Splitting behaviour of primes in $\textbf Q(\sqrt{-p_1 \cdots p_n})$ Let $K = \textbf Q(\sqrt{-p_1 \cdots p_n})\;, m = -p_1 \cdots p_n \;$ where $p_i$ are distinct prime numbers and $n \gt 1$
What are all the primes that ramify in $K$ ?
Maybe I am completly missing something but as far as I know $p$ ramifies in $K$ if $p$ divides the discriminant $d_k$.
So all primes $p_i$ ramify in $K$ ? I feel stupid for asking this.
 A: Yes, all those primes ramify. Or at least the ideals generated by those primes ramify. As numbers they might actually be irreducible, especially if you're specifically referring to imaginary quadratic rings.
To take a simple example, consider $K = \mathbb Q(\sqrt{-210}) = \mathbb Z[\sqrt{-210}]$. The primes 2, 3, 5, 7 as numbers are all actually irreducible in this ring, since the norms $-7, -5, -3, -2, 2, 3, 5, 7$ are impossible in this ring. However, none of those numbers are prime in this ring, since they all divide $(\sqrt{-210})^2$, yet none of them divide $\sqrt{-210}$, nor does that number divide any of them.
Thus, as ideals, we see that $\langle 2 \rangle = \langle 2, \sqrt{-210} \rangle^2$, $\langle 3 \rangle = \langle 3, \sqrt{-210} \rangle^2$, $\langle 5 \rangle = \langle 5, \sqrt{-210} \rangle^2$ and $\langle 7 \rangle = \langle 7, \sqrt{-210} \rangle^2$. To verify the first of these, note that any number in this ring with even norm is one of these forms: $2a$, $b \sqrt{-210}$ or $2a + b \sqrt{-210}$, where $a$ and $b$ are arbitrary numbers in $\mathbb Z[\sqrt{-210}]$.
Hope this helps clarify things for you.
