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Let $F,F'$ be two fields. Is there anything that can be said about the number of isomorphisms that can exist? In particular can there be more than one? What if $F$ is the complex numbers $\mathbb C$?

Context: I am trying to establish a bijection between characters in Banach algebras and maximal ideals in the algebra using the Gelfand Mazur theorem.

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  • $\begingroup$ There are lots of automorphisms of $\Bbb C$. If you want to fix $\Bbb R$, then there's only one: $a+bi \mapsto a-bi$, but if you just want it to be an automorphism as a ring there are uncountably many (this requires the axiom of choice, but I'm a firm believer in a set's right to choose). See for instance this article by Paul Yale. $\endgroup$ – user98602 Mar 31 '14 at 9:09
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There are $2^{2^{\aleph_0}}$ automorphisms of $\mathbb C$, a mind-boggling cardinal, and of those only the identity and conjugation can be explicitly described.
The others depend on set-theoretical wizardry (axiom of choice).

What is much more interesting is to take a field extension $k\subset K$ and to study the group $Aut(K/k)$ of automorphisms $\sigma: K\to K$ fixing $k$ pointwise: $\sigma(q)=q $ for all $q\in k$.
This study is called Galois theory, although in a strict sense Galois theory is restricted to the case where the extension $k\subset K$ is Galois, i.e. algebraic, separable and normal.
The results obtained in this set-up are then much more reasonable.

For example:
You may see $K$ as a vector space over $k$ (a point of view introduced, I think by Emil Artin).
If the dimension of that vector space $K$ over $k$ is finite and equal to $d$, then $Aut(K/k)$ is a finite set containing $\leq d$ elements, independently of whether the extension is Galois or not.
The simplest illustration is $Aut(\mathbb C/\mathbb R)$, a group consisting of just two elements, namely the conjugation and the identity of $\mathbb C$.
Contrast this with the absurdly large automorphism group of $\mathbb C$ evoked at the beginning!

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    $\begingroup$ To be fair the first group you describe is $\mathrm{Aut}(\mathbb{C}/\mathbb{Q})$, isn't it? $\endgroup$ – Najib Idrissi Mar 31 '14 at 9:38
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    $\begingroup$ Dear @nik: yes, that's absolutely correct: +1. In general the absolute automorphism group $Aut(K)$ of a field is equal to the relative automorphism group $Aut(K/k)$ where $k$ is the prime field of $k$, the smallest subfield of $K$ . That prime field $k$ is isomorphic to $\mathbb Q$ or $\mathbb F_p$ according as $K$ has characteristic $0$ or $p$. $\endgroup$ – Georges Elencwajg Mar 31 '14 at 9:50

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