Is every Polish ring topology on $\mathbb{C}$ defined by an absolute value?

Question

There is a unique up to isomorphism algebraically closed field of characteristic 0 and cardinality of the continuum. Let's call it $K$. We usually call it $\mathbb{C}$, but by this we impose a topological structure* on $K$. In fact there are many other possible ring topologies on $K$ (by a ring topology we mean a topology for which the addition and multiplication are continuous maps $K\times K \to K$). There seems to be a zoo of those**. I care more for better behaved ones.

It is natural to consider separable topologies (ie admitting a countable dense subset) such that the uniform structure imposed by the additive group structure is complete. In fact, I care mostly for Polish topologies (separable and defined by a complete metric). For these the uniform structure is automatically complete.

A famous Polish topology on $K$ is $\mathbb{C}_p$ which is obtained by completing the algebraic closure of $\mathbb{Q}_p$ with respect to the unique absolute value on it extending the $p$-adic absolute value on $\mathbb{Q}_p$ (absolute value = multiplicative norm). Here is another example: consider $k$, a countable field (eg $\mathbb{Q}$ or its algebraic closure), and take the degree valuation on $k(t)$. Complete to $k((t))$, extend (uniquely) the absolute value to the algebraic closure and complete again (use Krasner's Lemma). Get a Polish topology on $K$ for which $k$ is discrete.

All the examples of topologies on $K$ above, and all the examples I am aware of are defined by absolute values. Hence the question in the title: Is every Polish ring topology on $\mathbb{C}$ defined by an absolute value?

Remark: Given a topology on $K$ there is a unique minimal closed, algebraically closed subfield. I am willing to assume $K$ is minimal if it helps.

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Note: for locally compact fields it is always the case that the topology is defined by an absolute value. Indeed, one can use the modular function applied to multiplication operator on the additive group in order to construct an absolute value on the field.

$*$ There are plenty of ways to identify $K$ with $\mathbb{C}$. Given a topology on $K$ we can conjugate it by a field automorphism and define another. We consider here topologies up to conjugation.

$**$ related: https://mathoverflow.net/a/106454/89334

Answer 1

No. Note that if field $F$ has topology induced by $[0,\infty)$-valued absolute value, then either $\{x \in F : |x|<1\} = \{0\}$ so that the topology is discrete, or else there is $x$ with $0 < |x| < 1$, so that $$ \exists x,\qquad 0 < |x| < 1, \qquad \lim_{n\to\infty} x^n = 0 . \tag{1} $$ We will provide a counterexample: a Polish topological field, algebraically closed, cardinal $\mathfrak c$, not discrete, where $(1)$ fails.

This will be (a subfield of) a field $\mathbb C[[t^G]]$ of Hahn series. References: nLab , Wikipedia .

We begin with $\langle G, +, < \rangle$: an abelian, totally ordered, group. This is all that is needed for construction of $\mathbb C[[t^G]]$. [The "valuation group" here is $G$.]

In this example, we take $G = \mathbb Q^{(\omega)}$ with lexicographic order: That is, $G$ consists of all infinite sequences of rationals $q = \langle q_0,q_1,\cdots\rangle$ with $\exists K \in \mathbb N, \forall k \ge K, q_k = 0$. The order is: $q<p$ iff $$ \exists K \in \mathbb N, \quad (\forall k > K, q_k=p_k)\quad \& \quad(q_K < p_K) . $$

$G$ is countable.

$G$ is divisible.

Let $e_k = \langle 0,\cdots,0,1,0\cdots\rangle$ with $1$ in position number $k$, then $e_k \ll e_{k+1}$ in the sense that $n e_k < e_{k+1}$ for $n=1,2,3,\cdots$.

If $q \in G$, then there exists $p \in G$ with $p \gg q$. In fact, we may choose $p = e_{K+1}$ where $K$ is chosen so that $q_k = 0$ for all $k > K$.

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Properties of $F = \mathbb C[[t^G]]$.

$\bullet $ cardinal. Since $G$ is countable, we have $$\mathfrak c = \operatorname{card}(\mathbb C) \le \operatorname{card} F \le (\operatorname{card} \mathbb C)^{\operatorname{card}(G)} = \mathfrak c^{\aleph_0} = \mathfrak c,$$ so $\operatorname{card} F = \mathfrak c$.

$\bullet $ Since $\mathbb C$ is algebraically closed and $G$ is divisible, $F$ is algebraically closed.

$\bullet $ Hahn field $F$ comes with a valuation $v : F\setminus\{0\} \to G$. A base for the neighborhoods of $0$ are $V_g := \{x \in F : x = 0\text{ or } v(x) > g\}$ for $g \in G$. In this case, $G$ is countable, so $F$ is first-countable. Since $F$ is a topological group, $F$ is metrizable.

$\bullet $ Topological field $F$ is complete (as a uniform space) for general $G$, but in this case it is complete metric.

Unfortunately this $F$ is not separable. So we take a subfield. For $e_k \in G$ as above, choose $\alpha_k \in F$ with $v(\alpha_k) = e_k$. Then let $$ F_1 = \mathbb Q(\alpha_0,\alpha_1,\dots) $$ so $F_1 \subseteq F$ is a countable field. Let $F_2$ be the algebraic closure of $F_1$ in $F$. So $F_2$ is countable. Finally, let $F_3$ be the topological closure of $F_2$ in $F$. Then $F_3$ is separable, closed, and thus Polish. And $F_3$ is still algebraically closed. LINK

$(1)$ fails in $F_3$. Indeed, if $x \in F_3$, there exists $k$ with $e_k \gg v(x)$. So $v(x^n)= n v(x) < e_k$ for all $n$, and thus the sequence $x^n$ does not converge to $0$.

Cardinal of $F_3$ is $\mathfrak c$. Indeed, any Polish space with no isolated points contains a homeomorphic copy of the Cantor set. $\alpha_k \to 0$ so $F_3$ has no isolated points.