TITLE: Can we turn $\mathbb{R}^n$ into a field by changing the multiplication?
QUESTION [8 upvotes]: 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? $\mathbb{R}^n$?

REPLY [7 votes]: 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.