Timeline for Algebraic integers on the unit circle
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 1, 2010 at 14:42 | answer | added | Dror Speiser | timeline score: 3 | |
| Oct 1, 2010 at 13:35 | answer | added | Cam McLeman | timeline score: 4 | |
| Oct 1, 2010 at 13:31 | comment | added | S. Carnahan♦ | @Dror: I mentally replaced the article "a" with "the" in the first sentence. You are quite correct. | |
| Oct 1, 2010 at 12:56 | history | edited | Vagabond | CC BY-SA 2.5 |
deleted 62 characters in body; deleted 13 characters in body
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| Oct 1, 2010 at 12:27 | comment | added | Gerry Myerson | I think there's more confusion than question here, so I'm voting to close. The third paragraph is particularly confusing, as when one talks about units in a number field one is automatically talking about algebraic integers - but maybe "extension field" doesn't mean "number field" - but we shouldn't have to guess what the question means. | |
| Oct 1, 2010 at 11:54 | comment | added | Dror Speiser | @Scott: I'm not sure what's going on. Isn't ${1,-1}$ a set of algebraic integers which lie on the unit circle that generate a multiplicative subgroup of the circle and consists of the roots of a quadratic polynomial? | |
| Oct 1, 2010 at 10:15 | comment | added | S. Carnahan♦ | It should be pretty clear that such a group must be infinitely generated, since it is divisible. In other words, there is no polynomial whose roots generate the group. It seems pretty likely that the structure of the abstract group is $\mathbb{Q}/\mathbb{Z} \oplus \mathbb{Q}^{\oplus \infty}$ | |
| Oct 1, 2010 at 9:41 | history | asked | Vagabond | CC BY-SA 2.5 |