Homomorphism Between a Geometric Algebra and its Field of Scalars? Given a geometric algebra defined over a real vector space, is is possible to construct a homomorphism between the elements of the geometric algebra and the reals?
I was pondering an example of this: constructing an algebra homomorphism between the complex numbers and the real numbers. I have a hunch that it isn't not possible, but I'm not sure why it wouldn't be?
 A: I assume you mean algebra (or just ring) homomorphism, since it's trivial to find vector space homomorphisms.
If you are mainly interested in the Clifford algebras for nondegenerate forms, then the classification theorems explain when this happens.
Basically they say that Clifford algebras (for nondegenerate real or complex forms) are of the form $R$ or $R\times R$ where $R$ is a matrix ring over a division ring.
Such a ring has, respectively, two or four ideals that could serve as the kernel of the homomorphism you're talking about, and the resulting quotient ring would have to be $1$ dimensional over $\mathbb R$, so we see that the only two possibilities are either just $\mathbb R$ itself, or $\mathbb R\times \mathbb R$.
So $CL_{0,0}(\mathbb R)$ and $CL_{1,0}(\mathbb R)$ are the only ones that permit a ring homomorphism onto $\mathbb R$. 
An algebra between $\mathbb R$ and $\mathbb C$ is impossible simply for the reason that $\mathbb C$ has no nontrivial ideals. The image of any homomorphism out of $\mathbb C$ will be $\{0\}$ or isomorphic to $\mathbb C$.
A: An algebra-homorphism is (among other things) a vector-space morphism, so a linear map. In particular, if you map from a $m$-dimensional space to a $1$-dimensional space it should have a $(m-1)$-dimensional kernel. Let's call this subspace $K$ and the full algebra $A$. Now we use the other face of algebra homomorphism, namely that it respects multiplication. Let $k$ be an element of $K$ and let $a$ be a random other element of $A$. Let $\phi$ be the homomorphism. We have $\phi(ak) = \phi(a)\phi(k) = \phi(a)*0 = 0$ so $\phi$ maps $ak$ to $0$. But being mapped to zero under $\phi$ is the definition of lying in the kernel of $\phi$, so we see that $ak \in K$ for all $a$ in the algebra and all $k \in K$. 
In other words: $K$ is an ideal of the algebra. (I show here that it is a left ideal, but by the same reasoning we can show it is a two-sided ideal.)
$\mathbb{C}$ doesn't have any non-trivial ideals because it is a field: the fact that every element has an inverse prevents the '$ak \in K$ for every $a \in A, k \in K$'-behavior. The reason is simple: if $ak \in K$ for every $a$ then in particular $1 = k^{-1}k \in K$ and hence $a = a1 \in K$ for every $a$ in the algebra. But since $K$ was defined as the kernel of $\phi$ this would mean that $\phi$ maps EVERY element of $\mathbb{C}$ to zero. Ok, strictly speaking this is a homomorphism, but probably not one of the kind you are interested in.
A bit harder to show is that other geometric algebras such as $Mat(2, \mathbb{R})$ do not have non-trivial two-sided ideals (in particular because it does have left-ideals and right-ideals) and hence no homomorphism to the ground field. Perhaps google for Wedderburn Structure Theorem to learn more about this.
By contrast: here is an example of an algebra that does have a homomorphism to the ground field: the algebra of all $n \times n$ upper triangular matrices. In fact it has (at least, but my gut feeling is exactly) $n$ such homomorphisms.
