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Let $K/\mathbb{Q}$ a finite (i.e. algebraic and finitely generated) extension. Let $x \in K$, such that $||x||=1$ for all normalized absolute values of $K$ but at most one. Then $x$ is a root of unity.

By the product formula I get $||x||=1$ for all normalized absolute values of $K$. Hence $x \in S^1$, seen as the embedding $K \subset \mathbb{C}$.

But now I am a bit confused. Let $x = \frac{3}{5} + \frac{4}{5}i$ shows that the conditions $x$ algebraic and $x \in S^1$ are not sufficient for $x$ to be a root of unity, hence I will have to use that the $p$-adic absolutes are $1$ as well. However, I feel a bit clueless on how to go on. My intuition would tell me that in the example above something will go wrong for $p_i | (5)$, $p_i$ a prime ideal in $\mathbb{Q}(x)$, then taking the $p_i$-valuation. I don't really know if this is true though, as I don't know any statements about absolute values in fields like the given $K$, but the existence, definition and product formula. Also, understanding the example still doesn't give me a solution.

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  • $\begingroup$ For intuition, it may help to think of the converse, which is much easier. Note that for any root of unity, and any absolute value (e.g. a $p$-adic one) certainly $|x| = 1$ (because $|x|^n = 1$ and $|x|$ is a positive real number.) So what goes wrong in your example is that the $5$-adic valuation (or better, as you say, the $p_i$-adic valuation) is negative, and this property can't change if we take powers. $\endgroup$ – hunter Dec 17 '13 at 17:35
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    $\begingroup$ As you say, we can get rid of the superfluous "all but one" using the product formula. To proceed from there is a classic result. The easiest way is in terms of adeles; I am leaving this as a comment since you may not know adeles yet. $K$ sits discretely in $\mathbb{A}_K$. On the other hand, the product $S^1 \times \prod_p \mathcal{O}_{K, p}$ is compact. Therefore their intersection is finite. Taking units on both sides gives us that the set of elements with this property is a finite subgroup of $K^\times$ and hence is a group of roots of unity. $\endgroup$ – hunter Dec 17 '13 at 17:48
  • $\begingroup$ @YACP: Yes, sorry. Q are the rational numbers. I might better change to the mathbb script. $\endgroup$ – Louis Dec 17 '13 at 22:01
  • $\begingroup$ @hunter: we hadn't talked about adeles yet. But thanks for your comment! $\endgroup$ – Louis Dec 17 '13 at 22:01
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The archimedean places of $K$ correspond to the usual absolute value on $\Bbb C$, after first applying the embeddings $K\hookrightarrow\Bbb C$. Thus if $|x|=1$ for all archimedean places, we know that all of $x$'s conjugates have (the usual) absolute value $1$. Consider the minimal polynomial of $x$. In my Stewart & Tall, the following lemma is used in proving Dirichlet's unit theorem:

Lemma. If $p(t)\in\Bbb Z[t]$ is a monic polynomial all of whose roots have absolute value $1$, then all of its roots are roots of unity. The proof proceeds as follows:

  • Say $p(t)=(t-a_1)\cdots(t-a_k)$ and define $p_\ell(t)=(t-a_1^\ell)\cdots(t-a_k^\ell)$
  • As symmetric polynomials in $a_1,\cdots,a_k$, the coefficients of $p_\ell(t)$ are integers
  • $t^j$ coeff. of $p_\ell$ is bounded: $|e_{k-j}(a_1^\ell,\cdots,a_k^\ell)|=|\sum\square|\le\sum|\square|=\sum1=\binom{k}{j}$ (Vieta's)
  • Only finitely many $f(t)\in\Bbb Z[t]$ satisfying such bounds, so $p_1,p_2,p_3,\cdots$ has a repeat
  • Say $p_\ell(t)=p_m(t)$. So $\exists\pi\in S_k$ such that $a_j^\ell=a_{\pi(j)}^m$ (for each $j=1,\cdots,k$)
  • Thus $a_j^{\ell^2}=(a_{\pi(j)}^m)^\ell=(a_{\pi(j)}^\ell)^m=a_{\pi^2(j)}^{m^2}\Rightarrow\cdots\Rightarrow a_j^{\ell^{\large k!}}=a_j^{m^{\large k!}}\Rightarrow a_j^{\ell^{\large k!}-m^{\large k!}}=1$ for each $j$

I am not sure if this is desirable for a homework exercise but it's a fun proof nonetheless.

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Here's what I finally did (no guarantee for correctness):

First note that as the localizations $R_\mathfrak{p}$ are given by $\{y \in K\mid ||y||_\mathfrak{p} \leq 1\}$, $x$ is contained in all the $R_\mathfrak{p}$ hence also in $R$ (for $R$ being the integral closure of $\mathbb{Z}$ in $K$).

Let $r_1$ be the number of real embeddings of $K$, let $2r_2$ be the number of its complex embeddings into $\mathbb{C}$.

Let $\mathfrak{L}: K^\ast \rightarrow \mathbb{R}^{r_1+r_2}, y \mapsto (\log(|\sigma_1(y)|),...,\log(|\sigma_{r_1+r_2}(y)|)$

Then we had seen as a Lemma for Dirichlet's unit theorem:

Let $B \subset \mathbb{R}^{r_1+r_2}$ compact. Then $\mathfrak{L}^{-1}(B) \cap R$ is finite.

Now in our case: take $B = 0 \in \mathbb{R}^{r_1+r_2}$.

As for all $n \in \mathbb{N}$: $x^n \in \mathfrak{L}^{-1}(B) \cap R$, we are done.

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