Defining (|R)^n in a Field

Date: 03/27/2001 at 22:36:23
From: Bernardo Schiffrin G.
Subject: Algebra (Fields)

What multiplication operation would define (|R)^n  in a field?

Thanks a lot.   
Bernardo Schiffrin G.

Date: 03/29/2001 at 09:27:42
From: Doctor Rob
Subject: Re: Algebra (Fields)

Hi Bernardo.

(R^n,+) can be made into a field for any integer n > 0.

Let n be any natural number. Consider R^n and R. Each is an 
infinite-dimensional vector space over Q, the rational numbers. 
Let b = {e(i): i in I} be a basis of R over Q. Let u(j), 1 <= j <= n, 
be the unit vectors in R^n. Let B = {e(i)*u(j): i in I and 
1 <= j <= n}.  This is a basis of R^n. 

b has infinite cardinality. The cardinality of B is n times that, and 
n is a finite natural number, so b and B have the same cardinality.  
That means that there is a one-to-one correspondence between b and B.  
Fix one of these, f.  Let E(i) = f[e(i)]. Then B = {E(i): i in I}
is another way of writing the basis B.

Now define a Q-linear transformation T: R^n -> R via

   T( SUM  a[i]*E[i]) =  SUM  a(i)*e(i),
     i in I             i in I

where all a(i) are in Q, and all but a finite number of them are zero.  
This is one-to-one and onto. Now define addition by

   a + b = T^(-1)(T[a]+T[b]).

T maps a and b into R. You do the addition in the usual way there, 
and then apply T^(-1) to map back into R^n. Similarly define 
multiplication by

   a * b = T^(-1)(T[a]*T[b]).

This makes T an isomorphism between R^n and R. This makes R^n into a 
field, and the addition defined in this way is the same as the usual 
component-wise addition in R^n, because each E(i) is nonzero in only 
one of its components.

Thus R^n is a field, and is isomorphic as a field to R.  This holds 
true for any positive integer n.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/