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I was recently reading through Jacobson's Basic Algebra. I got to the section on Clifford algebras, and have the following question.

Let $Cl_\omega$ be the Clifford algebra with bilinear symmetric form $\omega$ on a vector space $V$.

I hear that the Jacobson radical $\text{rad}(Cl_\omega)$ is generated by $\ker\omega$, but it's not immediately clear to my why it is. How can one see this? Thanks.

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    Darn, is nobody interested in my question? :( Is there a way to make it better?2012-04-23
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    Personally I do not know what you mean by "radical." Do you mean the Jacobson radical?2012-04-23
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    @QiaochuYuan I do, I will add that, thanks.2012-04-23
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    This question has already been asked (and answered!) in this site. Search for it.2012-04-23
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    Hopefully you found it: http://math.stackexchange.com/questions/131953/quotient-of-a-clifford-algebra-by-its-radical-is-a-clifford-algebra2012-04-26
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    Thanks @rschwieb. I did find it, but I guess this is good for any who sees this question in the future.2012-05-03

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