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I consider an algebraically closed field of characteristic zero $F$ as a vector space over a real closed field $R \subset F$. I would like to define a norm on $F$ in an invariant fashion, i.e. if we consider another real closed subfield $R' \subset F$ then the norm would still be equivalent to the norm defined taking $R$.

More precisely, for instance for $\mathbb{C}$ we have the norm $N(z)=a^2 +b^2$ where $z=a+ib$, $a,b \in \mathbb{R}$. If we consider $R' \subset \mathbb{C}$ then the same $z \in \mathbb{C}$ will be written as $z=a'+jb'$, $a', b' \in R'$ and $R'$ is a real closed field of index 2 in $\mathbb{C}$. Is the norm $N'(z)=a'^2+b'^2$ equivalent to $N$?

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I don't think so. IIRC $\mathbb{C}$ has lots of wild automorphisms and most of them aren't even measurable, so it would be surprising to me if they did anything reasonable to the norm. – Qiaochu Yuan Feb 8 '14 at 5:16
up vote 3 down vote accepted

For any norm of a quadratic extension, you can specify the Galois-fixed elements as those $x$ for which $N(x) = x^2$. If you had an invariant norm, it would imply the real-closed field is unique. However, (assuming choice) the complex numbers have automorphisms that don't commute with complex conjugation, i.e., that take copies of $\mathbb{R}$ to different subfields.

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