Timeline for Zariski closure of set of units in a number ring
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 19, 2019 at 21:30 | comment | added | Leah | I now understand! This is a very helpful answer. Thanks! | |
| Jul 19, 2019 at 21:30 | vote | accept | Leah | ||
| Jul 19, 2019 at 14:23 | comment | added | Will Sawin | @Leah When doing the equation we fix some $\sigma$, which gives us a relation among the $e_i$s. Then, because we can fix any $\sigma$, the same equation must hold for any $\sigma$. | |
| Jul 19, 2019 at 13:56 | comment | added | Leah | I'm having a trouble following the paragraph starting "This can only be no[n] constant if...". Would you mind adding a few details? What might be confusing me is that I might be misinterpreting the previous paragraph, so let me say a bit about how I am interpreting it. First, when you say "complex conjugation element of the Galois group" I assume that you mean "Galois group of $\overline{\mathbb{Q}}$", right? It then looks to me like you are fixing some such $\sigma$, and the overline in your big equation is the action of this $\sigma$. | |
| Jul 19, 2019 at 13:09 | comment | added | Venkataramana | Thank you. That is helpful | |
| Jul 19, 2019 at 11:06 | comment | added | Will Sawin | @Venkataramana There is no CM subfield, so the Zariski closure of units is the norm one torus. | |
| Jul 19, 2019 at 6:40 | comment | added | Venkataramana | Take $F={\mathbb Q}(2^{1/3})$. What does it say about Zariski closure of units? | |
| Jul 18, 2019 at 22:57 | history | answered | Will Sawin | CC BY-SA 4.0 |