Timeline for What are the possible sets of degrees of irreducible polynomials over a field?
Current License: CC BY-SA 2.5
10 events
| when toggle format | what | by | license | comment | |
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| Sep 14, 2017 at 11:44 | comment | added | Watson | If $k$ has characteristic $0$, if $p>2$ is an odd prime and if $p r \in I$ for some $r \geq 1$, then for any $n \geq 1$, there is an integer $r_n \geq 1$ such that $r_n p^n \in I$ (Prop. 9.8.6. in Tauvel's book on Galois theory). | |
| Dec 16, 2010 at 0:52 | comment | added | Qiaochu Yuan | @Andres: thanks for the link; I think I saw that paper in connection to a different MO problem, but it is certainly relevant here as well. | |
| Dec 16, 2010 at 0:11 | comment | added | Andrés E. Caicedo | @Qiaochu: Have you seen the nice paper by J. Shipman, "Improving the fundamental theorem of algebra"? It is in the Math. Intelligencer, 29 (2007), no. 4, 9-14. It is precisely in the direction you want. | |
| Dec 15, 2010 at 19:41 | answer | added | Gjergji Zaimi | timeline score: 15 | |
| Mar 2, 2010 at 3:00 | comment | added | Pete L. Clark | @QY: This is a very interesting question. I'm sort of surprised that no one else has tried to answer it. If by tomorrow morning no one else has solved it, I will start thinking seriously about what kind of degree sets one can get by starting with Q and making a "Merkurjev-type construction", i.e., (infinitely) repeatedly applying any of the following basic operations: adjoin to $K$ all roots of irreducible polynomials of degree $d$. | |
| Mar 1, 2010 at 23:49 | answer | added | Pete L. Clark | timeline score: 18 | |
| Mar 1, 2010 at 20:44 | comment | added | Pete L. Clark | One more remark: all of the sets of degrees referred to above can be realized via algebraic extensions of $\mathbb{R}((t))$. | |
| Mar 1, 2010 at 20:41 | history | edited | Qiaochu Yuan | CC BY-SA 2.5 |
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| Mar 1, 2010 at 20:40 | comment | added | Pete L. Clark | I have only time at the moment to record my guess: let S be a set of prime numbers. Then there exists a field F (of characteristic 0) such that there exists a degree d field extension iff d is divisible only by primes in S. If 2 is not in S, we can build another field such that there exists a degree d field extension iff d or 2d is in S. I suspect the converse is also true. | |
| Mar 1, 2010 at 20:29 | history | asked | Qiaochu Yuan | CC BY-SA 2.5 |