Timeline for Where does the principal ideal theorem (from CFT) go?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 30, 2011 at 17:41 | vote | accept | Jonah Sinick | ||
| Apr 30, 2011 at 17:41 | vote | accept | Jonah Sinick | ||
| Apr 30, 2011 at 17:41 | |||||
| Apr 30, 2011 at 7:43 | answer | added | Franz Lemmermeyer | timeline score: 15 | |
| Apr 30, 2011 at 7:17 | answer | added | Kevin Ventullo | timeline score: 10 | |
| Apr 30, 2011 at 7:15 | answer | added | Charles Matthews | timeline score: 6 | |
| Apr 30, 2011 at 4:37 | comment | added | KConrad | Consider the sequence of fields K_0, K_1, K_2, ... where K_0 = K and, for i > 0, K_i is the Hilbert class field of K_{i-1}. These fields are called the class field tower over K, since each K_i is in K_{i+1}. One can show that K can be embedded in some number field with class number 1 iff the class field tower over K is a finite extension of K, in which case the top field in the tower is an example of a finite extension of K with class number 1. The proof of this iff statement does not require the principal ideal theorem, although that theorem does motivate the result! | |
| Apr 30, 2011 at 0:42 | history | edited | Jonah Sinick | CC BY-SA 3.0 |
added reference to dror comment
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| Apr 29, 2011 at 20:36 | comment | added | Dror Speiser | The smallest illustrative example is probably the following: $\mathbb{Q}(\sqrt{-5})$ has class number two, generated by the ideal above the ramified prime $2$. Taking, as Milne suggests, the square root of $2$ makes for an extension as required, but this is not the Hilbert class field, and is in fact ramified at 2. The Hilbert class field is $\mathbb{Q}(\sqrt{-5},i)$. | |
| Apr 29, 2011 at 19:57 | history | asked | Jonah Sinick | CC BY-SA 3.0 |