Assume that $K/\Bbb Q$ is a cyclic Galois extension, and $\mathfrak{p}$ a prime ideal of $K$ and $\sigma$ an element of the Galois group. What can we say about the classes $[\mathfrak{p}]$ and $[\sigma(\mathfrak{p})]$ in the ideal class group? Let $\left \langle [\mathfrak{p}] \right \rangle$ be the subgroup of the ideal class group generated by $[\mathfrak{p}]$. Do we have that necessarily $[\sigma(\mathfrak{p})] \in \left \langle [\mathfrak{p}] \right \rangle$? I cannot prove this. Also, I do not have any idea how to find a counter example.



Action of the Galois group on the ideal class group
This is true if $K$ is quadratic, since then $[\sigma(\mathfrak p)] = [\mathfrak p^{-1}]$.


It should be false for every other degree.


The only relation that the Galois action should satisfy is (for cyclic fields of degree $n$) that $ \prod\_{i=0}^{n-1} [\sigma^{i}(\mathfrak p)] $ is trivial in the class group, since it's the class of the norm of $\mathfrak p$, which is an ideal of $\mathbb Q$ and thus is principal.


For example, for cubic fields, if this were true then you could never have $p$ congruent to $2$ mod $3$ dividing the order of the class group, as the Galois automorphism of order $3$ would have to act on elements of order $p$ by scalar multiplication by an element of $\mathbb F\_p^\times$, and thus would have to be trivial, which is impossible since it would violate the above identity. So it suffices to find a cubic field whose class number is a multiple of $2,$ or $5,$ or ...


Searching quickly on the LMFDB, I found [this counterexample](https://www.lmfdb.org/NumberField/3.3.26569.1), whose class number is $4$.


