Timeline for How natural is the reciprocity map?
Current License: CC BY-SA 2.5
10 events
| when toggle format | what | by | license | comment | |
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| Mar 20, 2010 at 6:54 | vote | accept | abcdxyz | ||
| Jan 20, 2010 at 21:41 | comment | added | abcdxyz | It seems possible that there is a structure in which the suitable replacement of the multiplicative group and idelic class group will correspond to some abelian extension of that structure. However, this abelian extension is still not the maximal one. | |
| Jan 20, 2010 at 21:38 | comment | added | abcdxyz | My use of the word accident in the question might have been a bit misleading but I can not find a more suitable word. I did not mean accident in the sense close to being ad hoc. I agree that these maps are what we will get at the end. I was just wondering whether there is some hidden path in between. The use of the word accident in the later paragraph had yet another meaning. It seems a priori not clear why multiplicative group or idelic class group has anything to do with maximal abelian even though they are abelian. | |
| Jan 20, 2010 at 21:27 | comment | added | abcdxyz | For the third paragraph. I am convinced that this showed that the local reciprocity map is what we must get at the end. This explanation is still not satisfying in the same manner that deriving local class field theory from global one is not very satisfying. | |
| Jan 20, 2010 at 21:21 | comment | added | abcdxyz | However, once these choice are made the identification of $O^{times}_K$ and the Galois group of the chosen extension is canonical. So I don't think the non-canonical fact will give a very serious problem here either. | |
| Jan 20, 2010 at 21:18 | comment | added | abcdxyz | I don't quite agree with the second paragraph. Even though, the factorization $K^{\times}= \Z \times O^{\times}_K $ is not canonical, this can be fixed by looking at the exact sequence $ O^{\times}_K \rightarrow K^{times} \rightarrow Z$ where the second map is given by the additive valuation. This non-canonical problem won't affect much. The construction of Lubin Tate formal group not only depends on the uniformizer but also on the choice of the power series. The resulting maximal totally ramified abelian extensions of K depends on the uniformizer. | |
| Jan 20, 2010 at 21:06 | comment | added | abcdxyz | About the first part of your answer. I feel fortunate that you post this comment. I tried to find something like that for a while. In Cassels & Frohlich, Serre wrote that the change of base from $K^{\times} \rightarrow L^{\times} $ induced the transfer from $G^{ab}_K \rightarrow G^{ab}_L $. I was scared since I don't know much about group theory. Your comment above gave me a better feeling. However, this only say that if such a map exists it is natural but not yet the natural existence of such a map. | |
| Jan 20, 2010 at 20:56 | comment | added | abcdxyz | Thank you for the reply. I learned a lot from your answer but still I am not convinced. I will post my objections in different comments. | |
| Jan 19, 2010 at 19:40 | history | edited | Emerton | CC BY-SA 2.5 |
added 120 characters in body
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| Jan 19, 2010 at 16:15 | history | answered | Emerton | CC BY-SA 2.5 |