# divisors of zero

Discussion in 'Symbolic Algebra' started by Clifford Nelson, Apr 30, 2006.

1. ### Clifford NelsonGuest

ecd086c689/e365d3da1264633d?lnk=st&q=4D+field&rnum=12#e365d3da1264633d

If you can get the URL above to work you can see that Robin Chapman and
John Rickard found divisors of zero for B_5 numbers in about a day. I
don't know how they found them. How would you find them with
Mathematica?

Here are some URLs for background.

7fcb3e2ea0/788d899ebbf4cbeb?lnk=st&q=B+numbers&rnum=8#788d899ebbf4cbeb

http://bfi.org/node/574
http://mathworld.wolfram.com/SynergeticsCoordinates.html
http://library.wolfram.com/infocenter/MathSource/600/

Cliff Nelson

Dry your tears, there's more fun for your ears, "Forward Into The
Past" 2 PM to 5 PM, Sundays, California time, at: http://www.kspc.org/
Don't be a square or a blockhead; see:
http://bfi.org/node/574

Clifford Nelson, Apr 30, 2006

2. ### Robert IsraelGuest

I don't know what they did, either. But here's what I might do in Maple.
This is the formula for (a,b,c,d)*(e,f,g,h), quoted in Chapman's article:
b*h), 1/5*(-a*e + (a + b + c + d)*f - d*g -
b*(-e - f - g - h) - c*h),
1/5*(-b*e - a*f + (a + b + c + d)*g -
c*(-e - f - g - h) - d*h),
1/5*(-c*e - b*f - a*g - d*(-e - f - g - h) +
(a + b + c + d)*h)];

Since the components are all homogeneous of degree 2, we might try for
a=e=1. And to cut down the number of variables some more, try c=d and
g=h (OK, I cheated and noticed that Chapman's and Rickard's solutions
One of the results Maple gives is

{h = 1/2*RootOf(_Z^2+_Z-1,label = _L1),
f = -1/2-1/2*RootOf(_Z^2+_Z-1,label = _L1),
d = -1/2-1/2*RootOf(_Z^2+_Z-1,label = _L1),
b = 1/2*RootOf(_Z^2+_Z-1,label = _L1)}

Applying allvalues to this gives one solution as

h = -1/4+1/4*5^(1/2), b = -1/4+1/4*5^(1/2), f = -1/4-1/4*5^(1/2),
d = -1/4-1/4*5^(1/2)

Robert Israel
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada

Robert Israel, May 1, 2006

3. ### Robert H. LewisGuest

Here is another way to go about finding solutions to this system of polynomial equations:

L:= [1/5*((a + b + c + d)*e - d*f - c*g - a*(-e - f - g - h) -
b*h), 1/5*(-a*e + (a + b + c + d)*f - d*g -
b*(-e - f - g - h) - c*h),
1/5*(-b*e - a*f + (a + b + c + d)*g -
c*(-e - f - g - h) - d*h),
1/5*(-c*e - b*f - a*g - d*(-e - f - g - h) +
(a + b + c + d)*h)];

Since we expect infinitely many solutions (given the original problem) the u-resultant idea is worth trying. We add a new equation involving some of the original variables and some new coefficient variables, such as
co1 * b + co2 * c + co3 * d + co4 * f + co5 * g - 1. (Actually first I tried just
co1 * b + co2 * c - 1, then the above. It made no real difference). Given the homogeneity and the precise form of the equations, we set a = 1 and e = 1 with no loss of generality. We run the Dixon resultant method and get an interesting factor of the resultant, c^2 + (b + 2)c - b^2 + b + 1. We want this to be 0. Obviously there are many possibilities. Keeping it simple let's try b = 0. That yields c^2 + 2c + 1 = 0, or c = -1. Plugging this in and starting over (without the auxiliary first equation) yields a factor of the resultant, d^2 - d - 1 = 0.
So now we have that d is the golden ratio, which I'll call rt. Plugging that in and starting over yields a system for f, g, and h that happens to be linear so is trivially solved. The solution is
f = anything, h = (-rt+1) - (rt-1)f, g = -1 - (1-rt)f. All together this takes about 1 second of cpu time and less than an hour of human messing-around time.

Robert H. Lewis
Fordham University

Robert H. Lewis, May 2, 2006
4. ### Dave RusinGuest

This product formula shows that when we allow this algebra to act on
itself by left-multiplication, we get a linear embedding of the
algebra into the 4x4 matrix ring. (That is, it's an embedding of
vector spaces; such a product formula defines an embedding of rings
iff the associative law holds.) Then x = (a,b,c,d) is a zero-divisor
iff the corresponding linear map has determinant zero. Well, the
determinant is a homogeneous polynomial of degree 4 in a,b,c,d
(it's actually monic in each variable, after multiplying by 125).
So setting it equal to zero defines a projective surface; you just
want to know whether there are any points on that surface.

If the algebra is defined over an algebraically closed field like C,
then the answer is clearly yes (pick any a,b,c and solve for d).
If the base field is the reals this is not so obvious but some
random experimentation finds plenty of points (e.g. it's easy to
make the discriminant w.r.t. d vanish, and then find points).
Most of the simple things I tried let to the golden ratio or 5th
roots of unity in some way.

I looked for a little while to find a rational point on this surface
but was not successful. It may be that there is a simple modular
restriction disallowing this; with the previous paragraph in mind
I guess I might try to show there are no 5-adic points if I
thought there were no rational ones. (E.g. this works to show there
are no points with a=d.) But I didn't spend much time on this.

dave

Dave Rusin, May 2, 2006
5. ### Clifford NelsonGuest

Is my memory playing tricks on me or do I remember getting and verifying
answers in Mathematica of LinearSolve[M,v] for n by n matrices M and
length n vectors v even when M did not have an inverse and had a
determinant of zero?

Cliff Nelson

Dry your tears, there's more fun for your ears, "Forward Into The
Past" 2 PM to 5 PM, Sundays, California time, at: http://www.kspc.org/
Don't be a square or a blockhead; see:
http://bfi.org/node/574

Clifford Nelson, May 2, 2006
6. ### Robert IsraelGuest

It may not be playing tricks. Yes, you can sometimes solve equations
M x = v when M is not invertible. The general solution will be of the
form x_p + x_h where x_p is one particular solution and x_h is an
arbitrary solution of M x = 0. I don't know why you think this has
anything to do with the question at hand. Dave is just using the
fact that if M is an n x n matrix (over some field k), there are
nonzero solutions of M x = 0 (with x in k^n) if and only if
det(M) = 0. In the case at hand, with M corresponding to left
multiplication by (a,b,c,d) in your algebra, the existence of
such a solution x = (e,f,g,h) means that (if (a,b,c,d) is nonzero)
it is a zero-divisor.

Robert Israel
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada

Robert Israel, May 2, 2006