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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

One Answer

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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