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I can easily see how to construct non trivial automorphisms of an algebraic extension of $\mathbb{Q}$, for example in $\mathbb{Q}(\sqrt{5})$, the isomorphism $\phi$ fixing the rationals and sending $\sqrt{5}$ to $-\sqrt{5}$, coming from the fact that $\phi (5) = \phi (\sqrt{5} \sqrt{5}) = \phi (\sqrt{5})^2 = 5$ and hence, $\phi (\sqrt{5}) = - \sqrt{5}$ is a possibility.

But I'm having troubles seeing a non trivial automorphism in a transcendental one, like $\mathbb{Q}(\pi)$. Where should it send $\pi$ ? any help ?

user26857
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LeLoupSolitaire
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    There are tons of examples, using linear-fractional transformations. Send $\pi$ to $-\pi$ or to $\pi +1$ or to $1/\pi$, or more generally to $(a\pi+b)/(c\pi +d)$ where $a,b,c,d$ are rational with $ad-bc\not= 0$. The field $\mathbf Q(\pi)$ is isomorphic to the rational function field $\mathbf Q(x)$, and its field automorphisms are precisely the linear-fractional changes of variable $f(x) \mapsto f((ax+b)/(cx+d))$ for rational $a,b,c,d$ with $ad-bc \not= 0$. – KCd Oct 09 '16 at 16:49
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    With a more general field of coefficients $K$, the field automorphisms of $K(x)$ that fix each element of $K$ are the linear-fractional changes of variable $f(x) \mapsto f((ax+b)/(cx+d))$ where $a,b,c,d \in K$ with $ad-bc\not= 0$. (So $\text{Aut}(K(x)/K) \cong \text{PGL}_2(K)$.) When $K= \mathbf Q$ the field automorphisms of $\mathbf Q(x)$ automatically fix each number in $\mathbf Q$, so we did not have to specify the constants are fixed in that case. But in general it's needed, e.g., there are field automorphisms of $\mathbf C(x)$ not fixing $\mathbf C$, if you believe the axiom of choice. – KCd Oct 09 '16 at 17:00
  • Related: https://math.stackexchange.com/questions/415847, https://math.stackexchange.com/questions/1842130 – Watson Oct 09 '16 at 17:11

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