Primes that ramify in a cyclotomic extension Let $F$ be a number field and consider the cyclotomic extension $E = F(\zeta_{10})$ where $\zeta_{10}$ is a primitive 10th root of unity. Why is it true that the only primes of $F$ that ramify in in $E$ lies above 2 and 5?
 A: In fact, the only primes that can ramify in $E/F$ are primes above 5.  
This is because $E$ is the compositum of $F$ and $\mathbb{Q}(\zeta_{10})=\mathbb{Q}(\zeta_5)$ (this is true more generally for $n\equiv 2\pmod{4}$), and only 5 ramifies in $\mathbb{Q}(\zeta_5)/\mathbb{Q}$.

Added in response to the comment:  
Here's the general principle in play:  If $K$ is a number field, and $F$ and $G$ are two extensions of $K$ with compositum $E=FG$, and $\mathfrak{p}$ is a prime of $F$ above $p$ in $K$, then there is the following relationship between the ramification indices:
$$
e_{\mathfrak{p}}(E/F)\leq e_p(G/K).
$$
In your example (with $K=\mathbb{Q}$ and $G=\mathbb{Q}(\zeta_5)$) gives 
$$
e_{\mathfrak{p}}(F(\zeta_5)/F)\leq e_p(\mathbb{Q}(\zeta_5)/\mathbb{Q})=1
$$
for all $p\neq 5$.
The "$\leq$" occurs because $F/K$ can eat up/absorb/render redundant some of the ramification that occurs in $L/K$.  Most notably, you could take $F=\mathbb{Q}(\zeta_5)$ itself when $p=5$ to get an obvious counter-example to equality.  
