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.

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

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    $\begingroup$ About the note: in $\mathbf{C}$, the modular fonction yields $\Delta(2)=4$, while $|2|=2$. $\endgroup$ – YCor May 13 '16 at 22:22
  • $\begingroup$ By "comes from an absolute value", do you mean, given by a complete (multiplicative) norm? $\endgroup$ – YCor May 13 '16 at 22:23
  • $\begingroup$ @YCor, thanks for your comments. An edit was made in order to clarify things. $\endgroup$ – Uri Bader May 14 '16 at 15:31
  • $\begingroup$ Your absolute value has values in $[0,+\infty)$? Maybe for a counterexample try the Levi-Civita field (extended by adding $i$). Maybe we get a larger value group. $\endgroup$ – Gerald Edgar Apr 1 at 15:17
  • $\begingroup$ @GeraldEdgar My question regards indeed $[0,\infty)$ valued absolute values, but I certainly welcome more generous interpretation of it, eg: "Is every Polish ring topology on C defined by a valuation?". I do not understand your suggestion regarding the Levi-Civita field - Isn't it defined by a valuation a priori? $\endgroup$ – Uri Bader Apr 2 at 6:28

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 $k > K$.

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.

  • $\begingroup$ Thanks for the answer! I understand that $F$ is metrizable. Could you please elaborate on why is it completely so? Why do you insist on having $\mathbb{C}$ as your residue field rather than taking a countable algebraically closed field instead? Do you need this for completeness? I doubt it. If you insist on using $\mathbb{C}$, I find your argument for having $F_3$ algebraically closed insufficient. Note that Krasner's Lemma assumes the existence of an absolute value. Could you please elaborate here as well? Thanks. $\endgroup$ – Uri Bader Apr 3 at 20:54
  • $\begingroup$ A countable algebraically closed field could be used to start with instead of $\mathbb C$. But I still have to go to a subfield go make it separable. The field of Hahn series is always complete (in the uniform sense compatible with group translation). If it is also metrizable, then use an invariant metric, and metric completeness follows from general completeness. BUT If $F_3$ algebraically closed is known only for value group contained in $\mathbb R$, then I do have a problem. I may have to extend $F_1 \subset F_2 \subset F_3 \subset \cdots$ countably many steps. $\endgroup$ – Gerald Edgar Apr 3 at 21:40
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    $\begingroup$ I am convinced now that $F$ is an example of a field $\simeq\mathbb{C}$ which is endowed with a completely metrizable topology, but does not admit a compatible absolute value. This is the first example of this sort that I am aware of - thanks for giving it! However, at the moment I am not convinced that it has a closed subfield which is both separable and algebraically closed. $\endgroup$ – Uri Bader Apr 3 at 22:04

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