Timeline for Relative Leopoldt defect
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Jun 27, 2016 at 18:13 | comment | added | znt | You assume that Leopoldt is true for $F$! So the rank of the global units (from $F$ to $M$) goes up by $[F:\mathbf{Q}]=([M:\mathbf{Q}]-1)-([F:\mathbf{Q}]-1)$, and the rank of the topological closure in the local units might not change (it might go up by 0), so this instantly gives $[F:\mathbf{Q}]-0=[F:\mathbf{Q}]$ as a bound for the defect in this situation without any Waldschmidt. I don't know Waldschmidt's techniques but perhaps they might do better in this case. | |
| Jun 27, 2016 at 14:39 | comment | added | Adel BETINA | The rank of the global units of $M$ is $[M:\mathbb{Q}]-1$ and the rank of the product of the local units of $M$ is $[M:\mathbb{Q}]$, hence the Leopoldt defect of $M$ is less than than $[M:\mathbb{Q}]$, but the result of Walshmidt is more precise. When you said the unit rank goes up by ... you speak about local or global ? and the both case your answer is wrong. | |
| Jun 27, 2016 at 6:52 | comment | added | znt | I want to stress that the result you have referred to several times as "a result of Waldschmidt" is in this case trivially true -- just look at how the rank can change. Because it's trivial, perhaps Waldschmidt's methods give you something better in this case? | |
| Jun 26, 2016 at 23:34 | comment | added | Adel BETINA | In my question I take number fields with some assumptions in the aim to know if there is a better bound than the result of Waldshmidt in this case, and I know that Leopoldt conjecture holds for abelian extensions and cyclic extension of a quadratic complex extension (Baker-Brumer) | |
| Jun 26, 2016 at 9:30 | comment | added | znt | I think $[M:\mathbf{Q}]/2$ is $[F:\mathbf{Q}]$ and my comment was supposed to mean that this is a bound (my comment is an answer) and unless I'm mistaken the proof is trivial: the unit rank goes up by $[F:\mathbf{Q}]$ and so now you just imagine that the topological closure doesn't go up at all. I don't see why Leopoldt should hold for $M$ if it holds for $F$. It's known to hold for $F/\mathbf{Q}$ abelian but as far as I know it's not known for quadratic extensions of abelian extensions. | |
| Jun 25, 2016 at 23:55 | comment | added | Adel BETINA | We assume that F is finite over $\mathbb{Q}$ and I know that a theorem of waldschmidt implies that the leopoldt defect of $M$ is $[M:\mathbb{Q}]/2$, but I ask if there is a better bound or if Leopoldt conjecture holds for this case. | |
| Jun 25, 2016 at 23:50 | history | edited | Adel BETINA | CC BY-SA 3.0 |
added 8 characters in body
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| Jun 25, 2016 at 15:05 | comment | added | znt | $[F:\mathbb{Q}]$? $\ \ \ $ | |
| Jun 24, 2016 at 23:50 | history | asked | Adel BETINA | CC BY-SA 3.0 |