Defining (|R)^n in a FieldDate: 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/ |
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