Unramification of Ideals in Pure Cubic fields

Question

I need some explanation for this .Let $K=\mathbb Q{\sqrt[3]{m}} $ be a pure cubic field with non square element $\alpha $ in $K$ such that ideal $(\alpha) $ is an ideal square in K. Let $ L=K(\sqrt{\alpha})$ be a quadratic extension of K then why only primes above 2 and $\infty $ will ramify. Thank you in advance.

Answer 1

Let $\alpha$ be such an element. $(\alpha) = \mathfrak p^2$, and the hypothesis on $\alpha$ says that $\mathfrak p$ represents a nontrivial element of order $2$ in the ideal class group of $K$.
The extension $L/K$ may be ramified at the primes above $2,\infty$, and at the primes present in the factorisation of $\mathfrak p$.

Now, in the ideal class group, each ideal class is represented by infinitely many prime ideals. So we can find an ideal $\mathfrak q$ coprime with $\mathfrak p$ such that $\mathfrak {pq} = (a)$ for some $a \in K$

Now, we have $ (\alpha)\mathfrak q^2 = (a^2)$, so setting $\beta = a^2/\alpha$, we have $\mathfrak q^2 = (\beta)$, $L = K(\sqrt{\alpha}) = K(\sqrt{\beta})$, but $K(\sqrt{\beta})$ can only be ramified above the primes at $2$ and $\infty$, and at $\mathfrak q$.

Since $\mathfrak p$ and $\mathfrak q$ are coprime, $L$ can only be ramified at the primes above $2$ and $\infty$.