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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 $K=\mathbb{Q}(\mu_{p^\infty})$ the rational number adjoining all the pth power roots of unity.

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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 $\mathbb{Q}$... – Keenan Kidwell Feb 11 '12 at 23:17
Oh yeah, and for p=2 we have [\mu_8(\mu_8^2-1)]^2=2. – Dustin Clausen Feb 11 '12 at 23:35
Yes, in general a quadratic field of discriminant $D$ embeds in $\mathbb{Q}(\zeta_D)$. – Keenan Kidwell Feb 11 '12 at 23:39
So does it mean we do not have $Gal(K(\sqrt[p^\infty]{p})/K)\cong \mathbb{Z}_p$ for general $p$? – user21330 Feb 11 '12 at 23:48

1 Answer 1


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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Always Chevalley.... – Filippo Alberto Edoardo Apr 9 '12 at 0:39

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