Let $d$ be an odd prime and $d \equiv 1 \pmod{4}$. For quadratic extensions $\mathbb{Q}(\sqrt{d})/\mathbb{Q}$, the only prime that is ramified is $d$, and it is also the conductor of the extension. So the conductor of the extension is same as the conductor defined on the Dirichlet characters. How do I prove this?
Also, if I have $\mathbb{Q}(d^{1/p})/\mathbb{Q}$ where $p$ is some odd prime, how do I find the conductors? I am trying to use the discriminant-conductor formula for cubic extensions to try and estimate what conductors are using the discriminant but I am unable to make any progress. So how do I show this?
Since the extension $(L=\mathbb{Q}(d^{1/p}))/{K}$, (where $K$ contains the $p$-th roots of unity) is abelian so $\chi(1)=1$ but what are the values of the conductors? My understanding says that for $p=3$ there will be three Artin conductors defined at $1, \omega, \omega^2$ but what is their value? I know that only primes ramified in the extension $L/K$ in this case of $p=3$ are $(1-omega), d$ but then how is the conductor $3d$?
So how are these conductors related to the conductor of the extension and the ramified primes?
I would really appreciate the help.