Example of an algebra finite over a commutative subalgebra with infinite dimensional simple modules - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T01:18:51Z http://mathoverflow.net/feeds/question/28784 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/28784/example-of-an-algebra-finite-over-a-commutative-subalgebra-with-infinite-dimensio Example of an algebra finite over a commutative subalgebra with infinite dimensional simple modules Bedini 2010-06-19T22:59:17Z 2010-06-21T15:18:16Z <p>Let \$A\$ be an algebra over an algebraically closed field \$k.\$ Recall that if \$A\$ is a finitely generated module over its center, and if its center is a finitely generated algebra over \$k,\$ then by the Schur's lemma all simple \$A\$-modules are finite dimensional over \$k.\$ </p> <p>Motivated by the above, I would like an example of a \$k\$-algebra \$A,\$ such that:</p> <p>1) \$A\$ has a simple module of infinitie dimension over \$k,\$ </p> <p>2) \$A\$ contains a commutative finitely generated subalgebra over which \$A\$ is a finitely generated left and right module. </p> <p>Thanks in advance.</p> http://mathoverflow.net/questions/28784/example-of-an-algebra-finite-over-a-commutative-subalgebra-with-infinite-dimensio/28816#28816 Answer by Bugs Bunny for Example of an algebra finite over a commutative subalgebra with infinite dimensional simple modules Bugs Bunny 2010-06-20T07:58:40Z 2010-06-21T15:18:16Z <p>Doc, this is a stinker. Your condition (2) forces your algebra to be finitely generated PI, and every little hare knows that simple modules over such algebras are finite-dimensional. See 13.4.9 and 13.10.3 of McConnell-Robson...</p>