Can we turn $\mathbb{R}^n$ into a field by changing the multiplication? Of course $\mathbb{R}$ is a field with usual addition and multiplication.  When we move up a dimension into $\mathbb{R}^2$, however, there is not a clear way to multiply two vectors together to get something useful.  In fact, if we define multiplication of two vectors component-wise (as is arguably the most natural way), we get something that isn't even an integral domain.  However, if we implement the multiplication
$$ (a, b)(c, d) \mapsto (ac - bd, ad + bc), $$
then we obtain a copy of $\mathbb{C}$, which is again a field.  Can we do this for higher dimensions? That is, is there some clever multiplicative structure on $\mathbb{R}^3$ that produces a field? $\mathbb{R}^n$?
 A: It is a consequence of the theory of characteristic classes of vector bundles that the existence of a bilinear product $\mathbb{R}^n \times \mathbb{R}^n \rightarrow \mathbb{R}^n$ without zero divisors implies that $n$ is a power of 2. In fact it is only possible for $n=1,2,4,8$ where in dimensions 4 and 8 we get quaternions and octonions both of which are not fields because commutativity fails for both and associativity of the product fails for the octonions.
A: No, there is no field extension of $\mathbb R$ of degree $n$ other than for $n=2$, and we get something isomorphic to the complex numbers. 
At least with hindsight, this is not so surprising, if not "obvious", because $\mathbb C$ is algebraically closed (from Liouville's theorem, a corollary of Cauchy's theorems).
Yes, in some ways this is unfortunate, since certain scenarios are precluded. An example of a way to circumvent these obstacles is Hamilton's creation of "quaternions", a four-dimensional $\mathbb R$-vectorspace, which unfortunately (and surprisingly, tittilatingly, in Hamilton's time) produces a non-commutative ... thing.
That is, $\mathbb R$ is "so close" to being algebraically closed that it admits only something isomorphic to $\mathbb C$ as literal field extension. And, perhaps even-more-disappointingly, only $\mathbb H$ as division-algebra extension.
A: As an abelian group $\mathbb R^n$ is isomorphic to $\mathbb R$, so you just need to fix an isomorphism and use it to transport the structure.
A: A clifford algebra gives a natural way to talk about products of vectors, using the geometric product.  The geometric product of vectors is the associative and distributive.  If $a, b, c$ are vectors, then
$$(a + b) c = ab + ac, \quad a(b+c) = ab + ac, \quad (ab)c = a(bc) = abc$$
These products of several vectors, along with their linear combinations, are referred to as multivectors.  The clifford, or "geometric", algebra thus forms a ring.  However, you might notice that I left out commutativity of multiplication.  The geometric product is not in general commutative, so while a field cannot in general be built this way, the ring structure of a geometric algebra can be quite useful and generalizes to all dimensions.
