On an implication of Kronecker-Weber therorem Assuming Kronecker-Weber therorem:

Any abelien extension $K/\mathbb{Q}$ is a subextension of $\mathbb{Q}(\zeta_n)/\mathbb{Q}$

Then we have a surjective homomorphism
$$(\mathbb{Z}/n\mathbb{Z})^*=\mathrm{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q}) \to \mathrm{Gal}(K/\mathbb{Q})$$
I want to show that if $p$ is a prime not dividing $n$, then $K/\mathbb{Q}$ is not unramified at $p$. But I don't see how to use the "$p$ not dividing $n$" condition correctlly.  Could you give me some hints? Thanks in advance.
 A: Your phrase "is not unramifed at $p$" should be "is not ramified at $p$" or "is unramified at $p$", but not what you wrote.
As for the question itself, here is a general result that is quite useful: if $K$ is a number field and we can write $K = \mathbf Q(\alpha)$ where $\alpha$ is an algebraic integer that is the root of a polynomial $f(x) \in \mathbf Z[x]$, then for each prime $p$ such that $f(x) \bmod p$ is separable, $p$ is unramified in $K$.  Apply this to your situation with $K = \mathbf Q(\zeta_n)$ and $f(x) = x^n - 1$.
Twp things worth noting: (1) we are not requiring $f(x)$ to be the minimal polynomial of $\alpha$ over $\mathbf Q$, although the hypothesis on $f(x)$ above does pass to that minimal polynomial since it is factor of $f(x)$ in $\mathbf Z[x]$, and (2) there is no claim that the converse is true. It is definitely false in general that when $f(x) \bmod p$ is not separable that $p$ has to be ramified.  For example, take $K = \mathbf Q(\sqrt{5})$ and $f(x) = x^2 - 5$. We have $x^2 - 5 \equiv (x-1)^2 \bmod 2$, but $2$ is unramified in $K$. The ring of integers of $K$ is not $\mathbf Z[\sqrt{5}]$, but is a larger ring.
