Concrete example of non-invertible element in a Clifford algebra?

Question

Are there simple examples of non-invertible elements (i.e. zero divisors) in a real Clifford algebra $C(V)$ other than $0$?

Notation: $V$ has a positive definite metric $q$, and $uv+vu=-2q(u, v)$. Therefore, if $V$ is $1$-dimensional, $C(V)$ is isomorphic to ${\mathbb C}$; if $V$ is $2$-dimensional, $C(V)$ is isomorphic to ${\mathbb H}$. Thus we have to search $C(V)$ with $V$ of dimension at least $3$...

Answer 1

Let $e_1,\dotsc,e_n$ be the standard (orthonormal) basis. If $n$ is at least 3, then $e_1$ and $e_2e_3$ commute and both square to $-1$ so $$ (e_1 + e_2e_3)(e_1 - e_2e_3) = e_1^2 - (e_2e_3)^2 = 0 $$ so both $e_1\pm e_2e_3$ are not invertible.

By the same logic $$ (1 + e_1e_2e_3)(1 - e_1e_2e_3) = 0. $$

Answer 2

Let $V$ be $3$-dimensional. Then $ (e_1e_2e_3-1)(e_1-e_2e_3)=0. $