Timeline for Can you make any prime ideal ramify in some algebraic extension?
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| Oct 12, 2011 at 20:52 | answer | added | Charles Matthews | timeline score: 1 | |
| Oct 12, 2011 at 17:52 | comment | added | KConrad | J, you are missing something trivial: see Milne's Remark 3.39(b), where he indicates how to remove the freeness assumption by using localization. | |
| Oct 12, 2011 at 17:47 | comment | added | KConrad | J, you are referring in your parenthetical remark to www.math.uconn.edu/~kconrad/blurbs/gradnumthy/notfree.pdf, but this issue isn't relevant. A prime in a number field ramifies in a finite extension iff it divides the discriminant ideal of the extension, and there is no need for an initial hypothesis that the top ring of integers is a free module over the bottom ring of integers. (However, in the proof of the theorem one may localize at the prime of interest and the localized ring of integers becomes a PID, so leading to a free module as a technical convenience, not a hypothesis.) | |
| Oct 12, 2011 at 16:18 | comment | added | Tommaso Centeleghe | Also, if p ramifies in L/Q and K is linearly disjoint from L, then p ramifies also in LK/K, right? Look at Marcus book "Number Fields"! Or Lang. To construct ramified extensions of K at p is always good to keep the p-th power roots of unity in mind! If the power of p is big enough you will get plenty of ramification above p! | |
| Oct 12, 2011 at 16:15 | comment | added | Tommaso Centeleghe | Ramification of primes is a local phenomenon. If you complete, then you will get that the integers of the bigger extension are a free module over that of the smaller one. If that bothered you. Dedekind rings with finitely many primes have trivial class number by the approx theorem (Corps Locaux, by Serre has all of it) | |
| Oct 12, 2011 at 15:21 | comment | added | user3860 | Thanks, that makes sense. But why would one generally require the freeness when they state the ramification criterion? For example, Milne's Algebraic Number Theory (Thm 3.35) asserts this condition. | |
| Oct 12, 2011 at 15:19 | comment | added | Qing Liu | Take an element $f$ of $O_K$ which generates $\mathfrak p$ in the local ring $O_{K, \mathfrak p}$ and consider the extension $L=K(\sqrt{f})$. It is quadratic and ramifies at $\mathfrak p$ (look at the ramification index). The ramification is a local phenomena and doesn't have anything to do with the freeness of $O_L$ over $O_K$. | |
| Oct 12, 2011 at 15:15 | comment | added | Torsten Ekedahl | Just take the square root of an element of $K$ with valuation $1$ at the prime. | |
| Oct 12, 2011 at 15:04 | history | asked | user3860 | CC BY-SA 3.0 |