Is a field with archimedean absolute value with compact unit sphere complete?

Question

By the unit sphere of a valued field $(K,|\cdot|)$ I mean $\{x\in K:|x|=1\}$.

We know that if the unit closed ball $\{x\in K:|x|\le 1\}$ of a valued field $K$ with nontrivial absolute value is compact, then $K$ is a local field: for every $x\in K$, the set $\{x'\in K:|x'-x|\le 1\}$, being homeomorphic to the unit closed ball, is a compact neighborhood of $x$. From this question one learns that the compactness of the unit sphere implies the compactness of the unit closed ball for a nonarchimedean field. As a result, if the unit sphere of nonarchimedean field with nontrivial absolute value is compact, then is a local field.

I'm interested in the archimedean case: if $|\cdot|$ is an archimedean absolute value on $K$ with compact unit sphere, what can be said about $K$? Must it be complete, in which case it must be isomorphic to $\mathbb{R}$ or $\mathbb{C}$ by one of the two Ostrowski's theorems?

Edit: Every archimedean field can be embedded into $\mathbb{C}$, so we can suppose that $K\subset\mathbb{C}$. I have been able to show that $\{x\in K:|x|=1\}$ is compact if and only if it is equal to either $\{x\in \mathbb{C}:|x|=1\}$ or $\{\pm 1\}$ (see Note below). Of course, $\{x\in K:|x|=1\}=\{x\in \mathbb{C}:|x|=1\}$ if and only if $K$ contains the field generated by $\mathbb{Q}$ and $\{x\in \mathbb{C}:|x|=1\}$. So the question becomes how to determine the subfields $K$ of $\mathbb{C}$ that satisfy $\{x\in K:|x|=1\}=\{\pm 1\}$. Apart from subfields of $\mathbb{R}$, fields of the form $\mathbb{Q}(\omega\sqrt[3]{D})$ or $\mathbb{Q}(\omega^2\sqrt[3]{D})$ are also examples, where $\omega=\dfrac{1+\sqrt{3}{\rm i}}{2}$ is a primitive 3rd root of unity and $D\in\mathbb{Q}^+$ is not a cube.

Note: A subgroup of the group of unit circle is either discrete or dense, with the proof being similar to the subgroup of $\mathbb{R}$ case. If $\{x\in K:|x|=1\}$ is compact, it is either the whole unit circle or discrete. In the latter case there is a maximum $N\in\mathbb{N}^*$ such that $\exp\left(\dfrac{\pi\mathrm{i}}{N}\right)\in K$. If $N\ge 2$, then $\dfrac{r+\exp\left(\dfrac{\pi\mathrm{i}}{N}\right)}{r+\exp\left(\dfrac{\pi\mathrm{i}(N-1)}{N}\right)}\in K$ would have modulus $1$ and argument $2\arctan\dfrac{\sin\dfrac{\pi}{N}}{r+\cos\dfrac{\pi}{N}}\in \left(0,\dfrac{\pi}{N}\right)$ whenever $r\in(1,+\infty)\cap K$, which is a contradiction.

Edit 2: It should be clarified that, the question now is to find the subfields $K$ of $\mathbb{C}$, such that $\{x\in K:|x|=1\}=\{\pm 1\}$. Of course we are interested only in those that are not contained in $\mathbb{R}$.

The condition of $K$ is equivalent to: for every $x\in K\setminus(\mathbb{R}\cup\mathrm{i}\mathbb{R})$, one has $\overline{x}\notin K$. "$\Longleftarrow$": suppose that $x=\exp(\mathrm{i}\theta)\in K$, then $\overline{x}=x^{-1}\in K$, so $x\in\{1,\mathrm{i},-1,\mathrm{-i}\}$, and $x=\pm\mathrm{i}$ is clearly impossible (for example, $K$ contains $a+b\mathrm{i}$ and its conjugate whenever $a,b\in\mathbb{Q}$, $ab\neq 0$). $\Longrightarrow$": if $x\in K\setminus(\mathbb{R}\cup\mathrm{i}\mathbb{R})$, then $\frac{\overline{x}}{x}$ has norm $1$, but is not equal to $\pm 1$.

@EricWofsey Wow, the example is so easy! Silly me! If you don't mind, I'm going to wait to see if there are any examples apart from the proper subfields of $\mathbb{R}$ :) — Jianing Song, Jul 28 '23 at 03:27
All archimedian fields are isomorphic to subfield of $\mathbb R$ :) — mihaild, Jul 28 '23 at 09:08
The two examples you exhibit might not be subfields of $\mathbb R$ the way they are written, but they are isomorphic to subfields of $\mathbb R$. So the natural conjecture is that the fields you are looking for are exactly the fields which have an embedding into $\mathbb R$ (e.g. for number fields, the ones that are not totally imaginary). — Torsten Schoeneberg, Jul 28 '23 at 16:52
I changed the title and wording to avoid confusion with archimedean orders. -- Do you mind sharing your proof on why the "unit sphere" is either $\pm1$ or the full complex unit circle? — Torsten Schoeneberg, Jul 29 '23 at 19:37

Torsten Schoeneberg · Answer 1 · 2023-08-09T14:05:46.420

Not a full answer, just remarks. I call $U =\{x \in \mathbb C : \lvert x \rvert =1\}$ the complex unit circle.

Annoyingly, and contrary to what I naively believed in earlier comments, the property you look for is not invariant under field isomorphism, i.e. there can be subfields $K \simeq L$ of $\mathbb C$ such that $K\cap U = \{\pm1\}$ but $L \cap U$ is dense in $U$. For an algebraic example, take $K = \mathbb Q(s)$ with $s$ a Salem number, and let $L=\mathbb Q(\alpha)$ with $\alpha$ one of the non-real roots of the minimal polynomial of $s$.
This applies "even more so" to transcendental extensions, as the rational function field $K=\mathbb Q(T)$ can be embedded into $\mathbb R$, but of course we can alternatively send $T$ to any transcendental element of the unit circle in $\mathbb C$.
On the positive side, the argument you outline generalizes to the following characterization.

Proposition: Let $K \subset \mathbb C$ be a subfield. Then $K \cap U$ is dense in $U$ if and only if $K$ contains a non-real field $K_0 \not \subset \mathbb R$ which is stable under the complex conjugation $\mathbb C \rightarrow \mathbb C, x \mapsto \bar x$.

Proof: If $K \cap U$ is dense in $U$, take any $u \in U \setminus \{\pm1\}$; since $\bar u = u^{-1}$, $K_0 := \mathbb Q(u) \subset \mathbb C$ will do. Conversely, by assumptions $K_0$ and hence $K$ contains some pair $\bar x \neq x$, so it also contains all $$u_r := \dfrac{r+x}{r+\bar x} $$ for all $r\in \mathbb Q$, and these generate a dense subgroup of $U$.

How helpful this characterization is, is up to debate. At least it gives us some sufficient criteria, which are invariant under field isomorphisms:

Corollary: If $K$ is a field which contains a CM field, or a non-real Galois extension of $\mathbb Q$, then every embedding $K \hookrightarrow \mathbb C$ contains a dense subgroup of the unit circle.

(This applies e.g. to all $p$-adic fields which, according to axiom of choice, can be embedded into the complex numbers, and which (regardless of choice) contain e.g. many CM fields.)

In earlier comments, I had speculated that the number fields $K$ with dense $K\cap U$ might be exactly the totally complex ("totally imaginary") ones. This is false. The examples in comment 1 (cf. also the answers to Are all algebraic integers with absolute value 1 roots of unity? as well as K. Conrad's https://kconrad.math.uconn.edu/blurbs/galoistheory/numbersoncircle.pdf) are fields $K$ with dense $K\cap U$ that have ("also") real embeddings; on the other hand, e.g. $K := \mathbb Q(X)/(X^4+2)$ is totally complex but neither of its embeddings into $\mathbb C$ is stable under conjugation (if they were, they would be Galois). These examples also show that the sufficient criterion in the corollary above is not necessary.
In the short note 1, the above proposition is the main step to prove a characterization of CM-fields which singles them out from the more general fields we are after here (and which distinguishes them from the fields generated by Salem numbers or similar). Galois-theoretically, Galois CM fields are the ones among the above where complex conjugation is (not just contained in but) central in their Galois group, cf. https://mathoverflow.net/q/203012/27465. Also, it follows from https://mathoverflow.net/q/131475/27465 that CM fields are exactly the totally complex fields for which $\iota(K)$ is stable under conjugation for all embeddings $\iota: K \hookrightarrow \mathbb C$.

1 Blanksby, P. E.; Loxton, J. H., A note on the characterization of CM-fields, J. Aust. Math. Soc., Ser. A 26, 26-30 (1978). ZBL0413.12002.Blanksby, P., & Loxton, J. (1978). A note on the characterization of CM-fields. Journal of the Australian Mathematical Society, 26(1), 26-30. doi:10.1017/S1446788700011460

Thanks for your elaboration! :) – Jianing Song Aug 09 '23 at 13:49 — Jianing Song, Aug 09 '23 at 13:49

Is a field with archimedean absolute value with compact unit sphere complete?

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