Timeline for Collecting proofs that finite multiplicative subgroups of fields are cyclic
Current License: CC BY-SA 4.0
Post Revisions
40 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 1, 2025 at 20:34 | answer | added | Mohammad Ali Karami | timeline score: 0 | |
| Apr 1, 2025 at 5:23 | answer | added | Kan't | timeline score: 3 | |
| Mar 11, 2024 at 21:52 | answer | added | Taras Banakh | timeline score: 1 | |
| Feb 14, 2023 at 18:54 | answer | added | XYC | timeline score: 0 | |
| Feb 14, 2023 at 14:25 | comment | added | Nick S | If I am not mistaken, there is no need to introduce the Euler $\phi$ function in the Serre argument. All you have to do is compare the equation $x^n=1$ in your field to the number of elements in $\mahbb Z/n \mathbb Z$ of order $d|n$. | |
| Feb 13, 2023 at 22:03 | comment | added | Marc van Leeuwen | @LSpice I don't agree. Embedding an integral domain in a field is pointless for proving that a polynomial equation cannot have more solutions that its degree. No division is necessary for this, all that matter is that one cannot make a product zero without making at least one factor zero. | |
| Feb 13, 2023 at 18:04 | comment | added | LSpice | @MarcvanLeeuwen, re, how might a proof go that used field-ness rather than just domain-ness go? Even if you only knew the fact for fields, you could always embed the domain in its fraction field; and, even if you want to avoid saying that, just about any fact about domains (such as your fact about the number of roots of a polynomial) is probably going to be most easily provable by embedding in a field, whether or not you explicitly say you're doing so. | |
| Feb 13, 2023 at 17:43 | answer | added | Marc van Leeuwen | timeline score: 3 | |
| Feb 13, 2023 at 16:25 | comment | added | Marc van Leeuwen | Here is a meta-question: does any of these proofs use the assumption that $F$ is a field, rather than just an integral domain? I haven't found one below that clearly does (and cannot be trivially modified to avoid it). The one place it serves over and over again is that $X^n-1$ cannot have more than $n$ roots in $F$, which is because $F$ is an integral domain, and (therefore) $\deg(ab)=\deg(a)\deg(b)$ for $a,b\in F[X]$ (so decomposing $X^n-1$ in any way into monic non-constant irreducibles involves at most $n$ factors; absence of $X-r$ for some root $r$ contradicts $F$ integral after $X:=r$). | |
| Jan 12, 2023 at 5:47 | answer | added | Kenta Suzuki | timeline score: 4 | |
| Nov 21, 2022 at 12:56 | review | Close votes | |||
| Nov 23, 2022 at 14:15 | |||||
| Aug 1, 2022 at 23:34 | history | edited | Michael Hardy | CC BY-SA 4.0 |
added 1 character in body
|
| Nov 12, 2018 at 14:15 | answer | added | Charles Rezk | timeline score: 3 | |
| Nov 11, 2017 at 13:05 | comment | added | Watson | See also : math.stackexchange.com/questions/59903 | |
| Apr 5, 2017 at 7:15 | comment | added | Włodzimierz Holsztyński | Emil Artin comes instantly to my mind. | |
| Apr 5, 2017 at 6:45 | answer | added | Fedor Petrov | timeline score: 15 | |
| Dec 3, 2016 at 1:09 | answer | added | Jeff Strom | timeline score: 15 | |
| Dec 2, 2016 at 14:51 | answer | added | Todd Trimble | timeline score: 10 | |
| Dec 2, 2016 at 9:51 | answer | added | Octavian teodor Popp | timeline score: 2 | |
| Apr 19, 2016 at 21:54 | history | edited | Amir Sagiv | CC BY-SA 3.0 |
added finite groups
|
| Apr 19, 2016 at 21:49 | answer | added | Paul Pedersen | timeline score: 5 | |
| Apr 21, 2015 at 0:04 | history | edited | Gerry Myerson |
edited tags
|
|
| Apr 20, 2015 at 20:40 | comment | added | Fan Zheng | @DavidFeldman The Euler $\phi$ function is not at all an "extrinsic idea". It is the number of generators of that cyclic group, which is clearly something to the point. Also, the existence of generators is then equivalent to the fact that the Euler function is positive, which is very clear if you write out its Euler product. Plus, do you think it requires more mathematical maturity to understand the Euler $\phi$ function (the count of numbers between 1 and $n$ that is coprime to $n$) than to understand what a field is? | |
| Dec 25, 2014 at 16:42 | comment | added | Todd Trimble | For what it might be worth, the nLab has a proof here: ncatlab.org/nlab/show/root+of+unity#over_a_field | |
| Dec 25, 2014 at 16:18 | answer | added | Geoff Robinson | timeline score: 1 | |
| Feb 8, 2011 at 23:38 | comment | added | David Feldman | @Pete, cont. My pedagogical principle with students who lack mathematical maturity aims at postponing the "I never could have thought of that in a million years moments," in order to foster a sense that proofs of cornerstone theorems really would emerge given sufficient time and thought. | |
| Feb 8, 2011 at 23:38 | comment | added | David Feldman | @Pete I will teach Euler's $\phi$ and circle back to this theorem. But here's my pedagogical axe (and MO might not be the right place for this discussion...but where?) My students lack mathematical maturity and thus don't relish proofs that depend on extrinsic ideas. I aim to get them used to all that, but using examples where extrinsic ideas are essential. But with this proof, I think $\phi$ enters as a mere bookkeeping device. The mysterious stranger will seem like the protagonist in a short mysterious tale, which misleads beginners. | |
| Feb 8, 2011 at 18:36 | answer | added | inkspot | timeline score: 8 | |
| Feb 8, 2011 at 15:12 | answer | added | paul Monsky | timeline score: 13 | |
| Feb 8, 2011 at 14:25 | answer | added | KConrad | timeline score: 50 | |
| Feb 8, 2011 at 10:16 | answer | added | awllower | timeline score: 10 | |
| Feb 8, 2011 at 10:12 | history | edited | Pete L. Clark |
edited tags
|
|
| Feb 8, 2011 at 10:02 | comment | added | Pete L. Clark | One more comment: don't you want to introduce Euler's $\varphi$ function in a number theory course? As a number theorist, I would defend introducing it even in a pure algebra course, but in a number theory course it seems almost mandatory. | |
| Feb 8, 2011 at 10:00 | comment | added | Pete L. Clark | By the way, having just looked at my copy of Serre's book, I can say that "Serre's proof...runs a full page" is an exaggeration: it is about 2/3 of a page, with generous spacing. If we are talking pedagogy, then I recommend against optimizing the argument for length: better to have a medium length argument with all the details spelled out than a relatively cryptic short argument. | |
| Feb 8, 2011 at 9:45 | answer | added | Andrea Ferretti | timeline score: 64 | |
| Feb 8, 2011 at 9:40 | answer | added | Pete L. Clark | timeline score: 7 | |
| Feb 8, 2011 at 9:32 | answer | added | Qiaochu Yuan | timeline score: 19 | |
| Feb 8, 2011 at 9:03 | comment | added | Tobias Kildetoft | Well, it depends on what you want to use. The argument you give can be shortened somewhat by using that the order of any element in a finite abelian group divides the maximal order (and then noting that because of this, if m is the maximal order, then all elements in the subgroup are roots of a polynomial of degree m) | |
| Feb 8, 2011 at 8:59 | comment | added | S. Carnahan♦ | It looks like your short argument uses a nontrivial assertion about the exponent of a non-cyclic abelian $p$-group, and another assertion about the number of roots of a polynomial over a field. In particular, I don't see why this argument is substantially simpler than the corresponding fact for the case of $n$ with more than one prime divisor (which would allow you to eliminate the last sentence). | |
| Feb 8, 2011 at 8:30 | history | asked | David Feldman | CC BY-SA 2.5 |