Let $K$ be a number field generated by adjoining a root $α$ of a polynomial $f(x)$.
How to prove that:
If $\text{Gal}(\mathbb{Q}[α]/\mathbb{Q})=C_{p}$ acting regularly on the $p$ roots of $f$ ($p$ is a prime), then the discriminant of $O_K$ is a (p-1)th power?
If $\text{Gal}(\mathbb{Q}[α]/\mathbb{Q})=C_{3}\times C_3$ acting regularly on the $9$ roots of $f$, then the discriminant of $O_K$ is a 6th power?
I tried to generalize the proof of the following fact:
- If $\text{Gal}(\mathbb{Q}[α]/\mathbb{Q})$ is a subgroup of $A_n$, then the discriminant of $O_K$ is a square
which considers the action of the Galois group on $\prod_{(a_i,a_j)}(a_i-a_j)$, where $(a_i,a_j)$ run over all unordered pairs of roots.
The (absolute value of) discriminant $\mathcal{O}_K$ is the norm of different ideal $\mathcal{D}_{K}$. For every prime $\mathfrak{q}$ in $K$ lying above $q\in \mathbb{Z}$, exponent of $\mathfrak{q}$ in $\mathcal{D}_K$ is $$\sum_{k\geq 0} (|G_k|-1)$$ where $G_k$ are ramification group of $\mathfrak{q}/q$, with $G_0$ equals inertia group.
When $K/\mathbb{Q}$ has Galois group $C_p$, $|G_i| = p$ or $1$. Hence exponent of $\mathfrak{q}$ in $\mathcal{D}_K$ are divisible by $p-1$, so discriminant is a $(p-1)$-th power.
When $K/\mathbb{Q}$ has Galois group $C_p \times C_p$, a ramified prime $q$ must have ramification index $p$ (can be proved via Kronecker-Weber). Hence exponent of $\mathfrak{q}$ in $\mathcal{D}_K$ are divisible by $p-1$, but in this case there are either $p$ primes lying above $q$ or one with residual degree $p$, in both cases, the discriminant is a $p(p-1)$-th power.