Archimedean and Euclidean implies trivial automorphism group: Difference between revisions

Revision as of 22:57, 14 May 2009

Suppose $K$ is a Euclidean Archimedean field: a field that is both Euclidean and Archimedean. In other words, there exists a total ordering $\leq$ on $K$ making $K$ into an ordered field, such that $0\leq a$ if and only if $a$ is a square. Then, the automorphism group of $K$ is trivial.

Any field automorphism of $K$ must send squares to squares.
Any field automorphism of $K$ is also an automorphism of $K$ as an ordered field, i.e., it preserves the total ordering.
We now use the fact that the field automorphism fixes every rational number, combined with the fact that the rational numbers are dense in $K$ .

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 {{field property implication|
-stronger = quadratically closed Archimedean field|
+stronger = Euclidean Archimedean field|
 weaker = field with trivial automorphism group}}
 ==Statement==
-Suppose <math>K</math> is a [[quadratically closed Archimedean field]]: a field that is both [[fact about::quadratically closed field|quadratically closed]] and [[fact about::Archimedean field|Archimedean]]. In other words, there exists a total ordering <math>\le</math> on <math>K</math> making <math>K</math> into an [[ordered field]], such that <math>0 \le a</math> if and only if <math>a</math> is a square. Then, the [[automorphism group]] of <math>K</math> is trivial.
+Suppose <math>K</math> is a [[Euclidean Archimedean field]]: a field that is both [[fact about::Euclidean field|Euclidean]] and [[fact about::Archimedean field|Archimedean]]. In other words, there exists a total ordering <math>\le</math> on <math>K</math> making <math>K</math> into an [[ordered field]], such that <math>0 \le a</math> if and only if <math>a</math> is a square. Then, the [[automorphism group]] of <math>K</math> is trivial.
 ==Proof==