# Variant of Leopoldt's conjecture

Let $K$ be a number field with $[K : \mathbf{Q}] = d$, and let $p$ be a prime. Let $\sigma_1, \dots, \sigma_d$ be the embeddings of $K$ into $\mathbf{C}_p$. Let $u_1, \dots, u_k$ be a basis of $\mathcal{O}_K^\times$ modulo torsion, so $k = r_1 + r_2 - 1$ where $r_1$ and $r_2$ are the number of real and complex infinite places.

Consider the $d \times k$ matrix $M$ whose $i, j$ entry is $\log \sigma_i(u_j)$. Leopoldt's conjecture claims that $M$ has full rank, i.e. it has some $k \times k$ minor which is non-singular.

I'm wondering about the following conjecture:

If $K$ does not contain a CM field, then all $k \times k$ minors of $M$ are nonsingular.

Obviously this is stronger than Leopoldt's conjecture, so there is no hope of proving it; but does anyone know if such questions have been studied? Are counterexamples known?

(The statement is definitely false if $K$ contains a CM field $E$ with $[E : \mathbf{Q}] > 4$.)

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Curious: have you checked in any examples? –  Ramsey Jan 17 '13 at 20:53
Yes, I have checked a few examples (although only ones where p splits completely in K). –  David Loeffler Jan 17 '13 at 21:35
David, when you say that the statement is definitely false if $K$ is CM and $d>4$ do you really need CM or totaly imaginary is enough? The argument I figure out is that for every unit $u$, the complex aboslute value of its norm is $1$ and if we decompose this as the product over all embeddings $\tau_j$ in $\mathbb{C}$ of the absolute value of $\tau_j(u)$ we get $|\tau_j(u)|=|\tau_i(u)|$ if the two are conjugate. And this is enough to kill many minors (if there is enough room, i.e. $d>4$), right? –  Filippo Alberto Edoardo Jan 18 '13 at 1:57
Filippo: No, that does not work, because there is no uniquely defined notion of "conjugate" unless K is CM. –  David Loeffler Jan 18 '13 at 7:24
I think what David meant was that CM fields are characterized by the property that, under any embedding into the complex numbers, the image is preserved by complex conjugation and the resulting involution on the field is independent of the embedding. –  Ramsey Jan 18 '13 at 15:43