# Why is a semisimple (Wedderburn) ring von Neumann regular?

In exercise 6 of chapter 5 Passman's "A Course in Ring Theory" he asks us to prove that every semisimple (Or Wedderburn as he calls it) ring is von Neumann regular. It is easy to show that a direct product of von Neumann regular rings remains von Neumann regular, so by the Artin-Wedderburn theorem I just need to show that a full matrix ring over a division ring is von Neumann regular.

I believe I can do this by proving that $\operatorname{End}_DV$, where $V$ is a right module over a divison ring $D$ , is von Neumann regular, and then using the fact that $M_n(D)\cong \operatorname{End}_DD^n$. However, this way seems rather long, and I was wondering if there is a slicker way of showing that all semisimple rings are von Neumann regular?

• "... In particular, any module over a semisimple ring is injective and projective. Since "projective" implies "flat", a semisimple ring is a von Neumann regular ring." $\$ en.wikipedia.org/wiki/Semisimple_module#Semisimple_rings – Berci May 16 '17 at 22:25
• The result you want to prove is corollary 4.10 in Rotman's book "An introduction to homological algebra". Even more, the result used in rschwieb's answer is theorem 4.9. – Xam May 17 '17 at 1:34

A ring $R$ is von Neumann regular if and only if each principal left ideal is generated by an idempotent. In a semisimple ring, every left ideal is a direct summand.

• This is exactly what I was looking for. Thank you for the elegant solution – K.Power May 16 '17 at 22:30
• I am now looking at the converse. If a ring is Artinian and von Neumann regular then it must be semisimple. I'm trying to use contradiction by assuming the Jacobson radical is non-empty, but the contradiction is not showing itself. Will this approach work? – K.Power May 17 '17 at 0:04
• @K.Power You should just say "since it is right Artinian, it is right Noetherian, and all right ideals are f.g. By regularity, these are all summands. Thus all right ideals are summands, and the ring is semisimple." – rschwieb May 17 '17 at 1:27
• @K.Power It's easy to show that von Neumann regular rings all have Jacobson radical zero. If $a=axa\in J(R)$, then (by one definition of the Jacobson radical) $1-ax$ must be a unit of $R$. But $(1-ax)a=0$, so if $1-ax$ is a unit, $a=0$. – rschwieb May 17 '17 at 1:31
• @rschwieb thanks so much. That's an easier argument than the one I was pursuing. – K.Power May 17 '17 at 9:55

Another one, since the path of least resistance is already in egreg's answer.

All right $R$ modules are projective, hence they are flat, and this is equivalent to von Neumann regularity.

• Thank you for the alternative answer. I have not yet encountered flatness, as I will only manage to get to the chapter on tensor products in a few weeks. I will return to fully appreciate this answer then. – K.Power May 16 '17 at 22:40