Timeline for Polynomials for the alternating group $A_n$
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 31, 2020 at 3:10 | comment | added | ReverseFlowControl | @KConrad that works: people.math.gatech.edu/~mbaker/pdf/Coleman_GaloisNewton.pdf contains the proof in a very approachable way. | |
| Aug 31, 2020 at 3:02 | comment | added | KConrad | Just google "schur exponential galois" and you'll find a math.stackexchange page that answers your question. | |
| Aug 31, 2020 at 1:22 | comment | added | ReverseFlowControl | @KConrad: any available references to that work? | |
| Aug 31, 2020 at 1:18 | comment | added | KConrad | Schur proved that the truncated exponential polynomials $1 + x + x^2/2! + \cdots + x^n/n!$ are irreducible over $\mathbf Q$ for all $n \geq 1$ and their splitting field over $\mathbf Q$ has Galois group $A_n$ when $4 \mid n$. When $n$ is not divisible by $4$, the Galois group is $S_n$. The cases of $n$ being or not being divisible by $4$ corresponds to the discriminant of the polynomial being or not being a square, which is the minimial kind of information needed to know if a Galois group of an $n$th degree irreducible polynomial is or is not contained in $A_n$. | |
| Aug 31, 2020 at 0:43 | comment | added | ReverseFlowControl | @YCor: thank you for pointing that out. Non-uniqueness is not an issue, just need one, or more, family of functions parametrized by $n$ with $A_n$ as the Galois Group. | |
| Aug 31, 2020 at 0:40 | comment | added | ReverseFlowControl | @KConrad: the other question is more specific. I posted a generalized version here in the hopes that I may be pointed to some research on the topic. I will include a reference if I post again in a situation similar to this one. | |
| Aug 31, 2020 at 0:37 | history | edited | ReverseFlowControl | CC BY-SA 4.0 |
Factual update.
|
| Aug 31, 2020 at 0:30 | comment | added | KConrad | The question was asked recently at math.stackexchange.com/questions/3808650/…. Generally speaking, when posting the same question on both sites this should be mentioned to avoid duplicate answers/work. | |
| Aug 30, 2020 at 23:51 | comment | added | R. van Dobben de Bruyn | In some sense the simplest polynomial with Galois group $S_n$ is the polynomial $\sum_{i=0}^n (-1)^i \sigma_ix^i$ over $\mathbf C(\sigma_1,\ldots,\sigma_n) = \mathbf C(x_1,\ldots,x_n)^{S_n}$, where $\sigma_i$ is the $i^{\text{th}}$ elementary symmetric polynomial. Replacing $S_n$-invariants with $A_n$-invariants gives an analogous polynomial for $A_n$ (possibly over a field that is not pure transcendental ― I'm not sure). | |
| Aug 30, 2020 at 23:43 | answer | added | Gerry Myerson | timeline score: 3 | |
| Aug 30, 2020 at 23:29 | comment | added | YCor | Note that there are distinct $n,m$ with $(Z/nZ)^\times$ and $(Z/mZ)^\times$ isomorphic (for instance, $m,n=1,2$, or $m,n=3,4$), so it's not a unique "simplest" polynomial. | |
| Aug 30, 2020 at 23:26 | history | edited | YCor | CC BY-SA 4.0 |
removed capitals from title
|
| Aug 30, 2020 at 22:51 | comment | added | David E Speyer | See kconrad.math.uconn.edu/blurbs/galoistheory/cyclotomic.pdf , for example. | |
| Aug 30, 2020 at 22:49 | comment | added | ReverseFlowControl | @DavidESpeyer: I will update the question. Could you please provide a reference/proof/sketch-of-the-proof for that fact? Thank you. | |
| Aug 30, 2020 at 22:42 | comment | added | David E Speyer | The Galois group of $x^n-1$ (over $\mathbb{Q}$) is $(\mathbb{Z}/n \mathbb{Z})^{\times}$, the group of mulitiplicative units modulo $n$, not $C_n$. | |
| Aug 30, 2020 at 22:41 | comment | added | vidyarthi | a related question here, seems like a part of the inverse galois problem | |
| Aug 30, 2020 at 22:28 | history | asked | ReverseFlowControl | CC BY-SA 4.0 |