Let $K=\mathbb{Q}(\sqrt{-26})$.

(a) Show that $\mathcal{O}_{K}=\mathbb{Z}[\sqrt{-26}]$ and that the discriminant $d_{K}$ is equal to $-104$.

(b) Show that 2 ramifies in $\mathcal{O}_{K}$ by showing that $[2]=\mathfrak{p}_{2}^{2}$, and that $\mathfrak{p}_{2}$ is not a principal ideal. Show further that $[3]=\mathfrak{p}_{3} \overline{\mathfrak{p}}_{3}$ with $\mathfrak{p}_{3}=[3,1-\sqrt{-26}]$. Deduce that neither $\mathfrak{p}_{3}$ nor $\mathfrak{p}_{3}^{2}$ is a principal ideal, but $\mathfrak{p}_{3}^{3}=[1-\sqrt{-26}]$.

(c) Show that 5 splits in $\mathcal{O}_{K}$ by showing that $[5]=\mathfrak{p}_{5} \overline{\mathfrak{p}}_{5}$, and that

$$N_{K / \mathbb{Q}}(2+\sqrt{-26})=30$$

Deduce that $\mathfrak{p}_{2} \mathfrak{p}_{3} \mathfrak{p}_{5}$ has trivial class in the ideal class group of $K$. Conclude that the ideal class group of $K$ is cyclic of order six.

[You may use the fact that $\left.\frac{2}{\pi} \sqrt{104} \approx 6.492 .\right]$