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For a field $K$ let $\text{Aut}(K)$ denote the group of all automorphisms $f:K\to K$.

How can one show that $\text{Aut}(\mathbb{Q})$ is the trivial group and how to calculate $\text{Aut}(\mathbb{Q}(\sqrt{2}))$ ?

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    $\begingroup$ For the first, you can show that as $1\mapsto 1$ under any automorphism, then any automorphism fixes each $n\in \mathbb{N}$ by additivity, then $1/n$ by reciprocity and then $n/m$ by multiplicity. $\endgroup$ – James Oct 6 '14 at 21:47
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Hint: If $f$ is an automorphism of $\mathbb{Q}$, $f$ has to map every element to itself. (why?)

If $f$ is an automorphism of $\mathbb{Q}(\sqrt{2})$, $f$ fixes every element of $\mathbb{Q}$, meaning $f$ is determined by what it maps $\sqrt{2}$ to.

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Suppose $\varphi$ is an automorphism of $\mathbb{Q}$: we know that it has to fix 0 and 1, and so it fixes the integers by linearity, and hence it fixes $\mathbb{Q}$ by the fact that it's a field homomorphism.

Now, if $\sigma$ is an automorphism of $\mathbb{Q}(\sqrt{2})$, it has to fix $\mathbb{Q}$ by the above, and so as $\mathbb{Q}(\sqrt{2})$ is a $\mathbb{Q}$-vector space, $\sigma$ is determined by what it sends $\sqrt{2}$ to. Bear in mind that $\sigma(\sqrt{2})^2 = 2$...

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