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Let $K$ be a number field and $N$ a positive integer. Prove that if the absolute discriminant of $K$ is coprime to $N$, then $K \cap \mathbb{Q}[\zeta_{N}]=\mathbb{Q}.$

This is something that the Childress book on CFT leaves kind of vague when proving Artin's Lemma. I would like to see a proof if possible.

Thanks in advance!

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    @Fredrik +1 for making explicit why the intersection is $\mathbb{Q}$. In particular, it is not always true in the relative setting ($L$ and $K$ extensions of a number field $F$).2011-12-14

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If $L$ is an intermediate extension of $E$ over $F$, and $\Delta_{N/M}$ denotes the discriminant of $N$ over $M$, then $\Delta_{E/F} = N^{L}_F(\Delta_{E/L})\Delta_{L/F}^{[E:L]},$ where $N^L_F(\cdot)$ is the norm map from $L$ to $F$.

In particular, the discriminant of $L/F$ divides the discriminant of $E/F$.

Let $L = K\cap\mathbb{Q}(\zeta_N)$. Then the discriminant of $L$ over $\mathbb{Q}$ has to divide the discriminant of $K$, and also has to divide the discriminant of $\mathbb{Q}(\zeta_N)$.

The discriminant of $\mathbb{Q}(\zeta_N)$ is $(-1)^{\varphi(N)/2}\left(\frac{N^{\varphi(N)}}{\prod\limits_{p|N}p^{\varphi(N)/(p-1)}}\right).$ In particular, it can only be divisible by primes that divide $N$.

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    Thanks for the clear answer!2011-12-11