Proving R^n Can Be a Field: The n>2 Case

[dreamtheater]

Is there a proof that shows if R^n can be turned into a field, for specific n?

Obviously, n=1 is a field, and n=2 can be made into a field (which is just the complex plane.)

So what about n>2?

 

[Peeter]

What invertable, closed, multiplication and division operation are you going to define for n>2?

Even for R^2, your complex numbers, that's a space that's isomorphic with R^2, and is not R^2 itself. You can introduce invertable vector product for R^n (I like the geometric/clifford algebra product for this). But to get invertable and closed with that product you have to combine components of such vector products (ie: complex numbers, and quaternions, or other generalizations of these get by adding grade 0 and 2 components from this larger algebraic space).

 

[SiddharthM]

it is a theorem that for dimensions three or higher euclidean space cannot be turned into a field, or if we want to be pedantic: for n >/= 3 there is no field isomorphic to R^n.

I saw a proof of the fact in a complex analysis class sometime ago so I don't remember it but it uses (very) elementary methods and as such should be found easily.

 

[matt grime]

I have a small technical problem with the explanation given in 3: in what sense are you asserting that there are no fields isomorphic to R^n for n>=3? Or more accurately, in what category?

I think that it might be better to say - if F is a field, and F is in VECT(R) - cat of real vector spaces - then F has dimension 1 or 2.