Timeline for What is the Galois group of a polynomial over a finite field? [closed]
Current License: CC BY-SA 2.5
7 events
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| Apr 15, 2010 at 23:51 | comment | added | S. Carnahan♦ | I suppose one could make that interpretation, but I think jsmith12 should have given a precise definition in the statement of the question. Assuming the more generous interpretation of "Galois group", the automorphism group of an etale algebra over a finite field is a product of wreath products of cyclic groups by symmetric groups. | |
| Apr 15, 2010 at 21:40 | history | closed |
S. Carnahan♦ user350 Mariano Suárez-Álvarez Qiaochu Yuan Gjergji Zaimi |
too localized | |
| Apr 15, 2010 at 16:30 | comment | added | Dror Speiser | @Scott: that doesn't contradict anything in the question. What jsmith might mean is what is the automorphism group of $\mathbb{F}_7 [x]/(f)$, and that need not be a cyclic group. If he means the automorphism group of the splitting field, then while the answer is cyclic, it's size is related to the factors and not just the total degree. | |
| Apr 15, 2010 at 15:19 | comment | added | S. Carnahan♦ | Voting to close. Finite extensions of finite fields only admit cyclic galois groups. See en.wikipedia.org/wiki/Finite_field | |
| Apr 15, 2010 at 15:09 | comment | added | JBL | Indeed -- this sort of thing is well-suited to the appropriate forum at Art of Problem Solving: in this case, that's artofproblemsolving.com/Forum/index.php?f=8 | |
| Apr 15, 2010 at 14:52 | comment | added | Martin Brandenburg | I think that this is elementary algebra and does not belong to MO. please check out the faq (mathoverflow.net/faq). | |
| Apr 15, 2010 at 14:46 | history | asked | jsmith12 | CC BY-SA 2.5 |