Timeline for Do Grothendieck universes matter for an algebraic geometer?
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37 events
| when toggle format | what | by | license | comment | |
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| Jun 12, 2021 at 8:13 | comment | added | Joel David Hamkins | The question may be a duplicate of mathoverflow.net/q/24552/1946. | |
| May 19, 2019 at 3:31 | comment | added | Will Sawin | 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. | |
| May 19, 2019 at 3:04 | comment | added | user138661 | @WillSawin that is interesting. I have heard he does not post at all, under his name or otherwise. | |
| May 18, 2019 at 20:01 | comment | added | Will Sawin | 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. | |
| May 18, 2019 at 18:51 | answer | added | Nath Rao | timeline score: 15 | |
| May 17, 2019 at 8:04 | comment | added | Noah Schweber | 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 ... | |
| May 17, 2019 at 8:03 | comment | added | Noah Schweber | 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) | |
| May 17, 2019 at 8:00 | comment | added | Noah Schweber | 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. | |
| May 17, 2019 at 7:59 | comment | added | Noah Schweber | @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) | |
| May 14, 2019 at 21:21 | comment | added | Timothy Chow | @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. | |
| May 14, 2019 at 21:02 | comment | added | Robert Furber | @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. | |
| May 14, 2019 at 20:58 | comment | added | Timothy Chow | (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. | |
| May 14, 2019 at 20:58 | comment | added | Robert Furber | @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: 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. | |
| May 14, 2019 at 20:46 | comment | added | user138661 | @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, 2019 at 20:39 | comment | added | Timothy Chow | @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). | |
| May 14, 2019 at 19:44 | comment | added | François Brunault | @schematic_boi Doesn't the Stacks project follow this kind of strategy? (See just after Part 3, Lemma 54.20.3) | |
| May 14, 2019 at 19:26 | comment | added | François Brunault | @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. | |
| May 14, 2019 at 18:47 | comment | added | user138661 | @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, 2019 at 16:59 | comment | added | François Brunault | 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. | |
| May 14, 2019 at 7:09 | comment | added | user138661 | @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: arxiv.org/abs/1708.06494). Can you explain what are "the ordinary questions" in algebraic geometry? | |
| May 14, 2019 at 3:15 | comment | added | Timothy Chow | I think David Roberts is talking about McLarty's later paper, "The large structures of Grothendieck founded on finite order arithmetic." arxiv.org/pdf/1102.1773.pdf | |
| May 14, 2019 at 2:17 | comment | added | David Roberts♦ | 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. | |
| May 13, 2019 at 23:25 | comment | added | Noah Schweber | @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). | |
| May 13, 2019 at 22:51 | comment | added | Timothy Chow | 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" jstor.org/stable/pdf/20749620.pdf | |
| May 13, 2019 at 18:57 | comment | added | Noah Schweber | To eliminate sentences which fail in $L$, like $\neg$CH, we need to work a bit harder - *forcing* 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. | |
| May 13, 2019 at 18:51 | comment | added | Noah Schweber | 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) | |
| May 13, 2019 at 18:49 | comment | added | Noah Schweber | 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) | |
| May 13, 2019 at 18:38 | comment | added | Timothy Chow | I believe that Vidhyanath Rao has looked at this question before. See mathforum.org/kb/… 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. | |
| May 13, 2019 at 18:36 | comment | added | Timothy Chow | 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. | |
| May 13, 2019 at 18:21 | comment | added | user44143 | 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 (en.wikipedia.org/wiki/…), $\Pi^1_2$ sentences are provable in $ZF$ iff they are provable in $ZFC$. The sentences in question include the invariant subspace problem (mathoverflow.net/questions/169033/…), the formalized Millenium Prize Problems, and all ordinary questions in algebraic geometry. | |
| May 13, 2019 at 17:53 | history | became hot network question | |||
| May 13, 2019 at 17:33 | comment | added | Noah Schweber | 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." | |
| May 13, 2019 at 17:31 | comment | added | Noah Schweber | @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. | |
| May 13, 2019 at 17:29 | comment | added | arsmath | The Stacks Project is deliberately avoiding universes. Instead they are using the reflection principle that Noah mentions in his answer. | |
| May 13, 2019 at 17:26 | answer | added | Noah Schweber | timeline score: 23 | |
| May 13, 2019 at 17:05 | comment | added | Noah Schweber | "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.) | |
| May 13, 2019 at 16:46 | history | asked | user138661 | CC BY-SA 4.0 |