# Action of the Galois group on the ideal class group

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.

• you could have a look at [[math/9812171] Perfect forms and the Vandiver conjecture](arxiv.org/abs/math/9812171) Jan 26, 2021 at 20:48
• Look up genus theory. Jan 26, 2021 at 21:02
• @Niels I can not see how the results in that paper are related to my question. Could you please give an exposition on it? Jan 26, 2021 at 21:20
• @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. Jan 26, 2021 at 21:23
• looks like you got interested: mathoverflow.net/questions/382930/… Feb 3, 2021 at 15:13

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, whose class number is $$4$$.

• 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? Jan 27, 2021 at 9:54
• @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 $p$. Jan 27, 2021 at 14:50
• Thank you for your response. Jan 27, 2021 at 14:53