QUESTIONS AND PROBLEMS 459Ideal complex numbers. Numbers of the form m + çù, where ù = -1/2 + v^-3/2,
a primitive cube root of unity, satisfies the equation ù^2 = — 1 — u>, have properties
similar to the Gaussian integers. For this system the number 3 is not a prime, since
3 = (2 + ù)(1 -ù). Numbers of the form m + çù do have a unique factorization into
primes, but Ernst Eduard Kummer (1810-1893) discovered in 1844 that "complex
integers" of the form m + nia + n2Ct^2 -\ hnA-2«A~^2 , where 1+á + ···+áë_1 =0
and ë is prime, do not necessarily have the property of unique factorization.^21 This
fact seems to have a connection with Fermat's last theorem, which can be stated
by saying that the number
xX + yX = {x + y)(x + ay)(x + a^2 y) •••{x + ax~ly)is never equal to zx for any nonzero integers x, y, z. For that reason, Klein (1926,
p. 321) asserted that it was precisely in this context that Kummer made the dis-
covery. But Edwards (1977, pp. 79-81) argues convincingly that the discovery
was connected with the search for higher reciprocity laws, analogs of the quadratic
reciprocity discussed in Section 1 of Chapter 8.
However that may be, what Kummer made of the discovery is quite interesting.
He introduced "ideal" numbers that would divide some of the otherwise irreducible
numbers, just as the imaginary number 2 + i divides 5, and (he said) just as one
introduces ideal chords to be held in common by two circles that do not intersect
in the ordinary sense. By his definition, a prime number ñ that equals 1 modulo
ë has ë — 1 factors.^22 These factors may be actual complex integers of the form
stated. For example, when ë = 3, ñ — 13, we have 13 = (4 +ù)(3—ù). If they were
not actual complex numbers, he assigned an ideal factor of ñ to correspond to each
root î of the congruence îë î 1 mod ñ, thereby obtaining ë — 1 nonunit factors.
His rationale was that if f(a) divided ñ in the ordinary sense, then /(î) would
be divisible by ñ in the ordinary sense. More generally, a complex integer Ö(á)
was to have the ideal factor corresponding to î if Ö(î) = 0 mod p. Then if Ö(á)
was divisible by p, it would be divisible by all of the factors of p, whether actual
complex numbers or ideal complex numbers. When these ideal complex numbers
were introduced, unique factorization was restored.
Questions and problems
15.1. Prove that if every polynomial with real coefficients has a zero in the complex
numbers, then the same is true of every polynomial with complex coefficients. To
get started, let p(z) = zn + a\zn~l + · • • + an-\z + an be a polynomial with
complex coefficients áú,..., a„. Consider the polynomial q(z) of degree 2n given by
q(z) = p(z)p(z), where the overline indicates complex conjugation. This polynomial
has real coefficients, and so by hypothesis has a complex zero ZQ.15.2. Formulate Cauchy's 1812 result as the following theorem and prove it: Let ñ
be a prime number, 3 < ñ < ç. If a subgroup of the symmetric group on ç letters
contains all permutations of order p, it is either the entire symmetric group or the
alternating group.(^21) The first prime for which unique factorization fails is ñ = 23, just in case the reader was hoping
to see an example.
(^22) This relation between the primes ñ and ë speaks in favor of Edwards' argument that Kummer's
goal had been a higher reciprocity law.