Is it possible to make R^n a field for n>2 I know that real line is a field. And we can make $\mathbb R^2$ a field by defining complex operations. But can we extend this process? Is there any way to  make $\mathbb R^n$ field, at-least for some positive integer '$n$'? 
In particular, using similar constructions, which we used to get complex field from $\mathbb R^2$, can we make $\mathbb C^2$ field? If yes, this means we converted $\mathbb R^4$ in to a field.
 A: A division algebra is a ring in which every non-zero element has a multiplicative inverse (here multiplication is not necessarily commutative like in a field).
According to http://mathworld.wolfram.com/DivisionAlgebra.html the only finite-dimensional real division algebras occur for dimensions $n=1,2,4,8$.
If $n=4$ (quaternions) or $n=8$ (octonions) multiplication is noncommutative and there is no field structure. The proof of these facts are not simple. 
However there is a short proof that if $n>1$ and $\mathbb{R}^n$ is a division algebra (or a field) then $n$ is not odd. 
In fact, by the linearity of the multiplication, we can consider the map
$$L:\mathbb{R}^n\to \mbox{GL}(n,\mathbb{R})\cup \{0\}$$
such that $v\cdot w=L(v)w$. We have that $L(v)=0$ iff $v=0$. Moreover $L$ is linear and therefore continuous ($\mbox{GL}(n,\mathbb{R})$ is finite-dimensional).
Let $u$ be the neutral element of the multiplication and take a continuous path $\varphi:[-1,1]\to \mathbb{R}^n$ from $\varphi(1)=u$ to $\varphi(-1)=-u$ such that $\varphi(t)\not=0$ for any $t\in [-1,1]$ (use the fact that $n>1$).
Notice that $L(\pm u)=\pm I$ where $I$ is the $n\times n$ identity matrix.
Let $f(t)=\det(L(\varphi(t))$ then $f:[-1,1]\to \mathbb{R}$ is continuous, $f(1)=1$, $f(-1)=(-1)$ and by the Intermediate Value Theorem there is $t_0\in (-1,1)$ such that $f(t_0)=0$ that is $L(\varphi(t_0))\not\in \mbox{GL}(n,\mathbb{R})$. Contradiction.
