Are there any nontrivial unramified extensions between two cyclotomic fields? Fix $m$ and let $H$ be the Hilbert class field of $\mathbb{Q}(\zeta_m)$. I'm trying to show that $H\cap \mathbb{Q}(\zeta_n)=\mathbb{Q}(\zeta_m)$ for any $n$ such that $m\mid n$. To do this, I think that it suffices to show that there are no nontrivial unramified subextensions of $\mathbb{Q}(\zeta_n)/\mathbb{Q}(\zeta_m)$. Why is this true?
Here is what I've been thinking so far. Let $K$ be a nontrivial subextension of $\mathbb{Q}(\zeta_n)/\mathbb{Q}(\zeta_m)$. I'd like to show that at least one prime ramifies in $K/\mathbb{Q}(\zeta_m)$. Fixing a prime $p$, we can write $m=p^am'$ and $n=p^bn'$, for some $m'$ and $n'$ with $p\nmid m'n'$. In this case, the ramification indices for $p$ in $\mathbb{Q}(\zeta_m)$ and $\mathbb{Q}(\zeta_n)$ are $\phi(p^a)$ and $\phi(p^b)$, respectively. Thus, if $\frak{P}\mid \frak{p}\mid$ $p$ in $\mathbb{Q}(\zeta_n)/\mathbb{Q}(\zeta_m)/\mathbb{Q}$ then $e(\frak{P}\mid \frak{p})=$ $p^{b-a}$. We can choose $p$ such that $b-a>0$, so it follows that, if $\frak{p'}$ lies over $\frak p$ in $K$ then $e(\frak{p'}|\frak{p})$ is either 1 or a power of $p$. If $e(\frak{p'}|\frak{p})=1$ for all primes then...? Not sure where to go from here.
I think there is probably an easier way to do this. Advice or suggestions would be appreciated... 
 A: Writing $F_m= \mathbb{Q}(\zeta_m)$ throughout. By Kronecker-Weber, the fact that all unramified extensions between cyclotomic fields are trivial is equivalent to:

If $K/F_m$ is unramified and $K/\mathbb{Q}$ is abelian, then $K=F_m$.

We prove it by induction on number of prime factors of $m$.
If $m=p^n$, let $E$ denote the inertia field of $K/\mathbb{Q}$ for $p$ (independent of choice of prime lying above $p$). Then $p$ is unramified in $E$. For prime $q\neq p$, it is unramified in $F_{p^n}$, so unramified in $K$ by assumption, so unramified in $E$. Therefore $E/\mathbb{Q}$ is unramified everywhere, so $E=\mathbb{Q}$ (Minkowski's theorem), this says $p$ is totally ramified in $K$, hence $K=F_m$. (Note that the argument fails if $K/\mathbb{Q}$ is merely Galois)
For general case, let $m = p_1^{r_1}\cdots p_k^{r_k}, r = m/p_1^{r_1}$. Fix a prime $\mathfrak{p}$ in $F_r$ lying above $p_1$, let $E$ denote the inertia field of $K/F_r$ for $\mathfrak{p}$. Suppose $\mathfrak{p}_1 = \mathfrak{p}, \cdots, \mathfrak{p}_g$ are primes of $F_r$ lying above $p_1$. Let $\mathfrak{q}$ be any prime of $F_r$ not lying above $p_1$, $\mathfrak{q}$ is unramified in $F_m/F_r$, thus unramified in $K/F_r$, so unramified in $E/F_r$. Therefore $E/F_r$ is unramified except at possibly $\mathfrak{p}_2,\cdots,\mathfrak{p}_g$. However, $E/\mathbb{Q}$ is abelian (since $K$ is abelian), so acting on $\mathfrak{p}_1$ by various automorphisms shows that $E/F_r$ is unramified everywhere. Induction hypothesis implies $E = F_r$, thus $\mathfrak{p}$ is totally ramified in $K$, so $K = F_m$, completing the proof.
