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I recently learned that some parts of SGA require axioms beyond ZFC. I am just a simple algebraic geometer so I am trying to understand how can this fact impact my life (you may have engaged in a similar game with the axiom of choice at some point of your mathematical career). I think this was only needed in the volume developing cohomology of topoi, correct me if I am wrong.

The question: is there some statement about schemes not involving the word "topoi" that you know how prove using the additional axioms, but do not know how to prove without them? I believe this question has nothing to do with the notion of completeness in mathematical logic (because I am not asking if something can be proved in principle, just if there is an obvious argument). If there is some logical argument showing that the answer is negative, I would like to learn about that too.

Bonus points if the proof of the statement in your answer relies on a Weil cohomology theory, rather than something random (I do not know if this is even possible). To be more precise, a Weil cohomology theory is only defined for smooth projective schemes over a field, for which all of this should probably be irrelevant, so I mean that the cohomology theory you use restricts to a Weil cohomology theory for smooth projective schemes over a field.

P.S. There was similar discussion in the context of Fermat's theorem, and if I understand correctly, the user BCnrd insists that universes are useless for etale cohomology without giving a reference. I believe there are few people on this planet who know etale cohomology better than BCnrd so that's some useful information. The question is not, however, limited to etale cohomology.

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    "without giving a reference" That may be because no reference exists: I suspect this is the sort of thing where everyone familiar with the relevant arguments knows that they only need very concrete facts about etale cohomology, but nobody's bothered to write it up because it would be both straightforward and tedious. (Confession: I actually have no idea, being a complete doofus in the relevant topic, but this is what I've been told by people who aren't doofoi.) – Noah Schweber May 13 '19 at 17:05
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    The Stacks Project is deliberately avoiding universes. Instead they are using the reflection principle that Noah mentions in his answer. – arsmath May 13 '19 at 17:29
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    @arsmath How do they handle arguments which naively use >1 universes? At a glance, the reflection heuristic becomes much less convincing (per the end of my answer), although I may be missing something. – Noah Schweber May 13 '19 at 17:31
  • Supporting my initial comment, Pete Clark mentions in the comments below the linked answer that "as far as [he knows] there is no published complete proof of the relevant results that bypasses SGA IV so in that sense it is not an established fact." – Noah Schweber May 13 '19 at 17:33
  • If any algebraic geometers "have engaged in a similar game with the axiom of choice", the answer is probably no for them: By Shoenfield absoluteness (https://en.wikipedia.org/wiki/Absoluteness#Shoenfield's_absoluteness_theorem), $\Pi^1_2$ sentences are provable in $ZF$ iff they are provable in $ZFC$. The sentences in question include the invariant subspace problem (https://mathoverflow.net/questions/169033/is-the-invariant-subspace-problem-arithmetic), the formalized Millenium Prize Problems, and all ordinary questions in algebraic geometry. –  May 13 '19 at 18:21
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    In general, finding a "natural" mathematical statement that is unprovable in ZFC (and that makes no overt mention of uncountable sets) is very difficult, even if you expressly set yourself the task of doing so. So the chances that any statement you care about as an algebraic geometer is unprovable in ZFC are very small. However, if you want a meta-theorem that "automatically" eliminates universes, then my understanding is that there isn't one---not because of any fundamental obstacle, but because practitioners don't write in a way that makes it easy to formulate a suitable meta-theorem. – Timothy Chow May 13 '19 at 18:36
  • I believe that Vidhyanath Rao has looked at this question before. See http://mathforum.org/kb/thread.jspa?forumID=13&threadID=98167&messageID=493942#493942 for example. He wrote an even more informative post, but I can't locate it right now. I'll try emailing him to try to persuade him to post an answer here. – Timothy Chow May 13 '19 at 18:38
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    Adding to @MattF.'s comment, Shoenfield absoluteness does much more than that: it also, for example, eliminates any possible role of the Continuum Hypothesis. Specifically, Shoenfield says that if $M,N$ are models of ZF and $M$ is an inner model of $N$ (= a transitive subclass of $N$ containing all the $N$-ordinals), then every $\Delta^1_3$ sentence with real parameters from $M$ which is true in $M$ is also true in $N$ and (trivially) conversely. Eliminating CH (for example) can then be done as follows: (cont'd) – Noah Schweber May 13 '19 at 18:49
  • Suppose ZFC+CH proves some (parameter-free) $\Delta^1_3$ sentence $\varphi$ and $N\models$ ZF. $N$ has an inner model $L^N$ - its version of Godel's $L$ - and Godel's analysis of $L$ goes through in ZF alone, so in particular ZFC+CH is true in $L^N$. Since ZFC+CH proves $\varphi$, we have that $\varphi$ is true in $L^N$, which by Shoenfield lifts to $N$ itself. Since every model of ZF satisfies $\varphi$, Completeness tells us that ZF proves $\varphi$. (cont'd) – Noah Schweber May 13 '19 at 18:51
  • To eliminate sentences which fail in $L$, like $\neg$CH, we need to work a bit harder - [forcing*](https://en.wikipedia.org/wiki/Forcing_(mathematics)) is the key tool here, and getting the details right is technical (we need ill-founded models)*. But ultimately the "shape" is exactly the same. Note that all of this is a lot more constructive than one might expect (including the proof of Shoenfield absoluteness itself, in fact): while appeals to absoluteness as a black box may feel unsatisfying at first, they're really just concrete, if lengthy, combinatorial arguments in tidy packaging. – Noah Schweber May 13 '19 at 18:57
  • By the way, Colin McLarty has written a very relevant paper: "What does it take to prove Fermat's Last Theorem? Grothendieck and the logic of number theory" https://www.jstor.org/stable/pdf/20749620.pdf – Timothy Chow May 13 '19 at 22:51
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    @TimothyChow Some care is needed with that paper, in my opinion, in particular around what McLarty means by "uses universes." I don't think he makes any claim that universes are actually needed, merely that they served as a simplifying tool for introducing some large-scale concepts used in the proof. Personally, my read of that paper was as at least in part a defense of the use of unnecessarily powerful methods (with which I agree wholeheartedly). – Noah Schweber May 13 '19 at 23:25
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    McLarty has gone further, and shown that much less than ZFC is required for derived functor cohomology, and the less assumed on the schemes of interest (for instance, Noetherian, or countable), the weaker the axioms needed. – David Roberts May 14 '19 at 02:17
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    I think David Roberts is talking about McLarty's later paper, "The large structures of Grothendieck founded on finite order arithmetic." https://arxiv.org/pdf/1102.1773.pdf – Timothy Chow May 14 '19 at 03:15
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    @MattF. I am not sure I understand. I believe the statement "every non-empty affine scheme has a closed point" is equivalent to the axiom of choice (as claimed here: https://arxiv.org/abs/1708.06494). Can you explain what are "the ordinary questions" in algebraic geometry? –  May 14 '19 at 07:09
  • I'm not an expert of universes, but if you want to define etale cohomology without them, a standard way is to define the etale site of X using only X-schemes of cardinality less than some strong limit cardinal. Then one needs to prove that increasing this cardinal doesn't change the cohomology. – François Brunault May 14 '19 at 16:59
  • @FrançoisBrunault but is there a published reference using the approach you suggest? Personally to me, a confirmation from BCnrd means more than an article written by people I do not know, but still, the academic culture requires written references. I think Stacks project does not rely on universes though I did not carefully inspect the hypotheses they use. –  May 14 '19 at 18:47
  • @schematic_boi Unfortunately I don't know more, I only saw this briefly mentioned by Bhatt and Scholze in their article on the pro-étale topology. I would guess there are places where there are more (but maybe not all) details. – François Brunault May 14 '19 at 19:26
  • @schematic_boi Doesn't the Stacks project follow this kind of strategy? (See just after Part 3, Lemma 54.20.3) – François Brunault May 14 '19 at 19:44
  • @schematic_boi : "Ordinary questions," roughly speaking, are questions about specific geometric objects of interest, rather than very general statements about all geometric objects. For particular rings that you might be interested in, such as the ring of integers in a number field, or the coordinate ring of a projective variety, you don't need the full strength of the axiom of choice. "Every vector space has a basis" also needs AC in full generality, but in practice, we care about specific vector spaces for which we can get away with much less than AC (and sometimes no choice at all). – Timothy Chow May 14 '19 at 20:39
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    @TimothyChow OK, I am a bad logician, but here is a question. Is there some specific commutative unital ring for whom the existence of a maximal ideal is equivalent to AC? Or if no, is it possible to prove that no such ring exists? Because otherwise, I do not completely understand the point (I mean, I realize that we all have to hide behind some safe words to discuss this, since it seems very hard to actually describe exactly what statements are going to be true, so I do not really blame anyone, just to clarify). –  May 14 '19 at 20:46
  • @schematic_boi The proof of the equivalence of choice and every commutative unital ring having a maximal ideal works by explicitly constructing (under ZF) a ring. See here: https://academic.oup.com/jlms/article-abstract/s2-19/2/285/818317 However, I don't think there is one ring that has a maximal ideal iff the axiom of choice holds. Presumably for any given ring you can make choice fail for cardinals much bigger than it, while still keeping its maximal ideals intact. – Robert Furber May 14 '19 at 20:58
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    (continued) But this illustrates the difficulty with proving a metatheorem to eliminate "unnecessarily general" hypotheses. People tend to state and quote theorems in their general form because it's "cleaner" that way, even if they are only going to apply the theorem in a very special case for which the generality is unnecessary. If a paper is written in this manner then quoting a meta-theorem is typically not enough. Strictly speaking you would need to go through the paper and verify that indeed a less general theorem suffices. – Timothy Chow May 14 '19 at 20:58
  • @schematic_boi The version for the ultrafilter lemma is a little simpler -- $\mathcal{P}(X)/\mathcal{P}_{\mathrm{fin}}(X)$ has (as a Boolean ring) a maximal ideal iff $X$ has a nonprincipal ultrafilter. These are very ageometric rings, being far far away from finitely-presented rings over a field or a ring of algebraic integers. – Robert Furber May 14 '19 at 21:02
  • @RobertFurber : I think schematic_boi is asking a different question. Working over ZF, can we define a specific ring $R$ such that "Spec $R$ has a closed point" implies AC? That is, can we reduce AC to the existence of a closed point for a single affine scheme as opposed to all affine schemes? I think the answer is no, but I think to prove it requires a bit of work. One would need to show in ZF that from any ring we can form a "larger" ring for which it is consistent that the larger ring has no maximal ideal. – Timothy Chow May 14 '19 at 21:21
  • @TimothyChow The danger is that the definition of the ring in question could be something that itself refers to where choice fails (if it fails at all) - e.g. there is a single formula, the set defined by which can be well-ordered iff AC holds globally: the formula "Either $x=\emptyset$ and AC holds, or AC fails and $x=V_\alpha$ for the least $\alpha$ such that $V_\alpha$ is non-well-orderable." I don't see how to get that to work here - we can isolate the least rank of a ring with no maximal ideal (if such exists), but we can't pick one out of the set of all minimal-rank such rings (cont'd) – Noah Schweber May 17 '19 at 07:59
  • and I don't know a way to "fuse together" a set of rings $(R_i)_{i\in I}$ into a single ring $S$ such that from a maximal ideal in $S$ we can recover maximal ideals in each of the $R_i$s (whereas that's not an issue for the well-orderability example). But that does pose an issue for a negative answer to the question in its most direct phrasing. – Noah Schweber May 17 '19 at 08:00
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    That said, it is easy to show that any such definition really has to refer to the whole set-theoretic universe at once (and so be "non-concrete" in some sense). Specifically, suppose $\phi$ is a formula which ZF proves defines a ring. Take a model $M$ of ZFC, and build a symmetric extension $N$ of $M$ such that choice fails in $N$ but $M$ and $N$ agree up to rank $\alpha+2$ where $\alpha$ is the rank of the ring defined by $\phi$ in $M$. In passing from $M$ to $N$ we haven't added any new subsets of $\phi^M$, so $\phi^M$ still has a maximal ideal in $N$, which means $\phi^N\not=\phi^M$ (contd) – Noah Schweber May 17 '19 at 08:03
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    which is to say, $\phi$ "changed its mind" even though we only altered the universe well above $\phi$('s old position). So this does rule out a broad class of ways of defining rings. Whether it's satisfying, well ... – Noah Schweber May 17 '19 at 08:04
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    I've heard that, ironically, one of the reason BrnCnrd no longer posts on MO under that name (or any other approximation of his real name) is that he doesn't want people to believe what he says purely on the authority of his name - as you suggest you would (mostly quite reasonably!) do - but rather only on the basis of his arguments. – Will Sawin May 18 '19 at 20:01
  • @WillSawin that is interesting. I have heard he does not post at all, under his name or otherwise. –  May 19 '19 at 03:04
  • I haven’t seen him around in a while but that span of time is much less than the time since he last posted under his own name. – Will Sawin May 19 '19 at 03:31
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    The question may be a duplicate of https://mathoverflow.net/q/24552/1946. – Joel David Hamkins Jun 12 '21 at 08:13

2 Answers2

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It's inherently difficult to give a negative answer to a question like this, but here's a technical fact that pushes in that direction:

Let ZFC$_n$ be the subtheory of ZFC gotten by restricting Separation and Replacement to $\Sigma_n$ formulas. By the reflection principle,$^1$ for each $n$ the theory ZFC proves that there is an ordinal $\alpha_n$ such that $V_{\alpha_n}\models$ ZFC$_n$. That is: $$\mbox{For each $n\in\mathbb{N}$, ZFC proves Con(ZFC$_n$).}$$

We can think of the $V_{\alpha_n}$s as "approximate universes" which behave like universes for all "sufficiently simple" formulas, the point being that if you specify a complexity level ahead of time you can always assume you have an approximate universe appropriate to that complexity level.

Now the compactness theorem now naively suggests that - since we can only ever use finitely many sentences in a given proof - any argument with universes whatsoever can be replaced with one involving just approximate universes, and hence a proof in ZFC. This is of course false, but counterexamples have to be "global" as opposed to "local" - they need to at some point refer to the whole of the universe in question as a single completed object.

For exampe, the way ZFC + universes proves the consistency of ZFC is by showing that a universe $U$ is a model of ZFC. The statement "$U\models$ ZFC" is expressed in the language of set theory by talking about Skolem functions over $U$ (or something morally equivalent), and this takes place in the context of the powerset of $U$. But this sort of thing isn't to my knowledge how universes are applied in algebraic geometry - they instead use a universe to argue that a "sufficiently closed" object exists in that universe, and this "local" argument is exactly the sort of thing that the reflection principle tells us can generally be reduced to ZFC alone.

Basically, a candidate example needs to not just take place inside a universe, but rather over a universe.

That said, there is an obvious place to look for such: arguments using two (or $n$) universes. The larger universe does see the smaller universe as a completed object, so the coarse heuristic above suggests that we can replace only the larger universe with an approximate universe - that is, that arguments which are quickly phrased in terms of two universes can be directly translated to arguments involving only one universe. Now we can't cheat anymore - we nee actual arguments about algebraic geometry. My understanding is that we're still in a situation where universes are an unnecessary convenience, but now I'm far outside of my area of competence. Still, the above should give an indication of why a real essential use of universes in a concrete result (which will certainly only involve reference to a small fragment of the cumulative hierarchy) would be very surprising.

$^1$OK fine, the reflection principle is usually phrased for finite subtheories of ZFC. But $(i)$ that's not really any different as far as the heuristic is concerned, just more annoying to work with; and $(ii)$ the stronger version of reflection I've stated is also true (the point being that for each $n$, the schemes of $\Sigma_n$-Separation and -Replacement can be expressed in the language of set theory by a single sentence, which in turn can be proved from finitely many of the ZFC axioms which we can bash with the usual reflection hammer).

And on that note, it's worth pointing out two facts about reflection which help flesh out the picture:

  • First, given that ZFC proves the compactness theorem, we seem to be in tension with Godel's incompleteness theorem. What saves us is that "$\forall n$" and "ZFC proves" don't commute (unless ZFC is inconsistent of course): while ZFC does prove each specific instance of reflection, it can't prove the full version (unless, again, it's inconsistent).

  • It's also worth noting that a similar result holds for (first-order) Peano arithmetic (as does the analogous version of the previous bulletpoint), although of course we need to talk about mere consistent Henkinized complete theories as opposed to canonical-ish models. As a cute consequence, Kripke used this fact to give a purely model-theoretic proof of Godel's incompleteness theorem (in the absence of reflection, his argument would require the soundness of PA, similarly to how Godel's original argument assumed $\omega$-consistency rather than mere consistency).

Noah Schweber
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    In fact, I very much doubt that there's a single instance where Grothendieck universes are used where it wouldn't suffice to have a model of, say, ZFC with Replacement limited to $\Sigma_1$ formulas (let's keep full Separation to be sure); and for this, the $V_δ$ where $δ$ is a fixed point of $α\mapsto\beth_α$ provide a good supply. In what usual mathematical reasoning would such a $V_δ$ not suffice as a “universe”? (OK, maybe let $δ$ be of uncountable cofinality to get closure under sequences as well.) Who ever uses (uncountable) $\Sigma_2$ replacement outside of set theory? – Gro-Tsen May 13 '19 at 21:12
  • @Gro-Tsen Oh I would certainly bet large sums of money that that's true, but I can't actually claim meaningful confidence there (cf. my previous comment re: local doofositude). – Noah Schweber May 13 '19 at 21:47
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    @NoahSchweber : What I am most impressed by is that you used a word ("doofositude") that I understood the meaning of immediately, yet which gets zero Google hits. – Timothy Chow May 13 '19 at 22:36
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    @TimothyChow I'm also proud of "doofoi." – Noah Schweber May 13 '19 at 22:37
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    @NoahSchweber I'd have gone with "doofi" myself :) – Yemon Choi May 14 '19 at 00:15
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    Or doofopodes :-) – David Roberts May 14 '19 at 02:18
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    Remark regarding the first of the two bullet points at the end: I am fairly certain it is not even possible to express the full reflection principle in the language of ZFC, because of Tarski's undefinability theorem. $Sigma_n$-reflection is not a statement uniform in $n$. – Wojowu May 14 '19 at 09:39
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    @Wojowu That's true of course. – Noah Schweber May 14 '19 at 16:49
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    @YemonChoi is right. If doofoi were the plural, then the singular should be doofos. – Andreas Blass Jun 12 '21 at 16:22
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Tim Chow drew my attention to this thread, and asked if I cared to comment. Actually I wrote up a detailed version of my thoughts several years ago, On doing category theory within set theoretic foundations, and to my surprise, Solomon Feferman thought enough of it for that to appear in the book "What is Category Theory". Their website has disappeared, so here is the version I found in my back-ups.

As a homotopy theorist, I have no idea of what is need for Wiles' proof of FLT, or what is used in EGA or SGA. For homotopy theory, the crux of the matter comes down to two things. First, we construct towers iteratively (potentially of transfinite length) and then take its limit or colimit. This assumes that the whole tower is small, while all we know is that the length is small. This is an implicit use of replacement. [Granted, if the iterative step is predicative, ZFC is enough. But see the next paragraph] The earliest use of this I know of is the proof of Brown Representability Theorem dating back 1960. I recall the same kind of argument somewhere in Neeman's book on triangulated categories. So may be it also needed for derived categories.

Another similar situation: I have a functor F and a small subcategory D of the domain of F. I want to know that F(D) is small so that I can talk about its limit/colimit. This already happens with the Milnor short exact sequence for arbitrary generalized cohomology theories [and can enter into play when studying generalized cohomology of infinite dimensional spaces such as classifying spaces of groups.] If F is itself constructed as the (co)limit of a tower constructed by transfinite induction, trying to stay within ZFC proper is not trivial.

People seem to get hung up on indexed categories (just use fibered categories all the time) and functor categories. But there is another use of universes that is more troublesome: Prove a theorem about small categories (or quasicategories or Segal spaces or ...) and then apply to "the category of all spaces" (or ...). If the theorem about small categories was proved using only ZFC, then ZFC + global reflection does the trick. This was a suggestion of Feferman from the 60's, and my write-up was about getting around the second issue above. But published proofs ignore the subtleties: They implicitly assume Morse-Kelly. Then we need to invoke Grothendieck universes.

[If you don't understand the fuss in the third paragraph, think about this: $W$ is a set that satisfies all the ZFC axioms. For each natural number n, I construct an element $x_n$ of $W$. The details of the construction are hidden in a back box. Is the set $\{x_n\}$ an element of $W$? Or do you need to know the details of the construction?]

Timothy Chow
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Nath Rao
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