Ideal of Clifford algebra

Question

The left ideal of an algebra $A$ is a subspace $I$ of $A$ such that $ax\in I ~~\forall a\in A, x\in I$. A two-sided ideal is a subset of $A$ that is both a left and right ideal.

In Definition 2.4.1, $I=\{v\otimes v+||v||^2:v\in V\}$, but for $-1\in \mathbb R$ , $-v\otimes v-||v||^2$ can be presented by the form $v\otimes v+||v||^2$, what is wrong ?

Answer 1

The ideal $I$ is generated by $\{v\otimes v+||v||^2:v\in V\}$, but $I \neq \{v\otimes v+||v||^2:v\in V\}$. The ideal is much larger $I=\{T_1 \otimes (v\otimes v+||v||^2) \otimes T_2:v\in V, T_1, T_2 \in T(V)\}$, here $T(V)$ is the tensor algebra.

In particular it works for $T_2=-1$ or $T_1=-1$, here $T_{1,2}$ is a degree $0$ tensor: $$ (-1) \otimes (v\otimes v+||v||^2)=(v\otimes v+||v||^2)\otimes (-1)=-v\otimes v-||v||^2 \in I. $$ And in the same way for any $x \in \mathbb{R}$: $$ x\otimes (v\otimes v+||v||^2)=(v\otimes v+||v||^2)\otimes x=x(v\otimes v+||v||^2) \in I. $$