From: rusin@vesuvius.math.niu.edu (Dave Rusin)
Newsgroups: sci.math,sci.logic,sci.physics
Subject: Re: Extending the Fundamental Theorem of Algebra
Date: 27 Mar 1998 00:45:30 GMT
In article <6fdlgm$f56$1@nnrp1.dejanews.com>, wrote:
>The FTA states that any polynomial of degree n, with complex
>coefficients has n complex roots. This theorem is valid only
>within the complex normal field. When we expand the number
>system to e.g. four dimensions then the theorem will be:
>
>Any polynomial of degree n, with hypercomplex (4-D) coefficients
>has n^2 hypercomplex (4-D) roots.
In article <6fe8b3$npk$1@gannett.math.niu.edu>,
rusin@vesuvius.math.niu.edu (Dave Rusin) wrote:
>I don't know what definition of "hypercomplex" you're using, but
[I guessed M2(R) or H, for which the result is false.]
In article <6fekg3$b8t$1@nnrp1.dejanews.com>, wrote:
>The hypercomplex numbers that I am talking about are communitive.
>I am not concerned about non-communitive rings, but thanks for the
>directions.
>What is it that you would like to know about this 4-D algebra?
Sorry, I misunderstood. The four-dimensional _commutative_ real algebras
(with unity, that is, which are assumed to include the real numbers)
are more numerous. I forget the complete classification, although I can
describe them all if you intend them to be semisimple, which is the case
if the equation x^2=0 has a unique solution, x=0. If that's true, then
the ring is isomorphic to either C + C or R + R + R + R, that is,
you're really just talking about pairs of complex numbers (z1, z2) in
which both addition _and miultiplication_ are performed coordinate-wise,
or 4-tuples (x1, x2, x3, x4) of real numbers, again with coordinate-wise
addition and multiplication. (Note that I'm not claiming this is how
the operations are defined _in your presentation_, but rather that a
one-to-one correspondence can be set up between your ring and one of
these two, in a way which preserves addition and multiplication.)
Note that '1', the identity element for multiplication, is (1,1) or
(1, 1, 1, 1) respectively.
In the latter ring, the solution x^n=1 has 1 solution (x='1') if n is
odd, and has 16 solutions if n is even (namely each of the combinations
x=(+-1, +-1, +-1, +-1).) Thus what you wrote in your post is untrue in this
ring.
So now, IF your 'hypercomplex numbers' form a commutative ring with unity,
and are a four-dimensional vector space over the reals, and have no
nilpotent elements (i.e. x^2=0 => x=0 ) and you claim that for every
polynomial f of degree n there are n^2 roots (counted according to
multiplicity, whatever that means), THEN your ring is simply C + C, and
your claim is true and in fact trivially proved, since the solutions
to f(z) = 0 in this ring are precisely the pairs (z1, z2) of complex
numbers which are roots of f.
In the same way, yes, the number of roots of a polynomial of degree n
in a k-fold direct sum C + C + ... + C is n^k. (Here k can be
any integer; this is a real algebra of dimension 2k.)
These facts are pretty basic elements of commutative ring theory; see e.g.
http://www.math.niu.edu/~rusin/known-math/index/13-XX.html
for details on this field. These rings are rather unremarkable. The
situation is much more interesting when you allow nilpotence, e.g. the
ring { a + b u ; a and b lie in R } where multiplication is determined
by the rule u^2=0.
dave