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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?

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We have $V=V_1\oplus V_2$, where $V_2=\{x\in V\,|\,g(x,y)=0 \;\forall y\in V\}$. Assume momentarily that $V_2=\{0\}$ and hence $V=V_1$. Given $x\in V$, $x\ne0$, take $y\in V$ such that $g(x,y)\ne0$. Then in $Cl(V,g)$ we have $$ [x+y][x+y]=g(x+y,x+y)1=g(x,x)1+2g(x,y)1+g(y,y)1; $$ On the other hand, $$ [x+y][x+y]=[x][x]+[x][y]+[y][x]+[y][y]=g(x,x)1+[x][y]+[y][x]+g(y,y)1. $$ By subtracting, we find that $$ [x][y]+[y][x]=2g(x,y)1\ne0, $$ which proves the claim in our particular case.

If $V_1=\{0\}$ and $V=V_2$ (i.e., $g\equiv 0$), then $Cl(V,g)$ is the exterior algebra of $V$ and we know that the claim is true (assuming proprerties of the exterior algebra to be known). In the general case, $Cl(V,g)$ is the direct sum of two unital algebras $Cl(V,g)=Cl(V_1,g|_{V_1})\oplus Cl(V_2,0)$. For the summands, we already know the theorem. Now it is an easy exercise (left to the OP) to prove the claim for the direct sum.

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