I have a problem with part of the proof about dimensionality of Clifford algebra over finite dimensional vector space, but let me just state for reference what that is (Lang, Algebra XIX.4.):
Let $E$ finite dimensional vector space over a field $k$ and $g$ a symmetric form. Clifford algebra is a universal object $\rho : E \rightarrow C_g(E)$ in category of all linear maps $\psi : E\rightarrow L$ where $L$ is a unitary algebra, such that $\psi(x)^2 = g(x,x)*1_L$, i.e. such that there exists a unique algebra-homomorphism $\psi^* : C_g(E) \rightarrow L$ such that $\psi = \psi^*\rho$.
Anyway, I have a problem with part of the proof proving that its dimension is $\leq 2^n$. I understand clearly that if we find such $L$ so the dimension of $\psi(E)$ in $L$ is $2^n$ then (since $\psi^*$ is surjective) we will have $=2^n$.
Idea is that we take orthogonal basis of $E$: $\{v_i\}$ (always exists) and derive some identities about $e_i=\psi(v_i)$. The identites are: $$e_i^2=g(v_i,v_i)$$ and $$e_ie_j=-e_je_i$$
So the subalgebra of $L$ generated by $\psi(E)$ is generated as vector space over $k$ by elements $$e_1^{k_1}\ldots e_n^{k_n}\quad k_i\in\{0,1\}$$ and dimension of this algebra is obviously less than $2^n$, that is all very clear. But how do we get that $\dim C_g(E)\leq 2^n$. I know that Clifford algebra is generated by $E$ (or rather, by $\rho(E)$), so generated by $\prod_i \rho(v_i)$. The above identity implies $$\psi^*(\rho(v_i)\rho(v_j) +\rho(v_j)\rho(v_i))=0$$ but that doesn't mean commutativity, only that this identity is in the kernel of $\psi^*$. What's the trick here?
Here, as with universal enveloping algebras of Lie algebras, we put a filtration on the whole algebra: the "degree" of elements of the vector space is $1$, and the "degree" of any element is the smallest $d$ such that the element is a linear combination of monomials of vectors of degree at most $d$. For example, in a universal enveloping algebra of a Lie algebra, $xy-yx$ might seem to be of degree $2$, but it is $[x,y]$, so is of degree $1$.
In those universal enveloping algebras, and in Clifford algebras, vectors do not commute, but do nearly commute modulo lower-degree elements. In the enveloping algebra case, again, $xy=yx+[x,y]$. In a Clifford algebra, it's nearly anti-commuting: $Q(x+y)=Q(x)+xy+yx+Q(y)$, with the values of $Q$ being in the base field, shows that $yx=-xy+\hbox{lower-order}$.
Thus, in a Clifford algebra, given any monomial in vectors, switching two adjacent factors multiplies the whole by $-1$, modulo lower-degree terms.