Suppose that $f:\mathbb{C} \to \mathbb{C}$ is a ring homomorphism. Does $f$ necessarily fix the real axis?

Question

I suspect that only such ring homomorphisms are identity and conjugation, but I cannot see how any homomorphisms from $\mathbb{C}$ to itself fixes the real axis.

I have shown that $f$ should fix any rational numbers, but I cannot extend this result to irrational numbers. Normally, one would progress by showing that $f$ is actually continuous, and density of $\mathbb{Q}$ then implies the result.

But even if I restrict the domain to $\mathbb{R}$, since the codomain is $\mathbb{C}$, I cannot see how to show this. In fact, I even doubt if it actually is continuous. One certain way would be to show that range of $f$ is $\mathbb{R}$ when restricted to $\mathbb{R}$, but it seems to be impossible for me.

How should I progress?

There are many, many, maaaaany ring homomorphisms $\mathbb C\to\mathbb C$. Exactly two of them are continuous. See here or here — Mariano Suárez-Álvarez, Feb 09 '15 at 23:31
@mesel, well, no. If $k$ is not $1$, then $f(1)f(1)=k^2$ is not the same as $f(1\cdot 1)=k$. — Mariano Suárez-Álvarez, Feb 09 '15 at 23:35

score 4 · Answer 1 · answered Feb 09 '15 at 23:34

4

Not all automorphisms fix the real axis. For example, there is an automorphsm of $\mathbb C$ which maps $\sqrt[3]{2}$ to $\omega\sqrt[3]{2}$ with $\omega$ a primitive cubic root of unity.

answered Feb 09 '15 at 23:34

Mariano Suárez-Álvarez

135,076

score 0 · Accepted Answer · answered Feb 09 '15 at 23:41

0

Suppose $\pi$ is a homomorphism from $\mathbb{C}$ to $\mathbb{C}$. Then $\pi(\mathbb{C})$ must be a subring of $\mathbb{C}$ and the $\ker \pi$ an ideal. Since $\mathbb{C}$ is a field, it's only ideals are itself and $(0)$. So $\pi$ is either an automorphism, where $\ker\pi = (0)$, or trivial where $\ker \pi = \mathbb{C}$.

If $\pi$ is an automorphism (and continuous), then it is either the identity or complex conjugation. See this article here for proof.

answered Feb 09 '15 at 23:41

Eoin

5,809

1

This is not an answer for the actual question, is it? :-| – Mariano Suárez-Álvarez Feb 10 '15 at 01:15
@MarianoSuárez-Alvarez Possibly not. Hopefully it provided the OP with a different perspective to approach the problem, or just another resource to look at. – Eoin Feb 10 '15 at 02:46
1

The question is nicely sumarized in its title. What you wrote, while informative, has nothing to do with it :-) – Mariano Suárez-Álvarez Feb 10 '15 at 02:48

Suppose that $f:\mathbb{C} \to \mathbb{C}$ is a ring homomorphism. Does $f$ necessarily fix the real axis?

2 Answers2