This is a follow up question to this question on Clifford Algebras. As I understand it, if the associated bilinear form $\beta$ of a Clifford Algebra $C_{\ell}$ is non-degenerate, then $ker\beta$ is equal to $0$. This then implies that rad($Cl_{\ell}$) = {0}, since ) is the only thing that acts as zero in the Clifford Algebra.

But, if the bilinear form is degenerate, then the quotient $C_{\ell}/(rad(C_{\ell}))$ is not isomorphic to $C_{\ell}$, but rather to a smaller algebra. Can we say anything more than this?


Denote by $C\ell(V,\beta)$ the Clifford algebra of $V$ with bilinear form $\beta$.

Also define the set $\ker(\beta):=\{v\in V\mid \forall x\in V,\beta(v,x)=0\}$. It turns out that the ideal generated by $\ker(\beta)$ is exactly the Jacobson radical $C\ell(V,\beta)$, which I guess I'll denote as $rad(C\ell(V,\beta))$.

Now the bilinear form $\beta$ induces a bilinear form $\bar{\beta}$ on $\bar{V}:=V/\ker(\beta)$ in the obvious way: $\bar{\beta}(\bar{x},\bar{y})=\beta(x,y)$ for $x,y\in V$. You can easily show that $\bar{\beta}$ is nondegenerate on $\bar{V}$.

We can conclude that $C\ell(V,\beta)/rad(C\ell(V,\beta))\cong C\ell(\bar{V},\bar{\beta})$


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