Ideal of $TV$ which trivially intersects $V$ Let $V$ be a vector space over a fied $\mathbb{K}$ and let $TV = \bigoplus_{n \in \mathbb{N}} V^{\otimes n}$ be its tensor algebra. Let $b \colon V \times V \rightarrow \mathbb{K}$ be a bilinear form over V and let $I \subseteq TV$ the ideal of $TV$ generated by the elements of the form $v \otimes v + b(v,v)$.

I need to show that, identifing $V$ with $V^{\otimes 1}$ in $TV$, one has $V \cap I = \{ 0 \}$. It should be quite obvious, but I can't get it.

Edit: I have found a proof in Lawson, Michelsohn: Spin Geometry, but there's a passage which I don't understand:


  Let $\varphi = \sum a_i \otimes(v_i \otimes v_i + b(v_i,v_i)) \otimes b_i \in V \cap I$, where we may assume that the $a_i$'s and $b_i$'s are homogeneous. Let us now consider the sum $\overline\sum a_j\otimes (v_j \otimes v_j) \otimes b_j$, where this sum is taken over those indices with $\operatorname{deg} a_i + \operatorname{deg} b_i$ maximal. Since $\varphi \in V$, $\overline\sum a_j \otimes (v_j \otimes v_j) \otimes b_j = 0$. Then, by contraction with $b$, $\overline\sum b(v_j,v_j) ( a_j \otimes b_j ) = 0$. Proceeding inductively, we prove that $\varphi = 0$


Why do $\overline\sum b(v_j,v_j) ( a_j \otimes b_j ) = 0$? What do they mean by contraction with $b$?
Generators of $I$ are quadratic (not homogeneous because of the additive constant from b(v,v). ) So the ideal generated by them can not intersect lower homegeneous degree elements.
Generators of $I$ are quadratic (not homogeneous because of the additive constant from b(v,v). ) So the ideal generated by them can not intersect lower homegeneous degree elements.