ABSTRACT
The minimal polynomial of i over is t2+1, since i2+1=0 but So Now ξ is a zero of f over and f is irreducible by Eisenstein’s Criterion, Theorem
3.19. Hence f is the minimal polynomial of ξ over and Therefore,
3. We shall find the elements of the Galois group of By a direct check, or by Corollary 5.13, there is a -automorphism σ of K such that
and another, τ, such that
Products of these yield eight distinct -automorphisms of K, as follows:
Other products do not give new automorphisms, since σ4=1, τ2=1, τσ=σ3τ, τσ2=σ2τ, τσ3=στ. (The last two relations follow from the first three.)
Now any -automorphism of K sends i to some zero of t2+1, so i →±i; similarly, ξ is mapped to ξ, iξ, −ξ, or −iξ. All possible combinations of these (eight in number) appear in the above list, so these are precisely the -automorphisms of K.
