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Feb 16 at 13:58 comment added Jeff H @ChrisWuthrich You're right, I accidentally wrote down the Duke-Toth matrix for when a_p=2 rather than -2, but it doesn't affect the rest of my question -- in particular, it was not the source of my error. But thanks for catching that.
Feb 16 at 13:56 comment added Jeff H @re'emwaxman Yes, that's exactly the point I was somehow missing, but your comment made my false belief clear to me. Thanks so much again!
Feb 16 at 11:54 comment added re'em waxman The thing needed to be understood is that if $\mathfrak{p}$ and $\mathfrak{p}'$ are two primes above $p$ then $\text{Frob}_{\mathfrak{p}}$ and $\text{Frob}_{\mathfrak{p}'}$ are conjugate in $G$ but might be the same even if the primes are distinct.
Feb 16 at 11:15 comment added Chris Wuthrich Also your example doesn't have trace equal to $a_p=-2$.
Feb 15 at 21:37 comment added Jeff H @re'emwaxman : Thanks -- I figured my mistake had to be at that step. I'm still a bit confused about it, but your extreme example of an abelian extension is extremely helpful for convincing me that this is where I've goofed.
Feb 15 at 21:32 comment added GH from MO Please use a high-level tag like "nt.number-theory". I added this tag now.
Feb 15 at 21:31 history edited GH from MO CC BY-SA 4.0
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Feb 15 at 21:30 comment added re'em waxman I think your number theory is wrong. There is no correspondence between elements in $\mathcal{C}$ to primes above $p$, for example take $K/\mathbb{Q}$ abelian and any prime $p$ which does stay prime in $K$, then $|\mathcal{C}|=1$ and $g>1$.
Feb 15 at 19:51 history asked Jeff H CC BY-SA 4.0