Extension of $|\cdot|_\infty$ on $\mathbb R$ to $\mathbb C$

Question

Let $|\cdot|$ be the usual absolute value on $\mathbb C$. My question is:

Is the only extension of $|\cdot|$ on $\mathbb R$ to $\mathbb C$ $|\cdot|$ itself?

I'm not sure about the uniqueness. I want to show any two extensions of $|\cdot|$ to $\mathbb C$ induce same topologies, which means that one is a positive power of the other (and so are equal).
Is it true any two absolute values on $\mathbb C$ induce the same topology (as $\mathbb C$ is a finite dimensional vector space over $\mathbb R$)?

Thank you!

Answer 1

Yes, it is true that the usual absolute value $|\cdot|_\infty$ on $\Bbb{R}$ extends uniquely to $\Bbb{C}$, but no, it is false that all absolute values on $\Bbb{C}$ (or even on $\Bbb{R}$, for that matter) induce the same topology.

Here's a counterexample: with a bit of work one can prove that every $p$-adic absolute value $|\cdot|_p$ on $\Bbb{Q}$ extends (albeit non-uniquely) to an absolute value $|\cdot|_p$ on $\Bbb{C}$. Then observe that $|\cdot|_p$ cannot be equivalent to $|\cdot|_\infty$ because $|n|_p < 1$ for every $n \in \Bbb{Z}_{>0}$: in particular, every $|\cdot|_p$-ball of radius $>1$ contains the positive integers, but you can always find a positive integer outside of a $|\cdot|$-ball.

On the other hand, by one of Ostrowski's theorems we know that up to equivalence there is exactly one Archimedean valuation $|\cdot|_\infty$ on $\Bbb{Q}$ (you can find the proof on PlanetMath). Now, recall that $\Bbb{R}$ can be constructed as the completion of $\Bbb{Q}$ with respect to this absolute value. By the uniqueness of completions, this means that $|\cdot|_\infty$ is the unique Archimedean absolute value on $\Bbb{R}$. Finally, note that $\Bbb{C} \supset \Bbb{R}$ is an algebraic extension of degree $2$ and apply the following

Theorem. If $K$ is a complete field with respect to an absolute value $|\cdot|$, then $|\cdot|$ can be extended in a unique way to any given algebraic extension $L \supset K$. Furthermore, if $L$ is finite then $L$ is complete with respect to $|\cdot|$.

Proof. See Neukirch's Algebraic Number Theory, Theorem II.4.8.

Answer 2

Any absolute value on $\mathbf{C}$ restricting to the usual absolute value on $\mathbf{R}$ is in particular a norm on $\mathbf{C}$ as a two-dimensional real vector space. All norms on a finite-dimensional vector space induce the same topology. Thus your absolute value $|\cdot|$ on $\mathbf{C}$ restricts to a continuous homomorphism $S^1\to \mathbf{R}^+$, which must be trivial as $\mathbf{R}^+$ has no nontrivial compact subgroups, so you must have $|re^{i\theta}|=|r|$ for all $r$ and $\theta$.

Answer 3

You don't need all of Ostrowski's theorem to give that any extension of the absolute value of $\mathbb{R}$ is indeed the one used to. Here is an elementary argument. Let $||_1$ and $||_2$ be extensions of $||_{\infty}$ to $\mathbb{C}$ , where $||_1$ is the familiar extension (i.e the norm). Consider then the function $$f:\mathbb{C} \rightarrow \mathbb{R}$$ given by $f(z) = |z|_2/|z|_1.$ Then we note that since for $z= a+bi$ we have that $$|a+bi|_2 \leq |a|+|b| \leq \sqrt{2} \sqrt{a^2+b^2} = \sqrt{2} |a+bi|_1$$ we have that $$|a+bi|_2/|a+bi|_1 \leq \sqrt{2}.$$ But actually, $f(\alpha) \leq 1$ for all $\alpha$ as well, since if not, since the norm is multiplicative, we'd have that for $n$ large enough that $f(\alpha^n)=f(\alpha)^n > \sqrt{2}$ which is a contradiction. Thus, we see that $|\alpha|_2 \leq |\alpha|_1.$ This shows that the two valuations give the same topology, so the restriction of them to $\mathbb{R}$ shows that they're equal.