Timeline for A computation of ramification
Current License: CC BY-SA 2.5
11 events
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| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
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| Aug 8, 2010 at 16:00 | comment | added | H. Hasson | Qing's answer is saying exactly that just in a different language. The only thing I needed was that it is okay to use the same equation (which, a priori is only good enough to define the function field) to plug in a different y. The reason for that Qing explained in the comments to his answer. | |
| Aug 8, 2010 at 15:18 | vote | accept | H. Hasson | ||
| Aug 7, 2010 at 22:41 | comment | added | Dror Speiser | Still waiting on inverse-galois-solution article ;) Ok, cool. I get what you guys are saying. I think. So, in all specializations of $y$, except $1,-1,2,-2$, the constant coefficient is non-zero. Is this enough to prove no ramification, or do we need the machinery in Qing answer? | |
| Aug 7, 2010 at 22:01 | comment | added | H. Hasson | Ey, look who it is! How are you? If I understand your comment correctly, you're plugging in y=1 because you're saying the ramification above y=1 t=0 is the same as the ramification along t=0. This isn't true, though - the ramification along t=0 means generically (a horizontal branch divisor intersects t=0 at y=1). But it is a good idea to plug in a y not equal to one of 1,-1,2,-2. One should also be careful. The equation I gave is fine for defining the birational class, but it won't give a regular C((t))-curve, let alone define the normalization as a C[[t]]-curve. | |
| Aug 7, 2010 at 22:00 | comment | added | Qing Liu | But $y=\sqrt{1-t}$ is a horizontal ramification. | |
| Aug 7, 2010 at 21:58 | answer | added | Qing Liu | timeline score: 3 | |
| Aug 7, 2010 at 21:06 | comment | added | Dror Speiser | $\sqrt{1-t}$ doesn't really appear, so no need to fix choice. I haven't finished chapter 2 of Hartshorne, so I hope this isn't too wrong: the vertical divisor lives on the affine $y=1$, and using $\sqrt{2(2-t)} = 2-t/2+O(t^2)$, we get $z^4-t^2(27+O(t))$, showing that the ramification index is indeed 2. | |
| Aug 7, 2010 at 19:49 | comment | added | H. Hasson | No, it's supposed to be like that. | |
| Aug 7, 2010 at 19:47 | comment | added | KConrad | I think you left out an exponent of 3 on $sqrt(2(2-t))-y$. | |
| Aug 7, 2010 at 19:28 | history | asked | H. Hasson | CC BY-SA 2.5 |