Timeline for Why the roots of unity are the analogs of constants ?
Current License: CC BY-SA 2.5
6 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Jun 18, 2010 at 14:33 | vote | accept | user2330 | ||
| Jun 14, 2010 at 22:52 | comment | added | David E Speyer | Not always: Let $K= \mathbb{Q}(\sqrt{-5})$ and $L=K(i)$. Then $L$ is unrammified over $K$. | |
| Jun 14, 2010 at 21:00 | comment | added | Dror Speiser | @Pete: Doesn't adjoining roots of unity to a number field that weren't there before always result in a ramified extension? | |
| Jun 14, 2010 at 14:51 | comment | added | Pete L. Clark | Right. And, even more basically, if $K$ is a global field with constant field $\mathbb{F}_q$, then the roots of unity in $K$ are precisely the nonzero elements of $\mathbb{F}_q$. (There are also some differences: any constant field extension is everywhere unramified, whereas adjoining sufficiently many roots of unity to a number field gives a ramified extension.) | |
| Jun 14, 2010 at 14:43 | history | answered | David E Speyer | CC BY-SA 2.5 |