Re: New split complex numbers?

On Jan 27, 4:02 am, Narcoleptic Insomniac
<i_have_narcoleptic_insom...@xxxxxxxxx> wrote:
On Jan 26, 2008 9:50 PM CT, Narcoleptic Insomniac wrote:





On Jan 26, 2008 4:26 AM, Martin Hogbin wrote:

I just came across split complex numbers today, in
which sqrt(1) = j which is not equal to 1.

It is possible to extend this concept to define say
1^(1/3) =j,  or (-1)^(1/3)=j   where j is not a
complex number.
(also we could have a more general case
  (+-1)^(1/n)=j )

So, in the first case above, we then have j.j.j=1

we could then define  j.j=k  (k not a complex
number)

giving us    k.k=j

and k.j=1

So we could have numbers of the general form  a + bj
+ ck on which we could do algebra.

My question is, is this a new form of number or does
it turn out to form part of an existing algebra.
--
Martin Hogbin

The split complex numbers can be viewed as elements
from the quotient ring R[x] / <x^2 - 1>, where R[x] is
the set of polynomials over the real numbers and
<x^2 - 1> is the ideal generated by x^2 - 1.  It is
possible to "extend this concept" -- in fact, here you
are considering the quotient rings R[x] / <x^3 +- 1>.

Let I = x^3 - 1, then your first case boils down to
looking at the quotient ring R' = R[x] / <I>. Suppose
that f(x) is an arbitrary element of R[x], say

f(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + ...

In R' we have I = 0 which implies x^3 = 1.  Thus, the
quotient map sends f in R[x] to f' in R' where

f'(x) = a_0 + a_1 x + a_2 x^2 + a_3 + a_4 x + ...

= (a_0 + a_4 + ...) + (a_1 + a_5 + ...)x + (a_2 + a_6
+ ...) x^2.

Actually, in your post above you've set x = j and
x^2 = k to arrive at the relations

j.j = k [corresponding to x.x = x^2],

k.k = j [corresponding to x^2.x^2 = x^4 == x (mod I)],
and

k.j = 1 [corresponding to x^2.x = x^3 == 1 (mod I)].

Moreover, we can work out addition (+) and
multiplication (o) in R'. Let A and B be elements of R'
where

A = a_0 + a_1 x + a_2 x^2 and

B = b_0 + b_1 x + b_2 x^2.

Then we have...

A + B = (a_0 + b_0) + (a_1 + b_1) x + (a_2 + b_2) x^2,
and

A o B = (a_0 b_0 + a_1 b_2 + a_2 b_1) + (a_0 b_1 + a_1
b_0 + a_2 b_2) x + (a_0 b_2 + a_1 b_1 + a_2 b_0) x^2.

An important fact follows from this -- given any A in
R' there exists a B in R' such that A o B = 1.  (This
follows from the fact that there is a unique solution
to our three equations with three unknowns.)  In other
words, multiplicative inverses exist in R' which makes
it a field (i.e. R' is a division ring).  (Note this
also follows from the fact that I = <x^3 - 1> is
maximal.)

I don't know why I didn't think of this earlier, but
after reading Mariano's post it dawned on me.

We know x^3 - 1 = (x - 1)(x^2 + x + 1) and x^2 + x + 1 is
irreducible over R.  Now since

x^2 + x + 1 =

(x + 1/2 + i sqrt(3)/2)(x + 1/2 - i sqrt(3)/2)

..it follows that the splitting field for x^3 - 1 over R
is R(i sqrt(3)) = {r_0 + r_1 i sqrt(3) | r_0, r_1 \in R}.

Thus, R' = R[x] / <x^3 - 1> ~= R(i sqrt(3)), where the

No it isn't: the polynomial X^3-1 is a product of two
irreducibles, so in the ring R[x]/(x^3-1) there are
divisors of zero. In the field R(sqrt(-3)), of course,
there are no divisors of zero.

R[X]/(X^3-1) ~= R x C, as I states in my message.

-- m
.

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