I have been reading Lawson & Michelsohn's Spin Geometry, there is a sentence in the proof of Proposition $1.2$ that I don't understand. The proposition is the following:
For any quadratic form $q$, the associated graded algebra of $\text{Cl}(V,q)$ is naturally isomorphic to the exterior algebra $\Lambda^*V$.
In the proof they define a map $\bigotimes^rV\xrightarrow{\pi_r}\mathscr{F}^r\to\mathscr{F}^r/\mathscr{F}^{r-1}$, $v_{i_1}\otimes\cdots\otimes v_{i_r}\mapsto[v_{i_1}\cdots v_{i_r}]$, where $\pi_r$ is used in the definition of Clifford algebra, the quotient of the tensor algebra of $V$ by the ideal $\mathscr{I}_q$ which is generated by $v\otimes v+q(V)1$ for $v\in V$. The sentence I don't understand is
The kernel of $\bigotimes^r V\to\mathscr{F}^r/\mathscr{F}^{r-1}$ consists of the $r$-homogeneous pieces of elements $\varphi\in\mathscr{I}_q$ of degree $\leq r$.
(If we write $\varphi=\sum a_i\otimes(v_i\otimes v_i+q(v_i))\otimes b_i$, then the $r$-homogeneous part is of the form $\varphi_r=\sum a_i\otimes v_i\otimes v_i\otimes b_i$.)
Questions: Why is the kernel that? Why is it called the homogeneous part, isn't $q(v_i)$ of the same degree as $v_i\otimes v_i$?