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I have some trouble with an algebraic excercise about unitary mappings.

First part

If $f: \mathbb{R} \rightarrow \mathbb{R} $ is unitary, how to proof that $f$ can only be the identity?

My attempts so far

A unitary mapping $f:\mathbb{Z} \rightarrow \mathbb{Z}$ is the identity, because $$ f(n) = f(1+1+ \cdots +1) = f(1)+f(1)+ \cdots +f(1)=1+1+\cdots+1=n $$ This means that a unitary mapping $f: \mathbb{Q} \rightarrow \mathbb{Q}$ is the identity, as shown below. $$f(q)=f(ab^{-1})=f(a)f(b)^{-1}=ab^{-1}=q$$ I didn't know how to deal with $\mathbb{R}$, but I thought: $$x = x-\lfloor x \cdot 10^n \rfloor \cdot 10^{-n} + \lfloor x \cdot 10^n \rfloor \cdot 10^{-n}$$ The rightmost term is rational, so we can write $$f(x)=f(x-\lfloor x \cdot 10^n \rfloor \cdot 10^{-n}) + \lfloor x \cdot 10^n \rfloor \cdot 10^{-n} $$ We can act the same way by ceiling x. $$x = \lceil x \cdot 10^n \rceil \cdot 10^{-n} - (\lceil x \cdot 10^n \rceil \cdot 10^{-n}-x)$$

The leftmost term is rational, so $f(x) = \lceil x \cdot 10^n \rceil \cdot 10^{-n} - f(\lceil x \cdot 10^n \rceil \cdot 10^{-n}-x)$.

For large $n$ the expressions $\lceil x \cdot 10^n \rceil \cdot 10^{-n}-x$ and $x-\lfloor x \cdot 10^n \rfloor \cdot 10^{-n}$ will be small, but I don't know if $f(\lceil x \cdot 10^n \rceil \cdot 10^{-n}-x)$ and $f(x-\lfloor x \cdot 10^n \rfloor \cdot 10^{-n})$ behave the same way. Is this approach a good one or is there an easier one?

Second part

When this matter is proved, I can ask my second question: Is the only unitary mapping $f:\mathbb{C} \rightarrow \mathbb{C}$ the identity?

This is how far I came. We know that $f(a+bi) =f(a)+f(b)f(i)= a+bf(i)$, so if $f(i)=i$, we are done. But is this true? Intuitively I say yes, because we the rotating and stretching up the complex plane are not unitary, and a projection on the real axis is not a multiplacitive mapping, and I see no other options. How should this be done?

I am looking forward to your answers guys.

learner
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Koenraad van Duin
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    First part has a full answer here: http://math.stackexchange.com/questions/411459/ring-homomorphisms-mathbbr-to-mathbbr –  Jul 31 '13 at 10:11
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    For the second part (about $\mathbb C$), see http://math.stackexchange.com/questions/412010/wild-automorphisms-of-the-complex-numbers – azimut Jul 31 '13 at 10:49

3 Answers3

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As T. Bongers explained (+1), a ring homomorphism from reals to itself is the identity, because we can use squares to prove that it is order preserving, and hence continuous.

The purpose of my answer is to point out that in the case of complex numbers there are uncountably many isomorphisms! In fact the cardinality of the set of automorphisms of $\mathbb{C}$ is at least $2^{2^{\aleph_0}}$.

The catch is that if $f: \mathbb{C}\to\mathbb{C}$ is an automorphism, there is no need for $f$ to map reals to reals. A different approach is needed. Unfortunately it requires a bit more advanced machinery.

Let us pick a transcendence basis ${\cal B}$ for the extension $\mathbb{C}/\mathbb{Q}$. A cardinality argument shows that ${\cal B}$ must have continuum cardinality. Then obviously any permutation $\pi$ of ${\cal B}$ is an automorphism of $\mathbb{Q}({\cal B})$. Basically this is because the elements of ${\cal B}$ are algebraically independent variables, and we can permute them in an unconstrained manner and view $\pi$ as a field automorphism.

The next step is to observe that $\mathbb{C}$ is algebraic over $\mathbb{Q}({\cal B})$. As $\mathbb{C}$ is algebraically closed, any automorphism $\pi$ of $\mathbb{Q}({\cal B})$ can be extended to an automorphism of $\mathbb{C}$ (actually in infinitely many ways, but let's ignore that). Therefore $$ |Aut(\mathbb{C})|\ge |Sym({\cal B})|\ge 2^{2^{\aleph_0}}. $$ See the ever useful Basic Algebra I, by N. Jacobson for the missing details.

Jyrki Lahtonen
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For the first part:

If $r \in \mathbb{R}$ with $r \geq 0$, choose $s$ with $s^2 = r$. Then

$f(r) = f(s^2) = f(s)^2 \geq 0$

It follows that $f$ preserves the usual ordering on the reals; as a corollary, we can conclude that if $t > u$, then $f(t) > f(u)$.

Now given $r$, choose an increasing sequence of rational numbers $q_i$ tending to $r$ from below. Using what you've proven about $f$ fixing $\mathbb{Q}$, we see that

$$q_i = f(q_i) \leq f(r)$$

Letting $i \to \infty$, we see that $r \leq f(r)$. Do the same for a decreasing sequence from above.

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    +1 for handling the real case well. But you made a mistake assuming that in the complex case you could also conclude that $f(r)=r$ for all $r\in\mathbb{R}$. This is false, as it may happen that $f(\mathbb{R})\not\subseteq\mathbb{R}$. I'm afraid I cannot describe counterexamples, because Zorn's lemma (i.e. the axiom of choice) is used in a couple of spots while proving their existence :-) – Jyrki Lahtonen Jul 31 '13 at 10:49
  • But it is worth noting that if you don't believe in axiom of choice, then you may have all of them. Read the link given by azimut for more information. – Jyrki Lahtonen Jul 31 '13 at 10:54
  • thank you. That was clear and useful. – Koenraad van Duin Jul 31 '13 at 10:58
  • @JyrkiLahtonen Thanks very much for the correction. +1. –  Jul 31 '13 at 20:27
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  1. let $f:\mathbb{R}\to \mathbb{R}$ be a unitary ring homomorphism, then $f(1)=1$. But this fixes $\mathbb{Q}$ as you have seen, now you can use a density theorem and you have done it.

  2. use $f(i)=-i$, this is still unitary but not the identy. This is the complex conjugation.

Mebat
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