Let $K$ be a field (of characteristic $\neq 2$ if that matters) and $V$ a finite dimensional vector space over $K$, $dim(V)=n$. Let $g$ be a symmetric bilinear form on $V$.
By $Cl(V,g)$ I denote the clifford algebra associated to $V$ and $g$.
$Cl(V,g):=T(V)/I$ where $T(V)$ is the tensor algebra of $V$ and $I$ is the twosided ideal generated by $\{x\otimes x + g(x,x)1|\text{ } x\in V\}$.
My question regards the proof of the follwing claim
Claim: The linear map $i\colon V\rightarrow Cl(V,g)$ that maps $v\in V$ to $[v]\in Cl(V,g)$ is injective.
It is often mentioned that the Injectivity follows from the above construction of the clifford algebra, see e.g. http://en.wikipedia.org/wiki/Clifford_algebra. In "Dirac Operators in Riemannian Geometry" (Friedrich) the injectivity of the above map is stated as a corollary of the existence of $Cl(V,g)$ (but not proved).
This appears to me as if the injectivity would be a simple fact that is easily proven, however I don't know a "simple" proof. Am I missing something?
I know of two ways to prove the claim: One can use representation theory (see https://mathoverflow.net/questions/68378/clifford-algebra-non-zero), or one can first prove that the dimension of $Cl(V,g)$ is $2^{n}$, then show that the elements $[e_{i_1}]\cdot\ldots\cdot[e_{i_k}]$, $1\le i_1<\ldots <i_k\le n$, $1\le k\le n$ and $1$ generate $Cl(V,g)$ and therefore must be a basis and the injectivity of $i$ follows since it maps basis vectors to basis vectors.
So my question is:
Question: Are there easier (more elementary) ways to prove that $i$ is injective than those I mentioned?
