Timeline for Is there a better proof of this fact in number theory/formal group theory?
Current License: CC BY-SA 3.0
14 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Aug 4, 2014 at 3:55 | history | edited | Aaron Meyerowitz | CC BY-SA 3.0 |
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| May 28, 2014 at 0:26 | comment | added | Aaron Meyerowitz | But for $n=9$ I get $\gcd(9,0,4A_9,3A_8A9)=A_9$ and that is ok for $A_9=7$ but for $A_8,A_9=8,3$ or $5,3$ it fails. I might have overlooked something and maybe there is a further condition (or approach to binomial coefficients) which makes things work, but I don't know how interesting that is. | |
| May 28, 2014 at 0:22 | comment | added | Aaron Meyerowitz | An integral domain makes sense but I think $Z_6,Z_8$ can be made to work. In $Z_{12}$ with $I_k=k$ (up to $11$) $I_n=\prod_{d\mid n}A_d$ makes $A_k=k$ for $k=2,3,5,7,11$ but only forces $A_4 =2$ or $8.$ Similarly $A_6$ can be any of the four units, $A_8=2,3,8$ or $11$ and $A_9 =3$ or $7$ So the $A_k$ are not all forced (by that.) The binomial coefficients can't be defined via factorials. The $e_{n,m,k}$ approach make the main result look ok up to $n=7.$ Then we want $\gcd(8,4,8,A_8)=A_8$ (up to units) and that looks ok too. (cont) | |
| May 27, 2014 at 10:47 | comment | added | darij grinberg | Nice construction! As far as I remember, some answers about when cyclotomic polynomials have a gcd in the strong sense in $\mathbb{Z}\left[X\right]$ can be found in G. Dresden, Resultants of Cyclotomic Polynomials, home.wlu.edu/~dresdeng/papers/Res.pdf . | |
| May 27, 2014 at 10:41 | comment | added | darij grinberg | Your commutative ring should be an integral domain, I assume? | |
| May 27, 2014 at 10:41 | history | edited | darij grinberg | CC BY-SA 3.0 |
one more typo
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| May 27, 2014 at 9:27 | history | edited | darij grinberg | CC BY-SA 3.0 |
typo on line 1
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| May 27, 2014 at 9:11 | history | edited | Aaron Meyerowitz | CC BY-SA 3.0 |
Reworked the answer
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| May 27, 2014 at 9:11 | comment | added | Aaron Meyerowitz | I changed things and made more explicit note of this. | |
| May 25, 2014 at 7:46 | comment | added | KConrad | Your "obvious choice" $I_n = 1 + X + X^2 + \cdots + X^{n-1}$ is wrong if it is to be the product of $d$th cyclotomic polynomials for *all* $d$ dividing $n$. For that you need $I_n = X^n - 1$. | |
| May 25, 2014 at 7:18 | comment | added | მამუკა ჯიბლაძე | Very interesting! Has anyone considered a q-analog of the Lazard ring / universal formal group? | |
| May 25, 2014 at 6:05 | history | edited | Aaron Meyerowitz | CC BY-SA 3.0 |
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| May 25, 2014 at 5:59 | history | answered | Aaron Meyerowitz | CC BY-SA 3.0 |