Timeline for Does (the ideal class of) the different of a number field have a canonical square root?
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Jan 22, 2011 at 4:46 | answer | added | Emerton | timeline score: 9 | |
| Dec 21, 2010 at 17:14 | comment | added | Emerton | Dear Unknown, See the top of p.6. | |
| Dec 21, 2010 at 15:54 | comment | added | user4245 | Dear Emerton, I just open the file. It is quite long... which page should I start from ? | |
| Dec 21, 2010 at 15:44 | comment | added | Emerton | Dear Unknown, Did you look at the paper I linked to? | |
| Dec 21, 2010 at 15:22 | comment | added | user4245 | Dear Emerton: could you explain the parallel of integral domain and Hecke module as you mention in some detail, or suggest a good reference for it? thanks | |
| Oct 18, 2010 at 8:22 | answer | added | Stéphane Vinatier | timeline score: 11 | |
| Oct 3, 2010 at 3:15 | comment | added | Emerton | Dear Unknown, I have hand-written notes somewhere. Regards, Matthew | |
| Oct 3, 2010 at 2:50 | comment | added | user4245 | Hi,Emertion, do you have lecture notes of Dick Gross for the 1996 Harvard course in the reference of your artical? | |
| Sep 29, 2010 at 13:26 | answer | added | Chandan Singh Dalawat | timeline score: 8 | |
| Jul 25, 2010 at 16:18 | answer | added | Franz Lemmermeyer | timeline score: 9 | |
| Jul 24, 2010 at 4:34 | history | edited | Emerton | CC BY-SA 2.5 |
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| Jul 24, 2010 at 4:21 | history | edited | Chandan Singh Dalawat | CC BY-SA 2.5 |
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| Jul 24, 2010 at 2:53 | vote | accept | Emerton | ||
| Jul 24, 2010 at 2:44 | answer | added | user631 | timeline score: 28 | |
| Jun 22, 2010 at 9:59 | comment | added | naf | @KConrad: Thanks. I realised my mistake while trying to find an explicit example using sage. It is possible that the example was not a quadratic extension of Q but I don't really remember now. | |
| Jun 21, 2010 at 21:53 | comment | added | KConrad | Or maybe the error in Unknown's comment is that the extension of Q wasn't a quadratic field. | |
| Jun 21, 2010 at 21:51 | comment | added | KConrad | Unknown, there is a mistake in your comment: the different ideal in a quadratic field is always principal. More generally, if the ring of integers has the form Z[a] then the different ideal is (f'(a)) for f(x) the min. poly. of a over Q. In any quadratic field the ring of integers has the form Z[a] for some a. Thus whatever example you may have found couldn't possibly be quadratic over Q. Was your base field the quadratic field instead of Q (and some extension of it had a class group of order 4, etc.)? | |
| Jun 21, 2010 at 14:15 | history | edited | Emerton | CC BY-SA 2.5 |
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| Jun 21, 2010 at 14:01 | answer | added | Emerton | timeline score: 4 | |
| Jun 21, 2010 at 13:59 | history | edited | Emerton | CC BY-SA 2.5 |
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| Jun 21, 2010 at 10:40 | comment | added | naf | There exists a quadratic extension of Q with class group cyclic of order 4, the different the element of order 2, and such that the automorphism group acts non-trivially on the class group. It follows that there is no natural square root of the dfferent. (I had found such an example in response to a question of Fulton about the arithmetic Riemann-Roch theorem around 20 years ago by looking up some tables but unfortunately do not recall all the details.) | |
| Jun 20, 2010 at 16:25 | comment | added | Marty | I'd guess the answer has to be "no", if one wants a canonical construction. In the fn field case, there's a moduli space $S_g$ of "spin curves" (curves endowed with a sqrt of can. bundle), naturally a cover of $M_g$. Any natural construction of a sqrt of can. bundle would lead to a global section of the cover $S_g / M_g$. But this cover is provably nontrivial (for $g \geq 2$?). By analogy, I'd doubt the existence of a natural construction in the number field case as well. | |
| Jun 20, 2010 at 7:29 | comment | added | JBorger | Emerton: Are there any properties that you'd like the square-root to have? I ask because it's no doubt possible to give a completely ridiculous (but well-defined!) answer to your question by enumerating everything in sight and then picking the smallest ideal that does the job. (Unless I'm mistaken this can be done without invoking the axiom of choice---pick an enumeration of all possible finite presentations of rings, use this to make an ordering on all ideals in rings of algebraic integers, etc.) It would be nice to promote your question to a yes/no question without silly yes answers. | |
| Jun 19, 2010 at 23:29 | comment | added | BCnrd | Hi Marty. No, Serre just uses a fixed-point argument to make such a divisor class over the ground field. I haven't gone through the details, so I don't claim it leads to an illuminating construction. Just seemed a worthwhile thing to point out (since it is quite different from the character argument of Hecke). | |
| Jun 19, 2010 at 23:22 | comment | added | Marty | This, in the fn field setting, is the same as giving a spin structure on the curve, right? From a paper of Atiyah, I thought that the set of such spin structures was a torsor for a group of order $2^{2g}$, with $g$ the genus (in the $C$-analytic setting). Does Serre give a natural base point for this torsor? | |
| Jun 19, 2010 at 20:59 | comment | added | fherzig | For the class of the discriminant ideal being a square in the class group, see also Serre's very nice discussion of discriminants in Local Fields, section III.2. | |
| Jun 19, 2010 at 20:55 | comment | added | BCnrd | Comments on this paper at end of volume II of Serre's collected works give two proofs for fn fields over (quasi)finite fields. The first is Hecke's, and the second (given away from char. 2) is geometric: constructs divisor class on the curve (not just over alg. closure) which doubles to the canonical class. Since the different in fn field case for a sep'ble cover is difference of canonical classes (base pulled back to source), one gets construction in the fn field case. Probably no useful analogue for number fields or else Serre would have said something. Though could email Serre to ask him! | |
| Jun 19, 2010 at 20:36 | comment | added | KConrad | If this is possible, it likely will have to be very arithmetic. In a paper by Frohlich, Serre, and Tate (A different with an odd class, Crelle 209 (1962), 6–-7) they give a non-arithmetic example of an extension of Dedekind domains where the different is not a square in the class group. By comparison, if you look at discriminant ideals instead of different ideals then there is an algebraic construction of a square root: the Steinitz class. That is, if S/R is your Dedekind domain extension, decompose S as an R-module as S = R^{n-1}+M (direct sum) for an ideal M in R. Then [M]^2 = [disc(S/R)]. | |
| Jun 19, 2010 at 19:59 | history | asked | Emerton | CC BY-SA 2.5 |