Timeline for Serge Lang's proof of Brauer-Siegel theorem
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Mar 21, 2021 at 14:58 | comment | added | Melanka | @reuns I have one more question. Going through the proof I don't see why $1 - \epsilon/N $ is used instead of simply using $1 - \epsilon$. Have I missed something? | |
| Feb 10, 2021 at 20:46 | comment | added | 2734364041 | "Well, then Theorem 1 should have been '... for all fields normal over $\mathbb{Q}$ except for one possible exception...'" This is only necessary if you want $c_4(\epsilon)$ to be effective. If one does not care about that (Brauer-Siegel doesn't), then you can allow $c_4(\epsilon)$ to depend ineffectively on $k_0$. This is meaningful in many contexts, but not all. You can't simply "forget about this pathological $k_0$ and work with the rest" unless you are ok with the uncertainty of "am I working with $k_0$?" If $c_4(\epsilon)$ is ineffective, then it doesn't matter if $k=k_0$ or not. | |
| Feb 10, 2021 at 14:46 | comment | added | Melanka | @reuns Sorry for missing that. I have added an edit. | |
| Feb 10, 2021 at 14:46 | history | edited | Melanka | CC BY-SA 4.0 |
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| Feb 10, 2021 at 14:29 | comment | added | reuns | What is $\kappa(k)$, the distance between $1$ and the nearest zero of $\zeta_K(s)$, or something like that? | |
| Feb 10, 2021 at 14:12 | history | edited | Melanka | CC BY-SA 4.0 |
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| Feb 10, 2021 at 13:56 | comment | added | Melanka | I will also edit the post to include the resolution of case 2, in which I have some other clarifications. | |
| Feb 10, 2021 at 13:51 | comment | added | Melanka | @apolge Thanks for the explanation. But why cannot we just forget about this pathological $k_0$ and work with the rest? | |
| S Feb 10, 2021 at 7:57 | history | suggested | gmvh |
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| Feb 10, 2021 at 7:31 | review | Suggested edits | |||
| S Feb 10, 2021 at 7:57 | |||||
| Feb 9, 2021 at 22:54 | comment | added | alpoge | (Because the constant in the inequality, which is supposed to hold for all $k$, will depend on $k_0$! In other words either none of them vanish in the interval, in which case you’re good, or one of them does, in which case you get an inequality that works for every single one of the fields just in terms of that one zero.) | |
| Feb 9, 2021 at 20:33 | review | Suggested edits | |||
| Feb 9, 2021 at 21:14 | |||||
| Feb 9, 2021 at 19:55 | history | asked | Melanka | CC BY-SA 4.0 |