Why is the representation dimension of an Artin algebra never equal to 1? - MathOverflow 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 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 Answer by Hugh Thomas for Why is the representation dimension of an Artin algebra never equal to 1? 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>