Timeline for Proof of the result that the Galois group of a specialization is a subgroup of the original group?
Current License: CC BY-SA 2.5
6 events
| when toggle format | what | by | license | comment | |
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| Aug 5, 2010 at 16:25 | comment | added | damiano | Yes, but make sure that you check that $h(x)$ is not $x$! (This is a silly comment, but it is not implied by what you wrote that $x$ and $h(x)$ are coprime.) | |
| Aug 5, 2010 at 16:14 | comment | added | Adam | Thankyou both for your help! I hadn't even realized that this was basically the same result as the one I was asking about in my previous question, but using the ideal $(t-t_0)$ as opposed to $(p)$. Just to be sure, am I right in thinking that if in the last part of my question $k=1$ the result definitely holds? | |
| Aug 5, 2010 at 9:05 | comment | added | damiano | Thanks for the link: your answers both here and there are very interesting! | |
| Aug 5, 2010 at 1:43 | comment | added | KConrad | PARI says 3 factors in the quartic field as two primes with residue field degree 2, and that does correspond to a cycle structure in the Galois group (two disjoint 2-cycles). There is an example of a degree 6 number field generated by the root of some polynomial $f(x)$ such that neither the factorization type of $f(x) \bmod 3$ nor the way (3) factors in the number field correspond to the cycle structure of a permutation in the Galois group. See my answer to another question of Adam's at mathoverflow.net/questions/21247/… | |
| Aug 4, 2010 at 17:59 | history | edited | damiano | CC BY-SA 2.5 |
typos
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| Aug 4, 2010 at 17:37 | history | answered | damiano | CC BY-SA 2.5 |