divisors of zero

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

  1. http://groups.google.com/group/geometry.research/browse_frm/thread/23704a
    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.

    http://groups.google.com/group/geometry.research/browse_frm/thread/41a082
    7fcb3e2ea0/788d899ebbf4cbeb?lnk=st&q=B+numbers&rnum=8#788d899ebbf4cbeb

    http://bfi.org/node/574
    http://users.adelphia.net/~cnelson9/
    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
    #1
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  2. 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
    had that).
    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
    #2
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  3. 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
    #3
  4. Clifford Nelson

    Dave Rusin Guest

    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
    #4
  5. 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
    #5
  6. 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
    #6
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