Timeline for Splitting of prime ideals in non-Dedekind domains?
Current License: CC BY-SA 2.5
8 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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| Sep 8, 2010 at 22:05 | comment | added | BCnrd | Adam, concerning your pacifist tendencies, please go to the library (or some website of illegal scans, if the concept of library is too archaic) and compare two books on algebraic curves: the one by Fulton and the one by Chevalley. I suspect that a pacifist would become violent in rather different senses after looking at each of those books. :) | |
| Sep 8, 2010 at 18:57 | comment | added | Adam | Actually your suggestion to specialize $t$ into $\mathbb{Q}(s)$ gives a way to use Keith's method which I should have seen to start with: we can just consider $f$ as a polynomial in $\mathbb{Q}(s)[t]$, which is a Dedekind domain. I'm content to remain a pacifist for now! | |
| Sep 8, 2010 at 18:15 | comment | added | BCnrd | Adam, to paraphrase Hilbert, to do algebraic number theory without any algebraic geometry is like boxing without fists. (Strictly speaking, the content of my earlier comment can be done by pure algebra without any algebraic geometry. But that would be kind of silly, since it removes all of the insight which leads one to the argument in the first place.) | |
| Sep 8, 2010 at 17:36 | comment | added | Adam | So now I have to learn algebraic geometry as well? This problem is going to have taught me a lot of mathematics.... Thankyou Conrads! | |
| Sep 8, 2010 at 17:17 | comment | added | BCnrd | Dear Adam: For consistency with the notation in the earlier version, let's speak in terms of specialization of $t$ rather than $s$. Set $L = \mathbf{Q}(s)$ in the SGA1 argument that I gave in a comment to your question. That involves specializing $t$ into $\mathbf{Q}(s)$ rather than into $\mathbf{Q}$, but if you look at the proof you'll see that it provides an abundance of rational specializations of $t$ as well (in fact, all but finitely many). The argument is quite robust. | |
| Sep 8, 2010 at 17:16 | comment | added | KConrad | A domain other than a field in which nonzero proper ideals factor uniquely into a product of prime ideals is a Dedekind domain. In fact, a domain other than a field in which every nonzero ideal is a product of prime ideals is a Dedekind domain (so existence of prime ideal factorizations all the time implies uniqueness). The correct extension of Dedekind domains as far unique factn. is concerned is Krull domains, e.g., if A is Krull then A[x] is also Krull. | |
| Sep 8, 2010 at 16:58 | history | asked | Adam | CC BY-SA 2.5 |