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Let $L$ be a number field and $G=Aut(L\mid \mathbb{Q})$. Let $|\cdot|$ be the usual archimedean value on $\mathbb{C}$ and, by abuse of notation, its restriction to $\mathbb{Q}$. Then the archimedean values on $L$ are of the form $|w|_{\sigma,s}:=|\sigma(w)|^s$ for all $w\in L$, a positive $s$ and $\sigma$ an embedding of $L$ into $\mathbb{C}$. By taking non-equivalent such $\sigma$ (equivalent here meaning equal modulo composition with complex conjugation) we may identify the archimedean places on $L$ with all the $|\cdot|_{\sigma,1}$. and $w\in L$ defines an archimedean absolute value on $L$. Take two such non equivalent embeddings and the corresponding places $v_1$, $v_2$ and their completions $L_{v_1}$ and $L_{v_2}$. By Ostrowski's theorem both completions must be isomorphic to $\mathbb{R}$ or $\mathbb{C}$. Let us assume they are both isomorphic to $\mathbb{R}$, that is, $|w|_{v_1}=|\tau(w)|^s$ and $|w'|_{v_2}=|\tau'(w')|^t$ for some positive reals $s,t$ and all $w\in L_g$, $w'\in L_{g'}$ ($\tau$ and $\tau'$ are the aforementioned isomorphisms). We must have $s=t=1$ because $v_1$ and $v_2$ coincide with $|\cdot|$ on $\mathbb{Q}$. We must also have $\text{id}=\tau'\circ\tau^{-1}$ (because there are no nontrivial field automorphisms of $\mathbb{R}$) so for $w=w'\in L$ the above becomes \begin{equation}|w|_{v_1}=|w|_{v_2} \end{equation} which is a contradiction Question: where is the error?

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  • $\begingroup$ Why would both completions of $L$ be isomorphic to $R$ or to $C$? $\endgroup$ – Mariano Suárez-Álvarez Feb 2 '16 at 3:54
  • $\begingroup$ Any field that is complete with respect to an archimedean valuation must be. It's Theorem 4.2 on Neukirch's book. Of course, one could be isomorphic to R and the other to C, but I just wanted to take two embeddings such that the completions were both real (that might not be done for every $L$, but, say for $L$ the splitting field of an irreducible polynomial with 3 real roots and 2 complex conjugate roots). $\endgroup$ – Jaramillo Feb 2 '16 at 4:53
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The error is in writing $\tau'\circ \tau^{-1}= id$. This composition doesn't even make sense if you're considering both sides as maps $\mathbb{R}\to\mathbb{R}$, since $\tau$ does not have an inverse as a map $L\to\mathbb{R}$ (it only has an inverse when its codomain is restricted to $\tau(L)$). We can write that $\tau'\circ \tau^{-1}$ is a field isomorphism from $\tau(L)$ to $\tau'(L)$, but this isomorphism need not extend to an isomorphism $\mathbb{R}\to\mathbb{R}$ between their completions because it does not preserve the absolute value. So you don't get an automorphism of $\mathbb{R}$, and you can't conclude $\tau$ and $\tau'$ are the same.

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