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 watson December 26th, 2012 11:17 AM

If D is a division ring, why are D-mods semisimple?

Let D be a division ring. Why must every D-module be semisimple?

 Turgul December 27th, 2012 11:04 AM

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?

 Deveno December 27th, 2012 01:48 PM

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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