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Prove that every multivector which does not have an inverse has an idempotent for a factor.

Define an idempotent as a multivector $A$ with the property that $A^2=A$ and $A \neq 1$.

I can show it for specific cases, such as, $B = \beta + \mathbf b$, $\beta$ a scalar and $\mathbf b$ a vector, and $C = \langle C \rangle_0 + \langle C \rangle_1 + \langle C \rangle_2$. But I can't figure out how to show it in the general case.

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A geometric algebra over a nonsingular finite dimensional $F$-vector space is a semisimple ring, so it is von Neumann regular. As such, for every $a$ in the ring, there exists $x$ such that $axa=a$.

From this you can compute that $ax$ is an idempotent, and clearly it's a factor of $a$. If $ax=1$, then since the ring is Artinian it is also true that $xa=1$, and $a$ would be a unit. So if $a$ is not a unit, $ax\neq 1$.

rschwieb
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  • Ah. I guess my 1 semester of abstract algebra isn't enough for this question because I've never heard of semisimple rings (and it doesn't seem to even be in Artin's book which I have right here). So I'll just have to trust that this is correct and I'll accept it. –  Jun 30 '14 at 19:46
  • @BobDylan Which book are you looking at? There is more than one Artin, and there is more than one book by both Artins. – rschwieb Jun 30 '14 at 19:47
  • The book I used in my abstract algebra class was Algebra by Michael Artin. ----- I got this question from New Foundations for Classical Mechanics by Hestenes. He doesn't actually ask the reader to prove this, he just states it as a fact. So I thought I'd take a crack at it. –  Jun 30 '14 at 19:48
  • @BobDylan Thanks for letting me know. That's the fastest way I know to prove this: through a little abstract algebra. The basics of semisimple rings are worth picking up, if you have the time :) – rschwieb Jun 30 '14 at 19:53