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I can endow any field with a natural topology in the following way. Given a polynomial $f\in K[X]$, I denote by $\mathcal{O}(f)=\{x\in K\mid\exists y\in K^{\times}\ f(x)=y^2\}$, i.e. the set of elements $x\in K$ such that $f(x)$ is a non-zero square in $K$. I then consider the topology generated by the sets $\mathcal{O}(f)$.

This gives a unified description for a priori completely different topologies.

  • If $K$ is algebraically closed, one recovers the Zariski topology.
  • If $K=\mathbb{R}$, one recovers the Euclidean topology.
  • If $K=\mathbb{Q}_p$, one recovers the $p$-adic topology.
  • If $K$ is finite, one recovers the discrete topology.

The field $K$ together with this topology is in general not a topological field. What I can prove so far is that this topology is always T1 and finer than the Zariski topology. I should also mention that this gives a way to topologize the affine space $K^n$ by considering polynomials $f\in K[X_1,\dots,X_n]$.

I am wondering if people studied this topology. When is this topology Hausdorff? Do you know other fields where one recovers an interesting topology?

Bumblebee
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Jacques
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  • The notation $X^n$ when $X$ is a set is universally used to denote the Cartesian product of the set with itself $n$ times, you actually used it ($K^n$) so I suggest you don't use it to denote also the set ${x^n \mid x \in X}$ – jjagmath Nov 26 '23 at 11:08
  • @jjagmath Good point. I changed it. Thanks – Jacques Nov 26 '23 at 11:11
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    Why is it natural to consider squares exactly? – Martin Brandenburg Nov 26 '23 at 11:16
  • @MartinBrandenburg I don't have a rigourous argument for it. Let's just say that it is a topology that makes sense over any field. Further more it satisfies the following property: any field isomorphism is an homeomorphism. – Jacques Nov 26 '23 at 11:26
  • Fair enough. It is a good question, but maybe a bit too open-ended ("what can one prove ..."). More specific questions (which you can also split in several posts) are more suitable for being answered. For example the Hausdorff criterion. – Martin Brandenburg Nov 26 '23 at 12:45
  • Actually, $K$ can be any ring (here $K^\times$ is the group of units). What is the topology for $\mathbb{Z}$ then? Or $\mathbb{Q}$? – Martin Brandenburg Nov 26 '23 at 12:48
  • @MartinBrandenburg Right, one can make the same definition for a ring. What is the topology on $\mathbb{Q}$? I don't think we obtain the topology induced by the Euclidean distance. – Jacques Nov 26 '23 at 13:04
  • @jjagmath The set of non-zero squares of $\mathbb{Q}$ is an open set in $\mathbb{Q}$ for this topology. However, it is not an open set for the topology induced by the Euclidean distance. – Jacques Nov 26 '23 at 14:35
  • I find this incredibly interesting, if indeed one recovers all the topologies in the way you say. I have some recommendations for further questions, first of all, can this be generalised to rings? (I don't see why not, but I am not 100% sure I understand your construction) and crucially, are multiplication and addition continuous functions with this topology? – Carlyle Dec 12 '23 at 07:24
  • Could you also expand on the point "If $K$ is Algebraically closed this returns the Zariski topology" as far as I am aware the Zariski Topology has as its pointset the set of prime ideals of a ring, but this seems to only be defining a topology on a field (which doesn't have any non trivial ideals anyway) – Carlyle Dec 12 '23 at 07:52
  • @Carlyle Thank you for your enthusiasm! 1) You can indeed generalize this to any ring. The sets $\mathcal{O}(f)$ make indeed sense and you can take the topology generated by them. 2) If K is algebraically closed, then every non-zero element is a square. Hence $\mathcal{O}(f)$ is the preimage of $K^{\times}$ by $f$, i.e. the complement of $f^{-1}({0})$. This is the definition of the Zariski topology. – Jacques Dec 12 '23 at 15:17
  • @Carlyle 3) In general this topology does not give rise to a topological field. For instance for $K=\mathbb{C}$, consider the map $\phi:\mathbb{C}\times \mathbb{C}\to \mathbb{C}$, $(x,y)\to x-y$. Endow $\mathbb{C}\times \mathbb{C}$ with the product topology. If $\phi$ were continuous, then $\phi^{-1}({0})$ would be closed, but this is ${(x,x)\in\mathbb{C}\times\mathbb{C}}$. The latter is not closed in $\mathbb{C}\times\mathbb{C}$ because $\mathbb{C}$ is not Hausdorff for the Zariski topology. – Jacques Dec 12 '23 at 15:30

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