# 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?

## 1 Answer

$\newcommand\rad{\operatorname{rad}}$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.

• Thanks Mariano, I will give this some thought. However, do you mind expanding a bit on the first bullet point? This was my initial thought, but I was having difficulty just picturing how the induced map would act. – Jacqueline Pauwels Apr 15 '12 at 7:52
• The clifford algebra $C(\beta)$ is generated by the elements of $V$ and that of $C(\bar\beta)$ is generated by those of $V/\ker\beta$. There is an obvious way of mapping a generator of $C(\beta)$ to $C(\bar\beta)$! – Mariano Suárez-Álvarez Apr 15 '12 at 8:48
• Yikes! My mistake. So looking at the third bullet point, the elements of $\ker\beta$ map to zero in $C(\bar{\beta})$ since $\bar{\beta}$ is the induced form on $V/\ker\beta$, correct? Thus $I\subset\ker p$. But I'm still having trouble seeing why $I^n=0$ for some $n$. Could you please elaborate, if it's not a burden? – Jacqueline Pauwels Apr 17 '12 at 7:56
• The ideal is generated by finitely many nilpotent elements which are central. That is enough to show that the ideal is nilpotent. A typical element of the ideal is of the form $u=a_1v_1+\cdots a_kv_k$ with $\{v_1,\dots,v_k\}$ a basis of $\ker\beta$ and $a_1,\dots,a_k$ arbitrary elements of $C(\beta)$. Compute $u^{k+1}$ and show that it is zero. – Mariano Suárez-Álvarez Apr 18 '12 at 6:42
• I hope this is as simple as I think it is. In the expansion of $u^{k+1}$, each summand will have at least some factor $v_i^2$ for some $i$. But $v_i^2=0$, so $u^{k+1}=0$. Thanks again, I hope I didn't bother you. – Jacqueline Pauwels Apr 18 '12 at 7:16