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Let $k$ be a field. The usual motivation for the Zariski topology on affine space $\mathbb{A}^n(k)$ is that it is the coarsest topology for which the algebraic sets, the zero loci of polynomials, are closed.

This can be phrased in more topological or categorical terms: we may characterize the Zariski topology on $\mathbb{A}^n$ as the initial topology with respect to all regular maps into $k$, if we endow $k$ with the topology where the only nontrivial closed set is $0$. Call this topology $\tau,$ so $\tau =\{\varnothing,k^\times,k\}.$

There are other topologies $k$ could carry for which the Zariski topology is the initial topology for regular maps in to $k$, including for example the Zariski topology on $\mathbb{A}^1\cong k.$ But for the purposes of parsimony, and for non-circularity in motivating the definition of the Zariski topology, I prefer to use $\tau$, which is the coarsest topology on $k$ with this property.

Does this topological space $(k,\tau)$ occur in the literature, or have a name? Is there a natural or intrinsic algebraic justification for this topology (whatever that might mean)? It seems like an algebraic analogue of Sierpiński space, in that it classifies open sets in the regular category.

The only intrinsic topology that I know for an arbitrary ring is the $I$-adic topology. But the only ideal of a field is the zero ideal, and the $I$-adic topology for the zero ideal gives the discrete topology. So $\tau$ is not the $I$-adic topology.

I do not see any way to view $\tau$ as the Zariski topology on $k$, which, if it existed, should be the zero loci of the constant polynomials, hence the trivial topology. Actually that's not correct, that's not the Zariski topology on $k$, the Zariski topology is properly assigned to $\mathbb{A}^0=\text{pt}$, not to $k$, which is instead its coordinate ring. Anyway, $\tau$ is not the trivial topology.

We might identify $k$ with $\mathbb{A}^1,$ but the Zariski topology on $\mathbb{A}^1$ is usually the cofinite topology, which has $\lvert k\rvert$ many closed points, whereas $(k,\tau)$ has only one. So $\tau$ is not the Zariski topology either.

I'm hoping to provide an intrinsic motivation for this topology $\tau$ on $k$, to use in turn to motivate the definition of the Zariski topology from earlier principles, so even if we could view it as the Zariski (which, again, I don't see how we can), I'm hoping to hear a different justification.

ziggurism
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  • thanks to @Pierre-GuyPlamondon for your comments, they were very helpful to clarify my question. – ziggurism Dec 24 '17 at 16:50
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    The answer here by Slade seemed relevant https://math.stackexchange.com/a/863562/16490 – ziggurism Dec 24 '17 at 17:11
  • I think this answer by Vladimir Sotirov may contain hints of the answer – ziggurism Jan 15 '18 at 19:56
  • The closed sets of $\tau$ are $\varnothing,{0},$ and $k$. Which are also the zero loci of the constant functions $k\to k$ plus the identity map. I wonder whether this leads to an answer – ziggurism Apr 30 '20 at 23:25
  • If instead of just declaring our field to carry a topology, we also demand it to be a topological monoid/ring, then in fact it may have more closed sets. If ${0}$ is closed, then every singleton should be closed, since addition is continuous. Therefore instead of $\tau={\emptyset,k^\times,k}$, we actually do have the cofinite topology. – ziggurism Oct 24 '23 at 03:54

2 Answers2

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Another way could be to endow $k$ with the trivial valuation. Let $v:k\to \{0\}\cup\{\infty\}$ be the trivial valuation. The open unit ball is the elements with valuation bigger than $0$, which is just $0$ in this case. So $k^\times$ is a closed set. And clearly $k$ is also a closed set. And there are no other closed sets.

ugur efem
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  • I like this answer, because the spectrum of a DVR is literally Sierpinski space. But I'm confused. The spectrum of a field is a point, not Sierpinski space. I guess the usual definition of DVR does not allow the valuation to be trivial...? – ziggurism Dec 24 '17 at 17:20
  • Yes. Discrete valuation means the value group is isomorphic to $\mathbb Z$. Which we are violating in this case. – ugur efem Dec 24 '17 at 17:22
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    @ziggurism In the $\mathbb Z$-valuation on a DVR, the units have valuation $0$, and the non-units have positive valuation. Similarly here, the units have valuation $0$. – Dustan Levenstein Dec 24 '17 at 17:34
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    @DustanLevenstein so the assumption that your valuation is nontrivial means necessarily assuming our ring is not a field. I got it. – ziggurism Dec 24 '17 at 18:01
  • @ziggurism not quite... – Dustan Levenstein Dec 24 '17 at 18:03
  • @DustanLevenstein hmm let me try again. In a DVR the valuation must be surjective onto $\mathbb{Z}$, there are required to be some non-units of every positive valuation, while for our field $k$, our trivial valuation $\nu$ has no elements with $0<\nu(a)<\infty$? We only have non-units of valuation $\infty$? – ziggurism Dec 24 '17 at 18:13
  • And what's the maximal ideal in a DVR? Is it $\nu^{-1}(\mathbb{Z}\setminus 0),$ which is empty here? – ziggurism Dec 24 '17 at 18:15
  • @ziggurism Sorry didn't have the time to write an answer earlier. The valuation on a DVR extends to the field of fractions on a canonical way: The elements of $\mathfrak m^n$ which are not in $\mathfrak m^{n+1}$ have valuation $n$, for any $n \in \mathbb Z$ (where $\mathfrak m^n$ refers to the group operation on fractional ideals). A $\mathbb Z$-valuation on a field $k$ is precisely the same thing as a choice of DVR $D$ with $\operatorname{Frac}(D)=k$. This answer makes reference to the trivial valuation on $k$; there may be other valuations. – Dustan Levenstein Dec 24 '17 at 18:28
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    @DustanLevenstein ok I think I've got it now. $\mathfrak{m}=\nu^{-1}(n\geq1)=\bigcup_k^\infty \mathfrak{m}^k.$ Spec of a DVR has two points, the zero ideal $\nu^{-1}(\infty)$ and the maximal ideal. But our trivial valuation has only the zero ideal. – ziggurism Dec 24 '17 at 19:09
  • @ziggurism Correct. – Dustan Levenstein Dec 24 '17 at 19:21
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I think the usual way to describe this is that the inclusion $\mathbb{A}^1 \setminus \{ 0 \} \to \mathbb{A}^1$ is the universal standard open set.

It's universal in the following sense: the standard open subsets of any affine scheme $X$ are in bijective correspondence with the inverse images of $\mathbb{A}^1 \setminus \{0 \}$ under regular maps $X \to \mathbb{A}^1$.

Usually we write $D(f)$ for the standard open corresponding to $f : X \to \mathbb{A}^1$.

Incidentally, I don't think the phrase "universal standard open set" has common usage; it's just the natural description to give to the construction so described.

  • Perhaps this would be a good way to say it?Just as Sierpinski space $\top\to 2$ is the open subset classifier in Top, $\mathbb{A}^1\setminus 0\to\mathbb{A}^1$ is in Scm/$k$. – ziggurism Dec 24 '17 at 17:13
  • @ziggurism: It's not quite the classifier since there are open sets that aren't standard opens. For example, the open set $\mathbb{A}^2 \setminus { (0,0) }$ is not a standard open. (also, the classifying map wouldn't be unique; e.g. $D(f) = D(uf)$ for any unit $u$) –  Dec 24 '17 at 17:58
  • Hmm I see. To use that universal/univalent nomenclature, it's neither universal (exist exactly one), nor versal (exist one, not unique), nor univalent (exists at most one). It's the valent open set. – ziggurism Dec 24 '17 at 18:06
  • I think this answer was authored by Hurkyl, who has quit the site and deleted their account. – ziggurism Nov 18 '19 at 19:07