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Let $F$ be a totally real number field over $\mathbb Q$ of degree $m$ and let $\mathcal O_F$ be its ring of integers. By Dirichlet's unit theorem we know that $\mathcal O_F^* \cong \{\pm 1\} \times \mathbb Z^{m-1}$. Let $u \in \mathcal O_F^*$ be of infinite order. I'm interested in the group generated by $\{ \sigma_j(u) \mid 1\le j \le m \}$ where $\sigma_j$ are the $m$ real embeddings of $F$. Is this group always of rank $m-1$ (after dividing out potential torsion)?

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  • $\begingroup$ I don't the group of units so well. Why should we know the rank of the multiplicative group generated by $\{\sigma_j(u)\}_{j=1}^m$ ? $\endgroup$
    – reuns
    Apr 4, 2017 at 17:29
  • $\begingroup$ As explained in Franz Lemmermeyer's answer (+1) this is obviously false if $u$ happens to come from a smaller normal intermediate field. By normality all the conjugates $\sigma_j(u)$ then come from that same field, and there is no chance they could generate a full rank subgroup. $\endgroup$ Apr 4, 2017 at 20:43

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The answer is no. Take a totally real quartic field with a quadratic subfield, and let $u$ be the fundamental unit of the quadratic subfield. For fields of prime degree, google for Minkowski units and check out Narkiewicz's book on algebraic number theory.

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  • $\begingroup$ For quadratic fields the rank is always 1 since it can't be bigger because $\mathcal O_F^*$ has already rank 1 and it can't be smaller, because as $u$ is of infinite order so is $\sigma_j(u)$. I'm not interested in the torsion if this is your point. $\endgroup$ Apr 4, 2017 at 17:05
  • $\begingroup$ No - my point is that $u$ has infinite order and generates a subgroup of the unit group of the quartic field that does not have finite index. $\endgroup$
    – user23365
    Apr 4, 2017 at 17:39
  • $\begingroup$ Thanks a lot. I misread your answer: I read "quadratic" instead of "quartic" and was quite confused. $\endgroup$ Apr 5, 2017 at 8:01

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