Quotient of a Clifford algebra by its radical is a Clifford algebra?
I'm fumbling a bit in my reading on Clifford algebras. I'm hoping someone can shed some light on the following isomorphism.
Suppose you have a symmetric bilinear form $G$ over a vector space $V$, and let $\mathrm{Cl}_G(V)$ be the corresponding Clifford algebra. I'll denote it by $C_G$ for short when the vector space is clear.
Now $G$ induces a form $\hat{G}$ on $V/\ker(G)$, and apparently $C_G/\mathrm{rad}(C_G)\cong C_{\hat{G}}$, where $\mathrm{rad}(C_G)$ is the radical of $C_G$. I thought this will fall out easily from some application of the isomorphism theorems, but some epimorphism $C_G\to C_{\hat{G}}$ whose kernel conveniently happens to be $\mathrm{rad}(C_G)$.
However, I can't quite find such a map. Does anyone see what the trick is here? Thanks.
Later. I believe I now understand that there is a surjective map $p:C(\beta)\to C(\bar{\beta})$ induced by the quotient map $V\to V/\ker\beta$ described below. If $I$ is the ideal generated by $\ker\beta$, then since $p(\ker\beta)=0$ in $C(\bar{\beta})$, it follows that $I\subset\ker p$. However, I don't understand why $I$ is nilpotent. What is the explanation for this last bit?
$\newcommand\rad{\operatorname{rad}}$Let $\beta$ be an arbitrary symmetric form on a vector space $V$, let $\ker\beta$ be its kernel and let $\bar\beta$ be the induced non-degenerate form on $V/\ker\beta$.
Check that there is an surjective algebra map $p:C(\beta)\to C(\bar\beta)$ which is induced by the quotient map $V\to V/\ker\beta$ whose kernel is generaled by $\ker\beta$.
Notice that since the codomain of $p$ is a semisimple algebra (it is the Clifford algebra of a non-degenerate form), its kernel contains the radical $\rad C(\beta)$ of $C(\beta)$.
Now, every element of $\ker\beta$ is in the center of $C(\beta)$ and squares to zero. Check that the ideal $I$ generated by $\ker\beta$ in $C(\beta)$ is nilpotent and contained in $\ker p$. It follows that it is contained in the radical $\rad C(\beta)$.
Rejoice.