If $\zeta$ is a primitive $n$-th root of unity, prove that:
$$d(1, \zeta,…,\zeta^{\varphi(n)-1})=(-1)^{\varphi(n)/2}n^{\varphi(n)}\prod_{p\mid n} p^{-\frac{\varphi(n)}{p-1}}$$
Let $n=\prod_{i=1}^{m}p_i^{e_i}$. After looking it up in some books, I was able to understand why this is true for $m=1$. However, they all ignored the general case $m>1$ or simply stated that it could be done by induction on $m$, but I really can't see how it could be done.
The only interesting thing I could find out was that for $n,m$ with $\gcd(n, m)=1$, we get, on the right hand side of the equation:
$$(-1)^{\varphi(nm)/2}nm^{\varphi(nm)}\prod_{p\mid nm} p^{-\frac{\varphi(nm)}{p-1}}=$$
$$\left((-1)^{\varphi(n)/2}n^{\varphi(n)}\prod_{p\mid n} p^{-\frac{\varphi(n)}{p-1}}\right)^{\varphi(m)}\left((-1)^{\varphi(m)/2}n^{\varphi(m)}\prod_{p\mid m} p^{-\frac{\varphi(m)}{p-1}}\right)^{\varphi(n)}$$
That makes me think I'm getting somewhere, but I'm stuck with the problem of showing that $d(1, \zeta,…,\zeta^{\varphi(nm)-1})=[d(1, \zeta,…,\zeta^{\varphi(n)-1})]^{\varphi(m)}[d(1, \zeta,…,\zeta^{\varphi(m)-1})]^{\varphi(n)}$, which doesn't seem trivial at all. Any ideas? Thanks!
Best Answer
because we have a proposition that says
If $K, F$ are two number fields linearly disjoint over $\mathbb{Q}$, $KF$ their compositum, and their discriminants are coprime. then
$$\delta_{KL}=\delta_{K}^{[L:\mathbb{Q}]}\cdot\delta_{L}^{[K:\mathbb{Q}]}$$
and in our case we have $\mathbb{Q}(\zeta_{n})$ and $\mathbb{Q}(\zeta_{m})$ are linearly disjoint because $ gcd(n,m)=1$ , and their discriminants are coprime then
$$\delta_{\mathbb{Q}(\zeta_{mn})}=\delta_{\mathbb{Q}(\zeta_{n})}^{\phi(m)}\cdot\delta_{\mathbb{Q}(\zeta_{m})}^{\phi(n)}$$.
But the problem is $$\bigg( (-1)^{\phi(n)/2}n^{\phi(n)}\prod_{p\mid n}p^{\frac{-\phi(n)}{p-1}}\bigg)^{\phi(m)}\cdot\bigg( (-1)^{\phi(m)/2}n^{\phi(m)}\prod_{p\mid m}p^{\frac{-\phi(m)}{p-1}}\bigg)^{\phi(n)}=(-1)^{\phi(nm)}(nm)^{\phi(nm)}\prod_{p\mid nm}p^{\frac{-\phi(nm)}{p-1}}$$