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Given an abelian extension $L/\mathbb{Q}$, such that $L=\mathbb{Q}(\alpha)$ for a root $\alpha$ of an irreducible polynomial $f(x)$. By Kronecker-Weber theorem we know that there exists $m \in \mathbb{N}$, such that $L \subseteq \mathbb{Q}(\zeta_m)$. How can we find explicitly the least integer $m \in \mathbb{N}$, with this property? Could you please introduce me to a reference for this?

I know that my question is very short, but I am very curious about this algorithm after this comment. (Also, this question was very informative for me.)

Davood
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1 Answers1

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  • If $K/\Bbb{Q}$ is abelian then $K\subset \Bbb{Q}(\zeta_n)$. Clearly the $p$ ramified in $K$ must divide $n$.

  • Assuming that $K\subset \Bbb{Q}(\zeta_{rp^l}),p\nmid r$

    Since $\Bbb{Q}(\zeta_{rp^l})/\Bbb{Q}(\zeta_r)$ is cyclic and totally ramified at $p$, then $K\subset \Bbb{Q}(\zeta_{rp^{l-1}})$ iff $$e(p,K/\Bbb{Q})=e(p,K(\zeta_r)/\Bbb{Q}(\zeta_r))= [K(\zeta_r):\Bbb{Q}(\zeta_r)] \le [\Bbb{Q}(\zeta_{rp^{l-1}}):\Bbb{Q}(\zeta_r)]= \phi(p^{l-1})$$

(there is a subtlety for $p=2$ here as $\Bbb{Q}(\zeta_8)/\Bbb{Q}$ is not cyclic)

  • Whence $$K\subset \Bbb{Q}(\zeta_n), n = \prod_p p^{c_p}, \qquad {c_p = \inf \{ l,e(p,K/\Bbb{Q})\ |\ \phi(p^l)\}, \\ c_2=\ldots}$$ is the least possible $n$.

(the given expression for $c_p$ is only valid for $p$ odd, we need a special case for $p=2$ as $\Bbb{Q}(\sqrt2)$ has ramification index $2$ at $p=2$ but is contained in $\Bbb{Q}(\zeta_8)$ not $\Bbb{Q}(i)$.

reuns
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  • +1 Thanks a lot, dear reuns. Everything is clear and satisfying to me except this part $e(p,K/\Bbb{Q})=e(p,K(\zeta_r)/\Bbb{Q}(\zeta_r))= [K(\zeta_r):\Bbb{Q}(\zeta_r)]$. – Davood Dec 31 '20 at 13:44
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    The second equality is because it is a totally ramified extension. The first one is because $K(\zeta_r)/K$ is unramified (at $p$) – reuns Dec 31 '20 at 13:45
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    Sanity check: $K=\Bbb{Q}(\sqrt3)$ has ramification index $2$ at $p=2,3$ and $i\not \in K$ and $K\subset \Bbb{Q}(\zeta_{3^1 2^2})$. So my formula for $c_2$ is not correct. I don't know how to distinguish it from $K=\Bbb{Q}(\sqrt6)\subset \Bbb{Q}(\zeta_{3^1 2^3})$.. – reuns Dec 31 '20 at 13:47
  • Now the second equality is clear. I think the first equality should be also easy, But I can not see it. Why the fact that "$K(\zeta_r)/K$ is unramified" implies that "$e(p,K/\Bbb{Q})=e(p,K(\zeta_r)/\Bbb{Q}(\zeta_r))$"? – Davood Dec 31 '20 at 13:51
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    $p = \prod_j P_j^e$ in $O_K$ and $P_j = \prod_i Q_{i,j}$ in $O_{K(\zeta_r)}$ so $p=\prod_i Q_{i,j}^e$ in $O_{K(\zeta_r)}$ this gives $e(p,K/\Bbb{Q})=e(p,K(\zeta_r)/\Bbb{Q})$. Then redo the same starting with the factorization of $p$ in $\Bbb{Q}(\zeta_r)$ to obtain the factorization in $O_{K(\zeta_r)}$: $ p = \prod_j S_j = \prod_{i,j} T_{i,j}^{e'}$ thus $e=e'$. – reuns Dec 31 '20 at 13:55
  • Thank you so much for your patience, dear reuns. – Davood Dec 31 '20 at 13:56
  • So I should have said in $E/F/M$ when everything is Galois $e(p,E/M)=e(P_j,E/F)e(p,F/M)$. When it is not Galois it is still true replacing by $e(Q_{ij},E/M)=e(Q_{ij},E/F)e(Q_{ij}\cap F,F/M)$ where $e(Q_{ij},E/M)$ is the exponent of $Q_{ij}$ in the factorization of $(Q_{ij}\cap O_M) O_E$. – reuns Dec 31 '20 at 14:16