We study automorphisms of the field K = Q(i). Each element of K is of the form a + bi with a, b Q. The map

: Q(i) -> Q(i),    (a + bi) = a - bi

is an automorphism. This follows from the rules for complex conjugation.

The square of this map, · , is equal to the identity. In fact, the group of automorphisms coincides with <>.

In order to see this, note that, for each automorphism , we have

(a + bi) = (a) + (b) (i) = a + b (i).

The automorphism is fully determined by the image of (i). Now (i)² = (i²) = (-1) = -1. So (i) is a zero of X² + 1. Both zeros of X² + 1 correspond indeed to an automorphism: the identity and . So Aut(K) is a group of order two and hence isomorphic to Z/2Z. The possible automorphisms are apparently connected to the zeros of the polynomial X² + 1.

We study automorphisms of the field K = Q(x), where x = ³2.

Let be such an automorphism. If fixes x, then is obviously the identity.

If is not the identity, it must move x to another solution of X³ - 2. These solutions do not exist in K.

An intuitive way of seeing this runs as follows: the other solutions are e^2i/3x and e^4i/3x, and these are complex numbers, whereas K is a subfield of R.

Thus, the automorphism group of Q(x) is trivial. In particular, it is strictly smaller than the dimension of Q(x) over Q.