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Let $K$ be an algebraic number field. Let $A$ be the ring of integers in $K$. Let $L$ be an extension of $K$ of a prime degree $p$. Let $B$ be the ring of integers in $L$. Let $\mathfrak{D}_{L/K}$ be the different of $L/K$. Suppose every prime factor of $\mathfrak{D}_{L/K}$ does not divide $p$. Let $\alpha \in A$. Suppose $X^p - \alpha$ does not have a root in $K$. Let $\Gamma$ be an element of B such that $\Gamma^p = \alpha$.

My question: Is the following proposition true? If yes, how would you prove this?

Proposition $\Gamma$ is divisible by every prime factor of $\mathfrak{D}_{L/K}$. In particular, $\Gamma$ is not a unit.

Motivation This came from the following question.

Is there a purely imaginary unit in the cyclotomic number field of an odd prime degree?

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1 Answer 1

Note that $B$ contains the order $A[\Gamma] = A[X]/(X^p - \alpha),$ and so the different of $B$ (i.e. $\mathfrak D_{L/K}$) contains the different of $A[\Gamma]$, which equals $p\Gamma^{p-1}$. Thus every prime factor of $\mathfrak D_{L/K}$ divides $p \Gamma^{p-1}$. By assumption these are all prime to $p$, and so in fact every prime factor of $\mathfrak D_{L/K}$ divides $\Gamma^{p-1}$, and so divides $\Gamma$. Thus the first part of your proposition is true.

The second part is not true, because $\mathfrak D_{L/K}$ could be the unit ideal. E.g. take $p = 2$, $K = \mathbb Q[\sqrt{-5}]$, and $L = \mathbb Q[\sqrt{-5},i]$, with $\alpha = -1$ and $\Gamma = i$.

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Thanks. I forgot it could be the unit ideal. –  Makoto Kato Jul 27 '12 at 0:20

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