Given a ring homomorphism $f: \mathbb C \to \mathbb C$, prove it is either the identity mapping or complex conjugation

Question

I am struggling to round off this homework question and require a gentle push in the right direction.

Here's what I have so far:

If $f$ is a homomorphism from the complex numbers to the complex numbers, then I would have the following: $$ f(z+w)=f(z)+f(w) \; \forall (z,w) \in \mathbb{C}$$ $$f(zw)=f(z)f(w) \; \forall (z,w) \in \mathbb{C}$$

and so if I represent a complex number $x+yi$ as $(x+0i)+(0+yi)$ then I would arrive at the following: $$f(x+yi) = x+yf(i) = u(x,y)+v(x,y)i$$

The issue I have is that I am not sure now how to establish that the real component of this function is $x$. I attempted to look at particular cases such as $x=x , y=0$ and $x=0, y=y$ etc, but cannot establish even the real component is definitively $x$.

One such idea I had was to show that for certain, $f(i)$ is purely imaginary and so $yf(i)$ has no contribution to the real component, but again cannot arrive at a conclusion here.

Any help would be greatly appreciated!

Answer 1

Continuity is essential. There exist additive, multiplicative,
one-to-one discontinuous functions on $% %TCIMACRO{\U{2102} }% %BeginExpansion \mathbb{C} %EndExpansion !.$ However, continuity of $f$ can be replaced by the condition that $f$ maps $% %TCIMACRO{\U{211d} }% %BeginExpansion \mathbb{R} %EndExpansion $ into $% %TCIMACRO{\U{211d} }% %BeginExpansion \mathbb{R}$

We have $f(rz)=rf(z)$ for all rational $r.$ Since $(f(1))^{2}=f(1)$ we get $% f(1)=0$ or $1.$ If $f(1)=0$ then $f(r)=0$ for all $r$ rational, hence for all $r$ real. Note that $[f(i)]^{2}=f(-1)=-f(1)=0$ so $f(i)=0.$ We now get $% f(a+ib)=f(a)+f(i)f(b)=0.$ Thus $f\equiv 0$ in this case. Let $f(1)=1.$ Then $% [f(i)]^{2}=f(-1)=-f(1)=-1$ so $f(i)=i$ or $f(i)=-i.$ In the first case $% f(a+ib)=f(a)+f(i)f(b)=a+ib$ and in the second case $% f(a+ib)=f(a)+f(i)f(b)=a-ib.$ Conclusion: $f\equiv 0$ or $f(z)=z$ $\forall z$ or $f(z)=\overset{-}{z}$ $\forall z.$

Answer 2

This is quite false. Actually there are $2^{2^{\aleph_0}}$ ring automorphisms $\mathbb{C} \to \mathbb{C}$ other than the identity and complex conjugation, although they all require some choice to construct and they are all non-measurable. See, for example, this MO thread and this math.SE thread.

There are even ring homomorphisms that aren't automorphisms, which can be constructed as follows: $\mathbb{C}$ embeds into $\mathbb{C}(t)$ and hence into the algebraic closure $\overline{\mathbb{C}(t)}$. It's a general fact about algebraically closed fields of characteristic $0$ that an uncountable such field is determined up to isomorphism by its cardinality, so $\overline{\mathbb{C}(t)}$ is actually isomorphic to $\mathbb{C}$, and hence we have an injection $\mathbb{C} \to \mathbb{C}$ exhibiting $\mathbb{C}$ as a proper subfield of itself (in fact we have many such injections, acted on by a group at least as large as the absolute Galois group of $\mathbb{C}(t)$).

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What you can show about an arbitrary ring homomorphism $f : \mathbb{C} \to \mathbb{C}$ is that $f(i)^2 = f(i^2) = f(-1) = -1$, so $f(i) = i$ or $-i$. You can also show that $f(\mathbb{Q}) = \mathbb{Q}$ without much difficulty, so $f$ at least restricts to a map $\mathbb{Q}(i) \to \mathbb{Q}(i)$ which is either the identity or complex conjugation. The problem is that you have no control over the image $f(\mathbb{R})$ of the reals, and in particular no guarantee that it's $\mathbb{R}$ again. In fact the constructions above imply that $\mathbb{C}$ has a bunch of (non-measurable!) subfields abstractly isomorphic to $\mathbb{R}$ other than the obvious one.

This is despite the fact that (and this is a nice exercise) $\mathbb{R}$ itself has no nontrivial endomorphisms.