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 December 26th, 2012, 12:17 PM #1 Senior Member   Joined: Nov 2011 Posts: 100 Thanks: 0 If D is a division ring, why are D-mods semisimple? Let D be a division ring. Why must every D-module be semisimple?
 December 27th, 2012, 12:04 PM #2 Senior Member   Joined: Aug 2010 Posts: 195 Thanks: 5 Re: If D is a division ring, why are D-mods semisimple? If $k$ is a field, why must every $k$-module (ie vector space) be semisimple? What are the irreducible $k$-modules?
 December 27th, 2012, 02:48 PM #3 Senior Member   Joined: Mar 2012 Posts: 294 Thanks: 88 Re: If D is a division ring, why are D-mods semisimple? if D is a division ring, then a D-module is free (it satisfies, for example, all the usual axioms of a vector space). in particular, division rings have the invariant basis number property. we can thus regard any D-module M as a direct sum: $M= \bigoplus_{i \in I} D_i$ where: $D_i= D\ \forall i \in I$ since D is a division ring, it has no non-trivial proper D-submodules (which would necessarily be non-trivial proper ideals of D). this is because every element of D-{0} is a unit.

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