Let F be a totally real number field such that the Leopoldt conjecture holds at a prime number $p$ and $M$ be a quadratic totally real extension of $F$.
Is there a bound of the Leopoldt defect of $M$ ?
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$\begingroup$ $[F:\mathbb{Q}]$? $\ \ \ $ $\endgroup$– zntCommented Jun 25, 2016 at 15:05
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$\begingroup$ 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. $\endgroup$– Adel BETINACommented Jun 25, 2016 at 23:55
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1$\begingroup$ 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. $\endgroup$– zntCommented Jun 26, 2016 at 9:30
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$\begingroup$ 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) $\endgroup$– Adel BETINACommented Jun 26, 2016 at 23:34
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$\begingroup$ 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? $\endgroup$– zntCommented Jun 27, 2016 at 6:52
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