Timeline for Fields with trivial automorphism group
Current License: CC BY-SA 2.5
22 events
| when toggle format | what | by | license | comment | |
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| Aug 2, 2014 at 18:56 | comment | added | Pete L. Clark | @dke: Thanks, these are very interesting results. | |
| Apr 8, 2011 at 17:24 | comment | added | dke | For the question Pete points out, see the references in mathoverflow.net/questions/61058 | |
| Oct 3, 2010 at 15:22 | vote | accept | CommunityBot | ||
| Oct 3, 2010 at 15:22 | history | bounty ended | Vipul Naik | ||
| Sep 26, 2010 at 14:53 | history | bounty started | Vipul Naik | ||
| Sep 26, 2010 at 14:38 | history | edited | Vipul Naik | CC BY-SA 2.5 |
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| May 5, 2010 at 3:08 | answer | added | Dave Marker | timeline score: 7 | |
| May 3, 2010 at 8:11 | comment | added | K.J. Moi | One conceptual difference between $\mathbb{Q}(i)$ and $'\mathbb{Q}( \sqrt{2})$ is that a field in which $-1$ is a square cannot be ordered. Since $'\mathbb{Q}( \sqrt{2})$ has real embeddings it can obviously be ordered. There is a somewhat related invariant called the level of of the field, which if I remember correctly is the least number of summands needed to express -1 as a sum of squares. In the cases above the levels are 1 and $\infty$ respectively. If we had chosen $\sqrt{-2}$ instead the level would have been 2. | |
| Apr 29, 2010 at 6:09 | comment | added | Pete L. Clark | I just want to point out that one of the OP's questions remains wide open: can every field be embedded in a field with trivial automorphism group? I don't even myself know the answer for $\mathbb{C}$. | |
| Apr 28, 2010 at 23:36 | comment | added | Georges Elencwajg | Dear Kevin and Vipul: what you say is certainly true (and that's how I interpreted wzzx's sentence more or less) but I thought that the comment had to be made more precise for readers who didn't know these results beforehand. As for the result mentioned by Vipul, it is indeed due to Hilbert.It asserts irreducibility for infinitely many values of the parameters involved in the polynomial but I would be wary to claim that most (or the general) polynomials are irreducible. Anyway, I would encourage you to post a few lines on this powerful Hilbert irreducibility theorem. | |
| Apr 28, 2010 at 22:14 | answer | added | Pete L. Clark | timeline score: 33 | |
| Apr 28, 2010 at 22:10 | history | edited | Vipul Naik | CC BY-SA 2.5 |
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| Apr 28, 2010 at 22:10 | answer | added | Simon Thomas | timeline score: 16 | |
| Apr 28, 2010 at 22:06 | comment | added | Vipul Naik | I think wzzx means a polynomial of degree n whose Galois group is S_n, so adjoining one root of the polynomial does not adjoin any of the other roots. And I think he/she/ was referring to the base field itself being the rational numbers (or having trivial automorphism group). We do need the condition n > 2, though, for wzzx's statement to hold. There is a result (due to Hilbert?) that "most" polynomials over the rationals are "general" in the above sense. | |
| Apr 28, 2010 at 22:05 | comment | added | Kevin Buzzard | @Georges: by "general" I think he just means something like "irreducible and separable, of degree at least 3, and the Galois group of the splitting field is the full symmetric group". So for example over Q this produces a lot of examples. | |
| Apr 28, 2010 at 22:01 | comment | added | Kevin Buzzard | If F_un means the field with one element, then it probably doesn't count, because there's no field with 1 element: people talk about "schemes over the field with 1 element" (which is something like a scheme over Z with some extra properties) but the field with 1 element itself doesn't exist AFAIK. | |
| Apr 28, 2010 at 21:50 | comment | added | Steve Huntsman | Asked in utter ignorance: does $\mathbb{F}_{un}$ count? | |
| Apr 28, 2010 at 21:42 | comment | added | Georges Elencwajg | @ wzzx: Over $\mathbb C$ this is false for every polynomial, general (?) or not. For $\mathbb R$ I have no idea what a general polynomial is: for what you say to be true, a "general" real polynomial should have no complex root: hard to believe. | |
| Apr 28, 2010 at 21:26 | comment | added | Vipul Naik | Thanks. I think that's covered under my "self-normalizing subgroup" criterion for finite separable extensions. | |
| Apr 28, 2010 at 21:08 | answer | added | Robin Chapman | timeline score: 10 | |
| Apr 28, 2010 at 20:48 | comment | added | Ryan Thorngren | Extension by any single root of a general polynomial has trivial automorphism group. | |
| Apr 28, 2010 at 20:42 | history | asked | Vipul Naik | CC BY-SA 2.5 |