Timeline for Ramification and reduction
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 30, 2020 at 23:04 | comment | added | Will Sawin | @Macadam The issue is the morphism will not be well-defined at the vanishing points of $Q$ mod $\pi$ since $\pi P$ and $Q$ both vanish there. In general, for a morphism $a/b$ to be well-defined we want the ideal generated by $a$ and $b$ to be the unit ideal. | |
| Jul 30, 2020 at 22:51 | comment | added | Gabriel Soranzo | A last think (I promise) it's important for my (master) thesis: for me there was always unique a morphism $\Psi:\mathbb{P}^1_{\mathcal{O}_K}\to\mathbb{P}^1_{\mathcal{O}_K}$ such that $\psi$ was its extension of scalar with the trick $f=\pi P/Q$ as I said upper. But here you telled me that $\Psi$ may be not well defined: I can't see why because in all cases $\pi^k P$ and $Q$ are coprimes ($P$ and $Q$ are coprime and primitives so $\pi^k P$ ans $Q$ stay coprimes...) What's the problem? There are global sections generating the invertible sheaf $\mathcal{O}_{\mathbb{P}^1_{\mathcal{O}_K}}(d)$? | |
| Jul 30, 2020 at 20:51 | comment | added | Will Sawin | @Macadam If you look at proofs of Riemann-Hurwitz you will see the more abstract tools that are needed. | |
| Jul 30, 2020 at 20:43 | vote | accept | Gabriel Soranzo | ||
| Jul 30, 2020 at 20:43 | comment | added | Gabriel Soranzo | 1st pb: I'm agree if you mean that the image of $\Psi$ is closed: the points of $\mathbb{P}^1_\mathbb{K}$ are not closed in $\mathbb{P}^1_{\mathcal{O}_K}$ so the image of $\overline{\varphi}$ contains points of the closure of points of $\mathbb{P}^1_K$ so $\overline{\varphi}$ is not closed. For the 2nd: Ok! I'd so like to have better algebraic tools in disposal so I don't have to use equations for the morphism $f$... | |
| Jul 30, 2020 at 0:04 | comment | added | Will Sawin | @Macadam It is contradictory with the map being surjective in characteristic $0$ and constant in characteristic $p$ (then, what is the image? Is it closed?) The factors disappearing is OK, it just means the ramification points are going off to infinity. The field enlargement is to make a sufficient linear change of variables (rational linear transformation in $X$) that the points don't go off to $\infty$. | |
| Jul 29, 2020 at 21:27 | comment | added | Gabriel Soranzo | For the 1st pb: the fact that the image is closed is not contradictory with the constancy of the map, or I missed something? For the second pb: factors like $(cX-1)$ can reduced to $-1$ if $c\notin\mathcal{O}_K^*$ so disapear in $f'$. Maybe all this stuff work if the reduction is good? If we enlarge the field the $u/\pi$ becomes $u/\eta^q$ and the problem stay. Excuse my naive questions... | |
| Jul 29, 2020 at 20:45 | comment | added | Will Sawin | @Macadam If $\Psi$ is a well-defined morphism on $\mathbb P^1_{\mathcal O_K}$ then $\overline{\psi}$ cannot be constant (using that $\psi$ is finite, thus surjective.) For instance because $\overline{\psi}$ is a map from a projective variety, hence proper, so its image is closed. If the points are not in $\mathcal O_K$ (i.e. reduce mod $\pi$ to $\infty$) you get factors like $(cX -1)$ for $c \in \mathcal O_K$ and the same argument works. Alternately, you can extend to a larger field, which gives you enough freedom to change the variables so that none of the points reduce to $\infty$. | |
| Jul 29, 2020 at 20:00 | comment | added | Gabriel Soranzo | But I see two problems with that: if $q\neq 0$ the resulting reduced morphism is $(0,\overline{Q})$ or $(\overline{P},0)$ which is constant (should I add the hypothesis $\overline{\psi}$ not constant?). Second problem: what if the points are not in $\mathcal{O}_K$: they are in $\mathbb{P}^1_{\mathcal{O}_K}$ and not $\infty$ but they should be for example $[1:u/\pi]$ and so we can not reduce the linear factor $(X-u/\pi)$ in $f'$. | |
| Jul 29, 2020 at 20:00 | comment | added | Gabriel Soranzo | I need some details with the reasonning with the rationnal function $f$: I'm agree that we can write $f=\pi^k P/Q$ with $\pi$ an uniformizer of $\mathcal{O}_K$, $q\in\mathbb{Z}$, $P,Q\in\mathcal{O}_K[X]$ primives and prime between them so $\widetilde{\psi}$ has a sens (not as rationnal function but with $(\pi^k P,Q)$ or $(P,\pi^kQ)$ global section generating the invertible sheaf $\mathcal{O}(d)$). So it has a sens to reduce $f$ in the residue field $k$. | |
| Jul 29, 2020 at 14:30 | comment | added | Gabriel Soranzo | Ok and $3/2$ is $[2:3]$ so reduce to $[2:0]=[1:0]$ in $\mathbb{P}^1_{\mathbb{F}_3}$. So we have 3 points $0$, $3$ and $3/2$ which goes to $\overline{0}$ all with $e=2$ and $e(\overline{0})=4$ because the function becomes $z^4$ which is coherent with $1+(2-1)+(2-1)+(2-1)$. | |
| Jul 29, 2020 at 11:59 | comment | added | Will Sawin | @Macadam The derivative of $x^2 (x-3)^2 = 2 x (x-3)^2 + 2 x^2 (x-3) = 2x (x-3) (2x-3)$. There is a third ramification point $3/2$ which also has $e=2$. | |
| Jul 29, 2020 at 10:20 | comment | added | Gabriel Soranzo | Excuse my stupidity, but In a simple exemple, $P=x^2(x-3)^2$ in $\mathbb{Q}_3$, one hase $P_1=0$ and $P_2=3$ with $\overline{P_1}=\overline{P_2}=\overline{0}$ in $\mathbb{F}_3$, $e_1=2$, $e_2=2$ and the resulting ramification is $4$ because $\overline{P}=x^4$, correct? But I can't see it with your formula? | |
| Jul 28, 2020 at 22:39 | comment | added | Will Sawin | Any reference on Riemann-Hurwitz in characteristic $p$ should explain what you need here, because it's mostly the same ideas. | |
| Jul 28, 2020 at 22:21 | comment | added | Will Sawin | @Macadam In this case we can just define the Swan conductor as the order of vanishing of the derivative minus $e-1$, which is $0$ in the case of tame ramification and $>0$ in the case of wild ramification. | |
| Jul 28, 2020 at 22:12 | comment | added | Gabriel Soranzo | Thanks! Could you tell me in short words what is the Swan conductor or give me a reference? | |
| Jul 28, 2020 at 21:59 | history | answered | Will Sawin | CC BY-SA 4.0 |