From: wcw@math.psu.edu (William C Waterhouse) Subject: Re: Fields Date: 15 Oct 1999 21:33:57 GMT Newsgroups: sci.math Keywords: Why no 3-dimensional field extension of R? In article <7u2kah$4v3$1@nnrp1.deja.com>, wrote: > I've always heard it's impossible to construct a mathematical field > with dimension greater than three. So, the set of complex numbers is > the "largest" field you can have and that satisfies those algebraic > rules we find in the real number system for the addition and the > product operations. We can have vector spaces of any dimension, but we > can't, for instance, have a four-dimensional field. >... > Has anyone here ever seen that proof and could explain to me, even if > superficially, why we can't have a field of dimension over 3? The proof that there can't be a (commutative) field of finite dimension bigger than 3 over the real numbers rests basically on the fact that every real polynomial factors into factors of degree 1 or 2. (This is equivalent to saying that every complex polynomial has a complex root.) Here's an elementary argument from there. Let K be a finite-dimensional field over the reals. Take any b in K not itself a real number. The powers 1, b, b^2, b^3, ... cannot all be linearly independent, since K has finite dimension. Thus there is some equation a_0 b^n + a_1 b^{n-1} + ... + a_{n-1} b + a_n = 0, where the a_i are real. Dropping vanishing terms, we can assume a_0 is nonzero, and then we can divide by it and suppose a_0 = 1. This now means that b is a solution of the equation X^n + a_1 X^{n-1} + ... + a_n = 0. By assumption, we can factor this polynomial as a product of real polynomials f_j(X) of degree at most 2. Since K is a field, the product of the f_j(b) can only be zero when one factor is zero. Thus b satisfies an equation of degree 1 or 2. We can't have b satisfying an equation of degree 1, because then it would be real. So it satisfies an equation of the form X^2 + pX + q = 0 with p^2 - 4q negative. As usual, then, b-p/2 has square equal to p^2/4 - q. There is a real number r with r^2 = q-p^2/4, and we see that i = (1/r)b - p/2r has square equal to -1. Thus the span of 1 and b is a field that is just a copy of the complex numbers. Finally, I claim there are no other elements in K. If c were one outside the span of 1 and b, then the same proof shows that c satisfies some quadratic real equation X^2 + sX + t = 0. But that equation already has two solutions in the span of 1 and b, since a square root of the negative number s^2 - 4t is available there -- call those roots d and e. Then X^2 + sX + t = (X-d)(X-e), and we must have 0 = c^2 + sc + t = (c-d)(c-e). As we have a field, c is one of d or e and cannot actually lie outside the span of 1 and b. Note: the last step assumes that the element c commutes with d and e. As has been pointed out, there is a four-dimensional structure which is like a field except that multiplication is not commutative. William C. Waterhouse Penn State