There is an interesting issue in relation to these extension fields: automorphisms. These are self-bijections that satisfy the fields' operation properties. Thus, f(x+y) = f(x) + f
and f(x*y) = f(x) * f
. For natural numbers, integers, rational numbers, and real numbers, the only automorphism is the identity one. But let us consider complex numbers. f(x+y*i) = x + y*f(i) where we must find the possible values of f(i). Since f(i)
2 + 1 = 0, f(i) must equal either i or -i. We get both the identity automorphism and complex conjugation. Not surprisingly, these two automorphisms form a group: Z2.
Working with Q(sqrt(2)), one finds both the identity automorphism and reversing the sign of sqrt(2). Again, the automorphism group is Z2.
There is an interesting theorem about doing a sequence of polynomials. Start with field K1. Find its splitting field with respect to polynomial P1, giving splitting field K2. Do that again with polynomial P2, giving splitting field K3. Thus, K3 is the splitting field for some combined polynomial P12 over K1.
Let the automorphism group of K2 with K1 fixed be (K2:K1). This is the group of interchanges of roots of P1 that are not in K1, the polynomial's "Galois group". Yes, him who was killed in that duel. No, he communicated his work well before that duel.
Then, group (K3:K1) has (K2:K1) as a "normal subgroup", with "quotient group" (K3:K2).
This has implications for what polynomial equations can be solved with what radicals or nth roots, since doing so implies that the breakdown of the polynomial's Galois group must contain Z
and Z*
, where the latter is the group of multiplication of numbers relatively prime to n. So if a polynomial's roots can be found with square roots, its Galois group must break down to Z2's, while if one can use cube roots, that breakdown must include Z3's. Z*(2) = identity group and Z*(3) = Z2.
This means that the centuries-old problems of duplicating the cube and trisecting the angle cannot be solved by ruler-and-compass methods. That is because ruler-and-compass methods only do arithmetic and square roots, and because those two problems require cube roots. Archimedes used a neusis or marked ruler for cube roots, but most of his colleagues tried to avoid that.
Solution by radicals also has the general consequence that a polynomial's Galois group must be "solvable", with decomposition that is ultimately a chain of Z
's. That is the case for quadratic, cubic, and quadric (degree 4) equations. But for quintic (degree 5), the Galois group is, in general, not solvable. Thus, general quintics cannot be solved with radicals.