Timeline for Galois Groups of a family of polynomials
Current License: CC BY-SA 2.5
20 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 3, 2011 at 5:24 | vote | accept | Victor Miller | ||
| Nov 9, 2010 at 6:36 | answer | added | user631 | timeline score: 12 | |
| Nov 4, 2010 at 22:42 | answer | added | Dror Speiser | timeline score: 10 | |
| Nov 4, 2010 at 21:23 | comment | added | Victor Miller | @Homology: the slope of the segment in the Newton polygon is $-1/(p-2)$ so that every root of the factor corresponding to that has additive valuation $1/(p-2)$. So the ramification index of the extension of $\mathbb{Q}_p$ generated by a root is $p-2$ which is also the degree of the polynomial. | |
| Nov 4, 2010 at 20:50 | comment | added | Qiaochu Yuan | @Homology: in any case, irreducibility follows by the ideas described in the comments to mathoverflow.net/questions/18094/… . | |
| Nov 4, 2010 at 20:45 | comment | added | Homology | I don't get your proof of the irreducibility: over $\mathbb{Q}_p$, the Newton polygon only tells you that the polynomial has a linear factor, but it doesn't give a criterion for irreducibility. | |
| Nov 4, 2010 at 20:36 | answer | added | David E Speyer | timeline score: 6 | |
| Nov 4, 2010 at 20:30 | comment | added | Victor Miller | @David Speyer: sage says that the discriminant for 241 is $2^{240} 11^{478} 241^{238}$ | |
| Nov 4, 2010 at 20:28 | comment | added | tdnoe | For 241, the discriminant is 2^240 11^478 241^238 -- clearly a square. | |
| Nov 4, 2010 at 20:28 | comment | added | David E Speyer | We have either discovered a bug in Magma or Mathematica here. I vote that Mathematica is right, because computing a discriminant is a much less error prone operation than computing the full Galois group. | |
| Nov 4, 2010 at 20:28 | history | edited | Victor Miller | CC BY-SA 2.5 |
fixed result
|
| Nov 4, 2010 at 20:26 | comment | added | David E Speyer | Mathematica claims that the discriminant for 241 is a perfect square. The square root in question has 569 digits, so it won't fit in a comment box, but I can post it if anyone wants to see it. | |
| Nov 4, 2010 at 20:18 | comment | added | Victor Miller | Magma says that $7,17,49$ and $97$ are the only odd exceptions below 100. | |
| Nov 4, 2010 at 20:12 | comment | added | Dror Speiser | It seems to be true for all odd numbers of the form 2*n^2-1 | |
| Nov 4, 2010 at 20:10 | comment | added | Dror Speiser | Just calculate the discriminant - it is a square, and in fact, a square for all the discriminants in the series. Clearly we just need a nice closed form for the discriminant that is probably of the form a square times 2*q^2-1 | |
| Nov 4, 2010 at 20:02 | comment | added | Victor Miller | @David, I can do more calculations. I'm not sure that $f_n(x)$ is irreducible for non-prime $n$ (at least my proof of irreducibility doesn't work). @tdnoe: I'm trying the calculation for 241 now, but this may stress Magma to the limit! | |
| Nov 4, 2010 at 19:50 | comment | added | tdnoe | Your three exceptions 7, 17, 97 are the first three primes of the form 2*q^2-1 where q is prime, which is sequence A092057. Is the next exception 241? | |
| Nov 4, 2010 at 19:49 | comment | added | David E Speyer | Is there any reason not to expand your search to non-prime p? That would at least give us more data. | |
| Nov 4, 2010 at 19:43 | history | edited | Victor Miller | CC BY-SA 2.5 |
added results of further computations
|
| Nov 4, 2010 at 16:48 | history | asked | Victor Miller | CC BY-SA 2.5 |