Timeline for A quantitative version of Hensel's Lemma
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 29, 2016 at 12:20 | vote | accept | Daniel Loughran | ||
| Nov 28, 2016 at 12:49 | answer | added | Martin Bright | timeline score: 10 | |
| Nov 28, 2016 at 11:54 | history | edited | Daniel Loughran | CC BY-SA 3.0 |
deleted 5 characters in body
|
| Nov 27, 2016 at 18:45 | comment | added | Michael Stoll | Roughly speaking, since $X$ is smooth, the subset of $X({\mathbb Q}_p)$ consisting of points reducing mod $p$ to some specified point in $X({\mathbb F}_p)$ is analytically isomorphic to the corresponding set for affine $n$-space, for which the statement is pretty clear. (This is just an attempt at giving a geometric version of znt's previous comment.) | |
| Nov 27, 2016 at 16:08 | comment | added | znt | Smooth over Z_p means the completed local ring at a Z/pZ-point in the special fibre is Z_p[[X_1,X_2,...,X_n]] with n the rel dim of the morphism. Now count how many maps from that ring to Z/p^kZ. Any map from Spec(Z/p^k) to the scheme whose image is this point induces a map on the local rings and because Z/p^k is Artinian it will factor through the completion. Isn't this enough? Did I miss something? | |
| Nov 27, 2016 at 15:49 | comment | added | Daniel Loughran | @znt: I don't understand your point. Could you provide details or an answer? | |
| Nov 27, 2016 at 14:46 | comment | added | znt | It's smooth, so work with the completed local rings and then it's easy, no? | |
| Nov 27, 2016 at 12:28 | comment | added | kneidell | Check out N. Bourbaki, Commutative Algebra, III, S4.5, Corollary 3. I think the uniqueness in the corollary should be sufficient to prove that $\#X(\mathbb Z/p^k\mathbb Z)/\# X(\mathbb Z/p^{k-1}\mathbb Z)=p^n$, for all $k>1$, from which the statement follows (at least in the case where $X$ is affine). | |
| Nov 27, 2016 at 12:28 | comment | added | LSpice | Incidentally, I guess we could reduce to the case where $\#X(\mathbb Z/p\mathbb Z) = 1$, right? That might be easier to conceptualise. | |
| Nov 27, 2016 at 12:27 | comment | added | LSpice | Searching, as I'm sure you did. shows that Hensel's-lemma questions are quite popular here. I wonder if Wanderer's answer, on the implicit-function theorem, to PeteL.Clark's old question has any relevance? | |
| Nov 27, 2016 at 12:09 | history | asked | Daniel Loughran | CC BY-SA 3.0 |