This is another way to ask if Wiles's proof can be converted into a "purely number-theoretic" one. If there is no proof in Peano Arithmetic then there should be non-standard integers that satisfy the Fermat equation. I vaguely remember that most proofs in analytic number theory are known to be convertible into elementary ones, probably because some version of predicative analysis is a conservative extension of arithmetic. But Wiles's proof uses not as much analysis as high end algebraic geometry, so I am not sure. Is it convertible into elementary arithmetic, is that even known?
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Can you say precisely what you mean by "non-standard integers"? – Eric Wofsey Jul 28 '16 at 02:59
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@EricWofsey He means some model of Peano Arithmetic, other than the standard one, which would contain counterexamples to Fermat's Last Theorem. – Jul 28 '16 at 03:07
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2@MikeHaskel: That is the interpretation that would make the question make the most sense, but the inclusion of the nonstandard-analysis tag makes me unsure. – Eric Wofsey Jul 28 '16 at 03:08
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2@Eric Yes, that is what I meant. I think a copy of non-standard integers is embedded into hyperreals, hence the tag. – Conifold Jul 28 '16 at 03:13
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3I think of non-standard model of arithmetic as model of "true" arithmetic, in which case there can be no non-standard counterexamples. – André Nicolas Jul 28 '16 at 03:15
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5As far as I know this is still an open question (see http://mathoverflow.net/questions/39239/does-fermat-hold-in-non-standard-models, for instance). But my understanding is that experts strongly suspect the proof can be done in PA, just no one has actually undertaken to work out all the details. – Eric Wofsey Jul 28 '16 at 03:18
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8A recent (2016) paper by Glivicky and Kala constructs a model of Presburger arithmetic in which Fermat's Last Theorem fails (counterexamples exist). – hardmath Jul 28 '16 at 03:18
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There is an oft-mentioned result by Marc Krasner (circa 1940) that gives solutions to Fermat's equation using p-adic number systems. For example, I think Alf van der Poorten goes over this in his 13 Lectures on Fermat's Last Theorem, as it may be held out as evidence elementary proofs of FLT (e.g. via congruence and divisibility properties) are not possible. I'll check my copy for details. – hardmath Jul 28 '16 at 03:42
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@hardmath OP is asking about a the Peano Axioms and as I recall Presburger is strictly weaker. So I think its a stretch to say counterexamples exist. Unless Presburger is weaker in which case I apologize. – Jul 28 '16 at 14:42
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@dav11: I agree OP appears to be asking about Peano arithmetic, so I made a comment to put that request in perspective. Yes, Presburger arithmetic is weaker, so much so as to be decideable. It lacks multiplication, so what is constructed is a model of an extension of Presburger arithmetic. – hardmath Jul 28 '16 at 15:29
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@hardmath: Ah, I see :) – Jul 28 '16 at 22:18
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1@hard, you're conflating Ribenboim's 13 Lectures with van der Poorten's Notes. – Gerry Myerson Aug 08 '16 at 09:19
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@Gerry - Thanks, quite so. Now I have two missing books to look for! – hardmath Aug 08 '16 at 21:06
2 Answers
I am going to respond to two questions quoted below, which come from this comment. Here "TP" means "transfer principle". I've switched the order of the questions.
... why it does not prove that any sentence for which TP is preserved and which is true in the standard model, must be true in all models, and hence provable in PA.
Here is an informal summary of how the hyperreal construction works. If we begin with any structure $M$, with an associated formal language, and take an ultrapower, we will obtain a structure $M^*$. By Los's theorem, $M^*$ will satisfy the same sentences as $M$ in that formal language. In the cases of interest, $M$ has an internal real line $\mathbb{R}$ and an internal semiring of naturals $\mathbb{N}$. The ultraproduct $M^*$ with then have an internal "hyperreal" line $\mathbb{R}^*$ and an internal semiring of "hypernaturals" $\mathbb{N}^*$.
If we begin with different models $M_1$ and $M_2$ (e.g. if one of them is nonstandard), we simply obtain different models $M_1^*$ and $M_2^*$. Sentences from the appropriate language are true in $M_1$ if and only if they are true in $M_1^*$, and true in $M_2$ if and only if they are true in $M_2^*$, but if there was no connection between $M_1$ and $M_2$ to begin with then the ultrapower construction can't create one. The transfer principle only holds between each individual model $M$ and its ultrapower $M^*$.
At this point I am most curious as to why a naive transplantation of the TP argument to "nonstandard" embedding does not (indirectly) prove that FLT holds in all models of PA, even if it does not provide any means of (directly) converting Wiles's proof into PA.
The original question is right - "Wiles's proof uses not as much analysis as high end algebraic geometry". This is bad for nonstandard analysis, though, because when we make the hyperreal line we normally begin with a structure $M$ that is just barely more than the field of real numbers. This is good if we want to work with the kinds of constructions used in elementary analysis, but not as useful if we need to work with more general set-theoretic constructions.
Wiles' proof as literally read uses various set-theoretic constructions of "universes". So, to try to approach that in a nonstandard setting, we would want to start with a model $M$ that is much more than just a copy of the real line. That would not seem to help us show that FLT holds in every model of PA.
The thing about the proof is that the "literal" reading is too strong. I am no expert in the algebraic geometry used, but I followed the discussions in several forums, and here is the situation as I understand it. The proof relies on several general lemmas which, to be proved in utmost generality, were proved using very strong methods. However, in concrete cases such as FLT only weaker versions of the lemmas are needed, and experts seem to believe that the weaker versions should be provable in PA. But actually working out the proof would require a lot of effort to prove in PA the special, weaker cases of all the necessary lemmas and then combine them to get FLT.
This is very analogous in my mind to the fact that in logic we often prove things using strong axioms, but experts in logic recognize that these things are also provable, in concrete cases, in much weaker systems. We don't generally dwell on that, or even point it out, unless there is a specific reason. Indeed, the number of things which I know are provable in PA is much larger than the number for which I have ever written out a proof in PA.
Colin McLarty has published several papers on the axioms needed for FLT. You could look at What does it take to prove Fermat's Last Theorem? from the Bulletin of Symbolic Logic which is relatively accessible.
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Thanks, Carl. Could you add some preamble and reference to Mikhail's answer at the top of yours, because it is not clear what "it" or "TP argument" refers to in the quotes, and the context is buried at the bottom of a long comment thread under it. – Conifold Aug 13 '16 at 20:32
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No problen, I added a link to the comment, which should help people find it – Carl Mummert Aug 13 '16 at 21:09
The OP asks "Are there non-standard counterexamples to the Fermat Last Theorem?" The question can actually be interpreted in two different ways since the OP mentioned in a comment that "a copy of non-standard integers is embedded into hyperreals."
If the question is interpreted as applying to all possible models of PA then by the completeness theorem, if Wiles' theorem is not provable in PA then there should be models where the result fails.
If the question is interpreted as applying to models embeddable in the hyperreals, as the OP's comment suggests, then one could note the following. The theorem can be expressed by a first order formula, and so can Wiles' theorem (that there are no solutions). The transfer principle applies to such formulas. Therefore Wiles' theorem remains true over the hyperintegers, as well. Thus there are no nonstandard counterexamples, either.
This point was made in Mike Haskel's answer in a slightly different form but deleted presumably because Mike come to adopt the first interpretation. I am not sure if the OP can see the deleted answer (this depends on the reputation score). The point remains valid with regard to the second interpretation.
Fermat's theorem is expressible in PA. Therefore an elementary extension like Skolem's constructed in 1933 would also satisfy Wiles' theorem since Wiles did prove Fermat's last theorem, even though he used tools that go beyond PA. Therefore this provides an answer to the OP's question, who specifically mentioned the embedding of the nonstandard integers into the hyperreals in his first comment below the question.
With regard to your question as to the foundational status of the proof of FLT, a pretty much up-to-date account can be found here. A key name is McLarty.
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5I think it's clear from their focus on $PA$ that the OP is not just referring to the hyperintegers, but rather any nonstandard model of $PA$; and for such the question is much more complicated. – Noah Schweber Aug 01 '16 at 15:51
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1@NoahSchweber, Fermat's theorem is expressible in PA. Therefore an elementary extension like Skolem's constructed in 1933 would also satisfy Wiles' theorem since Wiles did prove Fermat's last theorem, even though he used tools that go beyond PA. Therefore this provides an answer to the OP's question. – Mikhail Katz Aug 03 '16 at 07:54
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1@NoahSchweber, what you are reacting to is mostly the discussion in the comments following the question. The OP is a historian rather than a logician as far as I can tell, and there is no reason to assume that by nonstandard integers he meant anything other than infinite numbers in nonstandard extensions of the ordinary integers. Furthermore, Conifold specifically mentioned the embedding of nonstandard integers into the hyperreals. – Mikhail Katz Aug 03 '16 at 13:17
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Sorry, I did not have a clear picture of the relation between general non-standard models of PA and the ones embeddable into hyperreals when asking the question, or even now. Is there a reference for that? Are the non-embeddable models particularly pathological somehow? – Conifold Aug 08 '16 at 21:02
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1@Conifold: every model of PA can be embedded in a similar way as the standard model; it's just that one would obtain nonstandard versions of the hyperreals by starting with a nonstandard model of PA. – Carl Mummert Aug 08 '16 at 23:48
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@Carl Thank you. Does this mean that in the answer "models embeddable in the hyperreals" stand for "models embeddable in the standard model of the hyperreals" (if there is such a thing)? Why wouldn't the transfer argument still work? – Conifold Aug 09 '16 at 02:40
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@Conifold The transfer argument doesn't work because PA isn't a complete theory. Consequently, merely knowing that a first-order sentence is true of the natural numbers doesn't mean it's true of any given model of PA. When people say "the hyperreals"/"the hyperintegers", they usually mean a model of true arithmetic—that is, a model the of the complete theory of the natural numbers. Your question is equivalent to asking whether F.L.T. follows from the axioms of Peano Arithmetic. That question is interesting and doesn't follow from any simple argument on general grounds. – Aug 09 '16 at 08:45
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@Conifold, I would explain this in a nontechnical fashion as follows. Constructing the hyperreals requires a stronger axiomatic background than that provided by PA. Typically the hyperreals are constructed in ZFC though there is current work on managing with much less. Thus, there are models of PA that cannot be models of ZFC, and moreover cannot be embedded in the hyperreals. Skolem's model is so embeddable because of the particular type of construction he used (definable sequences). – Mikhail Katz Aug 09 '16 at 11:55
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@CarlMummert, I can't see how you can say that every model of PA embeds into the hyperreals (even *nonstandard" ones whatever that means). Seems wrong to me. – Mikhail Katz Aug 09 '16 at 12:14
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3@Mikhail Katz: if we take any model of PA, which may be nonstandard, we can naturally extend it to a model of $\mathsf{ACA}_0$, which is a weakish system of second-order arithmetic that is able to define the real line (relative to the original model of PA). If we then take the usual ultraproduct of that model of $\mathsf{ACA}_0$, with respect to a nonstandard ultrafilter, we obtain a version of the hyperreals which embeds the original model of PA just as the standard hyperreals embed $\omega$. What in the normal ultraproduct construction that requires starting with the standard model of PA? – Carl Mummert Aug 09 '16 at 19:58
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2There has been a lot of work on nonstandard analogues of subsystems of second order arithmetic recently, actually. To name a few people, one could look for work by Keita Yokoyama, Sam Sanders, or Jerome Keisler. I am sure I am leaving out other important work here; these are just three I know have published in the area. – Carl Mummert Aug 09 '16 at 20:05
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1@Conifold: when we move to nonstandard models, the transfer principle becomes more interesting. The details are technical, but overall it is possible to get at least part of the transfer principle (certainly enough for $\Pi^0_1$ sentences like the conclusion of Fermat's Last Theorem). If anyone could show this leads directly to a proof of FLT from Peano Arithmetic, it would be a very significant feat. There is no obvious "golden ticket" in that situation that would suggest using nonstandard techniques to prove FLT. So it would be a very interesting if someone found a way to do it. – Carl Mummert Aug 09 '16 at 20:11
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1@Mikhail Katz: I meant "nonprincipal ultrafilter". The ultraproduct would be taken with a nonprincipal ultrafilter over the standard $\omega$. In some cases, I believe the ultrafilter may need to be sufficiently generic, in the sense of forcing; I vaguely remember that from some work of Yokoyama. In some cases it may be necessary to take the model of PA or $\mathsf{ACA}_0$ to be countable, IIRC. The details should be in https://www.math.tohoku.ac.jp/tmj/PDFofTMP/tmp34.pdf – Carl Mummert Aug 09 '16 at 20:12
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@Carl Mike's comment suggests to me that you and Mikhail are using "hyperreals" in two different senses: he means something like "standard interpretation(s)" in ZFC (although it is not clear to me if this presupposes also some "standard interpretation" of ZFC itself) with full transfer principle, and you use it more expansively for analogous ultraproducts that (maybe) satisfy a weaker version of it. Is that right? I am still confused as to why TP argument isn't the "golden ticket" if FLT is of the type for which TP is preserved. Or is that last part itself conjectural? – Conifold Aug 09 '16 at 21:34
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1@Conifold: yes, I think you have got the main issue. And FLT is of the type for which the transfer principle will be preserved in all reasonable cases. The issue is finding some way to use the nonstandard reals or nonstandard naturals to prove something about FLT. Is your question about the hyperreals specifically, or about why Wiles' proof can't easily be converted to an elementary proof? – Carl Mummert Aug 10 '16 at 01:50
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@Carl At this point I am most curious as to why a naive transplantation of the TP argument to "nonstandard" embedding does not (indirectly) prove that FLT holds in all models of PA, even if it does not provide any means of (directly) converting Wiles's proof into PA. Or for that matter why it does not prove that any sentence for which TP is preserved and which is true in the standard model, must be true in all models, and hence provable in PA. I guess I am missing the wedge that limits the scope of this argument, it seems like all true Goldbach sentences must be provable then, which can't be. – Conifold Aug 10 '16 at 02:54
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@CarlMummert, does $\text{ACA}_0$ prove the existence of a free ultrafilter? – Mikhail Katz Aug 10 '16 at 08:39
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2@Mikhail Katz: no, but that is irrelevant to the construction, which can be carried out in ZFC just like the construction of the standard hyperreals. We can take the ultraproduct of any structure with itself. If our original structure has its own internal notion of a real line (possibly nonstandard) then the ultraproduct will have its own internal version of a hyperreal line (again possibly nonstandard). Los' theorem applies to all ultraproducts, so we will have a transfer principle, exactly analogous to the usual hyperreals. Every model of $\mathsf{ACA}_0$ has a robust internal real line. – Carl Mummert Aug 10 '16 at 12:01
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@CarlMummert, in this model which presumably satisfies transfer, what does the collection of sets $A\subseteq\mathbb{N}$ such that $H\in A^\ast$ look like, where $H$ is a fixed infinite integer? – Mikhail Katz Aug 11 '16 at 07:55
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3@Mikhail Katz: The same usual ultrapower construction works. If $N$ is a number in the ground model ($N$ can be standard or nonstandard) and $S = (S_m : m \in \omega)$ is a set in the ultraproduct, where as usual each $S_m$ is a set in the ground model, then we have $N^* \in S$ if and only if ${ m : N \in S_m }$ is in the ultrafilter on $\omega$. The ground model has both a collection of "numbers" and a collection of "sets" of "numbers"; this is why we convert the model of PA to a model of $\text{ACA}_0$ before the ultrapower. The ultrapower uses an ultrafilter on the standard $\omega$. – Carl Mummert Aug 11 '16 at 11:13
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2If the original model $M$ is uncountable, it may give something more "hyperreal-like" to take an ultrafilter over $|M|$ instead of $\omega$. That makes it possible to construct a number in the ultrapower larger than every number in the ground model. – Carl Mummert Aug 11 '16 at 11:46