# Proof that there are only two automorphisms of $mathbb{Q}(sqrt d)$ fixing $mathbb{Q}$

Mathematics Asked by HyperPro on December 15, 2020

I have to proof that the only (field) automorphism of $$mathbb{Q}(sqrt d)$$ fixing $$mathbb{Q}$$ are $$id$$ and the conjugation $$sigma$$.

I know for every such automorphism $$tau$$ we have that $$tau(0)=0 \ tau(1)=1 \ tau(-a)=-tau(a) \ tau(a^{-1})=tau(a)^{-1}$$
and it is easy to see that satisfy all these properties. Any other automorphism I think about always violates any of these rules. But this does not mean that there could any very special and weird automorphism I just can not think about

An automorphism must fix $$mathbb Q$$, since it fixes $$1$$, and thus all integers, and thus all inverses of integers, and thus all products of integers and inverses of integers, which covers all rational numbers.

Now consider the polynomial $$f=X^2-d$$. If $$f(alpha)=0$$, then $$f(tau(alpha))=tau(f(alpha))=tau(0)=0$$, so any automorphism has to send roots of $$f$$ to roots of $$f$$. That is, $$sqrt d$$ is sent to itself (resulting in the identity automorphism), or to $$-sqrt d$$ (resulting in the conjugation).

Correct answer by Vercassivelaunos on December 15, 2020

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