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

Question

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.

2

Complex conjugation? – Ali Caglayan Mar 09 '14 at 10:06

score 4 · Accepted Answer · answered Mar 09 '14 at 10:10

4

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.

answered Mar 09 '14 at 10:10

Davide Giraudo

172,925

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? – Thiago Mar 09 '14 at 10:24
2

It's because $0=f(0)=f(z+(-z))=f(z)+f(-z)$. – Davide Giraudo Mar 09 '14 at 10:24
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But there are (assuming Axiom of Choice) infinitely many choices for e.g. $f(\pi)$. – Jyrki Lahtonen Mar 09 '14 at 19:54

score 2 · Answer 2 · answered Mar 09 '14 at 19:10

2

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.

answered Mar 09 '14 at 19:10

hmakholm left over Monica

286,031

One can say even more: the only measurable field isomorphisms are the identity and complex conjugation. – Elchanan Solomon Mar 09 '14 at 19:15
We can even say that the only Baire measurable field isomorphisms are the identity and complex conjugation. – Asaf Karagila Mar 09 '14 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. – Elchanan Solomon Mar 09 '14 at 19:27

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

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