Field automorphism: Difference between revisions
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The automorphisms of a given field ''K'' form a [[group (mathematics)|group]], the '''automorphism group''' <math>Aut(K)</math>. | The automorphisms of a given field ''K'' form a [[group (mathematics)|group]], the '''automorphism group''' <math>Aut(K)</math>. | ||
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 <math>Aut_L(K)</math> of the full automorphism group of ''K''. A [[field extension]] <math>K/L</math> of finite index ''d'' is ''[[normal]]'' if the automorphism group is of [[order (group theory)|order]] equal to ''d''. | 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 <math>Aut_L(K)</math> of the full automorphism group of ''K''. A [[field extension]] <math>K/L</math> of finite index ''d'' is ''[[normal extension|normal]]'' if the automorphism group is of [[order (group theory)|order]] equal to ''d''. | ||
==Examples== | ==Examples== | ||
* 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>. | ||
* The [[quadratic field]] <math>\mathbf{Q}(\sqrt d)</math> has a non-trivial automorphism which maps <math>\sqrt d \mapsto - \sqrt d</math>. The automorphism group is cyclic of order 2. | * The [[quadratic field]] <math>\mathbf{Q}(\sqrt d)</math> has a non-trivial automorphism which maps <math>\sqrt d \mapsto - \sqrt d</math>. The automorphism group is cyclic of order 2. | ||
A homomorphism of fields is necessarily [[injective function|injective]], since it is a [[ring homomorphism]] with trivial kernel, and a field, viewed as a [[ring theory|ring]], has no non-trivial [[ideal]]s. An [[endomorphism]] of a field need not be [[surjective function|surjective]], however. An example is the Frobenius map <math>\Phi: x \mapsto x^p</math> applied to the [[rational function]] field <math>\mathbf{F}_p(X)</math>, which has as image the proper subfield <math>\mathbf{F}_p(X^p)</math>. | A homomorphism of fields is necessarily [[injective function|injective]], since it is a [[ring homomorphism]] with trivial kernel, and a field, viewed as a [[ring theory|ring]], has no non-trivial [[ideal]]s. An [[endomorphism]] of a field need not be [[surjective function|surjective]], however. An example is the Frobenius map <math>\Phi: x \mapsto x^p</math> applied to the [[rational function]] field <math>\mathbf{F}_p(X)</math>, which has as image the proper subfield <math>\mathbf{F}_p(X^p)</math>.[[Category:Suggestion Bot Tag]] |
Latest revision as of 06:01, 16 August 2024
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 .