Timeline for Galois Groups of a family of polynomials
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
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| Nov 5, 2010 at 21:45 | comment | added | Dror Speiser | The Newton Polygons of these polynomials are relatively easy to study, and by some inspection, proving the assertion needed, prime > n/2, seems quite hard... These are just my calculations on page. Someone else confirm | |
| Nov 5, 2010 at 21:11 | comment | added | Victor Miller | @John Shareshian: I erred. The Galois group is of order 120, but it's not contained in $A_n$ since the discriminant is $-1$ times a square in that case. According to magma it's generated by the permutations $(1,5)(2,6)(3,4)$ and $(1,4,5,3,2,6)$. | |
| Nov 5, 2010 at 21:07 | comment | added | Victor Miller | @KConrad: Coleman cites a theorem of Camille Jordan which says that if $G$ is a transitive subgroup of $S_n$ which contains a $p$-cycle for some prime $p$ strictly between $n/2$ and $n-2$ then $G$ contains $A_n$. Coleman and Hajir both exploit this. Hajir points out that to find such a $p$ one need only find one which divides the denominator of some slope in any Newton Polygon. For the Laguerre polynomials he can always find this (with a small number of exceptions). I've been checking that criterion with these polynomials and it doesn't happen very often. | |
| Nov 5, 2010 at 20:17 | comment | added | KConrad | Victor, your "remark which needs to be checked" sounds quite similar to a theorem on conditions which guarantee a transitive subgroup of S_n is either S_n or A_n. See Theorems 2.1 and 2.2 at www.math.uconn.edu/~kconrad/blurbs/galoistheory/galoisSnAn.pdf | |
| Nov 5, 2010 at 16:41 | comment | added | Victor Miller | My proof was slightly different: Let $g_n(x)=(x−1)2f_n(x)=(x_n+1−x)−n(x−1)$. By differentiating and dividing by $(x−1)$ you get $2f_n(x)+(x−1)f_n′(x)=(n+1)(x^n−1)/(x−1)$. So Res$((x−1)f_n′(x))= $Res$((n+1)(x^n-1)/(x-1))$ ,which by one of the definitions of the resultant yields $(-1)^{n-1} (n+1)^{n-1} n^{n-2}$. Switching the order and attaching the $(-1)^{(n-1)(n-2)/2}$ for the discriminant and then dividing by $f_n(1)$ to get rid of the effect of $(x−1)$ gets the answer. – Victor Miller 0 secs ago | |
| Nov 5, 2010 at 15:31 | comment | added | John Shareshian | I am confused. As A_6 is simple of order 360, it cannot have a subgroup of order 120, as this would give a nontrivial homomorphism to S_3. | |
| Nov 4, 2010 at 23:11 | comment | added | Victor Miller | One more thing: the galois group of $f_7(x)$ is strictly contained in $A_6$. In fact it has order 120. | |
| Nov 4, 2010 at 23:00 | comment | added | Victor Miller | Very nice I was doing the same calculation in my head while driving home (probably not too safe). This combined with a remark in the two papers that I cited (which needs to be checked) about the lcm of the denominators of the slopes of all the segments in all Newton Polygons -- to the effect that if that lcm is divisible by a prime > n/2 then the galois group contains $A_n$ should finish it off. | |
| Nov 4, 2010 at 22:42 | history | answered | Dror Speiser | CC BY-SA 2.5 |