6
$\begingroup$

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.

  • 0
    Thanks @rschwieb. I did find it, but I guess this is good for any who sees this question in the future.2012-05-03

0 Answers 0