian if and only if the graph $E$ is finite and no cycle of $E$ has an exit. If $Q(E)=Q_K(E)$ denotes the regular algebra over $L(E),$ we prove that these conditions are further equivalent with any of the following: $L(E)$ contains no infinite set of orthogonal idempotents, $L(E)$ has finite uniform dimension, $L(E)$ is directly finite, $Q(E)$ is directly finite, $Q(E)$ is unit-regular, and a few more equivalences. In addition, if the involution on $K$ is positive definite, these conditions are equivalent with the following: the involution $\ast$ extends from $L(E)$ to $Q(E),$ $Q(E)$ is $\ast$-regular, $Q(E)$ is finite, $Q(E)$ is the maximal (total or classical) symmetric ring of quotients of $L(E),$ every finitely generated nonsingular $L(E)$-module is projective, and the matrix ring $M_n(L(E))$ is strongly Baer for every $n$. It may not be surprising that a noetherian Leavitt path algebra has these properties, but a more interesting fact is that these properties hold only if a Leavitt path algebra is noetherian. Using some of these equivalences, we give a specific description of the inverse of the isomorphism $V(L(E))\rightarrow V(Q(E))$ of monoids of equivalence classes of finitely generated projective modules for noetherian Leavitt path algebras. We also prove that two noetherian Leavitt path algebras are isomorphic as rings if and only if they are isomorphic as $\ast$-algebras. This answers in affirmative the Isomorphism Conjecture for the class of noetherian Leavitt path algebras: if $L_{\mathbb C}(E)$ and $L_{\mathbb C}(F)$ are noetherian Leavitt path algebras, then $L_{\mathbb C}(E)\cong L_{\mathbb C}(F)$ as rings implies $C*(E)\cong C*(F)$ as $\ast$-algebras.

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