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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!

Temoi
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RegimeBane
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    Why do you denote $ℂ$ by $(C)$? – k.stm Nov 20 '20 at 09:53
  • @k.stm it was merely a formatting error, I must have had extra parentheses on my \mathbb command – RegimeBane Nov 20 '20 at 09:54
  • A homomorphism of what? A group homomorphism, a ring homomorphism, …? – José Carlos Santos Nov 20 '20 at 09:55
  • @JoséCarlosSantos my apologies, it is a homomorphism of rings – RegimeBane Nov 20 '20 at 09:56
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    You need to preserve another structure. The axiom of choice implies that there are ring automorphisms of $\mathbb{C}$ that are neither the identity nor the conjugation. However, it is possible to show that $f(i)=\pm i$, since $f(i)^2+1=f(i^2+1)=0$. – Aphelli Nov 20 '20 at 09:57
  • @Mindlack What is it that you mean by "preserve another structure"? My area of expertise lies in real analysis right now, I'm working through elementary Abstract Algebra and would love to get more in depth with the theory as well! – RegimeBane Nov 20 '20 at 10:10
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    @Regime: for example, the statement is true if you require that $f$ is continuous or even just that it's measurable. Measurability implies that $f$ is $\mathbb{R}$-linear (see https://en.wikipedia.org/wiki/Cauchy%27s_functional_equation) and then it's pretty easy. – Qiaochu Yuan Nov 20 '20 at 10:10

2 Answers2

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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 $\mathbb{C}(t)$).

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.

Qiaochu Yuan
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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 $f(z)=\overset{-}{z}$ $\forall z.$

Kavi Rama Murthy
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  • Sure, that makes sense... is there any particular (rigorous) reason why $f(rz)=rf(z)$ is restricted first to $r \in \mathbb Q$ , then logically extended to $r \in \mathbb R$ ? – RegimeBane Nov 20 '20 at 10:33
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    @RegimeBane From additivity you get $f(nz)=nf(z)$ (by induction) .From this you can show that $f(\frac p q z)=\frac p q f(z)$ for any rational number $\frac p q$. But you cannot go beyond this unless you have some continuity condition on $f$. – Kavi Rama Murthy Nov 20 '20 at 11:36
  • Continuity of $f$ may not be replaced with the condition that $f$ maps $\Bbb R$ into $\Bbb R$. Zorn's-lemma-ing a counterexample is not difficult (e.g. take all partial homomorphisms $\Bbb C\to\Bbb C$ that map no real number to a non-real, and that also map $\pi$ to $\sqrt2$, ordered by extension). – Arthur Jun 16 '22 at 18:50