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It is well-known that any abelian extension of $\Bbb Q$ is contained within some cyclotomic field $\Bbb Q(\zeta_n)$. My question is about the more general statement:

If $L/K$ is an abelian extension, do we have $L \subset K(\zeta_n)$, where $\zeta_n \in \overline K$ is a root of $X^n-1$ (for some $n \in \Bbb N$) ?

For instance, any abelian extension of $\Bbb Q_p$ is contained within some cyclotomic field $\Bbb Q_p(\zeta_n)$. Morover, it holds when $K$ is a finite field (the Galois groups are cyclic, and any non zero element is a root of unity).

I believe that the statement is wrong, but I wasn't able to come up with a counterexample.

Thank you very much!

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  • $\begingroup$ Possibly related: math.stackexchange.com/questions/460397 $\endgroup$
    – Watson
    Sep 8, 2016 at 13:55
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    $\begingroup$ You might be interested in Hilbert's 12th problem, which asks for a generalisation of the Kronecker-Weber theorem to general number fields. Apart from the case of $\mathbb Q$, imaginary quadratic extensions (via the theory of complex multiplication) and a few other sporadic cases, this problem is wide open. $\endgroup$
    – Mathmo123
    Sep 11, 2016 at 9:51

2 Answers 2

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This is false for every number field $K$ other than $\mathbf Q$. The general proof uses ideas related to class field theory.

As a concrete example, consider the quadratic extension $\mathbf Q(\sqrt[4]{2})/\mathbf Q(\sqrt{2})$. Since $\mathbf Q(\sqrt{2})$ is inside a cyclotomic extension of $\mathbf Q$, namely it is in $\mathbf Q(\zeta_8)$, any cyclotomic extension of $\mathbf Q(\sqrt{2})$ is inside a cyclotomic extension of $\mathbf Q$. Therefore any cyclotomic extension of $\mathbf Q(\sqrt{2})$ is an abelian extension of $\mathbf Q$. Since $\mathbf Q(\sqrt[4]{2})$ is not abelian (or even Galois) over $\mathbf Q$, it is not contained in a cyclotomic extension of $\mathbf Q(\sqrt{2})$.

Many more counterexamples can be made by replacing $\mathbf Q(\sqrt{2})$ with other abelian extensions of $\mathbf Q$.

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No.

Consider the extension $\mathbb{Q}(t^{1/2})$ of the field $\mathbb{Q}(t)$. It is an extension of degree $2$, and is thus abelian. However, adding any root of unity $\zeta$ to $\mathbb{Q}(t)$ will give you the field $(\mathbb{Q}(\zeta))(t)$, in which $t$ has no square root.

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  • $\begingroup$ Thank you, Pierre-Guy! I see that in such a situation, the notion of cyclotomic extension is not suitable, while the notion of Kummer extension seems to be better: "When $K$ contains $n$ distinct $n$-th roots of unity, it states that any abelian extension of $K$ of exponent dividing $n$ is formed by extraction of roots of elements of $K$." $\endgroup$
    – Watson
    Sep 8, 2016 at 14:00
  • $\begingroup$ Perhaps in the same flavor, Artin Schreier extensions. $\endgroup$
    – peter a g
    Sep 8, 2016 at 14:06

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