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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 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.

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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.

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  • $\begingroup$ 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...? $\endgroup$ – ziggurism Dec 24 '17 at 17:20
  • $\begingroup$ Yes. Discrete valuation means the value group is isomorphic to $\mathbb Z$. Which we are violating in this case. $\endgroup$ – ugur efem Dec 24 '17 at 17:22
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    $\begingroup$ @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 $0$. $\endgroup$ – Dustan Levenstein Dec 24 '17 at 17:34
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    $\begingroup$ @DustanLevenstein so the assumption that your valuation is nontrivial means necessarily assuming our ring is not a field. I got it. $\endgroup$ – ziggurism Dec 24 '17 at 18:01
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    $\begingroup$ @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. $\endgroup$ – ziggurism Dec 24 '17 at 19:09
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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.

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  • $\begingroup$ 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/$k$. $\endgroup$ – ziggurism Dec 24 '17 at 17:13
  • $\begingroup$ @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 $u$) $\endgroup$ – Hurkyl Dec 24 '17 at 17:58
  • $\begingroup$ 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. $\endgroup$ – ziggurism Dec 24 '17 at 18:06

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