Timeline for Given $v,w$ primes of $k$, is there $K/k$ so $K_v\cap\Bbb Q^\text{cycl}=K_w\cap\Bbb Q^\text{cycl}=K\cap\Bbb Q^\text{cycl}$?
Current License: CC BY-SA 2.5
7 events
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| Jun 10, 2010 at 18:15 | history | edited | S. Carnahan♦ | CC BY-SA 2.5 |
giving up
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| Jun 10, 2010 at 17:28 | comment | added | BCnrd | Aaargh, the "unknown control sequence" thing above should be $O_ L$; I wish \mathcal didn't give such headaches on this system. Anyway, Scott, I recommend not spending more time on this unnatural question. I will now follow my own advice. :) | |
| Jun 10, 2010 at 17:25 | comment | added | BCnrd | Scott, 2nd attempt fails: no reason $[k'_v:\mathbf{Q}]$ finite. Any $\mathbf{Q}_ p$ contains finite ab. ext'ns $L/\mathbf{Q}$ of arb. large degree. Indeed, inside $\mathbf{Q}(\zeta_m)$ with $m = p^f - 1$ ($f > 0$), subfield $L$ of invariants under $p \bmod m$ embeds $\mathbf{Q}_ p$ since $p$ totally split in $L$. Clearly $[L:\mathbf{Q}] = \varphi(p^f - 1)/f$ and $\varphi(n) \ge \sqrt{n}$ for $n \ge 7$, so QED. (Cool tidbit: minimal number $r$ of $\mathbf{Z}$-algebra generators of $\mathca{O}_ L$ satisfies $p^r \ge \varphi(p^f - 1)/f$, so $r \rightarrow \infty$ as $f$ grows!) | |
| Jun 10, 2010 at 16:30 | history | edited | S. Carnahan♦ | CC BY-SA 2.5 |
added 282 characters in body
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| Jun 10, 2010 at 16:21 | history | edited | S. Carnahan♦ | CC BY-SA 2.5 |
Added Kronecker-Weber
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| Jun 10, 2010 at 6:15 | comment | added | BCnrd | Scott, why should $K$, $k_v[\zeta_n]$, and $k_w[\zeta_n]$ not have roots of unity apart from the $n$th roots of unity? For example, if $k$ is the maximal totally real subfield of the cylotomic field $\mathbf{Q}(\zeta_{15})$ then $\mu(k) = \{\pm 1\}$ but $k[\zeta_6] = \mathbf{Q}(\zeta_{15})$, so $\mu(k[\zeta_6])$ has order 30. You can now see the sort of subtlety which was missed in your calculations. | |
| Jun 10, 2010 at 4:44 | history | answered | S. Carnahan♦ | CC BY-SA 2.5 |