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Let $K$ be a number field, $\mathcal O_K$ the ring of integers of $K$ and let $\left \{ \varepsilon _1, \cdots, \varepsilon _{r_1+r_2-1} \right \}$ generator of $\mathcal O_K^\times$ mod $\mu(K)$ (here $\mu(K)$ denotes the roots of unity of $K$). Define the following mapping:

$\lambda : \mathcal O_K^\times \to \mathbb R^{r_1+r_2}$

$ \lambda(\varepsilon):=(\log |\sigma_1(\varepsilon )|, \cdots , \log |\sigma_{r_1}(\varepsilon )|, 2\log |\sigma_{r_1+1}(\varepsilon )|, \cdots , 2\log |\sigma_{r_1+r_2}(\varepsilon )| )$

where $\sigma_1, ..., \sigma_{r_1}$ are the real embeddings and $\sigma_{r_1+1}, ..., \sigma_{r_1+r_2}$ are the complex embeddings.

My question: Why is $\left \{ \lambda(\varepsilon_1) , \cdots \lambda(\varepsilon_{r_1+r_2-1}) \right \}$ a $\mathbb Z$-basis of $\lambda(\mathcal O_K^\times)$?

Thanks in advance.



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Hint: see… – Pete L. Clark Jun 12 '11 at 14:14
I know that the sum of all elements of any vector $\lambda(\varepsilon_i)$ is zero. But is this helpful for my problem? – David75 Jun 12 '11 at 15:01
You have a free abelian group $G$, a homomorphism $\lambda: G \rightarrow H$, and a basis $\{e_i\}$ of $G$, and you're wondering whether $\{ \lambda(e_i)\}$ is a basis for the image. This holds iff $\lambda$ is injective. So you want to check that if $\lambda(v) = 0$ then $v$ is a root of unity... – Pete L. Clark Jun 12 '11 at 15:08
Thanks for your answer. Now I have the following proof: Let $\lambda(\varepsilon)=0$. Then $\log| \sigma_i(\varepsilon )|=0$ for all $i=1,..,r_1+r_2$. Therefore $| \sigma_i(\varepsilon )|=1$ and $\varepsilon$ is a root of unity. – David75 Jun 12 '11 at 15:54
do you know how to prove that last assertion? It's not completely trivial. – Qiaochu Yuan Jun 12 '11 at 19:17

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