14
$\begingroup$

In a big topos whose objects are a kind of "space", it sometimes happens that when we define some "set" internally to the topos, the "topology" it automatically acquires coincides with the "correct" or "expected" one for the usual external definition. For instance, the real numbers object in the topos may be the set of real numbers with its usual "topology" (in whatever sense that topos represents "topology"). For example, this is the case for the Dedekind real numbers in the topos of sheaves on cartesian spaces $\mathbb{R}^n$, and for both Cauchy and Dedekind real numbers (which coincide) in Johnstone's topological topos.

Is there any topos in which this holds for the Zariski spectrum of a commutative ring? That is, if I start with an ordinary external commutative ring, map it into the topos as a "discrete" object, and then construct the object of prime ideals (or maybe filters -- whatever makes the most sense constructively) of that internal ring object, does the automatic intrinsic "topology" on the resulting object ever coincide with the usual Zariski topology?

One might naturally guess some topos of sheaves on algebraic spaces. But I don't think it's impossible that Johnstone's topological topos might also work; despite comments elsewhere that convergence isn't useful in the Zariski topology, Wikipedia tells me that the Zariski topology of a commutative Noetherian ring is a sequential topological space, and sequential spaces embed fully-faithfully in the topological topos.

$\endgroup$
15
  • 5
    $\begingroup$ Great question! This observation, due to Felix Cherubini and Thierry Coquand, feels relevant, even though it doesn't use the categories of sets as the base topos: Internally to the big Zariski topos with its structure sheaf $\mathbb{A}^1$, the constructively sensible versions of the spectrum construction actually turn out to have enough points (unexpectedly so, because constructively we usually lack enough points), at least for the spectrum of finitely presented $\mathbb{A}^1$-algebras. $\endgroup$ Feb 17, 2023 at 13:27
  • 7
    $\begingroup$ Or is this already an answer? Let $A$ be a finitely presented ring. The constant sheaf on $A$ in the big Zariski topos (of $\mathbb{Z}$) is from the internal point of view still just a ring, but it becomes an $\mathbb{A}^1$-algebra by tensoring $\cdot \otimes_{\mathbb{Z}} \mathbb{A}^1$. Then the internal spectrum of this algebra describes exactly the functor of points of $\mathrm{Spec}(A)$. It doesn't quite fit your bill because of the extra tensor product. Let me key in Matthias Hutzler, he will likely be able to contribute. $\endgroup$ Feb 17, 2023 at 13:44
  • 1
    $\begingroup$ @IngoBlechschmidt Yes, that's not quite what I had in mind, but it's definitely along the same lines. In addition to the tensor product, I was hoping for a topos that's a little more "topological". Is there a direct way to relate sheaves on the big Zariski topos to some more traditionally topological structure such as open sets or convergence spaces? And can you give a reference for the fact you mentioned? $\endgroup$ Feb 17, 2023 at 16:51
  • 2
    $\begingroup$ A reference for the observation on enough points is Lemma 3.1.9 in these notes by Felix, Matthias and Thierry. They even give a purely internal proof from the axiom of synthetic quasicoherence. Unfortunately I cannot offer any further insights right now. Every object $X$ of the big Zariski topos can be turned into a topological space, by the coend $\int^R X(R) \times \mathrm{Spec}(R)$. By the way, Felix told me that he posted this reference here, but the post got downvoted and was removed. I do not understand why this happened. $\endgroup$ Feb 17, 2023 at 20:42
  • 1
    $\begingroup$ Yes, the comment-answer should have been flagged for mod attention, not downvoted. $\endgroup$
    – David Roberts
    Feb 17, 2023 at 22:22

2 Answers 2

11
$\begingroup$

Let me give the condensed perspective: Regarding $A$ as a discrete condensed ring, I think the structure of the "internal spectrum" is codified by the functor that takes any extremally disconnected profinite set $S$ to the poset of sheaves of prime ideals in the constant sheaf on $A$ over $S$. (One could forget the poset structure and regard it only as a condensed set. I will comment below what structure this remembers.) Here, a "sheaf of prime ideals" is defined to be a sheaf of ideals $I$ of $A$ together with a sheaf of multiplicative subsets $M$ of $A$ such that the map $I\sqcup M\to A$ is an isomorphism (of sheaves of sets); I hope this is the correct way to talk about "internal prime ideals"?

I claim that this is the "correct" answer to this question. Recall that Makkai's conceptual completeness theorem as explained by Lurie in his course on categorical logic, or by Barwick-Glasman-Haine in their work on exodromy, gives a fully faithful embedding of the category of coherent locales (aka spectral spaces) to condensed posets. In one direction, this takes any coherent locale to the condensed category of points.

Summary: The spectrum of a ring is naturally a spectral space, i.e. coherent locale, so determined by its condensed poset of points. This is precisely the spectrum of $A$ as constructed internally in condensed sets.

Addendum: If one forgets the poset structure and only looks at the condensed set of prime ideals, one actually ends up getting a condensed set that is representable by a profinite set, which is precisely $\mathrm{Spec}(A)$ with its constructible topology.

$\endgroup$
4
  • $\begingroup$ Just so I understand completely: are you saying that the internal spectrum constructed in condensed sets does not get the condensed structure induced from the open-set Zariski topology in the naive way, but that it is instead the "correct" condensed structure in a different sense? $\endgroup$ Feb 28, 2023 at 20:11
  • $\begingroup$ Yes. The condensed structure it acquires gives the constructible topology of $\mathrm{Spec}(A)$. To see the actual Zariski topology, it is then sufficient to remember the specializations, recorded in the poset structure. $\endgroup$ Feb 28, 2023 at 20:14
  • 2
    $\begingroup$ As I discuss also in "The geometry of coherent topoi and ultrastructures", one should also acknowledge the work of Marmolejo on this topic, besides Makkai, Lurie and Barwick-Haine. $\endgroup$ Mar 1, 2023 at 10:20
  • 3
    $\begingroup$ I think the constructively correct way of constructing the prime spectrum of a ring is by looking at prime filters rather than prime ideals. A prime filter is a multiplicatively closed subset containing $1$ and such that if $a + b$ is a member, then at least one of $a$ or $b$ are members. Of course, this is classically equivalent to being the complement of a prime ideal, so I suppose in a context with enough points it is equivalent to speak of ideals whose complements are multiplicatively closed and contain $1$. $\endgroup$
    – Zhen Lin
    Mar 1, 2023 at 10:42
4
$\begingroup$

New answer, following up on Zhen Lin's comment for the "good" definition of prime filter. Sorry for the confusion!

A prime filter in a ring $A$ is a multiplicatively closed subset $S$ of $A$ (containing $1$) such that if $a+b\in S$, then $a\in S$ or $b\in S$. This definition makes sense internally in a topos. And it is almost a tautology that the category of sheaves on $\mathrm{Spec}(A)$ is precisely the classifying topos for prime filters in $A$ (where the universal prime filter is the sheaf of units of the structure sheaf).

In particular, if $X$ is any topological space, then continuous morphisms from $X$ to $\mathrm{Spec}(A)$ are equivalent to prime filters on the constant sheaf on $A$ on $X$.

In particular, working on the big site of topological spaces, the space of prime filters on the constant sheaf on $A$ gives precisely the topological space $\mathrm{Spec}(A)$.

(You could also work in pyknotic sets instead, and get the pyknotic space associated to $\mathrm{Spec}(A)$. Or work in Johnstone's topos, and get the sequential space...)

$\endgroup$
2
  • $\begingroup$ Is it really that simple? Maybe I missed a step. Fix a site $\mathcal{C}$. Suppose there were a general construction turning sheaves $\mathcal{C}^\textrm{op} \to \textbf{CRing}$ into sheaves $\mathcal{C}^\textrm{op} \to \textbf{Set}$ such that the value at $X$ is the prime spectrum of the ring at $X$, at least for some selected $X$. Then in particular we could do this for $\mathcal{C} = \textbf{CRing}^\textrm{op}$ and the identity functor to get a covariant (!) functor $\textbf{CRing} \to \textbf{Set}$ whose value at some selected $X$ is the prime spectrum of $X$... $\endgroup$
    – Zhen Lin
    Mar 2, 2023 at 10:06
  • $\begingroup$ The value of that thing at $X$ wouldn't be the spectrum of the ring at $X$; rather, the spectrum is some kind of total space over $\mathcal C$ and the value at $X$ are sections of it. These have the correct variance. $\endgroup$ Mar 2, 2023 at 13:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.