Timeline for Easy cases of Herbrand's theorem
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 18, 2018 at 3:28 | comment | added | user131093 | Short answer: No, there is no "easy" way to prove that $V_{p-2} = 0$. And yes, there is an easy way to prove $V_1 = 0$ --- as Sawin says, Kummer theory says any such Galois extension (without any ramification conditions) is of the form $\mathbf{Q}(\zeta_p, N^{1/p})$ where $N \in \mathbf{Q}$, at which point it is easy to see (using unique factorization) that all such extensions are ramified. | |
| Nov 16, 2018 at 1:59 | comment | added | Will Sawin | It should be possible to calculate one of $V_1$ or $V_{-1}$ (and therefore show it vanishes I guess) by Kummer theory. | |
| S Nov 16, 2018 at 1:40 | history | suggested | user130124 | CC BY-SA 4.0 |
nonzero should read zero (clearly, for instance $V$ is zero when $p$ is a regular prime).
|
| Nov 15, 2018 at 21:41 | review | Suggested edits | |||
| S Nov 16, 2018 at 1:40 | |||||
| Nov 15, 2018 at 20:17 | history | edited | David E Speyer | CC BY-SA 4.0 |
added 307 characters in body
|
| Nov 15, 2018 at 20:10 | history | asked | David E Speyer | CC BY-SA 4.0 |