most recent 30 from http://mathoverflow.net 2013-05-25T15:42:32Z http://mathoverflow.net/feeds/question/116525 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/116525/why-is-the-representation-dimension-of-an-artin-algebra-never-equal-to-1 Bernhard Boehmler 2012-12-16T14:01:47Z 2012-12-17T02:30:07Z

<p>Hi,</p> <p>in 1971 M.Auslander showed that the representation dimension of $A$ is $\neq 1$ for every Artin algebra $A$.</p> <p>Does anybody have a reference paper or book proving this? Is the proof easy and / or does it need many prerequisites?</p> <p>Thanks for the help.</p>

http://mathoverflow.net/questions/116525/why-is-the-representation-dimension-of-an-artin-algebra-never-equal-to-1/116558#116558 Hugh Thomas 2012-12-17T02:30:07Z 2012-12-17T02:30:07Z

<p>First of all, you have to assume that $A$ is non-semi-simple. For a semi-simple Artin algebra, the representation dimension is defined to be 1. </p> <p>For a non-semi-simple algebra, the representation dimension is, by definition, the smallest $d$ such that there exists $M$ an $A$-module which is both a generator and a co-generator, and such that the global dimension of the endomorphism ring of $M$ is $d$. </p> <p>To show the representation dimension of $M$ is not 1, we need to show that the endomorphism ring of $M$ is not hereditary. </p> <p>Since $M$ is a generator and a co-generator, it contains all the projective indecomposables and all the injective decomposables as direct summands. </p> <p>If $A$ is non-semi-simple, then it has a projective indecomposable module $P$ which is not simple. Let $Q$ be another projective which has a non-zero map to $P$. Suppose $P$ is the projective cover of the simple $S$, and let $I$ be its injective hull. Then the composition of the maps from $Q$ to $P$ and from $P$ to $I$ is zero. This shows that there are relations among the elements the endomorphism ring of $M$. It follows that the endomorphism ring is not hereditary. </p>