ramification in a cyclic extension of a cyclotomic field Let $F$ be a cyclotomic field generated by a primitive $p$-th root $\zeta$ of 1. Is it known that the prime ideal $(1-\zeta)$ in $\mathcal{O}_F$ ramifies in the extension $F(q^{\frac{1}{p}})$ where $q$ is a prime number distinct from $p$? I cannot find a reference that would explain the ramification behaviour of the above prime ideal. 
 A: A couple of preliminary remarks: first, the situation is purely local, at $p$, so we might as well take the base to be $\mathbb Q_p$, not $\mathbb Q$; second, the question has nothing to do with $q$ being prime, it can be any integer prime to $p$, and thus any $p$-adic unit $n$, though I’ll pretend that it’s in $\mathbb Z$. I’ll call $\zeta-1=\pi$, $F=\mathbb Q_p(\pi)$, minimal polynomial for $\pi$ is $X^p+(pX^{p-1}+p(p-1)/2)X^{p-2}+\cdots+(p(p-1)/2)X+p$, the intermediate coefficients being binomial numbers, as I’m sure you know. I’ll set $\lambda$ to be a $p$-th root of $n$.
The answer depends on the $p$-valuation of $n^p-n$, I’ll write it $v_p(n^p-n)$.
Case 1: $v_p(n^p-n)=1$. Here the extension $F(\lambda)\supset F$ is totally ramified of degree $p$, as I’ll now argue. Put $f(X)=X^p-n$, which has $\lambda$ for a root,
and if we look at $f(X+n)$, which has the shape (with $g(X)$ of degree at most $p-2$)  $X^p+pXg(X)+n^p-n$, what do we see? A polynomial with $\lambda-n$ for a root, and with Newton polygon with just the two vertices $(0,1)$ and $(p,0)$, using our hypothesis that $n^p-n$ is a unit times $p$. The slope is $-1/p$, which is not in our value group of $\frac1{p-1}\mathbb Z$, in fact to get a root, you need to make a ramified extension of degree $p$, and there you are for this case.
Case 2: $v_p(q^p-q)>1$, i.e. $n^p-n=p^2w$ for an integer $w$. Similar argument, it’s only a little more complicated. Again we put $f(X)=X^p-n$, and set $f_1(X)=f(X+n)=(X+n)^p-n$, which I’ll now write in the shape $X^p+pX^2g(X)+pn^{p-1}X+p^2w$. Here, $g$ is of degree at most $p-3$.
Again we look at the Newton polygon, but the vertices are at $(0,2+v_p(w))$, $(1,1)$, and $(p,0)$. The slopes of the segments are $-1-v_p(w)$ and $-1/((p-1)$, they’re distinct, so there is a vertex of the polygon at $(1,1)$, and the width-$1$ segment to its left corresponds to a linear factor of $f_1$. That is, $f_1$ has a root in $\mathbb Z_p[\pi]$, the same is true of $f$, and because the roots of unity are there, $f$ has all its roots in $\mathbb Z_p[\pi]$. Therefore, in this case, locally over $\mathbb Q_p$, there is no extension. What that means for the global situation is that $(\zeta-1)$ splits completely in the extension.
