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For any field $k$, let $\mu(k)$ denote the roots of unity in $k$. Now let $k$ be a number field and let $v, w$ be non-archimedean primes of $k$ with distinct residual characteristics. Does there exist a finite Galois extension $K/k$, with $v',w'$ primes of $K$ lying over $v,w$, such that $\mu(K)=\mu(K_{v'})=\mu(K_{w'})$?

For example, if $k=\mathbb{Q}$ and you're looking at the primes $3$ and $5$, then you can take $K=\mathbb{Q}(\zeta_{24})$ as your galois extension.

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Presumably the question is being asked for an arbitrary $k$. But why would one want to know this? It sounds like a strange (and very restrictive) condition. Is there some application in mind, or is this just an idle thought? – BCnrd Jun 10 '10 at 3:12
Just something that came to mind while I was thinking about the analogy between number fields and function fields. Number of roots of unity in number fields is something like the size of the constant field for function fields. If the stated condition held, you could potentially compare elements from these different "constant fields" because they would all global representatives. After the initial thought that I might want something like that, it seemed like a tricky question. – Joel Dodge Jun 10 '10 at 7:30

Okay, I'm giving up. I hadn't realized local fields had large subfields of cyclotomic extensions. Thanks to BCnrd - I learned something interesting today.

Second try (also wrong): Choose an embedding of $k$ in an algebraic closure $\overline{\mathbb{Q}}$ of $\mathbb{Q}$, and let $k' = k \cap \mathbb{Q}^{ab}$, where $\mathbb{Q}^{ab} \subset \overline{\mathbb{Q}}$ is the maximal abelian extension. By Kronecker-Weber, there exists a positive integer $m$ such that $k' \subset \mathbb{Q}[\zeta_m]$. Let ${k_v}'$ be the subfield of $k_v$ consisting of elements that are algebraic over $k$ and lie in $\mathbb{Q}^{ab}$ (under any choice of embedding), and let $m_v$ be a positive integer such that ${k_v}' \subset \mathbb{Q}[\zeta_{m_v}]$. Let $m_w$ be defined similarly. Let $n$ be the least common multiple of $m$, $m_v$, and $m_w$, and let $K = k[\zeta_n]$.

$K$ is an abelian Galois extension of $k$ and $\mu(K) = n$. The minimal polynomial of $\zeta_n$ over $k$ splits into (Galois conjugate) polynomials of equal degree over the subfield of $k_v \cap \overline{\mathbb{Q}}$, so $K \otimes_k k_v$ is isomorphic to a product of copies of $k_v[\zeta_n]$. Therefore, for any place $v'$ over $v$, $K_{v'} \cong k_v[\zeta_n]$ and $\mu(K_{v'}) = n$. Similarly, $\mu(K_{w'}) = n$ for any place $w'$ over $w$.

First Try (wrong - see BCnrd's comment): Let $n$ be the least common multiple of $\mu(k_v)$ and $\mu(k_w)$, let $K = k[\zeta_n]$, where $\zeta_n$ is a primitive $n$th root of unity, and let $v'$ and $w'$ be any chosen places over $v$ and $w$, respectively. $K$ is Galois over $k$, and if I'm not mistaken, we have $\mu(K_{v'}) = \mu(k_v[\zeta_n]) = n$ and $\mu(K_{w'}) = \mu(k_w[\zeta_n]) = n$. Is there a subtlety I'm missing?

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Scott, why should $K$, $k_v[\zeta_n]$, and $k_w[\zeta_n]$ not have roots of unity apart from the $n$th roots of unity? For example, if $k$ is the maximal totally real subfield of the cylotomic field $\mathbf{Q}(\zeta_{15})$ then $\mu(k) = \{\pm 1\}$ but $k[\zeta_6] = \mathbf{Q}(\zeta_{15})$, so $\mu(k[\zeta_6])$ has order 30. You can now see the sort of subtlety which was missed in your calculations. – BCnrd Jun 10 '10 at 6:15
Scott, 2nd attempt fails: no reason $[k'_v:\mathbf{Q}]$ finite. Any $\mathbf{Q}_ p$ contains finite ab. ext'ns $L/\mathbf{Q}$ of arb. large degree. Indeed, inside $\mathbf{Q}(\zeta_m)$ with $m = p^f - 1$ ($f > 0$), subfield $L$ of invariants under $p \bmod m$ embeds $\mathbf{Q}_ p$ since $p$ totally split in $L$. Clearly $[L:\mathbf{Q}] = \varphi(p^f - 1)/f$ and $\varphi(n) \ge \sqrt{n}$ for $n \ge 7$, so QED. (Cool tidbit: minimal number $r$ of $\mathbf{Z}$-algebra generators of $\mathca{O}_ L$ satisfies $p^r \ge \varphi(p^f - 1)/f$, so $r \rightarrow \infty$ as $f$ grows!) – BCnrd Jun 10 '10 at 17:25
Aaargh, the "unknown control sequence" thing above should be $O_ L$; I wish \mathcal didn't give such headaches on this system. Anyway, Scott, I recommend not spending more time on this unnatural question. I will now follow my own advice. :) – BCnrd Jun 10 '10 at 17:28

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