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.

$\begingroup$ you could have a look at [[math/9812171] Perfect forms and the Vandiver conjecture](arxiv.org/abs/math/9812171) $\endgroup$– NielsJan 26, 2021 at 20:48

1$\begingroup$ Look up genus theory. $\endgroup$– Franz LemmermeyerJan 26, 2021 at 21:02

$\begingroup$ @Niels I can not see how the results in that paper are related to my question. Could you please give an exposition on it? $\endgroup$– Tireless and hardworkingJan 26, 2021 at 21:20

3$\begingroup$ @FranzLemmermeyer Could you please introduce a specific book or note or ... to me about this subject? If I am not mistaken, probably genus theory would deal with the cases when the Galois group acts trivially on the ideal class group. $\endgroup$– Tireless and hardworkingJan 26, 2021 at 21:23

$\begingroup$ looks like you got interested: mathoverflow.net/questions/382930/… $\endgroup$– NielsFeb 3, 2021 at 15:13
1 Answer
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}^{n1} [\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, whose class number is $4$.

$\begingroup$ Your statements on quadratic extension and cyclic extensions are very clear. Also, I can check why your counterexample contradicts my claim (I can see it in another way), and I get the answer to my question. But I have another request, could you please tell more on the "Galois automorphism should have to act on elements of order $p$ by scalar multiplication by an element of $\mathbb F_p^\times$"? Can you write down the action of the Galois group? $\endgroup$ Jan 27, 2021 at 9:54

1$\begingroup$ @NeoTheComputer No, I just mean that if $\sigma([\mathfrak p]) \in \langle [\mathfrak p \rangle$ then $\sigma([\mathfrak p]) = [\mathfrak p]^a$ for some $a$ mod $p$, because $ [\mathfrak p]$ has order $p$. $\endgroup$ Jan 27, 2021 at 14:50

$\begingroup$ Thank you for your response. $\endgroup$ Jan 27, 2021 at 14:53