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I am trying to find a field isomorphism of $\mathbb{C}$ onto itself other than the identity map.

The isomorphism preserves the algebraic structures $f(x+y)=f(x)+f(y)$ and $f(xy)=f(x)f(y)$.

This means $f(i^2)=f(-1)=f(i)f(i)$. With this in mind, I have tried to come up with a bunch of different bijections $f$ like $f(z)=iz$, which fails that latter algebraic property, since:

$$ f(i)=i^2=-1 \Rightarrow f(i)f(i)=1 $$ $$ f(-1)=i $$

Could anyone give me some hints that would lead me to the solution? Because I'm not sure my approach is the righ one.

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    $\begingroup$ Complex conjugation? $\endgroup$
    – Ali Caglayan
    Mar 9, 2014 at 10:06

2 Answers 2

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We should have $f(0)=0$ hence $f(-z)=-f(z)$ for any $z\in\mathbb C$. In particular, $f(-1)=f(i)^2=-f(1)=-1$ (because $f(1)^2=f(1)$ and $f(1)\neq 0$).

We thus have $f(i)=i$ or $f(i)=-i$. In the first case, we get the identity map, and in the second one the complex conjugation.

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  • $\begingroup$ Why does f(0)=0 imply f(-z)=-f(z). Is it because f is odd? Does the concept of odd functions apply to complex functions? $\endgroup$
    – Thiago
    Mar 9, 2014 at 10:24
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    $\begingroup$ It's because $0=f(0)=f(z+(-z))=f(z)+f(-z)$. $\endgroup$
    – Davide Giraudo
    Mar 9, 2014 at 10:24
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    $\begingroup$ But there are (assuming Axiom of Choice) infinitely many choices for e.g. $f(\pi)$. $\endgroup$
    – Jyrki Lahtonen
    Mar 9, 2014 at 19:54
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The only continuous field isomorphisms $\mathbb C\to\mathbb C$ are the identity and complex conjugation.

Using the Axiom of Choice one can prove that additionally there exists infinitely many "wild" everywhere-discontinuous automorphisms of $\mathbb C$. Each of them acts either like the identity or like conjugation on all points with rational coordinates, but can do different things for irrational points.

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  • $\begingroup$ One can say even more: the only measurable field isomorphisms are the identity and complex conjugation. $\endgroup$
    – Elchanan Solomon
    Mar 9, 2014 at 19:15
  • $\begingroup$ We can even say that the only Baire measurable field isomorphisms are the identity and complex conjugation. $\endgroup$
    – Asaf Karagila ♦
    Mar 9, 2014 at 19:18
  • $\begingroup$ On the other hand, we can also say much less. The only smooth, harmonic field isomorphisms are the identity and complex conjugation. $\endgroup$
    – Elchanan Solomon
    Mar 9, 2014 at 19:27

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