Thread: Show element is invertible in a Clifford algebra

Here is the problem:

Let $\displaystyle F$ be a field and $\displaystyle V$ a quadratic space with an anisotropic form $\displaystyle q$ not representing $\displaystyle 1$ (i.e. $\displaystyle q(v)\neq1$ for all $\displaystyle v\in V$). Show that if $\displaystyle a\in F$ and $\displaystyle v\in V$ are not both zero, then $\displaystyle a+v$ is an invertible element in the Clifford algebra $\displaystyle C(V,q)$.

I'm not really sure how to start this problem. I don't fully understand Clifford algebras, so I don't even really know what invertible elements even look like. I understand the definition - $\displaystyle T(V)/I$, where I is the ideal generated by $\displaystyle v\otimes v-q(v)$ - but I don't understand the resulting algebra (except when $\displaystyle q$ is a binary or nonzero unary form).

Any help is appreciated.

Originally Posted by redsoxfan325

Here is the problem:

Let $\displaystyle F$ be a field and $\displaystyle V$ a quadratic space with an anisotropic form $\displaystyle q$ not representing $\displaystyle 1$. Show that if $\displaystyle a\in F$ and $\displaystyle v\in V$ are not both zero, then $\displaystyle a+v$ is an invertible element in the Clifford algebra $\displaystyle C(V,q)$.

I'm not really sure how to start this problem. I don't fully understand Clifford algebras, so I don't even really know what invertible elements even look like. I understand the definition - $\displaystyle T(V)/I$, where I is the ideal generated by $\displaystyle v\otimes v-q(v)$ - but I don't understand the resulting algebra (except when $\displaystyle q$ is a binary or nonzero unary form).

Any help is appreciated.

what do u mean by "not representing 1"? do u mean $\displaystyle q(v) \neq 1$ for all $\displaystyle v \in V$?

Originally Posted by NonCommAlg

what do u mean by "not representing 1"? do u mean $\displaystyle q(v) \neq 1$ for all $\displaystyle v \in V$?

Yes.

Originally Posted by redsoxfan325

Yes.

ok then, so $\displaystyle q(v) \neq a^2$ because otherwise $\displaystyle q(a^{-1}v)=1.$ let $\displaystyle b = a(a^2 - q(v))^{-1}$ and $\displaystyle c =-ba^{-1}$ and see that $\displaystyle (a+v)(b+cv)=1.$