# Field isomorphism of $\mathbb{C}$ onto itself

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.

• Complex conjugation? Mar 9, 2014 at 10:06

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.

• 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? Mar 9, 2014 at 10:24
• It's because $0=f(0)=f(z+(-z))=f(z)+f(-z)$. Mar 9, 2014 at 10:24
• But there are (assuming Axiom of Choice) infinitely many choices for e.g. $f(\pi)$. Mar 9, 2014 at 19:54

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.

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