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Field Automorphism


A field automorphism of a field F is a bijective map sigma:F->F that preserves all of F's algebraic properties, more precisely, it is an isomorphism. For example, complex conjugation is a field automorphism of C, the complex numbers, because

0^_=0
(1)
1^_=1
(2)
a+b^_=a^_+b^_
(3)
ab^_=a^_b^_.
(4)

A field automorphism fixes the smallest field containing 1, which is Q, the rational numbers, in the case of field characteristic zero.

The set of automorphisms of F which fix a smaller field F^' forms a group, by composition, called the Galois group, written Gal(F/F^'). For example, take F^'=Q, the rational numbers, and

F=Q(sqrt(2))
(5)
={a+sqrt(2)b:a,b in Q},
(6)

which is an extension of Q. Then the only automorphism of F (fixing Q) is sigma, where sigma(a+sqrt(2)b)=a-sqrt(2)b. It is no accident that sqrt(2) and -sqrt(2) are the roots of x^2-2. The basic observation is that for any automorphism sigma, any polynomial p with coefficients in F^', and any field element alpha,

 sigma(p(alpha))=p(sigma(alpha)).
(7)

So if alpha is a root of p, then sigma(alpha) is also a root of p.

The rational numbers Q form a field with no nontrivial automorphisms. Slightly more complicated is the extension of Q by 2^(1/3), the real cube root of 2.

 F=Q(2^(1/3))={a+2^(1/3)b+2^(2/3)c:a,b,c in Q}.
(8)

This extension has no nontrivial automorphisms because any automorphism would be determined by sigma(2^(1/3)). But as noted above, the value of sigma(2^(1/3)) would have to be a root of x^3-2. Since F has only one such root, an automorphism must fix it, that is, sigma(2^(1/3))=2^(1/3), and so sigma must be the identity map.


See also

Automorphism, Extension Field, Field, Galois Group

This entry contributed by Todd Rowland

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Cite this as:

Rowland, Todd. "Field Automorphism." From MathWorld--A Wolfram Web Resource, created by Eric W. Weisstein. https://mathworld.wolfram.com/FieldAutomorphism.html

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