From the calculation of the discriminant, I know that the extension $\Bbb Q (\sqrt {-535})/\Bbb Q$ ramifies only at $2,5,107$. ($\Delta=4\cdot(-535)=-4\cdot5\cdot 107$)
Since $\Bbb Q(\sqrt 5)/\Bbb Q$ is only ramified at $5$, my intuition is that adjoining $\sqrt5$ to $\Bbb Q (\sqrt{-535})$ will not add any ramification. But how do I prove it?
Attempt:
One way I know would be to compute the relative discriminant of $\Bbb Q (\sqrt{-535},\sqrt 5)/\Bbb Q (\sqrt{-535})$, but this is impractical. The other way would be to compute the absolute discriminant of $\Bbb Q (\sqrt{-535}, \sqrt 5)$. Is there a shortcut for this one that does not involve finding a basis for the ring of integers?
The field $F = \mathbb Q(\sqrt{-535},\sqrt{5})$ contains the field $\mathbb Q(\sqrt{-107})$, in which the prime 5 is not ramified. Therefore, 5 cannot be totally ramified in $F/\mathbb Q$.