Timeline for Algebraic theorems with no known algebraic proofs
Current License: CC BY-SA 4.0
23 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 26 at 2:40 | comment | added | Martin Brandenburg | Can we please stop pretending that in all these examples there is anything particular algebraic but non-topological, or the opposite? Things like formal power series, ordered fields, classes of slopes, quasi-isometries cannot be tagged with "algebra" XOR "topology". | |
| Nov 20 at 14:05 | comment | added | Wojowu | @JamesEHanson Someone has done the work of algebraically establishing the properties of the class of Archimedean fields to show there is a maximal one and has the usual completeness proeprty: drive.google.com/file/d/1fb1pNJVMuROmpLq-kDwXXeY2QxzCV0hy/view | |
| Nov 20 at 8:13 | comment | added | Jesse Elliott | On a related note, the ring of $p$-adic integers can be constructed algebraically as $\mathbb{Z}[[X]]/(p-X)$, and, just as $\mathbb{R}$ is isomorphic to the ring $\text{QEnd}(\mathbb{Z})$ of all quasi-endomorphisms of $\mathbb{Z}$, the ring of adeles over $\mathbb{Q}$ is isomorphic to $\text{QEnd}(\mathbb{Q})$. See studenttheses.universiteitleiden.nl/access/item%3A3596357/view | |
| Nov 20 at 1:54 | comment | added | Jesse Elliott | @TimothyChow Here is a purely algebraic definition of $\mathbb{R}$: arxiv.org/pdf/math/0301015. A purely algebraic definition of $\mathbb{C}$ is $\mathbb{C} = \mathbb{R}[X]/(X^2+1)$. It is a field because $X^2+1$ is irreducible over the field $\mathbb{R}$. The challenge is to give an algebraic proof that $\mathbb{C}$ is algebraically closed. | |
| Nov 19 at 23:14 | comment | added | Kevin Casto | @TimothyChow I would actually be pretty satisfied with a "purely algebraic" proof that "The degree two extension of any complete Archimedean ordered field is algebraically closed" and cordons off analysis purely to the existence statement. This seems hard though! | |
| Nov 19 at 10:26 | comment | added | Adam Epstein | @TimothyChow One possible criterion would be Galois invariance. | |
| Nov 19 at 9:28 | comment | added | Jannik Pitt | I think one should not conflate constructing an object and showing that the constructed object indeed coincides with the desired object, and deducing properties from an already defined object. The former is the business of the fundamental theorem of algebra: we construct $\mathbb{C}$ using analysis and show that it is algebraically closed as expected. The only well-posed question is thus, whether there is an alternative characterisation of $\mathbb{C}$ in "purely algebraic terms" not "explicitly encompassing" the algebraic closedness of $\mathbb{C}$. This does not seem very well-posed... | |
| Nov 19 at 8:55 | comment | added | PseudoNeo | Are there other algebraic characterizations of ℂ which don't explicitly say it is algebraically closed? | |
| Nov 19 at 8:27 | comment | added | Oliver | Maybe Artin's proof via Galois theory is algebraic enough. | |
| Nov 19 at 0:02 | comment | added | Timothy Chow | @LSpice Well, I guess I was implicitly thinking about a construction. I don't normally think of the fundamental theorem of algebra as stating that $\mathbb{C}$ is algebraically closed if it exists. | |
| Nov 18 at 19:23 | comment | added | LSpice | @JamesEHanson, re, sure, sorry; although the placement of my comment made it unclear, I was meaning to respond more to what I took to be @TimothyChow's implicit claim that there was no purely algebraic definition of $\mathbb R$ (which I took to mean 'characterisation', not 'construction', in the spirit of your definition/characterisation of $\mathbb C$). | |
| Nov 18 at 19:17 | history | made wiki | Post Made Community Wiki by Stefan Kohl♦ | ||
| Nov 18 at 19:01 | comment | added | James E Hanson | @LSpice Yes, it is the maximal Archimedean ordered field, but you do have to show that such a thing exists. (In particular, you would need to show that the category of Archimedean ordered fields is essentially small, posetal, and closed under directed colimits.) I think you could reason about that in a way that feels 'more algebraic,' but in some sense it's going to be very similar the normal construction of $\mathbb{R}$. Perhaps that's just part of what's so special about $\mathbb{R}$; its algebraic and topological properties are very tightly intertwined. | |
| Nov 18 at 18:52 | comment | added | LSpice | But isn't $\mathbb R$ easily singled out among ordered fields (as the order-complete, Archimedean ordered field), and can't ordered fields reasonably be considered part of algebra? | |
| Nov 18 at 18:40 | comment | added | James E Hanson | $\mathbb{R}$ is strange because it is 'algebraically unique' in the sense that it's rigid and therefore determined up to unique isomorphism, but there's seemingly no direct algebraic or categorical construction of it among rings or fields. (And this is true of all of the local fields of characteristic $0$ except $\mathbb{C}$.) | |
| Nov 18 at 18:33 | comment | added | user479223 | @TimothyChow Lol, fine. I get your point. | |
| Nov 18 at 18:33 | comment | added | Timothy Chow | @user479223 How do you define $\mathbb{R}$ in a purely algebraic manner? | |
| Nov 18 at 18:32 | comment | added | user479223 | Can't you define $\mathbb C$ as $\mathbb R^2$ with a strange multiplication. That seems purely algebraic to me. | |
| Nov 18 at 18:30 | comment | added | Timothy Chow | Yes, if $\mathbb{C}$ is defined to be an algebraically closed field, then the "theorem" that it's algebraically closed is trivial. In any case, I doubt that there can be a truly satisfying way to robustly define "algebraic definition" or "algebraic theorem" or "algebraic proof." | |
| Nov 18 at 18:25 | comment | added | Simon Henry | @JamesEHanson Well if you define $\mathbb{C}$ this way, then you can argue it is an algebraic definition (one can still debate over whether the cardinality requirement is is an algebraic notion of not), but then the fundamental theorem of algebra has a pretty obvious "algebraic" proof... | |
| Nov 18 at 18:20 | comment | added | James E Hanson | @TimothyChow I wonder if there's actually a robust way to define what a 'purely algebraic' definition is, because obviously $\mathbb{C}$ is the unique algebraically closed field with characteristic $0$ and transcendence degree $2^{\aleph_0}$, which doesn't strictly speaking refer to topology. Maybe in some sense the issue is that algebraic (or at least 'combinatorial') constructions of algebraically closed fields are in some sense highly non-canonical, whereas $\mathbb{C}$ of course has a specific topological construction. | |
| Nov 18 at 18:09 | comment | added | Timothy Chow | One could perhaps quibble that there is no "purely algebraic" definition of $\mathbb{C}$. | |
| Nov 18 at 18:01 | history | answered | J. W. Tanner | CC BY-SA 4.0 |