Field automorphism: Difference between revisions
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* The field '''Q''' of [[rational number]]s 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. | * The field '''Q''' of [[rational number]]s 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 number|prime]] order has only the identity automorphism. | * Similarly, a [[finite field]] of [[prime number|prime]] order has only the identity automorphism. | ||
* The field '''R''' of [[real number]]s 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 '''R''' of [[real number]]s 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 number]]s has two automorphisms, the identity and [[complex conjugation]]. | * The field '''C''' of [[complex number]]s has two automorphisms, the identity and [[complex conjugation]]. | ||
* A finite field '''F'''<sub>''q''</sub> of prime power order ''q'', where <math>q = p^f</math> is a power of the prime number ''p'', has the [[Frobenius automorphism]], <math>\Phi: x \mapsto x^p</math>. The automorphism group in this case is [[cyclic group|cyclic]] of order ''f'', generated by <math>\Phi</math>. | * A finite field '''F'''<sub>''q''</sub> of prime power order ''q'', where <math>q = p^f</math> is a power of the prime number ''p'', has the [[Frobenius automorphism]], <math>\Phi: x \mapsto x^p</math>. The automorphism group in this case is [[cyclic group|cyclic]] of order ''f'', generated by <math>\Phi</math>. |
Revision as of 14:51, 21 November 2008
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 form a group, the automorphism group .
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 of the full automorphism group of K. A field extension 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 is a power of the prime number p, has the Frobenius automorphism, . The automorphism group in this case is cyclic of order f, generated by .
- The quadratic field has a non-trivial automorphism which maps . 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 applied to the rational function field , which has as image the proper subfield .