240 8. Miscellaneous ProblemsE.69. Quaternions. A polynomial of degree n over a field has at most
n zeros. This result may no longer hold if instead of a field we choose a
structure for which not all the field axioms are valid. One such structure
was invented by the British mathematician William R. Hamilton in 1843.
Its elements are quaternions, generalized complex numbers of the forma+bi+cj+dkwhere a, b, c, d are real and i, j, k are distinct elements assumed to satisfy
the relationsi2 = j2 = k2 = -1, ij = k, jk = i, ki = j.We add, subtract and multiply quaternions in much the same way we
do complex numbers, and assume that we have access to associativity of
addition and multiplication and to the distributive law. For example, verify
that we must have ji = -k, ik = -j, kj = -i, ijk = -1 and
(a + bi + cj + dk)(p + qi + rj f sk) = (up - bq - cr - sk)
+(aq+bp+cs-dr)i+(ar+cp+dq-bs)j
+ (us + dp + br - cq) k.All the field axioms except commutativity of multiplication (Axiom M.2,
Section 1.7) hold. By considering the product (o + bi + cj + dk)(u - bi -
cj - dk), show that each element for which not all of a, b, c, d are zero
has a multiplicative inverse and determine what this inverse is. Do this for
several numerical examples and check your work.
An equation of the form ux = v where u and v are quaternions with
u # 0 has a unique solution. The situation for quadratic equations is more
interesting. Find all quaternion solutions to the equations x2 = 1 and
z2 = -1. Investigate the general quadratic equation ax2 + bx + c = 0.
Are there any quadratic polynomials irreducible over the quaternions?
Does the factor theorem hold? Is it true that a polynomial can always be
factored as a product of irreducibles? Is such a factorization unique up to
the order of the factors?
Investigate the equation x” = 1 for n 1 3.
Hints
Chapter 8
- Divide y2 + 1 by 3y - 5 to obtain x and clear fractions.
- The difference of the sides of the first equation can be factored.
- Factor (x + y + .z)~ - (x3 + y3 + ,r3).