Field structure for R^n Hi!
Is it possible to define a product on R^n for n>2 such that R^n can be made into a field?
R is a field in its own right with the standard operations and R^2 can be made into a field by introducing the product (a,b)*(c,d) = (ac-bd,ad+bc) i.e. the product of the complex numbers.
 A: No if you use the usual additive structure on $\mathbb{R}^{n}$ for the field addition; but if you give up commutativity of multiplication, you have the skew-field of hamiltonians, or quaternions, on $\mathbb{R}^{4}$, and if you then give up associativity of multiplication, you have the non-associative Cayley algebra, or octonians , on $\mathbb{R}^{8}$.  The Cayley-Dickson process builds the complex field from the real field, the skew-field of hamiltonians from the complex field, the non-associative Cayley algebra from the hamiltonians, and in general a $2^{n+1}$-dimensional involutive Cayley-Dickson algebra from the $2^{n}$-dimensional involutive Cayley-Dickson algebra.  A.A. Albert did much to articulate the state of affairs in the early part of the twentieth century, if memory serves.
A: It is not possible to do this beyond a certain point. R^2 can be made into a field the complex numbers, then after that some of the structure is lost with each doubling the quaternions do not have commutative multiplication, the octonions do not have commutative multiplication or associativity and finally after that after doubling octonions there is zero division. John Conway has a book about octonions that discusses this. There is a review here
A: No.
A: I guess it depends on what properties of $R^n$ you want. This is kind of a stupid answer but you can certainly pick a bijection $R^n \to R$ and define a field structure using that. It's not going to preserve anything nice about $R^n$ though.
A: No. $\mathbb{C}$ is algebraically closed, so any finite extension of $\mathbb{R}$ must be isomorphic to a subfield of $\mathbb{C}$, and therefore has dimension no more than $2$.
