| 7 Jul 2020 |
 solov-t | (can I ask some silly questions here? 😛) | 14:02:47 |
 Tim | that’s my favourite type of question! | 14:03:03 |
| solov-t | oh hey tim haha | 14:03:35 |
| solov-t | So... is there a relation between unramified primes and unramified morphisms? I don't know if the name is just a coincidence... But do we know when $Z \to O_K$ is unramified or something?
(I've just learned this stuff, so I know only the definitions) | 14:05:20 |
| Tim | there's a very short section in SGA1 on unramified morphisms that might have the answer | 14:08:29 |
| Tim | beyond that, I'm not too sure | 14:09:59 |
| Tim | wikipedia says that the morphism definition "corresponds" to the prime one, but doesn't give any further details | 14:10:23 |
| Tim | and I'm afraid I know basically nothing about unramified primes | 14:10:35 |
| solov-t | I wouldn't be surprised to see this on SGA1 i guess haha | 14:11:31 |
| solov-t | I would think something like O_K \to O_K is unramified iff all primes are unramified | 14:11:59 |
| solov-t | but this is not true, if I get this right: Z \to Z[x]/<x^2 - x - 1> is unramified but there are ramified primes | 14:12:33 |
 csaez | I think it should be the same. Spec O_K \to Spec Z should be unramified exactly when every prime of O_K is unramified. | 14:38:32 |
| csaez | However it's been a long time since I last thought about this things, so I might be wrong. | 14:38:58 |
| solov-t | I think this doesn't hold for $K = Q(\sqrt{5})$ | 14:42:39 |
| solov-t | the ring of integers is generated by \phi, the golden ratio | 14:42:57 |
| solov-t | so the cotangent bundle is generated by d\phi | 14:43:11 |
| solov-t | but \phi^2 = \phi + 1 | 14:43:18 |
| solov-t | so 2 d\phi = d\phi | 14:43:31 |
| solov-t | which implies d\phi = 0, so this is unramifed (r-right?) | 14:43:50 |
| solov-t | but i think 5 and 2 are ramified for this extension | 14:44:16 |
| csaez | I think you're right. Or at least I don't see the flaw in your argument. | 15:08:46 |
| csaez | Wait a moment, what you get is 2\phi d\phi = d\phi | 16:12:15 |
| csaez | Hence (2\phi - 1)d\phi = 0 | 16:12:33 |
| csaez | Or equivalently, \sqrt{5}d\phi = 0 | 16:12:53 |
| csaez | But this doesn't imply that d\phi=0, right? Because \sqrt{5} is not invertible in Z[\phi]. | 16:13:27 |
| solov-t | oh, that's right!! | 16:13:42 |
| solov-t | we're saved haha | 16:13:47 |
| csaez | Yeah | 16:13:55 |
| csaez | For a moment I was really puzzled because I remembered that ramification in ANT and AG essentially coincide | 16:14:29 |
| csaez | But as I said, it's been a long time since I thought about these kind of things, so I'm not sure of anything anymore | 16:14:56 |
