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In Kummer Theory, we always need the assumption that $a\not\in K^{\times n}$. For the local case, we can use the structure theorem of $K^{\times}$ to check this assumption. But in general, how can we check this condition?

For example, how to check $p\not\in K^{\times p}$, where $K=\mathbb{Q}(\mu_{p^\infty})$ is the rational number field with all the $p$th power roots of unity adjoined.

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    $\begingroup$ At least if $p$ is odd, adjoining a root of the irreducible $X^p-p\in\mathbb{Q}[X]$ to $\mathbb{Q}$ gives an extension which is neither totally real nor totally complex, and in particular, the extension can't be Galois. But every subfield of $\mathbb{Q}(\mu_{p^\infty})$ is Galois over $\mathbb{Q}$... $\endgroup$
    – Keenan Kidwell
    Commented Feb 11, 2012 at 23:17
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    $\begingroup$ Oh yeah, and for p=2 we have [\mu_8(\mu_8^2-1)]^2=2. $\endgroup$
    – Dustin Clausen
    Commented Feb 11, 2012 at 23:35
  • $\begingroup$ Yes, in general a quadratic field of discriminant $D$ embeds in $\mathbb{Q}(\zeta_D)$. $\endgroup$
    – Keenan Kidwell
    Commented Feb 11, 2012 at 23:39
  • $\begingroup$ So does it mean we do not have $Gal(K(\sqrt[p^\infty]{p})/K)\cong \mathbb{Z}_p$ for general $p$? $\endgroup$
    – user21330
    Commented Feb 11, 2012 at 23:48

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In Deux Théorèmes d'Arithmétique Chevalley shows the following related statement (Remarque p. 39): let $p$ be a prime, $K$ a field of characteristic different from $p$, and $\zeta$ a $p^e$-th primitive root of $1$ in some algebraic extension of $K$, where $e\geq 1$ is any integer. Assume moreover that $-1$ is a square in $K$ if $p=2$. Then if an element $x\in K(\zeta)$ is a $p^e$-th power, then it is already a $p^e$-th power in $K$.

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