Why is the radical of a Clifford algebra generated by the kernel of the associated symmetric form? [duplicate]
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 ...
While reading about the structure of Clifford algebra, there were two facts listed as bullet points about the center of Clifford algebra based on the parity of the dimension of the underlying vector ...