Field automorphism

In field theory, a field automorphism is an automorphism of the algebraic structure of a field, that is, a bijective function from the field onto itself which respects the fields operations of addition and multiplication.

The automorphisms of a given field K for a group, the automorphism group $$Aut(K)$$.

If L is a subfield of K, an automorphism of K which fixes every element of L is termed an L-automorphism. The L-automorphisms of K form a subgroup $$Aut_L(K)$$ of the full automorphism group of K. A field extension $$K/L$$ of finite index d is normal if the automorphism group is of order equal to d.

Examples

 * The field Q of rational numbers has only the identity automorphism, since an automorphism must map the unit element 1 to itself, and every rational number may be obtained from 1 by field operations. which are preserved by automorphisms.
 * Similarly, a finite field of prime order has only the identity automorphism.
 * The field R of real numbers has only the identity automorphism. This is harder to prove, and relies on the fact that R is an ordered field, with a unique ordering defined by the positive real numbers, which are precisely the squares, so that in this case any automorphism must also respect the ordering.
 * The field C of complex numbers has two automorphisms, the identity and complex conjugation.
 * A finite field Fq of prime power order q, where $$q = p^f$$ is a power of the prime number p, has the Frobenius automorphism, $$\Phi: x \mapsto x^p$$. The automorphism group in this case is cyclic of order f, generated by $$\Phi$$.
 * The quadratic field $$\mathbf{Q}(\sqrt d)$$ has a non-trivial automorphism which maps $$\sqrt d \mapsto - \sqrt d$$. The automorphism group is cyclic of order 2.

A homomorphism of fields is necessarily injective, since it is a ring homomorphism with trivial kernel, and a field, viewed as a ring, has no non-trivial ideals. An endomorphism of a field need not be surjective, however. An example is the Frobenius map $$\Phi: x \mapsto x^p$$ applied to the rational function field $$\mathbf{F}_p(X)$$, which has as image the proper subfield $$\mathbf{F}_p(X^p)$$.