This should be a very easy question, but the proof in Lawson/Michelson (Spin geometry) is wrong and I do not find a really correct and complete argument:


Let V be a nonzero real vector space with scalar product: Why is the Clifford-algebra (constructed from the tensor algebra by quotiening out an ideal) non-zero?



Clifford algebra non-zero
To show that an algebra constructed as a quotient of the tensor algebra of a vector space is nonzero, one of the main ways to go is to construct representations. We can do this for the Clifford algebra as follows.


Let $V$ be a vector space over a field $k$ and $(,):V\times V \to k$ a symmetric bilinear form on $V$. The Clifford algebra (for this form) is given by 
$$Cl(V)= T(V)/\langle v \otimes v - (v,v)\rangle.$$
We will construct a representation of the Clifford algebra on the exterior algebra $\bigwedge (V)$.


For $v \in V$, define two $k$-endomorphisms of $\bigwedge(V)$ by
$$ l\_v(x) = v \wedge x$$
and
$$ \delta\_v(x) = \sum\_{j=1}^k (-1)^{j-1}(v,x\_j) x\_1 \wedge \dots \wedge \widehat{x\_j} \wedge \dots \wedge x\_k$$
if $x = x\_1 \wedge \dots \wedge x\_k$.


Then check that $l\_v^2 = \delta\_v^2 = 0$, and moreover that $l\_v \delta\_v + \delta\_v l\_v = (v,v) \cdot \mathrm{id}$.


Extend the map linear $v \mapsto l\_v + \delta\_v$ to an algebra homomorphism from the tensor algebra $T(V)$ to $\mathrm{End}\_k(\bigwedge(V))$. By the previous remark, this descends to a map, let's call it $\phi$, from the Clifford algebra to $\mathrm{End}\_k(\bigwedge(V))$.


In particular, $\phi(v)1 = v$, so $V$ injects into the Clifford algebra.


Edit: I believe also that the map
$$ x \mapsto \phi(x)1$$
gives a linear isomorphism of the Clifford algebra with the exterior algebra.


A great reference for this stuff is Chevalley's monograph, The Algebraic Theory of Clifford Algebras and Spinors.


