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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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Prove the more general fact that if $K, L$ are number fields with coprime discriminants, then $K \cap L = \mathbb{Q}$. (To do this, recall the even more general fact that if $K \subset L$, then $D_K | D_L$.) – Qiaochu Yuan Dec 11 '11 at 20:33
Another proof: Recall that there are no unramified extensions of $\mathbb{Q}$ and that the primes that ramify are precisely those dividing the discriminant. If the discriminants in your example are coprime, then no primes in the intersection can ramify, so the intersection has to be $\mathbb{Q}$. – Fredrik Meyer Dec 11 '11 at 21:56
@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 $F$). – M Turgeon Dec 14 '11 at 1:20
up vote 6 down vote accepted

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! – Anna Dec 11 '11 at 20:55

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