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Do non-Hausdorff locally compact topological fields exist? If so, what is an example?

Edit: In light of Daniel Fischer's comment, I need to add the condition that topology not be the trivial $\{X,\emptyset\}$ topology.

Context:

X is locally compact if every point of X has a compact neighbourhood. (A neighbourhood of a point is a set containing an open set which contains the point.)

A topological group $G$ is Hausdorff if and only if the subgroup containing only the identity $\{e\}$ is closed. $G/\overline{\{e\}}$ is Hausdorff, so some authors just assume Hausdorff when discussing topological groups.

Commonly (wikipedia, Taibelson's book, etc) a local field is defined as a locally compact topological field, and it is proved that every local field has an absolute value (induced by the unique-up-to-multiplicative-factor property of Haar measure) and is therefore Hausdorff.

But the usual proofs of the existence of Haar measure require the locally compact topological group to be Hausdorff.

So it seems to me that the usual definition of local field assumes Hausdorff (otherwise I am not sure how to construct the Haar measure with the unique-up-to-multiplicative-factor property).

Thus I am wondering if non-Hausdorff locally compact topological fields exist. I looked in various books and googled around before asking.

Related: https://mathoverflow.net/questions/22800/must-a-locally-compact-group-be-hausdorff-in-order-to-possess-a-haar-measure

Daniel Fischer has pointed out that any field with the trivial (indiscrete, only open sets are whole space and empty set) topology is a non-Hausdorff compact topological field. He also pointed out that the $T_0$ axiom must be avoided to get non-Hausdorff.

Here's a candidate non-$T_0$ topology (from wikipedia)to put on the field $\mathbb{C}=\mathbb{R}^2$: Open sets are the Cartesian product of an open set in $\mathbb{R}$ and $\mathbb{R}$ itself, i.e., the product topology of $\mathbb{R}$ with the usual topology and $\mathbb{R}$ with the trivial topology; points $(a,b)$ and $(a,c)$ are not distinguishable.

RitterSport
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    There's always the (rather boring) indiscrete topology. Any field, endowed with the indiscrete topology is a topological field (unless one includes something like $T_0$ in the definition), and it is clearly quasicompact. – Daniel Fischer Jun 30 '17 at 18:19
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    @DanielFischer By quasicopmact, do you mean what most non-algebraic geoemeters call compact (every open cover has a finite subcover)? – RitterSport Jun 30 '17 at 18:23
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    I mean that. But I had the impression that algebraic geometers tend to call that compact, since the Zariski topology isn't Hausdorff. However, I tend to follow Bourbaki in this matter. – Daniel Fischer Jun 30 '17 at 18:28
  • @DanielFischer Your comment about the $T_0$ axiom is insightful. As soon as you have that, you have Hausdorff (because topological groups are completely regular and locally compact topological groups are normal) – RitterSport Jun 30 '17 at 18:35

1 Answers1

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Any field with the indiscrete topology is a non-Hausdorff locally compact topological field. However, this is the only example: if $K$ is a non-Hausdorff topological field, then the topology on $K$ must be indiscrete.

To prove this, note first that the intersection of all open sets containing $0$ is an additive subgroup $K_0$ of $K$ which is nontrivial since otherwise $K$ would be $T_0$ and hence Hausdorff. But for any nonzero $x\in K$, multiplication by $x$ is a homeomorphism $K\to K$ that fixes $0$, so $xK_0=K_0$. Since $K_0$ contains a nonzero element $y$, it contains $xy$ for all nonzero $x$, and every nonzero element of $K$ can be written in this form. Thus $K_0$ is actually all of $K$. That is, the only open set containing $0$ is $K$ itself; it follows that the topology is indiscrete.

Eric Wofsey
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  • Could you please explain why the intersection of all open sets containing $0$ is an additive subgroup of $K$? – RitterSport Jul 02 '17 at 15:49
  • Well, I don't actually use that fact in my proof. But $K_0$ is just the set of $x$ such that $0\in\overline{{x}}$, which is the same as the set of $x$ such that $-x\in\overline{{0}}$. But ${0}$ is a subgroup, and hence so is its closure. – Eric Wofsey Jul 02 '17 at 16:05
  • Oh, you're right. It's not used in the proof. – RitterSport Jul 02 '17 at 16:31
  • More generally, one has (Bourbaki, General Topology, Ch. III, §6, no. 4, Proposition 5): [Let $H$ be a dense subring of a topological ring $A$, and let $K$ be a subring (resp., an ideal) of $H$. Then the closure $\overline{K}$ of $K$ in $A$ is a subring (resp., an ideal) of $A$.] If we apply this result to a the closure of the zero ideal on a topological field $F$ (and noting that the closure of a point on a topological space equals the intersection of all neighborhoods of the point) we obtain that $F$ is either T₁ (whence Hausdorff) or has the trivial topology. – Elías Guisado Villalgordo Sep 30 '23 at 15:58